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Title: A term of commutative algebra
Description: Commutative algebra is a branch of algebra that deals with commutative rings, which are algebraic structures that satisfy the commutative property of multiplication. One important term in commutative algebra is the prime ideal. A prime ideal is a special type of ideal in a commutative ring, which has the property that whenever two elements in the ring multiply to an element in the prime ideal, at least one of the elements must be in the prime ideal. In other words, a prime ideal is an ideal that is "prime" in the sense that it cannot be further factored into smaller ideals. Prime ideals play a crucial role in algebraic geometry, number theory, and other areas of mathematics. They are used to define algebraic varieties, which are geometric objects that can be studied using algebraic techniques. The theory of prime ideals also connects to other important concepts in commutative algebra, such as maximal ideals, local rings, and polynomial rings.
Description: Commutative algebra is a branch of algebra that deals with commutative rings, which are algebraic structures that satisfy the commutative property of multiplication. One important term in commutative algebra is the prime ideal. A prime ideal is a special type of ideal in a commutative ring, which has the property that whenever two elements in the ring multiply to an element in the prime ideal, at least one of the elements must be in the prime ideal. In other words, a prime ideal is an ideal that is "prime" in the sense that it cannot be further factored into smaller ideals. Prime ideals play a crucial role in algebraic geometry, number theory, and other areas of mathematics. They are used to define algebraic varieties, which are geometric objects that can be studied using algebraic techniques. The theory of prime ideals also connects to other important concepts in commutative algebra, such as maximal ideals, local rings, and polynomial rings.
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A Term of
Commutative Algebra
By Allen ALTMAN
and Steven KLEIMAN
Version of September 1, 2013: 13Ed1
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0 Unported License
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edition number for publishing purposes
ISBN 978-0-9885572-1-5
Contents
Preface
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2
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4
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Rings and Ideals
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Radicals
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Exact Sequences
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6
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7
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8
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9
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10
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11
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12
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13
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14
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Noether Normalization
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16
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17
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18
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19
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20
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Appendix: Homogeneity
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Dimension
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Completion
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Discrete Valuation Rings
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Dedekind Domains
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Fractional Ideals
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Arbitrary Valuation Rings
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1
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2
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3
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4
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5
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6
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7
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10
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Localization of Rings
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13
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15
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17
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19
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21
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23
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25
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Localization of Modules
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Krull–Cohen–Seidenberg Theory
Noether Normalization
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Associated Primes
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Length
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Dimension
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Discrete Valuation Rings
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Fractional Ideals
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249
Disposition of the Exercises in [3]
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253
Bibliography
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Most books are monographs, with extensive coverage
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It is a clear, concise,
and efficient textbook, aimed at beginners, with a good selection of topics
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However, its age and flaws do show
...
Atiyah and Macdonald explain their philosophy in their introduction
...
It is designed to be read by students who have had a first elementary course in
general algebra
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The lecture-note origin of this book
accounts for the rather terse style, with little general padding, and for the condensed
account of many proofs
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” They endeavor “to build up to
the main theorems in a succession of simple steps and to omit routine verifications
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The present book also “grew out of a course of
lectures
...
Their book comprises eleven chapters, split into
forty-two sections
...
Atiyah and Macdonald “provided
...
” They
“provided hints, and sometimes complete solutions, to the hard” exercises
...
By
contrast, in the present book, the exercises are integrated into the development,
and complete solutions are given at the end of the book
...
In particular, it led to the addition of appendices on Fitting
Ideals and on Cohen–Macaulayness
...
)
There are 324 exercises below
...
The disposition of those
exercises is indicated in a special index preceding the main index
...
Here the exercises are tailored to provide a means for students to check, to
solidify, and to expand their understanding of the material
...
Rarely do they introduce new techniques,
although some introduce new concepts and many statements are used later
...
If they become stuck, then they should review the
relevant material; if they remain stuck, then they should change tack by studying
the given solution, possibly discussing it with others, but always making sure they
can eventually solve the whole exercise entirely on their own
...
Instructors are encouraged to examine their students, possibly orally at a blackboard, possibly via written tests, on a small, randomly chosen subset of all the
exercises that have been assigned over the course of the term for the students to
write up in their own words
...
Atiyah and Macdonald explain that “a proper treatment of Homological Algebra
is impossible within the confines of a small book; on the other hand, it is hardly
sensible to ignore it completely
...
— but
...
” Again, their philosophy is embraced and refined in the present
book
...
Also, projective modules are treated below, but not in their book
...
Thus they discuss the universal (mapping) property
(UMP) of localization of a ring, but provide an ad hoc characterization
...
Below,
the UMP is fundamental: there are many standard constructions; each has a UMP,
which serves to characterize the resulting object up to unique isomorphism owing
to one general observation of Category Theory
...
Atiyah and Macdonald prove the Adjoint-Associativity Formula
...
From it and the Left Exactness
of Hom, they deduce the Right Exactness of Tensor Product
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” More generally, as explained
below, this derivation shows that any left adjoint preserves arbitrary direct limits,
ones indexed by any small category
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Also, arbitrary direct sums are direct limits indexed
by a discrete category (it is not a directed set); hence, the general result yields that
Tensor Product and other left adjoints preserve arbitrary Direct Sum
...
Therefore,
the functor of localization of a module is canonically isomorphic to the functor of
tensor product with the localized base ring, as both are left adjoints of the same
functor, Restriction of Scalars from the localized ring to the base ring
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Since Localization is a left adjoint, it preserves Direct Sum
and Cokernel; whence, it is isomorphic to that tensor-product functor by Watts
Theorem, which characterizes all tensor-product functors as those linear functors
that preserve Direct Sum and Cokernel
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However, they do use the proof of Watts Theorem directly to show that,
under the appropriate conditions, Completion of a module is Tensor Product with
the completed base ring
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As such, Direct Limit is a left adjoint
...
Here the theory briefly climbs to a higher level of
abstraction
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The extra abstraction can be difficult, especially for beginners
...
They are closer to the kind of limits
treated by Atiyah and Macdonald
...
Further, they appear in the following lovely form of Lazard’s
Theorem: in a canonical way, every module is the direct limit of free modules of
finite rank; moreover, the module is flat if and only if that direct limit is filtered
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First, they study primary decompositions of ideals in rings
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The decompositions need not
exist, as the rings and modules need not be Noetherian
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Finally,
they prove that, when the rings and modules are Noetherian, decompositions exist
and the associated primes are annihilators
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Nowadays, associated primes are normally defined as prime annihilators
of elements, and studied on their own at first; sometimes, as below, irreducible
modules are not considered at all in the main development
...
First, the Noether Normalization Lemma is proved
below in a stronger form for nested sequences of ideals; consequently, for algebras
that are finitely generated over a field, dimension theory can be developed directly
without treating Noetherian local rings first
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Third, there is an elementary treatment
of regular sequences below and a proof of Serre’s Criterion for Normality
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The present book is a second beta edition
...
Thanks!
Allen B
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Kleiman
31 August 2013
This page is intentionally blank!
1
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Throughout this book, we emphasize universal mapping properties (UMPs); they are used
to characterize notions and to make constructions
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We close this section with a brief treatment
of idempotents and the Chinese Remainder Theorem
...
1) (Rings)
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Throughout this book, every ring has a multiplicative identity, denoted by 1
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As usual, the additive identity is denoted by 0
...
We allow 1 = 0
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A unit is an element u with a reciprocal 1/u such that u·1/u = 1
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The units form
a multiplicative group, denoted R×
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A ring homomorphism, or simply a ring map, φ : R → R′ is a map preserving
sums, products, and 1
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We call φ an isomorphism if it is
∼ R′
...
bijective, and then we write φ : R −→
We call φ an automorphism if it is bijective and if R′ = R
...
For example, the polynomial rings R[X] and R[Y ] in variables X and
Y are canonically isomorphic when X and Y are identified
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The notion
is psychological, and depends on the context
...
A subset R′′ ⊂ R is a subring if R′′ is a ring and the inclusion R′′ ,→ R a ring
map
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An R-algebra is a ring R′ that comes equipped with a ring map φ : R → R′ ,
called the structure map
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(1
...
— The simplest nonzero ring has two elements, 0 and 1
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Given any ring R and any set X, let RX denote the set of functions f : X → R
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For example, take R := F2
...
Then f (x) = 1
if x ∈ S, and f (x) = 0 if x ∈
/ S; in other words, f is the characteristic function
χS
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1
2
Rings and Ideals (1
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Further, χS + χT = χS△T , where S△T
is the symmetric difference:
S△T := (S ∪ T ) − (S ∩ T ) = (S − T ) ∪ (T − S);
here S − T denotes, as usual, the set of elements of S not in T
...
This
ring is canonically isomorphic to FX
2
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Clearly, FX
2 is Boolean
...
Consider the continuous functions f : X → F2
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Clearly, they form
a Boolean subring of FX
2
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25) asserts that every
Boolean ring is canonically isomorphic to the ring of continuous functions from a
compact Hausdorff topological space X to F2 , or equivalently, isomorphic to the ring
of open and closed subsets of X
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3) (Polynomial rings)
...
, Xn ] the polynomial
ring in n variables (see [2, pp
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268])
...
In fact, since π is a ring map, necessarily π is given by the formula:
(∑
) ∑
π
a(i1 ,
...
,in ) )xi11 · · · xinn
...
Similarly, let P ′ := R[{Xλ }λ∈Λ ] be the polynomial ring in an arbitrary list of
variables: its elements are the polynomials in any finitely many of the Xλ ; sum and
product are defined as in P
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Clearly, P ′ has
essentially the same UMP as P : given φ : R → R′ and given xλ ∈ R′ for each λ,
there is a unique π : P ′ → R′ with π|R = φ and π(Xλ ) = xλ
...
4) (Ideals)
...
Recall that a subset a is called an ideal if
(1) 0 ∈ a,
(2) whenever a, b ∈ a, also a + b ∈ a, and
(3) whenever x ∈ R and a ∈ a, also xa ∈ a
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If Λ = ∅, then this ideal consists just of 0
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Form the set a, or
Raλ , of all such linear
combinations; clearly, a is an ideal containing all aλ
...
Given a single element a, we say that the ideal ⟨a⟩ is principal
...
Similarly,
given ideals aλ of R, by the ideal∑
they generate, we mean the smallest
∑
ideal
aλ that contains
them
all
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Rings and Ideals (1
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They are clearly ideals
...
Further, for any ideal c, the distributive law holds: a(b + c) = ab + ac
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Then a = R if and only if 1 ∈ a
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It follows that a = R if and only if a contains a
unit
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If a ̸= R, then a is said to be proper
...
Let aR′ denote the ideal of R′ generated by φ(a);
we call aR′ the extension of a
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Clearly, the preimage
φ−1 (a′ ) is an ideal of R; we call φ−1 (a′ ) the contraction of a′
...
5)
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Set ae := φ(a)R′ and bc := φ−1 (b)
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(2) Then aece = ae and bcec = bc
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(4) If two extensions have the same contraction, then they are equal
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6) (Residue rings)
...
Recall its kernel Ker(φ)
is defined to be the ideal φ−1 (0) of R
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Conversely, let a be an ideal of R
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Recall that R/a inherits a ring structure, and is called the residue ring (or
quotient ring or factor ring) of R modulo a
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The element κx ∈ R/a is called the residue of x
...
Thus every ideal is a kernel!
Note that Ker(φ) ⊃ a if and only if φa = 0
...
Further, if ψ exists, then ψ is unique as κ is surjective
...
In addition, then ψ is injective if and only if a = Ker(φ)
...
Therefore, always
φ
$
∼ Im(φ)
...
6
...
Finally, R/a has, as we saw, this UMP: κ(a) = 0, and given φ : R → R′ such that
4
Rings and Ideals (1
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In other
words, R/a is universal among R-algebras R′ such that aR′ = 0
...
The UMP serves to determine R/a up to unique isomorphism
...
Then φ(a) = 0; so there is a unique
ψ : R/a → R′ with ψκ = φ
...
Then, as shown, (ψ ′ ψ)κ = κ, but 1 ◦ κ = κ where 1
7 R/a
κ
R
φ
/ R′
κ
ψ
1
ψ′
'
R/a
is the identity map of R/a; hence, ψ ′ ψ = 1 by uniqueness
...
Thus ψ and ψ ′ are inverse isomorphisms
...
There are many
more constructions to come, and each one has an associated UMP, which therefore
serves to determine the construction up to unique isomorphism
...
7)
...
, Xn ] the
polynomial ring
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, Xn ]
...
8)
...
Then
∼ R
...
Then π(F (X)) = b
...
Finally,
∼ R
...
6
...
9) (Nested ideals)
...
Given an ideal b ⊃ a, form the corresponding set of cosets of a:
b/a := {b + a | b ∈ b} = κ(b)
...
Also b/a = b(R/a)
...
Moreover, this correspondence preserves inclusions
...
Clearly, φ is surjective, and Ker(φ) = b
...
6), φ factors through
the canonical isomorphism ψ in this commutative diagram:
R −−−−−→ R/b
ψ y≃
y
/
R/a −
→ (R/a) (b/a)
Rings and Ideals (1
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10)
...
, Xn ] the polynomial ring
...
, am ∈ R
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, Xm − am ⟩
...
, Xn ]
...
11) (Idempotents)
...
Let e ∈ R be an idempotent; that is,
e2 = e
...
But Re is not a subring
of R unless e = 1, although Re is an ideal
...
Then e′ is idempotent and e · e′ = 0
...
Conversely, if two elements e1 , e2 ∈ R satisfy e1 + e2 = 1
and e1 e2 = 0, then they are complementary idempotents, as for each i,
ei = ei · 1 = ei (e1 + e2 ) = e2i
...
Let φ : R → R′ be a ring map
...
So the restriction of φ to Idem(R) is a map
Idem(φ) : Idem(R) → Idem(R′ )
...
12)
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The additive identity is (0, 0); the multiplicative
identity is (1, 1)
...
Then e and e′ are complementary
idempotents
...
Proposition (1
...
— Let R be a ring with complementary idempotents e and
e′
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Then φ is a ring isomorphism
...
Then φ′ is a ring map since
xye = xye2 = (xe)(ye)
...
So φ is a ring map
...
Also, φ is injective, since if xe = 0 and xe′ = 0, then x = xe + xe′ = 0
...
□
Exercise (1
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— Let R be a ring
...
Prove
(a) ab = a ∩ b and
(b) R/ab = (R/a) × (R/b)
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Prove a is also comaximal to bb′
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Prove am and bn are comaximal
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, an be pairwise comaximal
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Exercise (1
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— First, given a prime number p and a k ≥ 1, find the idempotents in Z/⟨pk ⟩
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Third, find the number
∏N
of idempotents in Z/⟨n⟩ where n = i=1 pni i with pi distinct prime numbers
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16)
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Show a = a′ × a′′ with a′ ⊂ R′ and a′′ ⊂ R′′ ideals
...
6
Rings and Ideals (1
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17)
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(See (10
...
)
(1) Set a := ⟨e⟩
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(2) Let a be a principal idempotent ideal
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(3) Set e′′ := e + e′ − ee′
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(4) Let e1 ,
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Show ⟨e1 ,
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(5) Assume R is Boolean
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2
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So we review the
basic theory
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We show maximal ideals are prime, and discuss examples
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Definition (2
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— Let R be a ring
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Denote
the set of zerodivisors by z
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A subset S is called multiplicative if 1 ∈ S and if x, y ∈ S implies xy ∈ S
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Exercise (2
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— Let a and b be ideals, and p a prime ideal
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(2
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— A ring is called a field if 1 ̸= 0 and if every nonzero
element is a unit
...
A ring is called an integral domain, or simply a domain, if ⟨0⟩ is prime, or
equivalently, if R is nonzero and has no nonzero zerodivisors
...
Conversely, any subring R of a field K,
including K itself, is a domain; indeed, any nonzero x ∈ R cannot be a zerodivisor,
because, if xy = 0, then (1/x)(xy) = 0, so y = 0
...
For example, the ring of integers Z is a domain, and Frac(Z) = Q ⊂ R ⊂ C
...
Then R[X]
is a domain too
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(2
...
1)
By induction, the polynomial ring in n variables R[X1 ,
...
, Xn ] = R[X1 ,
...
Hence the polynomial ring in an arbitrary set of variables R[{Xλ }λ∈Λ ] is a domain,
since any two elements lie in a polynomial subring in finitely many of the Xλ
...
So by induction, if f, g ∈ R[X1 ,
...
This reasoning can fail if R is not a domain
...
The fraction field Frac(R[{Xλ }λ∈Λ ]) is called the field of rational functions,
and is also denoted by K({Xλ }λ∈Λ ) where K := Frac(R)
...
4)
...
Exercise (2
...
— Let R := R′ × R′′ be a product of two rings
...
7
8
Prime Ideals (2
...
6) (Unique factorization)
...
We call
p prime if, whenever p | xy (that is, there exists z ∈ R such that pz = xy), either
p | x or p | y
...
We call p irreducible if, whenever p = yz, either y or z is a unit
...
In general, prime elements are irreducible; in a UFD, irreducible elements are
prime
...
398, p
...
18
...
297]
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7)
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If T is
multiplicative, then φ−1 T is multiplicative; the converse holds if φ is surjective
...
If T is multiplicative, then 1 ∈ S as φ(1) = 1 ∈ T , and
x, y ∈ S implies xy ∈ S as φ(xy) = φ(x)φ(y) ∈ T ; thus S is multiplicative
...
If φ is surjective, then every x′ ∈ T is of the form x′ = φ(x)
for some x ∈ S
...
□
Proposition (2
...
— Let φ : R → R′ be a ring map, and q ⊂ R′ an ideal
...
Proof: By (2
...
So the assertion
results from Definitions (2
...
□
Corollary (2
...
— Let R be a ring, p an ideal
...
Proof: By (2
...
So the assertion results
from the definition of domain in (2
...
□
Exercise (2
...
— Let R be a domain, and R[X1 ,
...
Let m ≤ n, and set p := ⟨X1 ,
...
Prove p is a prime ideal
...
11)
...
Show p is prime if and only if either p = p′ × R′′ with p′ ⊂ R′ prime or p = R′ × p′′
with p′′ ⊂ R′′ prime
...
12)
...
Assume ⟨x⟩ = ⟨y⟩
...
Definition (2
...
— Let R be a ring
...
Example (2
...
— Let R be a domain
...
10)
...
Moreover, ⟨X, Y ⟩ is maximal if and only if R is a field by (1
...
17)
below
...
15)
...
Proof: Suppose R is a field
...
Since R is a field, a ∈ R×
...
4) yields a = R
...
Take x ̸= 0
...
So ⟨x⟩ = R
...
4)
...
□
Prime Ideals (2
...
16)
...
Prove φ is injective
...
17)
...
Then m is maximal if and
only if R/m is a field
...
9)
...
15)
...
18)
...
Show that pR[X] and pR[X]+⟨X⟩ are prime ideals of R[X], and that if p is maximal,
then so is pR[X] + ⟨X⟩
...
19)
...
Show that every prime p is maximal,
and B/p = F2
...
20)
...
Assume that, given x ∈ R, there is n ≥ 2
with xn = x
...
Example (2
...
— Let k be a field, a1 ,
...
, Xn ] the
polynomial ring in n variables
...
, Xn − an ⟩
...
10); so m is maximal by (2
...
Exercise (2
...
— Prove the following statements or give a counterexample
...
(2) Given two prime ideals, their intersection is prime
...
(4) Given a ring map φ : R → R′ , the operation φ−1 carries maximal ideals of
R′ to maximal ideals of R
...
9), an ideal n′ ⊂ R/a is maximal if and only if κ−1 n′ ⊂ R is maximal
...
23)
...
, Xn ] the polynomial ring,
f ∈ P nonzero
...
(1) Let S ⊂ k have at least d + 1 elements
...
, an ∈ S with f (a1 ,
...
(2) Using the algebraic closure K of k, find a maximal ideal m of P with f ∈
/ m
...
24)
...
Proof: A field is a domain by (2
...
So (2
...
17) yield the result
...
25) (PIDs)
...
Examples include a field k, the polynomial ring k[X] in
one variable, and the ring Z of integers
...
12), p
...
18
...
291]
...
Then ⟨p⟩ is maximal; indeed, if ⟨p⟩ ⫋ ⟨x⟩,
then p = xy for some nonunit y, and so x must be a unit since p is irreducible
...
17) implies that R/⟨p⟩ is a field
...
26)
...
Example (2
...
— Let R be a PID, and p ∈ R a prime
...
Let
P := R[X] be the polynomial ring in one variable
...
Set m := ⟨p, g⟩
...
7) and (1
...
25); hence, m is maximal by (2
...
10
Prime Ideals (2
...
28)
...
Let P := R[X] be the polynomial ring in
one variable, and p a prime ideal of P
...
(2) Assume p is maximal
...
Proof: Assume p ̸= ⟨0⟩
...
Since p is prime, p contains
a prime factor f1′ of f1
...
Assume p ̸= ⟨f1 ⟩
...
Set K := Frac(R)
...
401], [8, Thm
...
15, p
...
So f1 and f2 are
relatively prime in K[X]
...
25) and (2
...
So c = g1 f1 + g2 f2 ∈ R ∩ p
...
But R ∩ p
is prime, and R is a PID; so R ∩ p = ⟨p⟩ where p is prime by (2
...
Set k := R/⟨p⟩
...
25)
...
Then
k[X]/q = P/p by (1
...
9)
...
Hence
q = ⟨g ′ ⟩ where g ′ is prime in k[X] by (2
...
Then q is maximal by (2
...
So p is
maximal by (1
...
Take g ∈ p with image g ′
...
□
Exercise (2
...
— Preserve the setup of (2
...
Let f := a0 X n + · · · + an be a
polynomial of positive degree n
...
Show that ⟨f ⟩ is not maximal
...
30)
...
Proof: Set S := {ideals b | b ⊃ a and b ̸∋ 1}
...
Given a totally ordered subset {bλ } of S, set b := bλ
...
Hence by
Zorn’s Lemma [11, pp
...
880, p
...
□
Corollary (2
...
— Let R be a ring, x ∈ R
...
Proof: By (1
...
Apply (2
...
□
3
...
Closely related to the nilradical is the radical of
a subset
...
In particular, we
study local rings; a local ring has only one maximal ideal, which is then its Jacobson
radical
...
Definition (3
...
— Let R be a ring
...
Proposition (3
...
— Let R be a ring, x ∈ R, and u ∈ R×
...
In particular, the sum of an
element of rad(R) and a unit is a unit
...
Let m be a maximal ideal
...
Since x ∈ m too, also u ∈ m, a contradiction
...
31)
...
Conversely, assume x ∈
/ rad(R)
...
So ⟨x⟩ + m = R
...
Then
u − xy = m ∈ m
...
31), or directly by (1
...
□
Exercise (3
...
— Let R be a ring, a ⊂ rad(R) an ideal, w ∈ R, and w′ ∈ R/a
its residue
...
What if a ̸⊂ rad(R)?
Corollary (3
...
— Let R be a ring, a an ideal, κ : R → R/a the quotient map
...
Then Idem(κ) is injective
...
Then
x3 = e3 − 3e2 e′ + 3ee′2 − e′3 = e − e′ = x
...
But κ(x) = 0; so x ∈ a
...
Hence 1 − x2 is a
unit by (3
...
Thus x = 0
...
□
Definition (3
...
— A ring A is called local if it has exactly one maximal ideal,
and semilocal if it has at least one and at most finitely many
...
6) (Nonunit Criterion)
...
Then
A is local if and only if n is an ideal; if so, then n is the maximal ideal
...
So, if n is an ideal,
then it is a maximal ideal, and the only one
...
Conversely, assume A is local with maximal ideal m
...
31)
...
Thus n is an ideal
...
7)
...
6) if both R′ and
R′′ are nonzero
...
11
12
Radicals (3
...
8)
...
Find its idempotents e
...
9)
...
Prove A is local
...
10)
...
A formal power series in the n variables
∑
X1 ,
...
, in ) with each ij ≥ 0
...
, 0) is called the constant term
...
, Xn ]]
...
, Xn ]] and a := ⟨X1 ,
...
Then
a(i) X1i1 · · · Xnin 7→ a(0)
is a canonical surjective ring map P → R with kernel a; hence, P/a = R
...
Then (1
...
A power series f is a unit if and only if its constant term a(0) is a unit
...
Conversely, if a(0)
2
is a unit, then f = a(0) (1 − g) with g ∈ a
...
Clearly f · f ′ = 1
...
Given a power series f ∈
/ n, its
initial term lies outside m, so is a unit by (2
...
So f itself is a unit
...
Thus (3
...
Example (3
...
— Let k be a ring, and A := k[[X]] the formal power
series ring
∑∞
in one variable
...
These series form a ring k{{X}}
...
∞
Set f := i=−m ai X i
...
Assume k is a field
...
Let a ⊂ A be a
nonzero ideal
...
Then X −m ∈ a
...
Then −m ≥ n
...
Then b ∈ A and f = bX n
...
Thus A is a PID
...
In fact, K = Frac(A) because any nonzero f ∈ K is of the
form f = u/X m where u, X m ∈ A
...
Define φ : A[Y ] → K by φ|A = ι and φ(Y ) : X −1
...
Set
m := Ker(φ)
...
17) and (1
...
So by (2
...
But m ∩ A = 0 as ι is injective
...
But XY − 1 belongs to m, and is
clearly irreducible; hence, XY − 1 = f u with u a/ unit
...
In addition,
/ ⟨X, Y ⟩ is /maximal
...
8), and so (3
...
However, ⟨X, Y ⟩ is not principal, as no nonunit
of A[Y ] divides both X and Y
...
28)
...
12)
...
Set S := {ideals b | b ⊃ a and b ∩ S = ∅}
...
Proof: Clearly, a ∈ S, and S is ∪
partially ordered by inclusion
...
Then b is an upper bound for {bλ } in S
...
20)
13
So by Zorn’s Lemma, S has a maximal element p
...
Take x, y ∈ R − p
...
So there
are p, q ∈ p and a, b ∈ R with p + ax ∈ S and q + by ∈ S
...
But pq + pby + qax ∈ p, so xy ∈
/ p
...
□
Exercise (3
...
— Let φ : R → R′ be a ring map, p an ideal of R
...
Exercise (3
...
— Use Zorn’s lemma to prove that any prime ideal p contains
a prime ideal q that is minimal containing any given subset s ⊂ p
...
15) (Saturated multiplicative subsets)
...
We say S is saturated if, given x, y ∈ R with xy ∈ S, necessarily x, y ∈ S
...
The group of units R×
and the subset of nonzerodivisors S0 := R − z
...
Further, let φ : R → R′ be a ring map, T ⊂ R′ a subset
...
The converse holds if φ is surjective
...
16)
...
Show that S is saturated
multiplicative if and only if R − S is a union of primes
...
17)
...
Define its
saturation to be the subset
S := { x ∈ R | there is y ∈ R with xy ∈ S }
...
(2) Show that R − S is the union U of all the primes
∪ p with p ∩ S = ∅
...
Show R − S = W
...
Given f, g ∈ R, show S f ⊂ S g if and only if ⟨f ⟩ ⊃ ⟨g⟩
...
18)
...
Show S is maximal
in the set S of multiplicative subsets T of R with 0 ∈
/ T if and only if R − S is a
minimal prime — that is, it is a prime containing no smaller prime
...
19) (Prime Avoidance)
...
, pn ideals such that p3 ,
...
If a ̸⊂ pj∪for all j, then there is an x ∈ a such that x ∈
/ pj for all j; or
n
equivalently, if a ⊂ i=1 pi , then a ⊂ pi for some i
...
If n = 1, the assertion is trivial
...
We may assume xi ∈ pi for every i, else we’re done
...
If n ≥ 3, then (x1 · · · xn−1 ) + xn ∈
/ pj for all j as,
if j = n, then xn ∈ pn and pn is prime, and if j < n, then xn ∈
/ pj and xj ∈ pj
...
20)
...
(1) Let P := k[X1 ,
...
Assume the
highest power of any Xi in f is less than d
...
, an ∈ S with f (a1 ,
...
(2) Let V be a k-vector space, and W1 ,
...
Assume r < d
...
31)
∪
Show i Wi ̸= V
...
Show W ∪
⊂ Wi for some i
...
, ar ideals with a ⊂ i ai
...
Exercise (3
...
— Let k be a field, R := k[X, Y ] the polynomial ring in two
variables, m := ⟨X, Y ⟩
...
(3
...
Then the radical of a is the
√ (Nilradical )
...
√√
√
√
Notice √ a = a
...
We call ⟨0⟩ the nilradical, and sometimes
denote it by nil(R)
...
Note that, if xn = 0 with n ≥ 1 and if m is any maximal ideal, then xn ∈ m and
so x ∈ m as m is prime by (2
...
Thus
nil(R) ⊂ rad(R)
(3
...
1)
We call R reduced if nil(R) = ⟨0⟩, that is, if R has no nonzero nilpotents
...
23)
...
In particular, take n = 12
...
24)
...
(1) Assume every ideal not contained in
nil(R) contains a nonzero idempotent
...
(2) Assume R
is Boolean
...
Exercise (3
...
— Let φ : R → R′ be a ring map, b ⊂ R′ a subset
...
√
√
Exercise (3
...
— Let e, e′ ∈ Idem(R)
...
Show e = e′
...
27)
...
Show there are complementary idempotents e1 and e2 with ei ∈ ai
...
28)
...
Assume a ⊂ nil(R)
...
Theorem (3
...
— Let R be a ring, a an ideal
...
(By convention, the empty
intersection is equal to R
...
Set S := {1, x, x2 ,
...
By (3
...
So x ∈
/ p⊃a p
...
∩
√
√
Conversely, take x ∈ a
...
Then x ∈ p
...
□
Exercise (3
...
— Let R be a ring
...
√
Proposition (3
...
— Let R be a ring, a an ideal
...
Radicals (3
...
Then
(
) i j
∑
(x + y)n+m−1 = i+j=m+n−1 n+m−1
xy
...
Thus x + y ∈ a
...
∩
Alternatively,
given any collection of ideals aλ , note that aλ is also an ideal
...
29)
...
32)
...
Assume
generated
...
√
a is finitely
Exercise (3
...
— Let R be a ring, q an ideal, p a finitely generated prime
...
Proposition (3
...
— A ring R is reduced and has only one minimal prime q
if and only if R is a domain
...
Then ⟨0⟩ is equal to the intersection of all the prime ideals p by (3
...
By (3
...
So ⟨0⟩ = q
...
The converse is obvious
...
35)
...
Assume R
∏is reduced and has finitely many
minimal prime ideals p1 ,
...
Prove φ : R → (R/pi ) is injective, and for each
i, there is some (x1 ,
...
Exercise (3
...
— Let R be a ring, X a variable, f := a0 + a1 X + · · · + an X n
and g := b0 + b1 X + · · · + bm X m polynomials with an ̸= 0 and bm ̸= 0
...
, an ⟩ = R
...
, an are nilpotent
...
, an are nilpotent
...
(4) Then f g is primitive if and only if f and g are primitive
...
37)
...
36) to the polynomial ring P := R[X1 ,
...
For (3), reduce to the case of one variable Y via this standard device: take d suitably
i
large, and define φ : P → R[Y ] by φ(Xi ) := Y d
...
38)
...
Show that
rad(R[X]) = nil(R[X]) = nil(R)R[X]
...
39)
...
Set
n
}
∑
m := M ∩ R and A :=
bn X n | bn ∈ a
...
The converse is false
...
Assume X ∈ M
...
Assume M is maximal
...
If a is finitely generated, then aR[[X]] = A
...
16
Radicals (3
...
40)
...
Then
every prime p of R is the contraction of a prime of R[[X]]
...
So by (3
...
In fact, a specific choice
for q is the set of series
an X n with an ∈ p
...
Plainly (R/p)[[X]] is a domain
...
39)(5) shows q may not be equal to pR[[X]]
...
Modules
In Commutative Algebra, it has proven advantageous to expand the study of rings
to include modules
...
We begin the expansion here by discussing residue modules, kernels, and
images
...
We also construct free modules, direct sums,
and direct products, and we describe their UMPs
...
1) (Modules)
...
Recall that an R-module M is an abelian
group, written additively, with a scalar multiplication, R × M → M , written
(x, m) 7→ xm, which is
(1) distributive, x(m + n) = xm + xn and (x + y)m = xm + xm,
(2) associative, x(ym) = (xy)m, and
(3) unitary, 1 · m = m
...
Moreover, a
Z-module is just an abelian group; multiplication is repeated addition
...
1), for any x ∈ R and m ∈ M , we have x · 0 = 0 and 0 · m = 0
...
For example, the ring R is itself an R-module,
and the submodules are just the ideals
...
Similar to (1
...
Given m ∈ M , we call the set of x ∈ R with xm = 0 the annihilator of m,
and denote it Ann(m)
...
Clearly, Ann(m) and Ann(M ) are
ideals
...
2) (Homomorphisms)
...
Recall that a
homomorphism, or R-linear map, is a map α : M → N such that:
α(xm + yn) = x(αm) + y(αn)
...
They are defined as subsets, but are obviously submodules
...
If so, then we
∼ N
...
So α is an isomorphism if and only if there is a set map β : N → M
such that βα = 1M and αβ = 1N , where 1M and 1N are the identity maps, and
then β = α−1
...
The set of homomorphisms α is denoted by HomR (M, N ) or simply Hom(M, N )
...
17
18
Modules (4
...
When α is the identity map 1M , we write
Hom(M, β) for Hom(1M , β); similarly, we write Hom(α, N ) for Hom(α, 1N )
...
3)
...
Consider the set map
ρ : Hom(R, M ) → M
defined by ρ(θ) := θ(1)
...
(4
...
— Let R be a ring, M a module
...
The module of endomorphisms Hom(M, M ) is
also denoted EndR (M )
...
Further, EndR (M ) is a subring of EndZ (M )
...
It is an endomorphism
...
(Thus we may view µR as representing R as a ring of operators on the abelian
group M
...
Conversely, given an abelian group N and a ring map
ν : R → EndZ (N ),
we obtain a module structure on N by setting xn := (νx)(n)
...
We call M faithful if µR : R → EndR (M ) is injective, or Ann(M ) = 0
...
(4
...
— Fix two rings R and R′
...
Let M ′ be an R′ -module
...
In other
words, the R-module structure on M ′ corresponds to the composition
φ
µ
′
R
→ EndZ (M ′ )
...
′
′
Indeed, R is an R -module, so an R-module by restriction of scalars; further,
(xy)z = x(yz) since (φ(x)y)z = φ(x)(yz) by associativity in R′
...
Then R′ has an
R-algebra structure that is compatible with the given R-module structure
...
Then φ(x)z = xz as (x · 1)z = x(1 · z)
...
Hence φ is a ring map,
because µR is one and µR′ is injective by (4
...
Thus R′ is an R-algebra, and
restriction of scalars recovers its given R-module structure
...
Then an R-module M has a compatible
′
R -module structure if and only if aM = 0; if so, then the R′ -structure is unique
...
6), so if and only if aM = 0; as EndZ (M ) may be noncommutative, we must
apply (1
...
Again suppose R′ is an arbitrary R-algebra with structure map φ
...
The subalgebra generated by
Modules (4
...
, xn ∈ R′ is the smallest R-subalgebra that contains them
...
, xn ]
...
, xn ) with
coefficients in R
...
, xn ]
...
, xn ∈ R′ such that R′ = R[x1 ,
...
(4
...
— Let R be a ring, M a module, M ′ ⊂ M a submodule
...
Recall that M/M ′ inherits a module structure, and is called the residue module,
or quotient, of M modulo M ′
...
Clearly κ is surjective, κ is linear, and κ has kernel M ′
...
Note that Ker(α) ⊃ M ′ if and only if α(M ′ ) = 0
...
Further, if β exists, then β is unique as κ is surjective
...
In addition, then β is injective if and only if M ′ = Ker(α)
...
In particular, always
∼ Im(α)
...
6
...
Finally, as we have seen, M/M ′ has the following UMP: κ(M ′ ) = 0, and given
α : M → N such that α(M ′ ) = 0, there is a unique homomorphism β : M/M ′ → N
such that βκα
...
(4
...
— Let R be a ring
...
If so, form α : R → M by x 7→ xm; then
∼ M as Ker(α) = Ann(m); see (4
...
1)
...
Conversely, given any ideal a, the R-module R/a is
cyclic, generated by the coset of 1, and Ann(R/a) = a
...
8) (Noether Isomorphisms)
...
First, assume L ⊂ M ⊂ N
...
Clearly α is surjective, and Ker(α) = M
...
6), α factors through
20
Modules (4
...
8
...
Clearly L + M is a submodule of N
...
Form the composition α′ of the inclusion map L → L + M and the quotient map
L + M → (L + M )/M
...
Hence owing
to (4
...
8
...
6
...
8
...
8
...
(4
...
— Let R be a ring, α : M → N a linear map
...
Note (4
...
Further, (4
...
1) becomes Coim(α) −→
(4
...
— Let R be a ring, Λ a set, M a module
...
Clearly,∑any submodule that contains them
all contains any (finite) linear combination
xλ mλ with xλ ∈ R
...
Thus N is the submodule generated by the mλ
...
Finally, the mλ are said to form a (free) basis of M if they
are free and generate M ; if so, then we say M is free on the mλ
...
We say M is free if it has a free basis
...
32)(2) or (10
...
For example, form the set of restricted vectors
R⊕Λ := {(xλ ) | xλ ∈ R with xλ = 0 for almost all λ}
...
It has a
standard basis, which consists of the vectors eµ whose λth component is the value
Modules (4
...
Clearly the standard basis is free
...
The free module R⊕Λ has the following UMP: given a module M and elements
mλ ∈ M for λ ∈ Λ, there is a unique homomorphism
α : R⊕Λ → M with α(eλ ) = mλ for each λ ∈ Λ;
(
) (∑
) ∑
namely, α (xλ ) α
xλ eλ = xλ mλ
...
(2) α is injective if and only if the mλ are linearly independent
...
Thus M is free of rank ℓ if and only if M ≃ Rℓ
...
11)
...
Then any two x, y in M are not
free; indeed, if x = a/b and y = −c/d, then bcx + ady = 0
...
Also M is not finitely generated
...
, mr /nr ∈ M , let d
be a common multiple of n1 ,
...
Then (1/d)Z contains every linear combination
x1 (m1 /n1 ) + · · · + xℓ (mℓ /nℓ ), but (1/d)Z ̸= M
...
Indeed, let p ∈ Z be any prime not
dividing n1 · · · nr
...
, mr /nr ]
...
12)
...
Let M be the
submodule of Frac(R) generated by 1, x−1 , x−2 ,
...
Prove that x−1 ∈ R, and conclude that M = R
...
13)
...
Theorem (4
...
— Let R be a PID, E a free module, {eλ }λ∈Λ a (free) basis,
and F a submodule
...
Proof: Well
⊕ order Λ
...
For all
µ, set Eµ := λ≤µ Reλ and Fµ := F ∩ Eµ
...
Choose fµ ∈ Fµ with πµ (fµ ) = aµ
...
Say µ∈Λ0 cµ fµ = 0 for some cµ ∈ R
...
Suppose
Λ1 ̸= ∅
...
Let µ∑
1 be the greatest element of Λ1
...
So πµ1 ( cµ fµ ) = cµ1 aµ1
...
But cµ1 ̸= 0 and
aµ1 ̸= 0, a contradiction
...
∪
Note F = λ∈Λ0 Fλ
...
Suppose
λ is
∑
least such that {fµ }µ∈Λλ does not generate Fλ
...
Then πλ (f ) = cλ
...
So cλ = bλ aλ for some bλ ∈ R
...
Then∑
g ∈ Fλ , and πλ (g) = 0
...
Hence g = µ∈Λν bµ fµ for some bµ ∈ R
...
Hence {fµ }µ∈Λλ generates Fλ
...
□
(4
...
— Let R be a ring, Λ a set, Mλ a module
for λ ∈ Λ
...
22
Modules (4
...
The direct sum of the Mλ is the subset of restricted vectors:
⊕
∏
Mλ := {(mλ ) | mλ = 0 for almost all λ} ⊂ Mλ
...
Clearly,
Mλ =
Mλ if Λ is finite
...
, λn }, then⊕ Mλ is also denoted by Mλ1 ⊕ · · · ⊕ Mλn
...
The direct product comes equipped with projections
(
)
∏
πκ :
Mλ → Mκ given by πκ (mλ ) := mκ
...
Often, α is denoted (αλ )
...
(4
...
1)
Clearly, this bijection is an isomorphism of modules
...
It is easy to see that it has this
⊕UMP: given homomorphisms βκ : Mκ → N , there is
a (unique) homomorphism
β
:
Mλ → N satisfying
βικ = βκ for all κ ∈ Λ; namely,
∑
∑
β (mλ ) =
βλ (mλ )
...
In other words, the
ικ induce this bijection of sets:
(⊕
)
∏
∼
Hom
Mλ , N −→
Hom(Mλ , N )
...
15
...
⊕
For example, if Mλ = R for all λ, then ∏Mλ = R⊕Λ by construction
...
15
...
3)
...
16)
...
Endow
∏
⊕
∏
Rλ and
Rλ with componentwise addition and ⊕
multiplication
...
Exercise (4
...
— Let R be a ring, M a module, and M ′ , M ′′ submodules
...
Exercise (4
...
— Let L, M , and N be modules
...
Prove that
M =L⊕N
and
α = ι L , β = πN , σ = ι N , ρ = πL
if and only if the following relations hold:
βα = 0, βσ = 1, ρσ = 0, ρα = 1, and αρ + σβ = 1
...
20)
23
Exercise (4
...
— Let L be a module, ⊕
Λ a nonempty set, Mλ a module for
λ ∈ Λ
...
Exercise
set, Mλ a module for λ ∈ Λ
...
20)
...
Prove a( Mλ ) = aMλ if a is finitely generated
...
Exact Sequences
In the study of modules, the exact sequence plays a central role
...
We introduce diagram
chasing, and prove the Snake Lemma, which is a fundamental result in homological
algebra
...
Finally,
we prove Schanuel’s Lemma, which relates two arbitrary presentations of a module
...
Definition (5
...
— A (finite or infinite) sequence of module homomorphisms
αi−1
α
i
· · · → Mi−1 −−−→ Mi −→
Mi+1 → · · ·
is said to be exact at Mi if Ker(αi ) = Im(αi−1 )
...
α
Example (5
...
— (1) A sequence 0 → L −
→ M is exact if and only if α is
injective
...
Dually — that is, in the analogous situation with all arrows reversed — a seβ
quence M −
→ N → 0 is exact if and only if β is surjective
...
” Dually, a sequence L −
→M −
→ N → 0 is exact if
and only if N = Coker(α) owing to (1) and (4
...
1)
...
3) (Short exact sequences)
...
If so, then the sequence is called short exact, and often we
regard L as a submodule of M , and N as the quotient M/L
...
Often, we identify L with ιL L and N with ιN N
...
4)
...
If every sequence is exact, then so are the two induced sequences
⊕ ′
⊕
⊕ ′′
∏ ′
∏
∏
Mλ →
Mλ →
Mλ and
Mλ → Mλ → Mλ′′
...
Proof: The assertions are immediate from (5
...
15)
...
5)
...
Set
M := M ′ ⊕ M ′′
...
2)(1) and (5
...
4), prove M/N = M ′ /N ⊕ M ′′
...
6)
...
Prove that, if M ′ and M ′′ are finitely generated, then so is M
...
7)
...
Set N := α (N ) and N ′′ := β(N )
...
24
Exact Sequences (5
...
□
(5
...
— We call a linear map ρ : M → M ′ a retraction of another α : M ′ → M if ρα = 1M ′
...
Dually, we call a linear map σ : M ′′ → M a section of another β : M → M ′′ if
βσ = 1M ′′
...
β
α
We say that a 3-term exact sequence M ′ −
→M −
→ M ′′ splits if there is an iso∼ M ′ ⊕ M ′′ with φα = ι ′ and β = π ′′ φ
...
9)
...
Then
the following conditions are equivalent:
(1) The sequence splits
...
(3) There exists a section σ : M ′′ → M of β, and α is injective
...
Then there exists φ : M −→
M
−1
and β = πM ′′ φ
...
Then plainly (2) and (3) hold
...
Set σ ′ := 1M − αρ
...
But ρα = 1M ′ as ρ is a
retraction
...
Hence there exists σ : M ′′ → M with σβ = σ ′ by (5
...
9)
...
Hence β = βαρ + βσβ
...
So β = βσβ
...
Thus 1M ′′ = βσ; that is, (3) holds
...
But βσ = 1M ′′ as (3) holds
...
But α is
injective, as ρ is a retraction of it
...
Thus (4
...
Assume (3)
...
□
Example (5
...
— Let R be a ring, R′ an R-algebra, and M an R′ -module
...
Define α : M → H by α(m)(x) := xm, and ρ : H → M by
ρ(θ) := θ(1)
...
Let β : M → Coker(α)
be the quotient map
...
9) implies that M is a direct summand of H with
α = ιM and ρ = πM
...
11)
...
Let N be
a submodule of M containing M ′ , and set N ′′ := N ∩ M ′′
...
Exercise (5
...
— Criticize the following misstatement of (5
...
Lemma (5
...
— Consider this commutative diagram with exact rows:
α
β
α′
β′
′
′′
M
→M
→0
−→ M
−
−
′
′′
γy
γ y
γ y
0−
→ N ′ −→ N −→ N ′′
It yields the following exact sequence:
φ′
ψ′
Ker(γ ′ ) −
→ Ker(γ) −
→ Ker(γ ′′ ) −
→ Coker(γ ′ ) −→ Coker(γ) −→ Coker(γ ′′ )
...
13
...
(
)
Proof:
Clearly
α restricts to a map φ, because α Ker(γ ′ ) ⊂ Ker(γ) since
(
)
α′ γ ′ Ker(γ ′ ) = 0
...
9), α′ factors through a unique
map φ′ because M ′ goes to 0 in Coker(γ)
...
Thus all the maps in (5
...
1) are defined except for ∂
...
16)
To define ∂, chase an m′′ ∈ Ker(γ ′′ ) through the diagram
...
By commutativity, γ ′′ β(m) = β ′ γ(m)
...
By exactness of the bottom row, there is a unique n′ ∈ N ′ such that
α′ (n′ ) = γ(m)
...
To see ∂ is well defined, choose another m1 ∈ M with β(m1 ) = m′′
...
Since β(m − m1 ) = 0, there
is an m′ ∈ M ′ with α(m′ ) = m − m1
...
So α′ γ ′ (m′ ) = α′ (n′ − n′1 )
...
So n′ and n′1 have the same image in
Coker(γ ′ )
...
Let’s show that (5
...
1) is exact at Ker(γ ′′ )
...
As in the
construction of ∂, take m ∈ M such that β(m) = m′′ and take n′ ∈ N ′ such that
α′ (n′ ) = γ(m)
...
Then the image of n′ in Coker(γ ′ ) is equal
′
′ ′
′
to 0; so there is m′ ∈ M ′ such that γ ′ (m′ ) = n′
...
So
′
′ ′
′
γα(m ) = α (n ) = γ(m)
...
Since β m − α(m ) = m′′ ,
clearly m′′ = ψ(m − α(m′ )); so m′′ ∈ Im(ψ)
...
Conversely, suppose m′′ ∈ Im(ψ)
...
So γ(m) = 0 and
′ ′
α (n ) = 0
...
Thus ∂(m′′ ) = 0, and so Im(ψ) ⊂ Ker(∂)
...
13
...
The other verifications of exactness are similar or easier
...
□
Exercise (5
...
— Referring to (4
...
15) (Five Lemma)
...
Via a chase, prove these two statements:
(1) If γ3 and γ1 are surjective and if γ0 is injective, then γ2 is surjective
...
Exercise (5
...
— Consider this commutative diagram:
0
0
0
y
y
y
′
0 −→ L
−−→
y
L
−−→
y
L′′ −→ 0
y
′
0−
→M
→M
→
−
−
y
y
′′
M
→0
−
y
′
0−
→N
−→
y
N′′ −
→0
y
0
N
−→
y
0
0
(5
...
1)
Exact Sequences (5
...
Applying the Snake
Lemma, prove that the first row is exact if and only if the third is
...
17)
...
Given n ∈ N and m′′ ∈ M ′′ with α′′ (m′′ ) = γ ′ (n),
show that there is m ∈ M such that α(m) = n and γ(m) = m′′
...
18) (Left exactness of Hom)
...
Then it is exact if and only if, for all modules N , the
following induced sequence is exact:
0 → Hom(M ′′ , N ) → Hom(M, N ) → Hom(M ′ , N )
...
18
...
Then it
is exact if and only if, for all modules M , the following induced sequence is exact:
0 → Hom(M, N ′ ) → Hom(M, N ) → Hom(M, N ′′ )
...
2)(2), the exactness of M ′ −
′′
that M = Coker(α)
...
18
...
So (5
...
1) is exact if and only if M ′′ has the
UMP of Coker(α), discussed in (4
...
Thus (1) holds
...
□
Definition (5
...
— A (free) presentation of a module M is an exact sequence
G→F →M →0
with G and F free
...
If M has a finite presentation, then M is said to be finitely presented
...
20)
...
α
Then there is an exact sequence 0 → K → R⊕Λ −
→M → 0 with α(eλ ) = mλ , where
α
→M → 0
...
10)(1), there is a surjection α : R⊕Λ →
→ M with α(eλ ) = mλ
...
Then 0 → K → R⊕Λ → M → 0 is exact by (5
...
Take a set of
generators {kσ }σ∈Σ of K, and repeat the process to obtain a surjection R⊕Σ →
→ K
...
□
Definition (5
...
— A module P is called projective if, given any surjective
linear map β : M →
→ N , every linear map α : P → N lifts to one γ : P → M ;
namely, α = βγ
...
22)
...
Theorem (5
...
— The following conditions on an R-module P are equivalent:
(1) The module P is projective
...
(3) There is a module K such that K ⊕ P is free
...
25)
(4) Every exact sequence N ′ → N → N ′′ induces an exact sequence
Hom(P, N ′ ) → Hom(P, N ) → Hom(P, N ′′ )
...
23
...
Proof: Assume (1)
...
So the sequence splits by (5
...
Thus (2) holds
...
By (5
...
Then (2) implies K ⊕ P ≃ R⊕Λ
...
Assume (3); say K ⊕ P ≃ R⊕Λ
...
Then the induced sequence
∏ ′
∏
∏
Nλ → Nλ → Nλ′′
...
4)
...
15), that sequence is equal to this one:
Hom(R⊕Λ , N ′ ) → Hom(R⊕Λ , N ) → Hom(R⊕Λ , N ′′ )
...
So owing to (4
...
2), the latter sequence is also equal to
Hom(K, N ′ ) ⊕ Hom(P, N ′ ) → Hom(K, N ) ⊕ Hom(P, N ) → Hom(K, N ′′ ) ⊕ Hom(P, N ′′ )
...
23
...
4)
...
β
Assume (4)
...
In other words, (5) holds
...
Then every α ∈ Hom(P, N ) is the image under Hom(P, β) of some
γ ∈ Hom(P, M )
...
Thus (1) holds
...
24)
...
Prove Hom(P, N ) is finitely generated, and is finitely presented if N is
...
25) (Schanuel)
...
To form θ, recall that P ′ is projective
( and
) α is surjective
...
Take θ := 10 π1
...
30)
Then θ has
( 1 −π )
0
1
29
as inverse
...
∼ L ⊕ P ′
...
Finally, take γ := θ ′ θ −1 and β := λ′ λ−1
...
26)
...
Prove M is finitely presented if and only if L is finitely generated
...
27)
...
infinitely many variables
...
] and M := P/⟨X1 , X2 ,
...
β
α
Proposition (5
...
— Let 0 → L −
→M −
→ N → 0 be a short exact sequence
with L finitely generated and M finitely presented
...
Proof: Let R be the ground ring, µ : Rm → M any surjection
...
Then the following diagram is commutative:
0−
→K
→
−
λy
ν
→N
→0
Rm −
−
µy
1N y
β
α
0 −→ L −→ M −→ N −
→0
∼ Ker µ
...
13) yields an isomorphism Ker λ −→
finitely generated by (5
...
So Ker λ is finitely generated
...
Hence
K is finitely generated by (5
...
Thus N is finitely presented by (5
...
□
β
α
Exercise (5
...
— Let 0 → L −
→M −
→ N → 0 be a short exact sequence with
M finitely generated and N finitely presented
...
α
β
Proposition (5
...
— Let 0 → L −
→M −
→ N → 0 be a short exact sequence
with L and N finitely presented
...
Proof: Let R be the ground ring, λ : Rℓ → L and ν : Rn →
→ N any surjections
...
Note Rn is projective by (5
...
Define µ : Rℓ ⊕ Rn → M by µ := γ + δ
...
13) yields an exact sequence
0 → Ker λ → Ker µ → Ker ν → 0,
and implies Coker µ = 0
...
26)
...
6)
...
26)
...
Appendix: Fitting Ideals
(5
...
— Let R be a ring, A := (aij ) an m × n matrix
with aij ∈ R
...
31
...
Let B := (bij ) be an r × r submatrix of A
...
For any
i, expansion yields det(B) = j=1 (−1)i+j bij det(Bij )
...
Thus
R = I0 (A) ⊃ I1 (A) ⊃ · · ·
...
31
...
Then det(U) is a unit, as U V = I yields
det(U ) det(V ) = 1
...
Thus Ir (U) = R for all r ≤ m
...
32)
...
(1) If α
is injective, then n ≤ m
...
Proof: For (1), assume n > m, and let’s show α is not injective
...
Note (5
...
1) yields In (A) = ⟨0⟩ as n > m and
I0 (A) = R
...
Then 0 ≤ r < n
...
If r = 0, set z := (x, 0,
...
Then z ̸= 0 and
α(z) = 0; so α is not injective
...
As x ̸= 0, also x ∈
/ Ann(Ir (A))
...
By renumbering, we may assume that B is the upper left r × r
submatrix of A
...
∑r+1
Let ci be the cofactor of a(r+1)i in det(C); so det(C) = i=1 a(r+1)i ci
...
So xcr+1 ̸= 0
...
, cr+1 , 0,
...
Then z ̸= 0
...
Denote by Ak the kth row of A, by D the matrix obtained
by replacing the (r + 1)st row of C with the first (r + 1) entries of Ak , and by z · Ak
the dot product
...
If k ≤ r, then D has two equal rows; so
z · Ak = 0
...
Thus α(z) = 0
...
Thus (1) holds
...
Thus (2) holds
...
33)
...
Then for all r
(1) Ir (AB) ⊂ Ir (A)Ir (B)
and
(2) Ir (UAV) = Ir (A)
...
Given sequences I := (i1 ,
...
, jr ) with 1 ≤ j1 < · · · < jr ≤ q, set
xi1 j1
...
xi1 n
...
...
...
...
x ir j r
xir 1
...
Set C := AB
...
, ir ) with
30
Appendix: Fitting Ideals (5
...
, kr ) with 1 ≤ k1 < · · · < kr ≤ p, note
( 1
)
r
det(CIK ) = det CIK
,
...
,
AjIr bjr kr
j1 =1
n
∑
jr =1
(
)
det AjI1 ,
...
=
j1 ,
...
, jr ) with
1 ≤ ji ≤ n
...
, AjIr = 0 as two columns are
equal
...
Then J is a permutation σ of H := (h1 ,
...
Denote the sign of σ by (−1)σ
...
, AjIr = (−1)σ det(AIH )
...
Hence
∑
det(CIK ) = H det(AIH ) det(BHK )
...
For (2), note that Ir (W ) = R for W = U, U −1 , V, V −1 by (5
...
So (1) yields
Ir (A) = Ir (U−1 UAVV−1 ) ⊂ Ir (UAV) ⊂ Ir (A)
...
Lemma (5
...
— Let R be a ring, M a module, r an integer, and
α
µ
Rn −
→ Rm −
→M →0
and
β
π
Rq −
→ Rp −
→M →0
presentations
...
Then Im−r (A) = Ip−r (B)
...
Set K := Ker(µ)
...
But Im(α) is generated by the
columns of A
...
So there’s a matrix C such that AC = B
...
Given k, denote by Ik the k × k identity matrix
...
Then, therefore, there is a block triangular matrix V := 0Iqn
such that
Iq
(A|B)V = (A|0mq )
...
So Is (A|B) = Is (A|0mq ) by (5
...
But Is (A|0mq ) = Is (A)
...
Similarly, Is (A|B) = Is (B)
...
Second, assume m = p and that there’s an isomorphism γ : Rm → Rp with
γα
π
πγ = µ
...
Then Rn −−→ Rp −
→ M → 0 is a presentation,
and GA represents γα
...
But G is
invertible
...
33)(2)
...
Third, assume that q = n( + t and)p = m + t for some t ≥ 1 and that β = α ⊕ 1Rt
and π = µ + 0
...
t
(
)
Given an s×s submatrix C of A, set D := 0Cts 0Ist
...
Thus Is (A) ⊂ Is+t (B)
...
If D includes part of the (m + i)th row of B, then D must also include
part of the (n + i)th column, or D would have an all zero row
...
38)
includes part of
( the0(n
)+ i)th column, then D must include part of the (m + i)th
hk
row
...
But det(D) = det(C)
...
But Ih (A) ⊂ Is (A)
by (5
...
2)
...
Thus Is+t (B) ⊂ Is (A)
...
Finally, in general, Schanuel’s Lemma (5
...
(5
...
— Let R be a ring, M a finitely presented module, r an
α
integer
...
It is independent of the choice of presentation by (5
...
By definition, Fr (M ) is finitely generated
...
31
...
(5
...
1)
Exercise (5
...
— Let R be a ring, and a1 ,
...
Set M := (R/⟨a1 ⟩) ⊕ · · · ⊕ (R/⟨am ⟩)
...
Exercise (5
...
— In the setup of (5
...
(1) Show that m is the smallest integer such that Fm (M ) = R
...
Assume
R is a domain
...
Theorem (5
...
— Let R be a PID, M a free module, N
a submodule
...
Then there exists a decomposition
M = M ′ ⊕ M ′′ and elements x1 ,
...
, an ∈ R such that
M ′ = Rx1 ⊕ · · · ⊕ Rxn ,
N = Ra1 x1 ⊕ · · · ⊕ Ran xn ,
⟨a1 ⟩ ⊃ · · · ⊃ ⟨an ⟩ ̸= 0
...
Proof: Let’s prove existence by induction on n
...
So M ′′ = M , and the displayed conditions are trivially satisfied
...
Assume n > 0
...
Then some cλ0 ̸= 0
...
Thus πλ0 (N ) ̸= 0
...
∪Partially order S by inclusion
...
Then b is an ideal
...
Then
b ∈ αλ (N ) for some λ
...
By Zorn’s Lemma, S has a maximal element,
say α1 (N )
...
Given any linear map β : M → R, set b := β(y1 )
...
Write c = da1 + eb for d, e ∈ R, and set γ := dα1 + eβ
...
But γ(y1 ) = c
...
But ⟨a1 ⟩ ⊂ ⟨c⟩
...
40)
33
maximality, ⟨a∑
1 ⟩ = γ(N )
...
Thus β(y1 ) = b ∈ ⟨a1 ⟩
...
Then πλ (y1 ) = cλ∑
...
Set x1 := dλ eλ
...
So α1 (y1 ) = a1 α1 (x1 )
...
So a1 α1 (x1 ) = a1
...
Thus α1 (x1 ) = 1
...
As α1 (x1 ) = 1, clearly Rx1 ∩ M1 = 0
...
Hence (4
...
Further, M1 is free by (4
...
Set N1 := M1 ∩ N
...
So N ⊃ Ra1 x1 ⊕ N1
...
Then α1 (y) = b, so b ∈ ⟨a1 ⟩
...
Thus N = Ra1 x1 ⊕ N1
...
If φ(a) = 0, then aa1 = 0 as α1 (x1 ) = 1,
and so a = 0 as a1 ̸= 0
...
Note N1 ≃ Rm with m ≤ n owing to (4
...
Hence N ≃ Rm+1
...
So (5
...
By induction on n, there exists a decomposition M1 = M1′ ⊕ M ′′ and elements
x2 ,
...
, an ∈ R such that
M1′ = Rx2 ⊕ · · · ⊕ Rxn , N1 = Ra2 x2 ⊕ · · · ⊕ Ran xn , ⟨a2 ⟩ ⊃ · · · ⊃ ⟨an ⟩ ̸= 0
...
Also ⟨a1 ⟩ ⊃ · · · ⊃ ⟨an ⟩ ̸= 0
...
Finally, consider the projection π : M1 → R with π(xj ) = δ2j for j ≤ 2 ≤ n and
π|M ′′ = 0
...
Then ρ(a1 x1 ) = a1
...
By maximality, ρ(N ) = α1 (N )
...
Thus ⟨a2 ⟩ ⊂ ⟨a1 ⟩, as desired
...
Thus M ′ is determined
...
37)(2) with M ′ /N for M , each ai is determined up to unit multiple
...
39)
...
(1) Then M can be generated by m elements if and only if Fm (M ) = A
...
Proof: For (1), assume M can be generated by m elements
...
10)(1)
α
and (5
...
So Fm (M ) = A by (5
...
For the converse, assume also M cannot be generated by m−1 elements
...
Then Fm−1 (M ) = A by (5
...
1)
...
6)
...
33)(2), we may assume a11 (= 1 and
) the other entries in the first row and
0
first column of A are 0
...
Then B defines a presentation As−1 → Am−1 → M → 0
...
Thus Fk (M ) ̸= A for k < m
...
In (2), if M is free of rank m, then there’s a presentation 0 → Am → M → 0;
so Fm (M ) = A and Fm−1 (M ) = ⟨0⟩ by (5
...
Conversely, if Fm (M ) = A, then
α
(1) and (5
...
10)(1) yield a presentation As −
→ Am → M → 0
...
35)
...
□
Proposition (5
...
— Let R be a ring, and M a finitely presented module
...
Set a := Ann(M )
...
34
Appendix: Fitting Ideals (5
...
10)(1) and (5
...
Say α has matrix A
...
35
...
So assume
r ≤ m and set s := m − r + 1
...
Note that this sequence is a presentation
...
Given an (s − 1) × (s − 1) submatrix B of A, enlarge it to an s × s submatrix B′
of (A|xIm ) as follows: say the ith row of A is not involved in B; form the m × s
submatrix B′′ of (A|xIm ) with the same columns as B plus the ith column of xIm
at the end; finally, form B′ as the s × s submatrix of B′′ with the same rows as B
plus the ith row in the appropriate position
...
By constuction,
det(B′ ) ∈ Is (A|xIm )
...
34)
...
Thus (1) holds
...
But
Fm (M ) = R by (5
...
1)
...
For the second inclusion, given any m × m submatrix B of A, say B∑= (bij )
...
Set mi := µ(ei )
...
Let C be the matrix of cofactors of B: the (i, j)th entry of C is (−1)i+j
times the determinant of the matrix obtained by deleting the jth row and the ith
column of B
...
Hence det(B)mi = 0 for all i
...
But Im (A) is generated by all such det(B)
...
Thus (2) holds
...
Direct Limits
Category theory provides the right abstract setting for certain common concepts,
constructions, and proofs
...
We elaborate
on two key special cases of direct limits: coproducts (direct sums) and coequalizers
(cokernels)
...
Further, we prove direct limits are preserved by left adjoints; whence, direct limits
commute with each other, and in particular, with coproducts and coequalizers
...
In fact, many statements are just
concise restatements in more expressive language; they can be understood through
a simple translation of terms
...
(6
...
— A category C is a collection of elements, called objects
...
We write α : A → B or A −
→ B to mean α ∈ HomC (A, B)
...
We say α is an isomorphism with inverse β : B → A if αβ = 1B and βα = 1A
...
Given categories C and C′ , their product C × C′ is the category whose objects
are the pairs (A, A′ ) with A an object of C and A′ an object of C′ and whose maps
are the pairs (α, α′ ) of maps α in C and α′ in C′
...
2) (Functors)
...
Namely, given
categories C and C′ , a (covariant) functor F : C → C′ is a rule that assigns to
each object A of C an object F (A) of C′ and to each map α : A → B of C a map
F (α) : F (A) → F (B) of C′ preserving composition and identity; that is,
(1) F (βα) = F (β)F (α) for maps α : A → B and β : B → C of C, and
(2) F (1A ) = 1F (A) for any object A of C
...
Note that a functor F preserves isomorphisms
...
For example, let R be a ring, M a module
...
A second example is the forgetful functor from
((R-mod)) to ((Sets)); it sends a module to its underlying set and a homomorphism
to its underlying set map
...
Namely, given two functors F, F ′ : C ⇒ C′ , a natural transformation θ : F → F ′ is a collection of maps
θ(A) : F (A) → F ′ (A), one for each object A of C, such that θ(B)F (α) = F ′ (α)θ(A)
35
36
Direct Limits (6
...
We call F and F ′ isomorphic if there are natural
transformations θ : F → F ′ and θ′ : F ′ → F with θ′ θ = 1F and θθ′ = 1F ′
...
For example, fix a module N ; then Hom(•, N ) is a contravariant
functor from ((R-mod)) to ((R-mod))
...
3)
...
2)(1) is equivalent to the
commutativity of the corresponding diagram:
(
)
HomC (B, C) −
→ HomC′ F (B), F (C)
y
y
(6
...
1)
(
)
HomC (A, C) −
→ HomC′ F (A), F (C)
(2) Given γ : C → D, show (6
...
4) (Adjoints)
...
We call (F, F ′ )
an adjoint pair, F the left adjoint of F ′ , and F ′ the right adjoint of F if, for
every pair of objects A ∈ C and A′ ∈ C′ , there is given a natural bijection
HomC′ (F (A), A′ ) ≃ HomC (A, F ′ (A′ ))
...
4
...
Indeed,
let F and G be two left adjoints of F ′
...
To see that θ(A) is natural in A, take a map α : A → B
...
6)
37
HomC′ (F (A), F (A)) ≃ HomC (A, F ′ F (A)) ≃ HomC′ (G(A), F (A))
y
y
y
HomC′ (F (A), F (B)) ≃ HomC (A, F ′ F (B)) ≃ HomC′ (G(A), F (B))
x
x
x
HomC′ (F (B), F (B)) ≃ HomC (B, F ′ F (B)) ≃ HomC′ (G(B), F (B))
Chase after 1F (A) and 1F (B)
...
So
both map to the same image in HomC′ (G(A), F (B))
...
So θ(B)G(α) = F (α)θ(A)
...
Similarly, there is a natural transformation θ′ : F → G
...
But, by naturality, the following diagram is commutative:
HomC′ (F (A), F (A)) ≃ HomC (A, F ′ F (A)) ≃ HomC (G(A), F (A))
y
y
y
HomC′ (F (A), G(A)) ≃ HomC (A, F ′ G(A)) ≃ HomC (G(A), G(A))
Chase after 1F (A)
...
Counterclockwise, its image is 1G(A) , owing to the definition of θ′
...
Similarly, θθ′ = 1F , as required
...
10),
Hom((R-mod)) (R⊕Λ , M ) = Hom((Sets)) (Λ, M )
...
3),
(
)
(
)
Hom((R-alg)) R[X1 ,
...
, Xn }, R′
...
5)
...
Let φA,A′ : HomC′ (F A, A′ ) −→
C
bijection, and set ηA := φA,F A (1F A )
...
We call the natural transformation A 7→ ηA the unit of (F, F ′ )
...
(3) Prove the natural map ηA : A → F ′ F A is universal from A to F ′ ; that is,
given f : A → F ′ A′ , there is a unique map f ′ : F A → A′ with F ′ f ′ ◦ ηA = f
...
Prove the equation in
(2) defines a natural bijection making (F, F ′ ) an adjoint pair, whose unit is η
...
4): the “free module” functor
and the “polynomial ring” functor
...
)
38
Direct Limits (6
...
6) (Direct limits)
...
Assume Λ is small; that is, its
objects form a set
...
(Note: given κ and µ, there may be more than one map κ → µ, and so
more than one transition map αµκ
...
As usual, universality implies that, once equipped with its insertions αµ , the limit
lim Mλ is determined up to unique isomorphism, assuming it exists
...
In any case,
−→
let us use lim Mλ to denote a particular choice
...
We say that C has direct limits if it
−→
has direct limits indexed by every small category Λ
...
Furthermore, whenever the corresponding two
direct limits exist, the maps F (αµ ) : F (Mµ ) → F (lim Mλ ) induce a canonical map
−→
ϕ : lim F (Mλ ) → F (lim Mλ )
...
6
...
At times, given
lim Mλ , we construct lim F (Mλ ) by showing F (lim Mλ ) has the requisite UMP
...
Then, given a natural transformation
from λ 7→ Mλ to λ 7→ Nλ , universality yields unique commutative diagrams
Mµ −
→ lim Mλ
−→
y
y
Nµ −→ lim Nλ
−→
To put it in another way, form the functor category CΛ : its objects are the
functors λ 7→ Mλ from Λ to C; its maps are the natural transformations (they form
a set as Λ is one)
...
−→
In fact, it is just a restatement of the definitions that the “direct limit” functor
lim is the left adjoint of the diagonal functor
−→
∆ : C → CΛ
...
(6
...
— Let C⨿be a category, Λ a set,
⨿ and Mλ an object of C for
each λ ∈ Λ
...
Direct Limits (6
...
Thus, given such a P , there
exists a unique map β :
Mλ → P with βιµ = βµ for all µ ∈ Λ
...
There are no µ in Λ, so no inclusions ιµ : Mµ → B, so no equations
βιµ = βµ to restrict β
...
For instance, suppose C =⨿
((R-mod))
...
For any Λ, the coproduct
Mλ is just the direct sum
Mλ (a convention if
Λ = ∅)
...
Then the empty
set
is
an
initial object
...
Note that the coproduct is a special case of the direct limit
...
Then lim Mλ = Mλ with the insertions equal to the inclusions
...
8) (Coequalizers)
...
Their
coequalizer is defined as the object of C universal among objects P equipped with
a map η : N → P such that ηα = ηα′
...
In particular, the coequalizer of α and 0 is just Coker(α)
...
Take the smallest equivalence relation ∼ on N with
α(m) ∼ α′ (m) for all m ∈ M ; explicitly, n ∼ n′ if there are elements m1 ,
...
Clearly, the coequalizer is the quotient N/∼ equipped with the quotient map
...
Indeed, let Λ be
the category consisting of two objects κ, µ and two nontrivial maps φ, φ′ : κ ⇒ µ
...
Then the coequalizer is lim Mλ
...
9)
...
Their pushout is defined as the object of C universal among objects P equipped
with a pair of maps γ : M → P and δ : N → P such that γα = δβ
...
Show that, in ((Sets)), the pushout is the disjoint union
M ⊔ N modulo the smallest equivalence relation ∼ with m ∼ n if there is ℓ ∈ L
with α(ℓ) = m and β(ℓ) = n
...
Lemma (6
...
— A category C has direct limits if and only if C has coproducts and
coequalizers
...
Proof: If C has direct limits, then C has coproducts and coequalizers because
they are special cases by (6
...
8)
...
Conversely, assume that C has coproducts and coequalizers
...
Let Σ be the set of all transition
maps
⨿
λ
αµλ : M⨿
λ → Mµ
...
Set M :=
σ∈Σ Mσ and
N := λ∈Λ Mλ
...
Correspondingly, there are two maps α, α′ : M → N
...
Given maps βλ : Mλ → P with βµ αµλ = βλ , there is a unique map β : N → P with
βιλ = βλ by the UMP of the coproduct
...
14)
through C by the UMP of the coequalizer
...
−→
Finally, if F : C → C′ preserves coproducts and coequalizers, then F preserves
arbitrary direct limits as F preserves the above construction
...
11)
...
Proof: The assertion follows from (6
...
7) and have coequalizers by (6
...
□
Theorem (6
...
— Every left adjoint F : C → C′ preserves direct limits
...
Given an object P ′ of C′ , consider all possible commutative diagrams
−→
F (ακ
µ)
F (αµ )
F (Mκ ) −−−−→ F (Mµ ) −−−−→ F (lim Mλ )
−→
β ′
β ′
′
y κ
y µ
yβ
1
(6
...
1)
1
P ′ −−−−−−−−→ P ′ −−−−−−−−−→ P ′
where αµκ is any transition map and αµ is the corresponding insertion
...
Say F is the left adjoint of F ′ : C′ → C
...
12
...
−→
□
Proposition (6
...
— Let C be a category, Λ and Σ small categories
...
Then the functor category CΛ does too
...
Then a map σ → τ in
Σ yields a natural transformation from λ 7→ Mσλ to λ 7→ Mτ λ
...
13
...
Hence, with λ fixed, the rule
σ 7→ Mσλ is a functor from Σ to C
...
So λ 7→ limσ∈Σ Mσλ is a functor from Λ to
−→
−→
C
...
Finally, the latter is the direct limit of the functor
−→
τ 7→ (λ 7→ Mτ λ ) from Σ to CΛ , because, given any functor λ 7→ Pλ from Λ to C
equipped with, for τ ∈ Σ, compatible natural transformations from the λ 7→ Mτ λ
to λ 7→ Pλ , there are, for λ ∈ Λ, compatible unique maps limσ∈Σ Mσλ → Pλ
...
17)
41
Theorem (6
...
— Let C be a category with direct limits
indexed by small categories Σ and Λ
...
Then
limσ∈Σ limλ∈Λ Mσ,λ = limλ∈Λ limσ∈Σ Mσ,λ
...
6), the functor limλ∈Λ : CΛ → C is a left adjoint
...
13), the
−→
□
category CΛ has direct limits indexed by Σ
...
12) yields the assertion
...
15)
...
Then functor lim : CΛ → C preserves coproducts and coequalizers
...
7) and (6
...
So (6
...
Exercise (6
...
— Let C be a category, Σ and Λ small categories
...
(2) Assume C has direct limits indexed by Σ and by Λ
...
−→
−→
−→
Exercise (6
...
— Let λ 7→ Mλ and λ 7→ Nλ be two functors from a small
category Λ to ((R-mod)), and {θλ : Mλ → Nλ } a natural transformation
...
−→
−→
−→
Show that the analogous statement for kernels can be false by constructing a
counterexample using the following commutative diagram with exact rows:
µ2
Z −−→
µ2
y
µ2
Z−
→ Z/⟨2⟩ −
→0
µ2
µ2
y
y
Z −−→ Z −
→ Z/⟨2⟩ −
→0
7
...
After making the definitions, we study an
instructive example where the limit is Q
...
We conclude that forming them preserves
exact sequences, and so commutes with forming the module of homomorphisms out
of a fixed finitely presented source
...
1) (Filtered categories)
...
Given a category C, we say a functor λ 7→ Mλ from Λ to C is filtered if Λ is
filtered
...
−→
For example, let Λ be a partially ordered set
...
Regard Λ as a category whose
objects are its elements and whose sets Hom(κ, λ) consist of a single element if
κ ≤ λ, and are empty if not; morphisms can be composed, because the ordering is
transitive
...
Exercise (7
...
— ∪
Let R be a ring, M a module, Λ a set, Mλ a submodule for
each λ ∈ Λ
...
Assume, given λ, µ ∈ Λ, there is ν ∈ Λ such that
Mλ , Mµ ⊂ Mν
...
Prove M = lim Mλ
...
3)
...
Exercise (7
...
— Show that every direct sum of modules is the filtered direct
limit of its finite direct subsums
...
5)
...
Then Mn = Q and Mm , Mn ⊂ Mmn
...
2)
yields Q = lim Mn where Λ is ordered by inclusion of the Mn
...
Thus we may view Λ as ordered by divisibility of the n ∈ Λ
...
Clearly,
βn is a Z-module isomorphism
...
5
...
Thus Q = lim
−→ n
where the transition maps are the µs
...
6)
...
5)
...
Show lim Nn = Q/Z
...
8)
43
Theorem (7
...
— Let Λ be a filtered category, R a ring, and C either ((Sets)) or
((R-mod)) or ((R-alg))
...
Define a relation
∼ on the set-theoretic disjoint union Mλ as follows: m1 ∼ m2 for mi ∈ Mλi if
there are transition maps αµλi : Mλi → Mµ such that αµλ1 m1 = αµλ2 m2
...
Set M :=
Mλ ∼
...
−→
Proof: Clearly ∼ is reflexive and symmetric
...
Given
mi ∈ Mλi for i = 1, 2, 3 with m1 ∼ m2 and m2 ∼ m3 , there are αµλi for i = 1, 2
and ανλi for i = 2, 3 with αµλ1 m1 = αµλ2 m2 and ανλ2 m2 = ανλ3 m3
...
1)(1)
yields αρµ and αρν
...
1)(2) yields ασρ
with ασρ (αρµ αµλ2 ) = ασρ (αρν ανλ2 )
...
Thus m1 ∼ m3
...
Given mi ∈ Mλi for i = 1, 2,
there are αµλi by (7
...
Set
αλ1 m1 + αλ2 m2 := αµ (αµλ1 m1 + αµλ2 m2 )
...
First, consider µ
...
Then (7
...
Possibly, αρµ αµλi ̸= αρν ανλi , but (7
...
In sum, we have this diagram:
λ1
/5 µ
λ2
)/
2, ρ
/ σ
/ τ
ν
Therefore, (ατσ ασρ αρµ )(αµλ1 m1 + αµλ2 m2 ) = (ατσ ασρ αρν )(ανλ1 m1 + ανλ2 m2 )
...
Second, suppose m1 ∼ m′1 ∈ Mλ′1
...
Thus addition is well defined on M
...
Then clearly M is an R-module
...
κ
κ
Finally,
βλ induce
⊔ let βλ : Mλ → N be maps with βλ αλ = βκ for all αλ
...
Suppose m1 ∼ m2 for mi ∈ Mλi ; that is, αµ1 m1 = αµλ2 m2
for some αµλi
...
So there is a unique map
β : M → N with βαλ = βλ for all λ
...
The proof is now complete
...
8)
...
7)
...
−→
(2) Given mi ∈ Mλi for i = 1, 2 such that αλ1 m1 = αλ2 m2 , there are αµλi such
that αµλ1 m1 = αµλ2 m2
...
Then given λ and mλ ∈ Mλ
such that αλ mλ = 0, there is αµλ such that αµλ mλ = 0
...
12)
Proof: The assertions follow directly from
⊔ (7
...
Specifically, (1) holds, since
lim Mλ is a quotient of the disjoint union Mλ
...
Finally, (3) is the special case of (2)
where m1 := mλ and m2 = 0
...
9)
...
−→
(1) Prove that R = 0 if and only if Rλ = 0 for some λ
...
Prove that R is a domain
...
Prove that R is a field
...
10)
...
For each λ, let
Nλ ⊂ Mλ be a submodule, and let N ⊂ M be a submodule
...
Definition (7
...
— Let R be a ring
...
, Xr ]/a for some variables Xi and finitely generated ideal a
...
12)
...
Given N ∈ C, form the map (6
...
1),
θ : lim Hom(N, Mλ ) → Hom(N, lim Mλ )
...
(2) The following conditions are equivalent:
(a) N is finitely presented;
(b) θ is bijective for all filtered categories Λ and all functors λ 7→ Mλ ;
(c) θ is surjective for all directed sets Λ and all λ 7→ Mλ
...
Then
the βµλ are the transition maps of lim Hom(N, Mλ )
...
−→
−→
For (1), let n1 ,
...
Given φ and φ′ in lim Hom(N, Mλ ) with
−→
θ(φ) = θ(φ′ ), note that (7
...
Then θ(φ) = αλ φλ and θ(φ′ ) = αµ φ′µ by
construction of θ
...
So αλ φλ (ni ) = αµ φ′µ (ni ) for all i
...
8)(2) yields λi and αλλi and αλµi such that αλλi φλ (ni ) = αλµi φ′µ (ni ) for all i
...
Indeed, given νi−1 and ανλi−1 and ανµi−1 , by
ν
(7
...
By (7
...
Set ανλi := ανρii αρλii αλλi
i
i
and ανµi := ανρii αρλii αλµi
...
Set ν := νr
...
Hence ανλ φλ = ανµ φ′µ
...
Similarly, φ′ = βν (ανµ φ′µ )
...
Thus θ is injective
...
Thus (1) holds
...
Assume (a)
...
, n′s
...
Then there are homogeneous linear
polynomials fj with fj (e1 ,
...
So fj (n1 ,
...
Filtered Direct Limits (7
...
Repeated use of (7
...
1)(1) yields λ and mλi ∈ Mλ with αλ mλi = mi for all i
...
, mλr )) = fj (m1 ,
...
, nr )) = 0
...
8)(2) and (7
...
, mλr )) = 0
...
10) and (4
...
Set ψ := βµ (φµ )
...
Hence θ(ψ)(ni ) = mi := φ(ni ) for all i
...
Thus θ is surjective
...
Thus (b) holds
...
Finally, assume (c)
...
Then N = lim Nλ by (7
...
However, θ is surjective
...
So (7
...
Hence αλ ψλ = θ(ψ)
...
So αλ is surjective
...
So Nλ = N
...
Say
n1 ,
...
Set F := Rr and let ei be the ith standard basis vector
...
Define κ : F → N by κ(ei ) := ni for all i
...
Then F/N ′ −→
Let’s show N ′ is finitely generated
...
Then
′
N = lim Nλ′ by (7
...
Set Nλ := F/Nλ′
...
17)
...
Since θ is surjective, there is ψ ∈ Hom(N, Nλ )
with θ(ψ) = 1N
...
8)(1) yields λ and ψλ ∈ Hom(N, Nλ ) with βλ (ψλ ) = ψ
...
So αλ ψλ = 1N
...
Then αµ ψµ = αλ ψλ = 1N for all µ
...
For all µ and i, let nµi be the image in Nµ of ei
...
Hence repeated use of (7
...
1)(1) yields µ such that
αµλ nλi = αµλ (ψλ αλ nλi ) for all i
...
But the nµi generate Nµ
for all i
...
So αµ : Nµ → N is an isomorphism
...
Thus N ′ is finitely generated
...
In the case C = ((R-alg)), replace F by a polynomial ring R[X1 ,
...
With these replacements, the above proof shows (a) implies (b)
...
The rest of the proof works after we
replace F by a polynomial ring, the ei by the variables, N ′ by the appropriate ideal,
and the Nλ′ by the finitely generated subideals
...
13) (Finite presentations)
...
The proof of (7
...
, Xr ]/a of R′ ,
where R[X1 ,
...
Similarly, for a finitely presented module M , that proof gives another
solution to (5
...
Theorem (7
...
— Let R be a ring, Λ a
filtered category
...
18)
objects are the 3-term exact sequences, and its maps are the commutative diagrams
L
−→
y
M
−→
y
N
y
L′ −
→ M′ −
→ N′
βλ
γλ
Then, for any functor λ 7→ (Lλ −−→ Mλ −→ Nλ ) from Λ to C, the induced sequence
β
γ
lim Lλ −
→ lim Mλ −
→ lim Nλ is exact
...
Then given ℓ ∈ lim Lλ , there is ℓλ ∈ Lλ with αλ ℓλ = ℓ
−→
by (7
...
By hypothesis, γλ βλ ℓλ = 0; so γβℓ = 0
...
For the opposite inclusion, take m ∈ lim Mλ with γm = 0
...
8)(1), there is
−→
mλ ∈ Mλ with αλ mλ = m
...
So by (7
...
So γµ αµλ mλ = 0 by commutativity
...
Apply αµ to get
βαµ ℓµ = αµ βµ ℓµ = αµ αµλ mλ = m
...
So Ker(γ) = Im(β) as asserted
...
15)
...
Assume αµλ aλ ⊂ aµ for each transition map αµλ
...
−→
If each aλ is prime, show a is prime
...
Exercise (7
...
— Let M := lim Mλ be a filtered direct limit of modules, with
−→
transition maps αµλ : Mλ → Mµ and insertions αλ : Mλ → M
...
Assume αµλ Nλ ⊂ Nµ for all αµλ
...
−→
Exercise (7
...
— Let R := lim Rλ be a filtered direct limit of rings
...
−→
Filtered Direct Limits (7
...
18)
...
Assume
−→
each ring Rλ is local, say with maximal ideal mλ , and assume each transition map
αµλ : Rλ → Rµ is local
...
Prove that R is local with maximal ideal
−→
m and that each insertion αλ : Rλ → R is local
...
19) (Hom and direct limits again)
...
Here is an alternative proof
that the map θ(N ) of (6
...
1) is injective if N is finitely generated and bijective if
N is finitely presented
...
3)
...
It induces the following commutative diagram:
0−
→ lim Hom(N, Mλ ) −
→ lim Hom(Rn , Mλ ) −
→ lim Hom(R⊕Σ , Mλ )
−→
−→
−→
θ(N )y
θ(Rn )y≃
θ(R⊕Σ )y
0−
→ Hom(N, lim Mλ ) −
→ Hom(Rn , lim Mλ ) −
→ Hom(R⊕Σ , lim Mλ )
−→
−→
−→
The rows are exact owing to (5
...
14), the
exactness of filtered direct limits
...
15),
and direct limit does so by (6
...
7); hence, θ(Rn ) is bijective, and θ(R⊕Σ )
is bijective if Σ is finite
...
Exercise (7
...
— Let Λ and Λ′ be small categories, C : Λ′ → Λ a functor
...
Assume C is cofinal; that is,
(1) given λ ∈ Λ, there is a map λ → Cλ′ for some λ′ ∈ Λ′ , and
(2) given ψ, φ : λ ⇒ Cλ′ , there is χ : λ′ → λ′1 with (Cχ)ψ = (Cχ)φ
...
Show that
limλ′ ∈Λ′ MCλ′ = limλ∈Λ Mλ ;
−→
−→
more precisely, show that the right side has the UMP characterizing the left
...
21)
...
8
...
We construct the product, and establish various properties: bifunctoriality,
commutativity, associativity, cancellation, and most importantly, adjoint associativity; the latter relates the product to the module of homomorphisms
...
We prove Watt’s Theorem; it
characterizes “tensor-product” functors as those linear functors that commute with
direct sums and cokernels
...
(8
...
— Let R a ring, and M , N , P modules
...
Denote the set of all these maps by BilR (M, N ; P )
...
(8
...
— Let R be a ring, and M , N modules
...
2
...
The above construction yields a canonical bilinear map
β : M × N → M ⊗ N
...
Theorem (8
...
— Let R be a ring, M , N modules
...
θ : HomR (M ⊗R N, P ) −→
R
(8
...
1)
Proof: Note that, if we follow any bilinear map with any linear map, then the
result is bilinear; hence, θ is well defined
...
Further, θ is injective since M ⊗R N is generated by the image of β
...
10) it extends to a map α′ : R⊕(M ×N ) → P ,
and α′ carries all the elements in (8
...
1) to 0; hence, α′ factors through β
...
□
Exercise (8
...
— Let R be a ring, R′ an R- algebra, and M an R′ -module
...
Define α : M → M ′ by αm := 1 ⊗ m, and ρ : M ′ → M by
ρ(x ⊗ m) := xm
...
48
Tensor Products (8
...
5) (Bifunctoriality)
...
Then there is a canonical commutative diagram:
α×α′
M
× N −−−→ M ′
× N′
β
′
y
yβ
α⊗α′
M ⊗ N −−−→ M ′ ⊗ N ′
Indeed, β ′ ◦ (α × α′ ) is clearly bilinear; so the UMP (8
...
Thus • ⊗ N
and M ⊗• are commuting linear functors — that is, linear on maps, compare (9
...
Proposition (8
...
— Let R be a ring, M and N modules
...
(commutative law)
(2) Then multiplication of R on M induces an isomorphism
R ⊗R M = M
...
2
...
Thus (1) holds
...
Clearly β is bilinear
...
Given a bilinear map α : R × M → P , define γ : M → P by
γ(m) := α(1, m)
...
Also, α = γβ as
α(x, m) = xα(1, m) = α(1, xm) = γ(xm) = γβ(x, m)
...
Thus b has the UMP, so (2) holds
...
7)
...
Set K := Frac(R)
...
(8
...
— Let R and R′ be rings
...
At times, we think of N as a left Rmodule, with multiplication xn, and as a right R′ -module, with multiplication nx′
...
A
(R, R′ )-homomorphism of bimodules is a map that is both R-linear and R′ -linear
...
Then M ⊗R N
is an (R, R′ )-bimodule with R-structure as usual and with R′ -structure defined
by x′ (m ⊗ n) := m ⊗ (x′ n) for all x′ ∈ R′ , all m ∈ M , and all n ∈ N
...
5) with α := µx and α′ := µx′
...
Then R′ is an (R, R′ )-bimodule
...
It is said to be obtained by extension of scalars
...
Further, given an R′ -module
P , it is easy to check that HomR′ (N, P ) is an (R, R′ )-bimodule under sourcewise
multiplication by elements of R
...
9)
...
Show there is a canonical R-linear map τ : M ⊗R N → M ⊗R′ N
...
Show K is equal to
Ker(τ ), and τ is surjective
...
50
Tensor Products (8
...
10)
...
Then there are two canonical (R, R′ )-isomorphisms:
M ⊗R (N ⊗R′ P ) = (M ⊗R N ) ⊗R′ P,
(
)
HomR′ (M ⊗R N, P ) = HomR M, HomR′ (N, P )
...
For each (R, R′ )-bimodule Q, call a map τ : M × N × P → Q trilinear if it is
R-bilinear in M × N and R′ -bilinear in N × P
...
It is, clearly, an (R, R′ )-bimodule
...
Thus
(
)
Tril(R,R′ ) (M, N, P ; Q) = Hom M ⊗R (N ⊗R′ P ), Q
...
Hence each of M ⊗R (N ⊗R′ P ) and (M ⊗R N ) ⊗R′ P is the universal target of a
trilinear map with source M × N × P
...
To establish the isomorphism of adjoint associativity, define a map
(
)
α : HomR′ (M ⊗R N, P ) → HomR M, HomR′ (N, P ) by
(
)
α(γ)(m) (n) := γ(m ⊗ n)
...
First, α(γ)(m) is R′ -linear, because given x′ ∈ R′ ,
γ(m ⊗ (x′ n)) = γ(x′ (m ⊗ n)) = x′ γ(m ⊗ n)
since γ is R′ -linear
...
(
)
Thus α(γ) ∈ HomR M, HomR′ (N, P )
...
)
To obtain an inverse to α, given η ∈ HomR M, HomR′ (N, P ) , define a map
ζ : M × N → P by ζ(m, n) := (η(m))(n)
...
Given x ∈ R, clearly (η(xm))(n) = (η(m))(xn); so
δ((xm) ⊗ n) = δ(m ⊗ (xn))
...
9) with Z for R and with R for R′
...
Finally, it is easy to verify that α(β(η)) = η and β(α(γ)) = γ, as desired
...
11)
...
First, let M be an
R-module, and P an R′ -module
...
(cancellation law)
(left adjoint)
′
Instead, let M be an R -module, and P an R-module
...
(right adjoint)
′
In other words, • ⊗R R is the left adjoint of restriction of scalars from R′ to R,
and HomR (R′ , •) is its right adjoint
...
3) and the unitary law
...
18)
51
Exercise (8
...
— In the setup of (8
...
Corollary (8
...
— Let R, R′ be rings, N a bimodule
...
Proof: By adjoint associativity, •⊗R N is the left adjoint of HomR′ (N, •)
...
12) and (6
...
□
Example (8
...
— Tensor product does not preserve kernels, nor even injections
...
Tensor it with N := Z/⟨2⟩, obtaining
µ2 : N → N
...
Exercise (8
...
— Let M and N be nonzero k-vector spaces
...
Exercise (8
...
— Let R be a ring, a and b ideals, and M a module
...
13) to show that (R/a) ⊗ M = M/aM
...
Exercise (8
...
— Show Z/⟨m⟩ ⊗Z Z/⟨n⟩ = 0 if m and n are relatively prime
...
18) (Watts)
...
Then there is a natural transformation θ(•) : • ⊗F (R) → F (•) with θ(R) = 1, and
θ(•) is an isomorphism if and only if F preserves direct sums and cokernels
...
But Hom(R, M ) = M by (4
...
Set
N := F (R)
...
Explicitly, θ(M )(m ⊗ n) = F (ρ)(n) where ρ : R → M is defined by ρ(1) = m
...
Either way, it’s not hard to see θ(M ) is natural in M and θ(R) = 1
...
13)
...
20)
...
18
...
If F preserves direct sums, then θ(R⊕Λ ) = 1N ⊕Λ
and θ(R⊕Σ ) = 1N ⊕Σ ; in fact, given any natural transformation θ : T → U , let’s
show that, if T and U preserve direct sums, then so does θ
...
But the UMP of direct sum says that,
given any N , a map
T⊕
(Mλ ) → N
⊕ is determined by its compositions with the
inclusions T (ιλ )
...
52
Tensor Products (8
...
Since • ⊗ N does too, the rows of (8
...
1) are
exact by (5
...
Therefore, θ(M ) is an isomorphism
...
19)
...
Show
that F always preserves finite direct sums
...
(8
...
— Let R be a ring, M a module, and form the diagram
δ
σ
M
M −−
→ M ⊕ M −−M
→M
where δM := (1M , 1M ) and σM := 1M + 1M
...
Then
σN (α ⊕ β)δM = α + β,
(8
...
1)
because, for any m ∈ M , we have
(σN (α ⊕ β)δM )(m) = σN (α ⊕ β)(m, m) = σN (α(m), β(m)) = α(m) + β(m)
...
Then F (α ⊕ β) = F (α) ⊕ F (β)
...
Hence F (α + β) = F (α) + F (β) by (8
...
1)
...
Conversely, every additive functor preserves finite direct sums owing to (8
...
However, not every additive functor is R-linear
...
Define F (M ) to be M , but with the scalar product of x ∈ C and m ∈ M to be xm
where x is the conjugate
...
Then F is additive, but not linear
...
21) (Equational Criterion for Vanishing)
...
Then any t ∈ M ⊗ N can be
written as a finite sum t = mλ ⊗ nλ with mλ ∈ M
...
σ xλσ mσ = mλ for all λ and
Proof: By (8
...
∑
It follows that t can be written as a finite sum t = mλ ⊗ nλ with mλ ∈ M
...
Then
)
)
∑ (
∑
∑
∑ (∑
m λ ⊗ nλ = λ
σ xλσ mσ ⊗ nλ =
σ mσ ⊗
λ xλσ nλ = 0
...
20), there is a presentation R⊕Σ −
→ R⊕Λ −
→ N → 0 with
α(eλ ) = nλ for all λ where {eλ } is the standard basis of R⊕Λ
...
13) the
following sequence is exact:
1⊗β
1⊗α
M ⊗ R⊕Σ −−−→ M ⊗ R⊕Λ −−−→ M ⊗ N → 0
...
So the exactness implies there is an element
∑
s ∈ M ⊗ R⊕Σ such that∑(1 ⊗ β)(s) =
mλ ⊗ eλ
...
Write β(eσ ) = λ xλσ eλ
...
∑
Since the eλ are independent, mλ = σ xλσ mσ , as asserted
...
26)
53
(8
...
— Let R be a ring, S and T algebras with structure maps
σ : R → S and τ : R → T
...
Now, define
S × T × S × T → U by (s, t, s′ , t′ ) 7→ ss′ ⊗ tt′
...
So it induces a bilinear map
µ: U × U → U
µ(s ⊗ t, s′ ⊗ t′ )(ss′ ⊗ tt′ )
...
In fact, U is an R-algebra
with structure map ω given by ω(r) := σ(r) ⊗ 1 = 1 ⊗ τ (r), called the tensor
product of S and T over R
...
Clearly ιS is an R-algebra homomorphism
...
Given an R-algebra V , define a map
γ : Hom((R-alg)) (S ⊗R T, V ) → Hom((R-alg)) (S, V ) × Hom((R-alg)) (T, V )
...
Conversely, given R-algebra homomorphisms θ : S → V
and ζ : T → V , define η : S × T → V by η(s, t) := θ(s) · ζ(t)
...
It is easy to see that the map
(θ, ζ) 7→ ψ is an inverse to γ
...
In other words, S ⊗R T is the coproduct of S and T in ((R-alg)):
: S
ιS
σ
R
τ
$
ιT
θ
$
S ⊗R T
:
ψ
*/
4 V
ζ
T
Example (8
...
— Let R be a ring, S an algebra, and X1 ,
...
Then there is a canonical S-algebra isomorphism
S ⊗R R[X1 ,
...
, Xn ]
...
, xn of T ,
there is an R-algebra homomorphism R[X1 ,
...
3)
...
22),
there is a unique S-algebra homomorphism S ⊗R R[X1 ,
...
Thus both
S ⊗R R[X1 ,
...
Xn ] possess the same UMP
...
, Ym , we obtain
R[X1 ,
...
, Ym ] = R[X1 ,
...
, Ym ]
...
24)
...
Let M [X] be
the
in X with coefficients in M , that is, expressions of the form
∑n set of polynomials
i
i=0 mi X with mi ∈ M
...
Exercise (8
...
— Let R be a ring, (Rσ′ )σ∈Σ a family of algebras
...
Prove that the
assignment J 7→ RJ′ extends to a filtered direct system and that lim RJ′ exists and
−→
is the coproduct of the family (Rσ′ )σ∈Σ
...
26)
...
Set k := Q(ω) and K := k[ 3 2]
...
9
...
First, we
study exact functors in general
...
Notably, we prove Lazard’s Theorem, which characterizes the flat modules as the
filtered direct limits of free modules of finite rank
...
Lemma (9
...
— Let R be a ring, α : M → N a homomorphism of modules
...
1
...
Proof: If the equations hold, then the second short sequence is exact owing to
∼ Im(α) by (4
...
the definitions, and the first is exact since Coim(α) −→
Conversely, given the commutative diagram with two short exact sequences, α′′
is injective
...
So M ′ = Ker(α)
...
So N ′ = Im(α)
...
Thus the equations hold
...
2) (Exact Functors)
...
Assume F is R-linear; that is, the associated map
HomR (M, N ) → HomR′ (F M, F N )
(9
...
1)
is R-linear
...
But M = 0 if
and only if 1M = 0
...
Thus if M = 0, then F M = 0
...
2
...
Call F exact if it preserves exact sequences
...
23)
...
When F is contravariant, call F left
exact if it takes cokernels to kernels
...
Call F right exact if it preserves cokernels
...
Proposition (9
...
— Let R be a ring, R′ an algebra, F an R-linear functor from
((R-mod)) to ((R′ -mod))
...
(2) F preserves short exact sequences
...
(4) F preserves cokernels and injections
...
Proof: Trivially, (1) implies (2)
...
2), clearly (1) yields (3) and (4)
...
Let 0 → M ′ → M → M ′′ → 0 be a short exact sequence
...
Thus (2) holds
...
54
Flatness (9
...
Given α : M → N , form the diagram (9
...
1)
...
Hence (9
...
α
β
Finally, assume (5)
...
Now, (5) yields Ker(F (β)) = F (Ker(β)) and Im(F (α)) = F (Im(α))
...
Thus (1) holds
...
4)
...
Assume F is exact
...
(2) An R-module M vanishes if F M does
...
α
β
Fα
Fβ
(4) A sequence M ′ −
→M −
→ M ′′ is exact if F M ′ −−→ F M −−→ F M ′′ is
...
5) (Flatness)
...
It is equivalent by (9
...
13)
...
We say an R-algebra is flat or faithfully flat if it is so as an R-module
...
6)
...
Further, M is faithfully flat if every Mλ is flat and at least one is faithfully flat
...
Then (8
...
18), taking T (M ) := M ⊗ N ′ and U (M ) := M ⊗ N
...
4)
...
⊕
Further, M ⊗ N = (Mλ ⊗ N ) by (8
...
So if M ⊗ N = 0, then Mλ ⊗ N = 0
for all λ
...
□
Proposition (9
...
— A nonzero free module is faithfully flat
...
Proof: It’s easy to extend the unitary law to maps; in other words, R ⊗ • = 1
...
Thus a nonzero free module is faithfully flat by (9
...
Every projective module is a direct summand of a free module by (5
...
6)
...
8)
...
Example (9
...
— In (9
...
Its converse needn’t
hold
...
By (9
...
But neither Ri is so, as R1 ⊗R2 = R1 ⊗(R/R1 ) = R1 /R12 = 0
...
10)
...
Show that M ⊗R N
is flat
...
11)
...
Show that
M ⊗R R′ is flat over R′
...
12)
...
Show
that M is flat over R
...
19)
Exercise (9
...
— Let R be a ring, R′ an algebra, R′′ an R′ -algebra, and M an
R′′ -module
...
Prove that
R′ is flat over R
...
14)
...
Assume R/a is flat
...
Exercise (9
...
— Let R be a ring, R′ a flat algebra
...
α
For every R-module M , the map M −
→ M ⊗R R′ by αm = m ⊗ 1 is injective
...
Every prime p of R is the contraction of some prime q of R′ , or p = φ−1 q
...
Every nonzero R-module M extends to a nonzero module, or M ⊗R R′ ̸= 0
...
16)
...
Assume M ′′ is flat
...
(2) Then M is flat if and only if M ′ is flat
...
20), there is an exact sequence 0 → K → R⊕Λ → N → 0
...
5), as M ′′ and R⊕Λ are flat by hypothesis
and by (9
...
So the rows and columns are exact, as tensor product is right exact
...
13), implies γ is injective
...
To prove (2), take an injection N ′ → N , and form this commutative diagram:
′
0−
→ M′
⊗ N′ −
→M⊗
→ M ′′
⊗ N′ −
→0
N −
′
′′
αy
αy
α y
0 −→ M ′ ⊗ N −→ M ⊗ N −→ M ′′ ⊗ N −→ 0
Its rows are exact by (1)
...
Then α is injective
...
Thus M ′ is flat
...
Then α′ is injective
...
Hence α is injective by the Snake lemma
...
Thus (2) holds
...
17)
...
Assume N ⊗ M ′ −−−→ N ⊗ M is injective for all N
...
Flatness (9
...
18)
...
Proposition (9
...
— A filtered direct limit of flat modules lim Mλ is flat
...
Then Mλ ⊗ β is injective for each λ since
Mλ is flat
...
14)
...
13)
...
□
−→
−→
Proposition (9
...
— Let R and R′ be rings, M an R-module, N an (R, R′ )bimodule, and P an R′ -module
...
(9
...
1)
Assume P is flat
...
Proof: The map θ exists by Watts’s Theorem, (8
...
Explicitly, θ(φ ⊗ p)(m) = φ(m) ⊗ p
...
So θ is bijective if M = Rn for any n, as
HomR (•, Q) preserves finite direct sums for any Q by (4
...
Assume that M is finitely generated
...
20), we obtain a presentation
R⊕Σ → Rn → M → 0, with Σ finite if P is finitely presented
...
The
right-hand vertical map is bijective if Σ is finite
...
□
Exercise (9
...
— Let R be a ring, R′ an algebra, M and N modules
...
Assume R′ is flat over R
...
Definition (9
...
— Let R be a ring, M a module
...
Proposition (9
...
— Let R be a ring, M a module, and (Rm , α) 7→ Rm the
forgetful functor from ΛM to ((R-mod))
...
−→
M
Proof: By the UMP, the α : Rm → M induce a map ζ : lim Rm → M
...
First, ζ is surjective, because each x ∈ M is in the image of
(R, αx ) where αx (r) := rx
...
By construction, (Rm ,α) Rm → lim Rm is surjec−⊕
→
tive; see the proof of (6
...
So y is in the image of some finite sum (Rmi , αi ) Rmi
...
Then
R = Rm
...
Then y is the image of some
y ′ ∈ Rm under the insertion ιm : Rm → lim Rm
...
So α(y ′ ) = 0
...
They
58
Flatness (9
...
So, by definition of direct limit, they have the same compositions
with the insertion ιm
...
Thus ζ is injective, so bijective
...
24) (Lazard)
...
Then the following
conditions are equivalent:
(1) M is flat
...
20
...
(3) Given a finitely presented module P and a map β : P → M , there exists a
γ
α
factorization β : P −
→ Rn −
→ M;
(4) Given an α : Rm → M and a k ∈ Ker(α), there exists a factorization
φ
α : Rm −
→ Rn → M such that φ(k) = 0
...
, kr ∈ Ker(α) there exists a factorization
φ
α : Rm −
→ Rn → M such that φ(ki ) = 0 for i = 1,
...
ρ
α
(6) Given Rr −
→ Rm −
→ M such that αρ = 0, there exists a factorization
m φ
n
α: R −
→ R → M such that φρ = 0
...
(8) M is a filtered direct limit of free modules of finite rank
...
Then (9
...
Assume (2)
...
There are γ1 ,
...
, xnn ∈ M
with β(p) =
γi (p)xi by (2)
...
, γn ), and let α : R → M
be given by α(r1 ,
...
Then β = αγ, just as (3) requires
...
Set P := Rm /Rk, and let κ : Rm → P denote
the quotient map
...
By (3), there is a factorization β : P −
→ Rn → M
...
Then
φ
β : Rm −
→ Rn → M is a factorization of β and φ(k) = 0
...
Set m0 := m and α0 = α
...
, r
such that φi · · · φ1 (ki ) = 0
...
Then (5) holds
...
Let e1 ,
...
Then α(ki ) = 0
...
Then φρ = 0, as required by (6)
...
Given (Rm1 , α1 ) and (Rm2 , α2 ) in ΛM , set m := m1 + m2 and
α := α1 + α2
...
Thus the first
condition of (7
...
Given σ, τ : (Rr , ω) ⇒ (Rm , α) in ΛM , set ρ := σ − τ
...
So (6)
φ
yields a factorization α : Rm −
→ Rn → M with φρ = 0
...
Hence the second condition of (7
...
Thus (7) holds
...
23)
...
Say M = lim Mλ with the Mλ free
...
5), and
−→
□
a filtered direct limit of flat modules is flat by (9
...
Thus M is flat
Exercise (9
...
— Prove that
∑ the Condition
(9
...
28)
xi ∈ R and yi ∈ M , there are xij ∈ R and yj′ ∈ M such that
∑
∑
′
j xij yj = yi for all i and
i xij xi = 0 for all j
...
25
...
26) (Ideal Criterion for Flatness)
...
a ⊗ N −→
∼ N with a⊗x 7→ ax
...
6)(2) implies R ⊗N −→
∼ aN
...
25)
...
, xn ⟩
...
21) yields (9
...
1)
...
□
Example (9
...
— Let R be a domain, and set K := Frac(R)
...
Indeed, (8
...
So K is flat by (9
...
Suppose K is projective
...
23)
...
So there is an x ∈ K with α(x) ̸= 0
...
Take any nonzero b ∈ R
...
Since R is a domain,
b · α(x/ab) = 1
...
Thus R is a field
...
3) yields R = K
...
28)
...
Prove (1) if M is flat, then for
x ∈ R and m ∈ M with xm = 0, necessarily m ∈ Ann(x)M , and (2) the converse
holds if R is a Principal Ideal Ring (PIR); that is, every ideal a is principal
...
Cayley–Hamilton Theorem
The Cayley–Hamilton Theorem says that a matrix satisfies its own characteristic
polynomial
...
” Using the Trick, we obtain various results, including the uniqueness of the
rank of a finitely generated free module
...
Then we turn to the important
notions of integral dependence and module finiteness for an algebra
...
We end
with a discussion of integral extensions and normal rings
...
1) (Cayley–Hamilton Theorem)
...
Let In be the n × n identity matrix, and T a variable
...
Let a be an ideal
...
The Cayley–Hamilton Theorem asserts that, in the ring of matrices,
pM (M) = 0
...
2) below; indeed, take M : Rn , take m1 ,
...
Conversely, given the setup of (10
...
Then φα = αΦ
...
Hence, if p(Φ) = 0, then p(φ) = 0 as α is surjective
...
2) are equivalent
...
2) (Determinant Trick)
...
, mn , and φ : M → M an endomorphism
...
Then pM (φ) = 0 in End(M )
...
Let ∆ stand for the matrix (δij φ − µaij ) with entries in the commutative subring
R[φ] of End(M ), and X for the column vector (mj )
...
Multiply
on the left by the matrix of cofactors Γ of ∆: the (i, j)th entry of Γ is (−1)i+j
times the determinant of the matrix obtained by deleting the jth row and the ith
column of ∆
...
But Γ∆ = det(∆)In
...
Hence det(∆) = 0
...
Thus pM (φ) = 0
...
3)
...
Then
M = aM if and only if there exists a ∈ a such that (1 + a)M = 0
...
Say m1 ,
...
Set M := (aij )
...
Set
a := a1 + · · · + an ∈ a
...
2) with φ := 1M
...
So M ⊂ aM ⊂ M
...
□
60
Cayley–Hamilton Theorem (10
...
4)
...
If φ is surjective, then φ is an isomorphism
...
By the UMP of
P , there is an R-algebra homomorphism µ : P → End(M ) with µ(X) = φ
...
4)
...
Since φ is surjective, M = aM
...
3), there is a ∈ a with (1 + a)M = 0
...
Then 1M + φq(φ) = 0
...
Then φψ = 1 and ψφ = 1
...
□
Corollary (10
...
— Let R be a nonzero ring, m and n positive integers
...
, vn of the free module Rn form a free basis
...
Proof: Form the surjection φ : Rn →
→ Rn taking the ith standard basis element
to vi
...
4)
...
10)(3)
...
Then Rn has m generators
...
The result is a free basis by (1); so it can contain no zeros
...
□
Exercise (10
...
— Let R be a nonzero ring, α : Rm → Rn a map of free modules
...
Show that m ≥ n
...
7)
...
Assume a is finitely generated
and idempotent (or a = a2 )
...
Exercise (10
...
— Let R be a ring, a an ideal
...
(2) R/a is flat over R, and a is finitely generated
...
(4) a is generated by an idempotent
...
Exercise (10
...
— Prove the following conditions on a ring R are equivalent:
(1) R is absolutely flat; that is, every module is flat
...
(3) Every finitely generated ideal is idempotent
...
Exercise (10
...
— Let R be a ring
...
Prove R is absolutely flat
...
Prove any quotient ring R′ is absolutely flat
...
Prove every nonunit x is a zerodivisor
...
Prove R is a field
...
11) (Nakayama)
...
Assume M = mM
...
Proof: By (10
...
By (3
...
Thus M = (1 + a)−1 (1 + a)M = 0
...
Say m1 ,
...
Then n ≥ 1 and m1 = a1 m1 + · · · + an mn with ai ∈ m
...
2), we may set
xi := (1 − a1 )−1 ai
...
Thus n = 0 and so M = 0
...
20)
Example (10
...
— Nakayama’s Lemma (10
...
For example, let A be a local domain, m the maximal ideal, and
K the fraction field
...
Then
any z ∈ K can be written in the form z = x(z/x)
...
Proposition (10
...
— Let R be a ring, m ⊂ rad(R) an ideal, N ⊂ M modules
...
(2) Assume M is finitely generated
...
, mn generate M if
and only if their images m′1 ,
...
Proof: In (1), the second hypothesis holds if and only if m(M/N ) = M/N
...
11) applied with M/N for M
...
, mn
...
Hence N = M if the m′i generate M/mM by (1)
...
□
Exercise (10
...
— Let R be a ring, a an ideal, and α : M → N a map of
modules
...
Show that α is surjective
...
15)
...
Let α, β : M ⇒ N
be two maps of finitely generated modules
...
Set γ := α + β
...
Exercise (10
...
— Let A be a local ring, m the maximal ideal, M a finitely
generated A-module, and m1 ,
...
Set k := A/m and M ′ := M/mM , and
write m′i for the image of mi in M ′
...
, m′n ∈ M ′ form a basis
of the k-vector space M ′ if and only if m1 ,
...
Exercise (10
...
— Let A be a local ring, k its residue field, M and N finitely
generated modules
...
(2) Show
that M ⊗A N ̸= 0 if M ̸= 0 and N ̸= 0
...
18) (Local Homomorphisms)
...
Then the following three conditions are equivalent:
(1) φ−1 n = m;
(2) 1 ∈
/ mB;
(3) mB ⊂ n
...
18
...
If (2) holds, then mB
lies is some maximal ideal, but n is the only one; thus (3) holds
...
If the above conditions hold, then we say φ : A → B is a local homomorphism
...
19)
...
Prove that M is faithfully flat over A if and only if M is flat over
A and nonzero
...
Proposition (10
...
— Consider these conditions on an R-module P :
(1) P is free and of finite rank;
(2) P is projective and finitely generated;
(3) P is flat and finitely presented
...
Cayley–Hamilton Theorem (10
...
22), and a projective module is
always flat by (9
...
Further, each of the three conditions requires P to be finitely
generated; so assume it is
...
Let p1 ,
...
Set F := Rn
...
Then the sequence splits by (5
...
So (5
...
Hence L is finitely generated
...
Assume P is flat and R is local
...
Then,
by (9
...
Now,
F ⊗ k = (R ⊗ k)n = k n by (8
...
Finally,
rechoose the pi so that n is minimal
...
16)
...
Assume P is finitely presented
...
26)
...
17)(1)
...
Thus (3) implies (1)
...
21)
...
Then R′ is said to be
module finite over R if R′ is a finitely generated R-module
...
(10
...
1)
Such an equation is called an equation of integral dependence of degree n
...
Exercise (10
...
— Let G be a finite group of automorphisms of a ring R
...
Show that every x ∈ R is integral over RG , in
fact, over the subring R′ generated by the elementary symmetric functions in the
conjugates gx for g ∈ G
...
23)
...
Then the following conditions are equivalent:
(1) x satisfies an equation of integral dependence of degree n;
(2) R[x] is generated as an R-module by 1, x,
...
Proof: Assume (1) holds
...
For any m, let Mm ⊂ R[x] be the R-submodule generated by 1,
...
For m ≥ n, clearly xm − xm−n p(x) is in Mm−1
...
So also xm ∈ Mm−1
...
Hence Mn−1 = R[x]
...
If (2) holds, then trivially (3) holds with R′′ := R[x]
...
Assume (4) holds
...
2), take φ := µx
...
Since M is faithful, p(x) = 0
...
□
Exercise (10
...
— Let k be a field, P := k[X] the polynomial ring in one
variable, f ∈ P
...
Using the free basis 1, X of P over R, find
an explicit equation of integral dependence of degree 2 on R for f
...
25)
...
Set R′ := P/a, and let x be the image of X in R′
...
Then the following conditions are equivalent:
64
Cayley–Hamilton Theorem (10
...
, xn−1 form a free basis of R′ over R;
(3) R′ is a free R-module of rank n
...
Then p(x) = 0 is an equation of integral dependence
of degree n
...
, xn−1 generate R′ by (1)⇒(2) of (10
...
Suppose
b1 xn−1 + · · · + bn = 0
with the bi ∈ R
...
Then q(x) = 0
...
Hence
q = f p for some f ∈ P
...
Hence q = 0
...
Trivially, (2) implies (3)
...
Then (3)⇒(1) of (10
...
Form the induced homomorphism ψ : P/⟨p⟩ → R′
...
Since (1) implies (3), the quotient P/⟨p⟩ is free of rank n
...
4)
...
Thus (1) holds
...
26)
...
Then M is a finitely generated R-module
...
Then clearly the products xi mj generate M over R
...
27) (Tower Law for Integrality)
...
If x ∈ R′′ is integral over R′ and if R′ is integral over R,
then x is integral over R
...
For m = 1,
...
, am ] ⊂ R′′
...
23)
...
26) and induction on m
...
So Rn [x] is module finite over Rn by (1)⇒(2)
of (10
...
Hence Rn [x] is module finite over R by (10
...
So x is integral over
R by (3)⇒(1) of (10
...
□
Theorem (10
...
— Let R be a ring, R′ an R-algebra
...
, xn ] with all xi integral over R;
(3) R′ is module finite over R
...
Assume (2) holds
...
Then R′′ is module finite
over R by (1)⇒(2) of (10
...
We may assume R′ is module finite over R′′ by
induction on n
...
26) yields (3)
...
23); so (1) holds
...
29)
...
, Rn be R-algebras, integral over R
...
Definition (10
...
— Let R be a ring, R′ an algebra
...
If
R ⊂ R′ and R = R, then R is said to be integrally closed in R′
...
Cayley–Hamilton Theorem (10
...
31)
...
Set R := Ri , set R := Ri′ , and set x := (x1 ,
...
Prove
(1) x is integral over R if and only if xi is integral over Ri for each i;
(2) R is integrally closed in R′ if and only if each Ri is integrally closed in Ri′
...
32)
...
Then R is an R-algebra, and is integrally closed in R′
...
Then the ring R[x, y] is integral over R by
(2)⇒(1) of (10
...
So ax and x + y and xy are integral over R
...
Finally, R is integrally closed in R′ owing to (10
...
□
Theorem (10
...
— A UFD is normal
...
Given x ∈ Frac(R), say x = r/s with r, s ∈ R
relatively prime
...
21
...
Then
rn = −(a1 rn−1 + · · · + an sn−1 )s
...
Hence s is a unit
...
□
Example (10
...
— (1) A polynomial ring in n variables over a field is a UFD,
so normal by (10
...
√
(2) The ring R := Z[ 5] is not a UFD, since
√
√
(1 + 5)(1 − 5) = −4 = −2 · 2,
√
√
and 1 + 5,
√ and 1 − 5 and 2 are irreducible, but not associates
...
” The ring Z[τ ] is known to be a PID; see
[12, p
...
Hence, Z[τ ] is a UFD, so normal by (10
...
On the other hand, τ 2 − τ − 1 = 0; hence, Z[τ ] ⊂ R
...
√
(3) Let d ∈ Z be square-free
...
Then R is the normalization Z of Z in K; see [2, pp
...
(4) Let k be a field, k[t] the polynomial ring in one variable
...
Then Frac(R) = k(t)
...
However, k[t] is normal by (1); hence, k[t] ⊃ R
...
Let k[X, Y ] be the polynomial ring in two variables, and φ : k[X, Y ] → R the
k-algebra homomorphism defined by φ(X) := t2 and φ(Y ) := t3
...
Set p := Ker φ
...
9), but not maximal by (2
...
Clearly p ⊃ ⟨Y 2 − X 3 ⟩
...
28) implies that p = ⟨Y 2 − X 3 ⟩
...
Exercise (10
...
— Let k be a field, X and Y variables
...
Prove that R is a domain, but not a field
...
Prove that k[t] is the integral closure of R in Frac(R)
...
Localization of Rings
Localization generalizes construction of the fraction field of a domain
...
The result is universal among algebras rendering all these elements units
...
We relate the ideals in the original ring to those in the localized ring
...
(11
...
— Let R be a ring, and S a multiplicative subset
...
This relation is an equivalence relation
...
As to transitivity, let (y, t) ∼ (z, r)
...
Then xturv = ysurv = ztvsu
...
Denote by S −1 R the set of equivalence classes, and by x/s the class of (x, s)
...
This product is well defined
...
Then there is v ∈ S such that yrv = ztv
...
Thus xy/st = xz/sr
...
Then, similarly, this sum is well defined
...
It is called
the ring of fractions with respect to S or the localization at S
...
Then φS is a ring map,
and it carries elements of S to units in S −1 R as s/1 · 1/s = 1
...
2)
...
Prove S −1 R = 0
if and only if S contains a nilpotent element
...
3) (Total quotient ring)
...
Then S0 is a saturated multiplicative subset, as noted in (3
...
The map
φS0 : R → S0−1 R is injective, because if φS0 x = 0, then sx = 0 for some s ∈ S, and
so x = 0
...
Let S ⊂ S0 be a multiplicative subset
...
Suppose R is a domain
...
Further, S −1 R is a domain by (2
...
Exercise (11
...
— Find all intermediate rings Z ⊂ R ⊂ Q, and describe each R
as a localization of Z
...
Theorem (11
...
— Let R be a ring, S a multiplicative subset
...
In
fact, given a ring map ψ : R → R′ , then ψ(S) ⊂ R′× if and only if there is a ring
map ρ : S −1 R → R′ with ρφS = ψ; that is, this diagram commutes:
R
φS
/ S −1 R
$
ψ
ρ
R′
Further, there is at most one ρ
...
66
Localization of Rings (11
...
Let s ∈ S
...
Hence
ψ(s)ρ(1/s) = ρ(s/1 · 1/s) = 1
...
Next, note that ρ is determined by ψ as follows:
ρ(x/s) = ρ(x/1)ρ(1/s) = ψ(x)ψ(s)−1
...
Set ρ(x/s) := ψ(s)−1 ψ(x)
...
Say x/s = y/t
...
Hence
ψ(x)ψ(t)ψ(u) = ψ(y)ψ(s)ψ(u)
...
Now, st = ts, so
ψ(t)−1 ψ(s)−1 = ψ(s)−1 ψ(t)−1
...
Thus ρ is well defined
...
Clearly, ψ = ρφS
...
6)
...
Then the
canonical map φS : R → S −1 R is an isomorphism if and only if S consists of units
...
Conversely, if S consists of units, then the identity map R → R has the UMP
that characterizes φS ; whence, φS is an isomorphism
...
7)
...
Consider R := R′ × R′′ and set
S := { (1, 1), (1, 0) }
...
Exercise (11
...
— Take R and S as in (11
...
On R × S, impose this relation:
(x, s) ∼ (y, t)
if
xt = ys
...
Exercise (11
...
— Let R be a ring, S ⊂ T a multiplicative subsets, S and T
their saturations; see (3
...
Set U := (S −1 R)×
...
(2) φ−1
S U = S
...
(4) S R = S −1 R
...
10)
...
(1) Show there’s a unique R-algebra map φST : S −1 R → T −1 R and φTU φST = φSU
...
∪ (3) Let Λ be a set, Sλ ⊂ S a multiplicative subset for all λ ∈ Λ
...
Assume given λ, µ ∈ Λ, there is ν such that Sλ , Sµ ⊂ Sν
...
Using (1), show lim Sλ−1 R = S −1 R
...
11)
...
(1) Show S0 is the largest multiplicative subset S with φS : R → S −1 R injective
...
(3) Suppose every element of R is either a zerodivisor or a unit
...
Definition (11
...
— Let R be a ring, f ∈ R
...
We call
the ring S −1 R the localization of R at f , and set Rf := S −1 R and φf := φS
...
13)
...
Then
/
Rf = R[X] ⟨1 − f X⟩
...
19)
/
Proof: Set R′ := R[X] ⟨1 − f X⟩, and let φ : R → R′ be the canonical map
...
5)
...
Then 1 − xφ(f ) = 0
...
So
φ(f n ) is a unit for n ≥ 0
...
Define
θ : R[X] → R′′ by θ|R = ψ and θX = ψ(f )−1
...
So θ factors
via a homomorphism ρ : R′ → R′′ , and ψ = ρφ
...
□
Proposition (11
...
— Let R be a ring, S a multiplicative subset, a an ideal
...
−1
(2) Then a∩S ̸= ∅ if and only if aS −1 R = S −1 R if and only if φ−1
R) = R
...
Then ax/s + by/t = (axt + bys)/st;
further, axt + bys ∈ a and st ∈ S
...
But the
opposite inclusion is trivial
...
As to (2), if a ∩ S ∋ s, then aS −1 R ∋ s/s = 1, so aS −1 R = S −1 R; whence,
−1
−1
φS (aS −1 R) = R
...
Then aS −1 R ∋ 1
...
So there exists a t ∈ S such that at = st
...
So a ∩ S ̸= ∅
...
□
Definition (11
...
— Let R be a ring, S a multiplicative subset, a a subset of
R
...
If a = aS , then we say a is saturated
...
16)
...
(1) Then Ker(φS ) = ⟨0⟩S
...
(3) Then aS is an ideal
...
Clearly, (2) holds as 1 ∈ S
...
□
Exercise (11
...
— Let R be a ring, S a multiplicative subset, a and b ideals
...
Exercise (11
...
— Let R be a ring, S a multiplicative subset
...
Proposition (11
...
— Let R be a ring, S a multiplicative subset
...
Then
−1 S
(a) φ−1
S b = (φS b)
and
−1
(b) b = (φ−1
R)
...
Then φ−1
R) = aS
...
Then
(a) p = pS
and
(b) pS −1 R is prime
...
23)
69
Proof: To prove (1)(a), take a ∈ R and s ∈ S with as ∈ φ−1
S b
...
Hence a ∈ φ−1
b
...
The opposite inclusion holds as 1 ∈ S
...
To prove (1)(b), take a/s ∈ b
...
So a ∈ φ−1
S b
...
Thus
b
⊂
(φ
b)(S
R)
...
Then a/1 ∈ b
...
Thus (1)(b) holds too
...
Then there is s ∈ S with as ∈ a
...
−1
−1
So a/1 ∈ aS −1 R
...
Now, take x ∈ φ−1
R)
...
14)(1)
...
So x ∈ aS
...
Thus (2) holds
...
Conversely, if sa ∈ p with s ∈ S ⊂ R − p,
then a ∈ p as p is prime
...
−1
As for (b), first note pS −1 R ̸= S −1 R as φ−1
R) = pS = p by (2) and (3)(a)
S (pS
−1
−1
and as 1 ∈
/ p
...
Then ab ∈ φ−1
R), and the
S (pS
S
latter is equal to p by (2), so to p by (a)
...
So either a/s ∈ pS −1 R or b/t ∈ pS −1 R
...
Thus (3) holds
...
20)
...
(1) Then a 7→ aS −1 R is an inclusion-preserving bijection from the set of all ideals
a of R with a = aS to the set of all ideals b of S −1 R
...
(2) Then p 7→ pS −1 R is an inclusion-preserving bijection from the set of all
primes of R with p ∩ S = ∅ to the set of all primes q of S −1 R
...
Proof: In (1), the maps are inverses by (11
...
Further, (1) implies (2) by (11
...
8), and by (11
...
□
Definition (11
...
— Let R be a ring, p a prime ideal
...
We call
the ring S −1 R the localization of R at p, and set Rp := S −1 R and φp := φS
...
22)
...
Then Rp is local with
maximal ideal pRp
...
Then φ−1
p b ⊂ p owing to (11
...
Hence (11
...
Thus pRp is a maximal ideal, and the only one
...
Suppose x/s is a unit
...
So there is a u ∈
/ p with xyu = stu
...
Hence x ∈
/ p
...
Then s/x ∈ Rp
...
Thus by (11
...
Hence (3
...
□
(11
...
— Let R be a ring, S a multiplicative subset, R′ an R-algebra
...
1) as follows
...
It is easy to check, as in (11
...
Denote by S −1 R′ the set of equivalence classes, and by x/s the class of (x, s)
...
We call S R the localization of R with respect to S
...
Then φ′S makes S −1 R′
70
Localization of Rings (11
...
Note that elements of S become units in S −1 R′
...
5), that S −1 R′ has the following UMP: φ′S is an algebra map, and elements
of S become units in S −1 R′ ; further, given an algebra map ψ : R′ → R′′ such that
elements of S become units in R′′ , there is a unique R-algebra map ρ : S −1 R′ → R′′
such that ρφ′S = ψ; that is, the following diagram is commutative:
R′
φ′S
/ S −1 R′
ψ
$
ρ
R′′
In other words, S −1 R′ is universal among R′ -algebras rendering the s ∈ S units
...
Then there is a commutative diagram of
R-algebra maps
τ
′
′′
R
−−−−−−−→ R
φ′
φS y
y S
S −1 τ
S −1 R′ −−−→ S −1 R′′
Further, S −1 τ is an S −1 R-algebra map
...
Then T is multiplicative
...
23
...
Exercise (11
...
— Let R′ /R be a integral extension of rings, S a multiplicative
subset of R
...
Exercise (11
...
— Let R be a domain, K its fraction field, L a finite extension
field, and R the integral closure of R in L
...
Show every element
of L can, in fact, be expressed as a fraction b/a with b ∈ R and a ∈ R
...
26)
...
Assume that R′ is a finitely generated R-algebra, and that L is a finite dimensional
K-vector space
...
Proposition (11
...
— Let R be a ring, S a multiplicative subset
...
Assume S ⊂ T
...
Proof: Let’s check (T ′ )−1 (S −1 R) has the UMP characterizing T −1 R
...
Next, let ψ : R → R′ be a map carrying T
into R′×
...
First, ψ carries S into R′× since S ⊂ T
...
Now, given r ∈ T ′ , write r = x/s
...
So x ∈ T
...
So ρ factors through a
unique map ρ′ : (T ′ )−1 (S −1 R) → R′
...
□
Localization of Rings (11
...
28)
...
Then Rp is the
localization of Rq at the prime ideal pRq
...
Set T := φ−1
S (T )
...
20)(2)
...
27) yields the assertion
...
29)
...
(1) Set T ′ := φS (T ) and assume S ⊂ T
...
(2) Set U := {st ∈ R | s ∈ S and t ∈ T }
...
Proposition (11
...
(— Let) R be a ring, S a multiplicative subset, X a variable
...
Proof: In spirit, the proof is like that of (1
...
□
Corollary (11
...
— Let R be a ring, S a multiplicative subset, X a variable, p
an ideal of R[X]
...
(
)
Then p is prime and p ∩ S = ∅ if and only if pR′ [X] is prime and pφ−1 pR′ [X]
...
30) and (11
...
□
Exercise (11
...
— Given a domain
R and a multiplicative subset S with 0 ∈
/ S
...
12
...
The result is a module
over the localized ring, and comes equipped with a linear map from the original
module; in fact, the result is universal among modules with those two properties
...
So Localization preserves direct
limits, or equivalently, direct sums and cokernels
...
Moreover, Localization is exact; so the localized
ring is flat
...
Proposition (12
...
— Let R be a ring, S a multiplicative subset
...
Proof: Assume M has a compatible S −1 R-structure, and take s ∈ S
...
So µs · µ1/s = µ(s/1)(1/s) = 1
...
So µs is bijective
...
Then µR : R → EndZ (M )
sends S into the units of EndZ (M )
...
5)
...
5)
...
2) (Localization of modules)
...
Define a relation on M × S by (m, s) ∼ (n, t) if there is u ∈ S such that
utm = usn
...
1), this relation is an equivalence relation
...
Then S −1 M is an S −1 R-module with addition given by m/s + n/t := (tm + sn)/st
and scalar multiplication by a/s · m/t := am/st similar to (11
...
We call S −1 M
the localization of M at S
...
Then S −1 a = aS −1 R by (11
...
Similarly,
−1
S (aM ) = S −1 aS −1 M = aS −1 M
...
23)
...
Clearly, φS is R-linear
...
1)
...
Similarly, if S = R−p for some prime ideal p, then we call S −1 M the localization
of M at p, and set Mp := S −1 M and φp := φS
...
3) (UMP)
...
Then S −1 M is universal among S −1 R-modules equipped with an R-linear
map from M
...
5): given an R-linear map ψ : M → N
where N is an S −1 R-module, it is easy to prove that ψ factors uniquely via the
S −1 R-linear map ρ : S −1 M → N well defined by ρ(m/s) := 1/s · ψ(m)
...
4)
...
Show that M = S −1 M if and only if M is an S −1 R-module
...
13)
73
Exercise (12
...
— Let R be a ring, S ⊂ T multiplicative subsets, M a module
...
Show T −1 M = T −1 (S −1 M ) = T1−1 (S −1 M )
...
6)
...
Show that S
becomes a filtered category when equipped as follows: given s, t ∈ S, set
Hom(s, t) := {x ∈ R | xs = t}
...
Define βs : Ms → S −1 M
∼ S −1 M
...
Show the βs induce an isomorphism lim Ms −→
−→
Exercise (12
...
— Let R be a ring, S a multiplicative subset, M a module
...
Prove the converse if M is finitely generated
...
8)
...
(1) Set S := 1 + a
...
(2) Use (1), Nakayama’s Lemma (10
...
7), but not the determinant
trick (10
...
3): if M = aM , then sM = 0 for an s ∈ S
...
9) (Functoriality)
...
Then φS α carries M to the S −1 R-module S −1 N
...
3) yields
a unique S −1 R-linear map S −1 α making the following diagram commutative:
φS
−1
M
−−→ S M
α
−1
y
yS α
φS
N −−→ S −1 N
The construction in the proof of (12
...
(12
...
1)
Thus, canonically, we obtain the following map, and clearly, it is R-linear:
HomR (M, N ) → HomS −1 R (S −1 M, S −1 N )
...
9
...
9
...
Thus S −1 (•) is a linear functor from ((R-mod)) to ((S −1 R-mod))
...
10)
...
Then the functor
S −1 (•) is the left adjoint of the functor of restriction of scalars
...
Then N = S −1 N by (12
...
9
...
And
(12
...
2) is natural in M and N by (6
...
Thus the assertion holds
...
11)
...
Then the
functor S −1 (•) preserves direct limits, or equivalently, direct sums and cokernels
...
10), the functor is a left adjoint
...
12); equivalently, it preserves direct sums and cokernels by (6
...
Exercise (12
...
— Let R be a ring, S a multiplicative subset, P a projective
module
...
Corollary (12
...
— Let R be a ring, S a multiplicative subset
...
74
Localization of Modules (12
...
11), the assertion
is an immediate consequence of Watts Theorem (8
...
Alternatively, both functors are left adjoints of the same functor by (12
...
11)
...
4)
...
14)
...
Show S −1 (M ⊗R N ) = S −1 M ⊗R N = S −1 M ⊗S −1 R S −1 N = S −1 M ⊗R S −1 N
...
15)
...
Show
Fr (M ⊗R R′ ) = Fr (M )R′
and
Fr (S −1 M ) = Fr (M )S −1 R = S −1 Fr (M )
...
16)
...
Given a submodule N , its saturation N S is defined by
N S := {m ∈ M | there is s ∈ S with sm ∈ N }
...
Proposition (12
...
— Let R be a ring, M a module, N and P submodules
...
(1) Then (a) N S is a submodule of M , and (b) S −1 N is a submodule of S −1 M
...
−1
−1
S
(3) Then φS (S N ) = N ; in particular, Ker(φS ) = 0S
...
(5) If N ⊂ P , then (a) N S ⊂ P S and (b) S −1 N ⊂ S −1 P
...
(7) Then (a) (N + P )S ⊃ N S + P S and (b) S −1 (N + P ) = S −1 N + S −1 P
...
Assertion (5)(b) follows by taking M := P
...
4) with M := S −1 M
...
16)(3) and (11
...
Assertions (4)(a) and (5)(a) can be proved as in (11
...
As to (6)(a), clearly (N ∩ P )S ⊂ N S ∩ P S
...
Then stn ∈ N ∩ P and st ∈ S
...
Thus (a) holds
...
As to (6)(b), since N ∩ P ⊂ N, P , using (1) yields S −1 (N ∩ P ) ⊂ S −1 N ∩ S −1 P
...
Hence utn/uts = usp/uts ∈ S −1 (N ∩ P )
...
As to (7)(a), given n ∈ N S and p ∈ P S , there are s, t ∈ S with sn ∈ N and
tp ∈ P
...
Thus (7)(a) holds
...
So (1)(b) yields S −1 (N + P ) ⊃ S −1 N + S −1 P
...
Thus (7)(b) holds
...
18)
...
α
(1) Let M1 −
→ M2 be a map of modules, which restricts to a map N1 → N2 of
submodules
...
α
β
(2) Let 0 → M1 −
→ M2 −
→ M3 be a left exact sequence, which resticts to a left
exact sequence 0 → N1 → N2 → N3 of submodules
...
Localization of Modules (12
...
19)
...
Set T S M := ⟨0⟩S
...
Prove the following:
(1) T S (M/T S M ) = 0
...
(3) Let α : M → N be a map
...
(4) Let 0 → M ′ → M → M ′′ be exact
...
(5) Let S1 ⊂ S be a multiplicative subset
...
Theorem (12
...
— Let R be a ring, and S a multiplicative subset
...
Proof: As S −1 (•) preserves injections by (12
...
11),
it is exact by (9
...
β
α
Alternatively, given an exact sequence M ′ −
→M −
→ M ′′ , for each s ∈ S, take a
copy Ms′ → Ms → Ms′′
...
6), make S into a filtered category, and make
these copies into a functor from S to the category of 3-term exact sequences; its
limit is the following sequence, which is exact by (7
...
The latter argument can be made more direct as follows
...
Hence Ker(S −1 β) ⊃ Im(S −1 α)
...
So β(tm) = 0
...
So (S −1 α)(m′ /ts) = m/s
...
Thus Ker(S −1 β) = Im(S −1 α), as desired
...
21)
...
Then S −1 R
is flat over R
...
20), and is isomorphic to S −1 R⊗R •
by (12
...
Thus S −1 R is flat
...
6), write S −1 R as a filtered direct limit of copies of R
...
7)
...
19)
...
22)
...
Then S −1 (M/aM ) = S −1 M S −1 (aM ) = S −1 M aS −1 M
...
20) and (12
...
□
Corollary (12
...
— Let R be a ring, p a prime
...
Proof: We have Frac(R/p) = (R/p)p = Rp /pRp by (11
...
22)
...
24)
...
(1) Let m1 ,
...
If M is finitely generated and if the mi /1 ∈ S −1 M
generate over S −1 R, then there’s f ∈ S so that the mi /1 ∈ Mf generate over Rf
...
Then there is h ∈ S such that Mh is a free Rh -module of rank n
...
Set C := Coker(α)
...
11)
...
Then S −1 α is surjective by (4
...
11)
...
In addition, assume M is finitely generated
...
Hence, (12
...
Hence αf is surjective
...
10)(1)
...
76
Localization of Modules (12
...
, mn /sn be a free basis of S −1 M over S −1 R
...
, mn /1 as the 1/si are units
...
Then (12
...
But S −1 α is
bijective
...
Since M is finitely generated, C is too
...
Then 0 → Kf → Rfn −−→ Mf → 0 is exact by (12
...
Take a
finite presentation Rp → Rq → M → 0
...
20), it yields a finite presentation
Rfp → Rfq → Mf → 0
...
26)
...
Then (12
...
But
−1
S K = 0
...
Set h := f g
...
Let x ∈ K
...
Hence there
is b such that f b g a x = 0 in K
...
Then hc x = 0
...
But
□
Cf = 0 implies Ch = 0
...
Proposition (12
...
— Let R be a ring, S a multiplicative subset, M and N
modules
...
Further, σ is injective if M is finitely generated, and σ is an isomorphism if M is
finitely presented
...
21) with R′ := S −1 R, since S −1 R is flat
by (12
...
13)
...
26)
...
Then M is
faithful since z ∈ S implies z · (1/2z) = 1/2 ̸= 0; thus, µR : R → HomR (M, M ) is
injective
...
So (12
...
On the other
hand, S −1 M = 0 as s · r/s = 0 for any r/s ∈ M
...
25)
is not injective
...
25)(2) can fail if M is not finitely generated
...
27)
...
Then S −1 Mn = 0 for all n as(∏
nm ≡) 0 (mod n) for all m
...
)/1 is nonzero in S −1
Mn as the kth component
m · (1, 1,
...
Thus
S
M
̸
=
(S −1 Mn )
...
)So (12
...
13)
⊕
⊕
yields Q ⊗
Mn = (Q ⊗ Mn )
...
28)
...
Set M := n≥2 Z/⟨n⟩ and
N := M
...
25) is not injective
...
Support
The spectrum of a ring is the following topological space: its points are the
prime ideals, and each closed set consists of those primes containing a given ideal
...
We relate the support to the closed set of
the annihilator
...
We end this section by proving that the following
conditions on a module ar equivalent: it is finitely generated and projective; it is
finitely presented and flat; and it is locally free of finite rank
...
1) (Spectrum of a ring)
...
Its set of prime ideals is denoted
Spec(R), and is called the (prime) spectrum of R
...
Let V(a) denote the subset of Spec(R) consisting of those
primes that contain a
...
Let b be a second ideal
...
Conversely, if
V(b) ⊂ V(a), then a ⊂ b, owing√to the Scheinnullstellensatz (3
...
Therefore,
√
V(a) = V(b) if and only if a = b
...
2) yields
V(a) ∪ V(b) = V(a ∩ b) = V(ab)
...
Finally, V(R) = ∅, and V(⟨0⟩) = Spec(R)
...
Given an element f ∈ R, we call the open set
D(f ) := Spec(R) − V(⟨f ⟩)
a principal open set
...
Further, f, g ∈
/ p if and only if f g ∈
/ p, for any f, g ∈ R and prime p; in other words,
D(f ) ∩ D(g) = D(f g)
...
1
...
′
−1
(13
...
2)
′
Notice φ (p ) ⊃ a if and only if p ⊃ aR ; so Spec(φ) V(a) = V(aR )
...
Thus Spec(•) is a contravariant functor from ((Rings)) to
((Top spaces))
...
1
...
9) and (2
...
Furthermore, the localization map
R → Rf induces a topological embedding
Spec(Rf ) ,→ Spec(R),
whose image is D(f ), owing to (11
...
77
(13
...
4)
78
Support (13
...
2)
...
Show that p is a closed
point — that is, {p} is a closed set — if and only if p is a maximal ideal
...
3)
...
Let X1 , X2 ⊂ X be
closed subsets
...
(2) There are complementary idempotents e1 , e2 ∈ R with V(⟨ei ⟩) = Xi
...
(4) There are ideals a1 , a2 ⊂ R with a1 ⊕ a2 = R and V(ai ) = Xi
...
Exercise (13
...
— Let φ : R → R′ be a map of rings, a an ideal of R, and b an
ideal of R′
...
Prove these two statements:
(1) Every prime of R is a contraction of a prime if and only if φ∗ is surjective
...
Is the converse of (2) true?
Exercise (13
...
— Let R be a ring, S a multiplicative subset
...
Set φ∗S := Spec(φS ) and S −1 X := Im φ∗S ⊂ X
...
Exercise (13
...
— Let θ : R → R′ be a ring map, S ⊂ R a multiplicative subset
...
Via (13
...
23), identify Spec(S −1 R) and Spec(S −1 R′ ) with their images S −1
X ⊂ X and
S −1 Y ⊂ Y
...
Exercise (13
...
— Let θ : R → R′ be a ring map, a ⊂ R an ideal
...
Let θ : R/a → R′ /b be the induced map
...
Set
∗
′
θ∗ := Spec(θ) and θ := Spec(θ)
...
1), identify Spec(R/a) and Spec(R
/b)
∗
∗−1
∗
with V(a) ⊂ X and V(b) ⊂ Y
...
Exercise (13
...
— Let θ : R → R′ be a ring map, p ⊂ R a prime, k the residue
field of Rp
...
Show (1) that θ∗−1 (p) is canonically homeomorphic
to Spec(R′ ⊗R k) and (2) that p ∈ Im θ∗ if and only if R′ ⊗R k ̸= 0
...
9)
...
Show that the image of
Spec(Rp ) in Spec(R) is the intersection of all open neighborhoods of p in Spec(R)
...
10)
...
Show
∩
Im Spec(θ) = Im Spec(φ) Im Spec(ψ)
...
11)
...
For each λ, let φλ : R′ → Rλ be a ring map with
φµ = αµλ φλ for all αµλ , so that φ := αλ φλ is independent of λ
...
Exercise (13
...
— Let A be a domain with just one nonzero prime p
...
Define φ : A → R by φ(x) := (x′ , x) with x′ the
residue of x
...
Show φ∗ is bijective, but not a homeomorphism
...
23)
79
Exercise (13
...
— Let φ : R → R′ be a ring map, and b an ideal of R′
...
Show (1) that the closure φ∗ (V(b)) in Spec(R) is equal to V(φ−1 b)
and (2) that φ∗ (Spec(R′ )) is dense in Spec(R) if and only if Ker(φ) ⊂ nil(R)
...
14)
...
Show that R′ is faithfully flat if and only if Spec(φ) is surjective
...
15)
...
Show that the induced map Spec(Rq′ ) → Spec(Rp ) is surjective
...
16)
...
Given f ∈ R, set Sf := {f n | n ≥ 0}, and
let S f denote its saturation; see (3
...
Given f, g ∈ R, show that the following
conditions are equivalent:
√
√
(1) D(g) ⊂ D(f )
...
(3) ⟨g⟩ ⊂ ⟨f ⟩
...
(5) g ∈ ⟨f ⟩
...
−1
−1
f
(7) there is a unique R-algebra map φg : S f R → S g R
...
Show that, if these conditions hold, then the map in (8) is equal to φfg
...
17)
...
(1) Show that D(f ) 7→ Rf is a well-defined
contravariant functor from the category of principal open sets and inclusions to ((Ralg))
...
−→
Exercise (13
...
— A topological space is called irreducible if it’s nonempty
and if every pair of nonempty open subsets meet
...
Set X := Spec(R)
and n := nil(R)
...
Exercise (13
...
— Let X be a topological space, Y an irreducible subspace
...
(2) Show that Y is contained in a maximal irreducible subspace
...
They are called its irreducible components
...
Show that its irreducible components
are the closed sets V(p) where p is a minimal prime
...
20)
...
Then X is quasi∪n
∪
compact: if X = λ∈Λ Uλ with Uλ open, then X = i=1 Uλi for some λi ∈ Λ
...
As X = λ∈Λ Uλ , then ∅ = V(aλ ) = V
aλ
...
Hence
there
are
λ
,
...
...
So R = aλi
...
Thus X = Uλi
...
21)
...
Show U is quasi-compact if and only if X − U = V(a) where a is finitely generated
...
22)
...
Set X := Spec(R)
...
Show m = 0
...
23)
...
Prove that the four
following conditions are equivalent:
(1) R/ nil(R) is absolutely flat
...
80
Support (13
...
(4) Every prime p of R is maximal
...
Prove that X is totally disconnected; namely, no two distinct
points lie in the same connected component
...
24)
...
Show a
subset U ⊂ X is both open and closed if and only if U = D(f ) for some f ∈ B
...
(Following Bourbaki, we shorten
“quasi-compact” to “compact” when the space is Hausdorff
...
25) (Stone’s Theorem)
...
Equivalently, show B is isomorphic to the ring R of open and
∼ R is given by f 7→ D(f )
...
26)
...
Its support is the set
Supp(M ) := { p ∈ Spec(R) | Mp ̸= 0 }
...
27)
...
∪
(1) Let 0 → L → M → N → 0 be
Supp(N ) = Supp(M )
...
Then Supp(L)
∪
(2) Let Mλ be submodules with
Mλ = M
...
(3) Then Supp(M ) ⊂ V(Ann(M )), with equality if M is finitely generated
...
For every prime p, the sequence 0 → Lp → Mp → Np → 0
is exact by (12
...
So Mp ̸= 0 if∪and only if Lp ̸= 0 or Np ̸= 0
...
In (2), Mλ ⊂ M
...
To prove the opposite
inclusion, take p ⊕
∈
/
Supp(Mλ )
...
By hypothesis,
the natural map
Mλ → M is surjective
...
11)
...
Alternatively,
given
m/s
∈
Mp , express m as a finite sum
p
∑
m=
m
with
m
∈
M
...
λ
λ
λ
∏
Set t := tλ
...
So m/s = 0 in Mp
...
Thus p ∈
/ Supp(M ), and so (2) holds
...
Let p be a prime
...
7), Mp = 0 if Ann(M
∩ ) ∩ (R − p) ̸= ∅,
and the converse holds if M is finitely generated
...
Thus (3) holds
...
28)
...
We say x is nilpotent
on a
√
module M if there is n ≥ 1 with xn m = 0 for all m ∈ M ; that is, x √
∈ Ann(M )
...
Proposition (13
...
— Let R be a ring, M a finitely generated module
...
∩
Proof: First, nil(M ) = p⊃Ann(M ) p by the Scheinnullstellensatz (3
...
But
□
p ⊃ Ann(M ) if and only if p ∈ Supp(M ) by (13
...
Proposition (13
...
— Let R be a ring, M and N modules
...
30
...
Proof: First, (M ⊗R N )p = Mp ⊗Rp Np by (12
...
30
...
The opposite inclusion follows from (10
...
□
Support (13
...
31)
...
Prove that
Supp(M/aM ) ⊂ Supp(M ) ∩ V(a),
with equality if M is finitely generated
...
32)
...
Prove
Supp(M ⊗R R′ ) ⊂ Spec(φ)−1 (Supp(M )),
with equality if M is finitely generated
...
33)
...
Prove
V(p) ⊂ Supp(M )
...
34)
...
Find Supp(M ), and show that it is not Zariski closed
...
35)
...
These conditions are
equivalent: (1) M = 0; (2) Supp(M ) = ∅; (3) Mm = 0 for every maximal ideal m
...
In particular, (2) holds
...
Finally, assume M ̸= 0, and take a nonzero m ∈ M , and set a := Ann(m)
...
Then, for all f ∈ R − m, we have f m ̸= 0
...
Thus (3) implies (1)
...
36)
...
Set S := R − 0 and
T (M ) := T S (M )
...
Prove M is torsionfree if and only if Mm is torsionfree for all maximal ideals m
...
37)
...
Assume
Mm = Nm for every maximal ideal m
...
First assume M ⊂ N
...
38)
...
Suppose
Mm = 0 for all maximal ideals m ⊃ a
...
Exercise (13
...
— Let R be a ring, P a module, M a submodule, and p ∈ P
an element
...
Show p ∈ M
...
40)
...
Show a = m aRm where
m runs through the maximal ideals and the intersection takes place in Frac(R)
...
41)
...
(2) S −1 R is reduced for all multiplicative subsets S
...
If Rm is a domain for all maximal ideals m, is R necessarily a domain?
Exercise (13
...
— Let R be a ring, Σ the set of minimal primes
...
n
(2) R = i=1 Ri where Ri is a domain if and only if Rp is a domain for any
prime p and Σ is finite
...
, pn } = Σ
...
43)
...
82
Support (13
...
20)
...
First Im(βm αm ) = 0
...
20) and (9
...
Hence Im(βα) = 0 by (13
...
So βα = 0
...
/
/
Set H := Ker(β) Im(α)
...
20) and (9
...
So Hm = 0 owing to the hypothesis
...
35), as required
...
44)
...
Prove elements mλ ∈ M
generate M if and only if, at every maximal ideal m, their images mλ generate Mm
...
45)
...
, mn its maximal ideals,
M, N finitely presented modules
...
Then M ≃ N
...
Then (12
...
However,
(2
...
Set
γ
:=
x
s
φi
...
Then αmj : Mmj −→
as
x
and
s
are
units
...
Then βj (Mmj ) ⊂ mj Nmj as xi ∈ mj for i ̸= j
...
∼ N
...
15)
...
43) implies γ : M −→
□
Proposition (13
...
— Let R be a ring, M a module
...
Proof: If M is flat over R, then M ⊗R Rm is flat over Rm by (9
...
But
M ⊗R Rm = Mm by (12
...
Thus Mm is flat over Rm
...
Let α : N ′ → N be an
injection of R-modules
...
43)
...
But that map is equal to (M ⊗ α)m by (12
...
So (M ⊗ α)m is injective
...
43)
...
Exercise (13
...
— Let R be a ring, R′ a flat algebra, p′ a prime in R′ , and p
its contraction in R
...
Exercise (13
...
— Let R be a ring, S a multiplicative subset
...
Show S −1 R is absolutely flat
...
Definition (13
...
— Let R be a ring, M a module
...
It is enough that an f exist for each maximal ideal m as every
p lies in some m by (2
...
Similarly, we define the properties locally finitely
presented, locally free of finite rank, and locally free of rank n
...
50)
...
(1) If M is locally finitely generated, then it is finitely generated
...
∪
Proof: By (13
...
, fn ∈ R with D(fi ) = Spec(R) and
n
finitely many mi,j ∈ M such that, for some ni,j ≥ 0, the mi,j /fi i,j generate Mfi
...
Given any maximal ideal m, there is i such that fi ∈
/ m
...
Then (12
...
Hence the mi,j /1 generate
Support (13
...
Thus (13
...
Assume M is locally finitely presented
...
So
there is a surjection Rk →
→ M
...
Then K is locally finitely
generated owing to (5
...
Hence K too is finitely generated by (1)
...
It yields the desired finite presentation Rℓ → Rk → M → 0
...
□
Theorem (13
...
— These conditions on an R-module P are equivalent:
(1) P is finitely generated and projective
...
(3) P is finitely presented, and Pm is free over Rm at each maximal ideal m
...
(5) P is finitely generated, and for each p ∈ Spec(R), there are f and n such
that p ∈ D(f ) and Pq is free of rank n over Rq at each q ∈ D(f )
...
20)
...
Then Rm is local by (11
...
If P is finitely presented, then Pm is finitely presented, because localization preserves direct sums and
cokernels by (12
...
Assume (2)
...
46), so free by (10
...
Thus (3) holds
...
Fix a surjective map α : M → N
...
So Hom(Pm , αm ) : Hom(Pm , Mm ) → Hom(Pm , Nm ) is surjective by (5
...
22)
...
25) as P is finitely presented
...
Hence Hom(P, α) is surjective by (13
...
Therefore, P
is projective by (5
...
Thus (1) holds
...
Given any prime p, take a maximal ideal m containing it
...
By (12
...
Thus (4) holds
...
Then P is locally finitely presented
...
50)(2)
...
Given q ∈ D(f ), let S be the image of R − q in Rf
...
5) yields Pq = S −1 (Pf )
...
Thus (5) holds
...
Finally, assume (5), and let’s prove (4)
...
Take a free basis p1 /f k1 ,
...
The pi define
a map α : Rn → P , and αp : Rpn → Pp is bijective, in particular, surjective
...
24)(1) provides g ∈ R − p such that αg : Rgn → Pg
is surjective
...
If also
q ∈ D(f ), then by hypothesis Pq ≃ Rqn
...
4)
...
Clearly, D(f ) ∩ D(g) = D(h)
...
1), D(h) = Spec(Rh )
...
Hence αh : Rhn → Ph is bijective owing to
(13
...
Thus (4) holds
...
52)
...
Exercise (13
...
— Let A be a semilocal ring, P a locally free module of rank
n
...
Exercise (13
...
— Let R be a ring, M a finitely presented module, n ≥ 0
...
14
...
We prove each prime has at least one prime lying over it —
that is, contracting to it
...
Further, one prime is maximal if and only if the other
is, and two overprimes cannot be nested
...
Lemma (14
...
— Let R ⊂ R′ be an integral extension of domains
...
Proof: First, suppose R′ is a field
...
Then 1/x ∈ R′ , so
satisfies an equation of integral dependence:
(1/x)n + a1 (1/x)n−1 + · · · + an = 0
with n ≥ 1 and ai ∈ R
...
Conversely, suppose R is a field
...
Then y satisfies an
equation of integral dependence
y n + a1 y n−1 + · · · + an−1 y + an = 0
with n ≥ 1 and ai ∈ R
...
Take n minimal
...
So dividing by −an y, we obtain
1/y = (−1/an )(y n−1 + · · · + an−1 ) ∈ R′
...
2)
...
We say p′ lies over p if p′ contracts to p
...
3)
...
Let p′ ⊂ q′ be nested primes of R′ , and a′ an arbitrary ideal of R′
...
Then p′ is maximal if and only if p is
...
Then p′ = q′
...
(4) (Going up) Suppose a′ ∩ R ⊂ p
...
Proof: Assertion (1) follows from (14
...
To prove (2), localize at R − p, and form this commutative diagram:
′
R
→ Rxp′
x −
R −→ Rp
84
Krull–Cohen–Seidenberg Theory (14
...
17)(1), and the extension is integral by (11
...
Here p′ Rp′ and q′ Rp′ are nested primes of Rp′ by (11
...
By the same token,
both lie over pRp , because both their contractions in Rp contract to p in R
...
22)
...
To prove (3), again we may replace R by Rp and R′ by Rp′ : if r′′ is a prime ideal
of Rp′ lying over pRp , then the contraction r′ of r′′ in R′ lies over p
...
Now, R′ has a maximal ideal r′
by 2
...
Thus r = p
...
□
Exercise (14
...
— Let R ⊂ R′ be an integral extension of rings, and p a prime
of R
...
Show (a) that p′ Rp′ is the only
maximal ideal of Rp′ , (b) that Rp′ ′ = Rp′ , and (c) that Rp′ ′ is integral over Rp
...
5)
...
Suppose R′ has at least two distinct primes p′ and q′ lying over p
...
Show that, in fact, if y lies in q′ , but not in
p′ , then 1/y ∈ Rp′ ′ is not integral over Rp
...
6)
...
Set R′ := k[X],
and Y := X 2 , and R := k[Y ]
...
Is Rp′ ′
integral over Rp ? Explain
...
7)
...
Suppose f = gh with g, h ∈ R′ [X] monic
...
Proof: Set R1 := R′ [X]/⟨g⟩
...
Then 1, x1 , x21 ,
...
25) as g is monic; hence, R′ ⊂ R1
...
Repeat
this
process,
extending
R
...
The xi and
yj are integral over R as they are roots of f
...
□
Proposition (14
...
— Let R be a normal domain, K := Frac(R), and L/K a
field extension
...
Then p ∈ R[X], and so p(y) = 0 is an equation of integral dependence
...
Write f = pq with q ∈ K[X]
...
7) the coefficients of p are integral over
R, so in R since R is normal
...
9) (Going down for integral extensions)
...
Then there is a prime p′ lying over p and contained in q′
...
Given y ∈ pRq′ ′ ∩ R, say y = x/s with
∑m
x ∈ pR′ and s ∈ R′ − q′
...
, xm ]
...
28) and xR′′ ⊂ pR′′
...
86
Krull–Cohen–Seidenberg Theory (14
...
1), and f (x) = 0 by the Determinant Trick (10
...
Set K := Frac(R)
...
By (14
...
Further, f ≡ X n (mod p)
...
Hence g and h have all nonleading coefficients in p
...
Then f is the minimal polynomial of x over K
...
So s satisfies the equation
sn + b1 sn−1 + · · · + bn = 0
with
bi := ai /y i ∈ K
...
9
...
So (14
...
1) is the minimal polynomial of s over K
...
8)
...
Then bi ∈ p as ai = bi y i ∈ p
...
So s ∈ q′ ,
a contradiction
...
Thus pRq′ ′ ∩ R ⊂ p
...
Thus pRq′ ′ ∩ R = p
...
13)
...
Set p′ := p′′ ∩ R′
...
20)(2), as desired
...
10)
...
Proof: Let R be the ring, p the minimal prime
...
20)(2)
...
Hence, given x ∈ p, there is s ∈ R − p with sxn = 0 for some n ≥ 1
...
Then sxn−1 ̸= 0, but (sxn−1 )x = 0
...
□
Theorem (14
...
— Let R be a ring, R′ a flat
algebra, p ⫋ q nested primes of R, and q′ a prime of R′ lying over q
...
Proof: The canonical map Rq → Rq′ ′ is faithfully flat by (13
...
Therefore,
Spec(Rq′ ′ ) → Spec(Rq ) is surjective by (13
...
Thus (11
...
Alternatively, R′ ⊗R (R/p) is flat over R/p by (9
...
Also, R′ /pR′ = R′ ⊗R R/p
by (8
...
Hence, owing to (1
...
By (3
...
Let’s show that p′ lies over ⟨0⟩
...
Then the multiplication map µx : R → R is injective
...
Hence, (14
...
□
Exercise (14
...
∪ — Let R be a reduced ring, Σ the set of minimal primes
...
div(R) = p∈Σ p and that Rp = Frac(R/p) for any p ∈ Σ
...
13)
...
Assume Σ is finite
...
∪
(2) z
...
∏
(3) K/pK = Frac(R/p) for each p ∈ Σ, and K = p∈Σ K/pK
...
17)
87
Exercise (14
...
— Let A be a reduced local ring with residue field k and finite
set Σ of minimal primes
...
Let P be a finitely
generated module
...
Exercise (14
...
— Let A be a reduced local ring with residue field k and a
finite set of minimal primes
...
Show that P is a free A-module of rank r if
and only if P ⊗ B is a free B-module of rank r
...
16) (Arbitrary normal rings)
...
If R is a domain, then this definition
recovers that in (10
...
32)
...
17)
...
, pr all its minimal primes, and K
the total quotient ring
...
(2) R is reduced and integrally closed in K
...
Assume the conditions hold
...
15
...
After proving this lemma, we derive
several versions of the Nullstellensatz
...
Then we study the (Krull) dimension: the maximal length of any chain of primes
...
Further, if the algebra is a domain,
then its dimension is equal to the transcendence degree of its fraction field
...
At
the same time, we prove it in significantly greater generality: for Jacobson rings
...
1) (Noether Normalization)
...
, xn ] a
finitely generated k-algebra, and a1 ⊂ · · · ⊂ ar a chain of proper ideals of R
...
, tν ∈ R such that
(1) R is module finite over P := k[t1 ,
...
, thi ⟩
...
Proof: Let R′ := k[X1 ,
...
Set a′0 := Ker φ and a′i := φ−1 ai for i = 1, · · · , r
...
10), and clearly (1) and (2) hold
...
The proof proceeds by induction on r (and shows ν := n works now)
...
Then t1 ∈
/ k because
a1 is proper
...
, tn ∈ R so that x1 is integral over
P := k[t1 , t2 ,
...
Then (10
...
Further, by the theory of transcendence bases [2, (8
...
526], [10, Thm
...
1,
p
...
, tn are algebraically independent
...
Then x = t1 x′ where x′ ∈ R ∩ Frac(P )
...
34) as P is a polynomial algebra
...
Thus (2) holds too
...
, tn , we are going to choose ℓi and set ti := xi − xℓ1i
...
Now, say t1 =
a(j) xj11 · · · xjnn with (j) := (j1 ,
...
Recall t1 ∈
/ k, and note that x1 satisfies this equation:
∑
a(j) xj11 (t2 + xℓ12 )j2 · · · (tn + xℓ1n )jn = t1
...
Take ℓ > max{ji } and ℓi := ℓi
...
Let e(j ′ ) be largest among the e(j) with a(j) ̸= 0
...
Thus x1 is integral over P , as desired
...
3)
89
Suppose k is infinite
...
Then P [x1 ] = R
...
, xn ; that is, Hi is a linear
combination of monomials of degree i
...
As k is infinite, (3
...
, an ) ̸= 0
...
Again since Hd is homogeneous,
we may replace ai by ai /a1
...
, an ) ̸= 0
...
, an ) is
the coefficient of xd1 in Hd (x1 , t2 + a2 x1 ,
...
So after we collect like
powers of x1 , the equation
Hd (x1 , t2 + a2 x1 ,
...
, tn + an x1 ) + t1 = 0
becomes an equation of integral dependence for x1 over P , as desired
...
We may assume a1 ̸= 0
...
The case n = 1 follows from the first case (but is
simpler) because k[x1 ] is a PID
...
By the first case, there
exist elements u2 ,
...
un are algebraically independent and
satisfy (1) and (2) with respect to R and t1 R
...
, tn
satisfying (1) and (2) with respect to k[u2 ,
...
, un ]
...
, tn ]
...
, un ] and the
latter is module finite over P , the former is module finite over P by (10
...
Thus
(1) holds, and so t1 ,
...
Further, by assumption,
a1 ∩ k[t2 ,
...
, th ⟩
for some h
...
So a1 ∩ P ⊃ ⟨t1 ,
...
∑d
Conversely, given x ∈ a1 ∩ P , say x = i=0 fi ti1 with fi ∈ k[t2 ,
...
Since
t1 ∈ a1 , we have f0 ∈ a1 ∩ k[t2 ,
...
, th ⟩
...
, th ⟩
...
, th ⟩
...
Finally, assume the lemma holds for r − 1
...
, un ∈ R be algebraically
independent elements satisfying (1) and (2) for the sequence a1 ⊂ · · · ⊂ ar−1 , and
set h := hr−1
...
, tn satisfying (1)
and (2) for k[uh+1 ,
...
, un ]
...
, tn ] = ⟨th+1 ,
...
Set ti := ui for 1 ≤ i ≤ h
...
, tn ]
...
, un ], and k[u1 ,
...
27)
...
, tn are algebraically
independent over k
...
Set m := hi
...
, tm ∈ ai
...
, vm ) and f(v) ∈ k[tm+1 ,
...
Then f(0)
lies in ai ∩ k[tm+1 ,
...
We are going to see the latter intersection is equal to
⟨0⟩
...
, un ], which is equal to ⟨0⟩
...
, tn ] = ⟨tm+1 ,
...
Thus f(0) = 0
...
, thi ⟩
...
, thi ⟩
...
Thus (2) holds, and the proof is complete
...
2)
...
Set f := X q Y − XY q and R := k[X, Y ] ⟨f ⟩
...
For every a ∈ k, show that R is not module finite over
P := k[y −ax]
...
1), no k-linear combination works
...
90
Noether Normalization (15
...
3)
...
Set
/
R := k[X, Y, Z] ⟨X 2 − Y 3 − 1, XZ − 1⟩,
and let x, y, z ∈ R be the residues of X, Y, Z
...
Show that x and y are integral over P for any a, b and that z is
integral over P if and only if b ̸= 0
...
4) (Zariski Nullstellensatz)
...
Assume R is a field
...
Proof: By the Noether Normalization Lemma (15
...
, tν ]
...
23)
...
1)
...
So P = k
...
Alternatively, here’s a short proof, not using (15
...
Say R = k[x1 ,
...
Set
P := k[x1 ] and K := Frac(P )
...
, xn ]
...
Suppose x1 is transcendental over k, so P is a polynomial ring
...
, xn ]
...
26) yields f ∈ P with Rf /Pf module finite,
so integral by (10
...
But Rf = R
...
1)
...
r
Set g := 1 + f
...
So 1/g = h/f for some h ∈ P and r ≥ 1
...
But f and g are relatively prime, a contradiction
...
Hence P = K, and K/k is finite
...
Thus R/k is too
...
5)
...
, xn ] an algebra-finite extension, and m a maximal ideal of R
...
Then there
are a1 ,
...
, xn − an ⟩
...
Then K is a finite extension field of k by the Zariski
Nullstellensatz (15
...
But k is algebraically closed
...
Let ai ∈ k be
the residue of xi , and set n := ⟨x1 − a1 ,
...
Then n ⊂ m
...
, Xn ] be the polynomial ring, and φ : R′ → R the k-algebra
map with φXi := xi
...
, Xn − an ⟩
...
But n′ is
maximal by (2
...
So n is maximal
...
□
Corollary (15
...
— Let k be any field, P := k[X1 ,
...
Then m is generated by n elements
...
Then K is a field
...
4)
...
If n = 0, then m = 0
...
Set R := k[X1 ] and p := m∩R
...
Set k1 := R/p
...
But K is a finite-dimensional k-vector space
...
So k ⊂ k1 is an integral extension by (10
...
Since k is a field, so is k1 by (14
...
Note P/pP = k1 [X2 ,
...
7)
...
So by
induction m/p is generated by n − 1 elements, say the residues of f2 ,
...
Then m = ⟨f1 ,
...
□
Theorem (15
...
— Let k be a field, and R a finitely
generated k-algebra
...
Then
∩
√
a = m⊃a m
where m runs through all maximal ideals containing a
...
12)
91
√
∩
Proof: We may assume
a = 0 by replacing R by R/a
...
√
Conversely, take f ∈
/ 0
...
2)
...
30)
...
Now, R is a finitely generated k-algebra
by hypothesis; hence, Rf is one too owing to (11
...
Therefore, by the weak
Nullstellensatz, Rf /n is a finite extension field of k
...
By construction, K is a k-subalgebra of Rf /n
...
So k ⊂ K is an integral extension by (10
...
Since k is a field, so is K by (14
...
Thus m √
is maximal
...
Hence f ∈
/ m
...
Thus 0 = m
...
8)
...
(So
K contains a copy of every finite extension field
...
, Xn ] be the
polynomial ring, and f, f1 ,
...
Assume f vanishes at every zero in K n of
f1 ,
...
, an ) ∈ K n and f1 (a) = 0,
...
Prove that there are polynomials g1 ,
...
Lemma (15
...
— Let k be a field, R a finitely generated k-algebra
...
Let p0 ⫋ · · · ⫋ pr be a chain of primes
...
degk K
...
Proof: By the Noether Normalization Lemma (15
...
, tν ] such that pi ∩ P = ⟨t1 ,
...
Set L := Frac(P )
...
degk L
...
23)
...
Hence ν = d
...
3)(2)
yields hi < hi+1 for all i
...
But hr ≤ ν and ν = d
...
If r = d, then r is maximal, as it was just proved that no chain can be longer
...
Then p0 = ⟨0⟩ since R is a domain
...
Further, pr is maximal since pr is contained in some maximal ideal and it is prime
...
3)(1)
...
Suppose there is an i such that hi + 1 < hi+1
...
, thi +1 ⟩ ⫋ (pi+1 ∩ P )
...
10), equal to k[thi +1 ,
...
34)(1)
...
Hence, the Going-down Theorem (14
...
, thi +1 ⟩
...
Thus hi + 1 = hi+1 for all i
...
Hence r = hr
...
Thus r = d, as desired
...
10) (Krull Dimension)
...
For example, if R is a field, then dim(R) = 0; more generally, dim(R) = 0 if
and only if every minimal prime is maximal
...
25)
...
11)
...
Prove that dim(R/p) < r
...
19)
Exercise (15
...
— Let R′ /R be an integral extension of rings
...
Theorem (15
...
— Let k be a field, R a finitely generated k-algebra
...
degk (Frac(R))
...
9)
...
14)
...
Suppose R is a domain
...
Proof: A chain of primes p0 ⫋ · · · ⫋ p ⫋ · · · ⫋ pr in R gives rise to a pair of
chains of primes, one in Rp and one in R/p,
p0 Rp ⫋ · · · ⫋ pRp
and
0 = p/p ⫋ · · · ⫋ pr /p,
owing to (11
...
9) and (2
...
But by (15
...
The first equation follows
...
□
Definition (15
...
— We call a ring catenary if, given any two nested primes,
all maximal chains of primes between the two have the same (finite) length
...
16)
...
Proof: Let R be the algebra, and q ⊂ p two nested primes
...
Then the proof of (15
...
□
Exercise (15
...
— Let k be a field, R a finitely generated k-algebra, f ∈ R
nonzero
...
Prove that dim(R) = dim(Rf )
...
18)
...
Set p := ⟨f ⟩ and R := Pp
...
Exercise (15
...
— Let R be a ring, R[X] the polynomial ring
...
(In particular, dim(R[X]) = ∞ if and only if dim(R) = ∞
...
Appendix: Jacobson Rings
(15
...
— We call a ring R Jacobson if, given any ideal a, its
radical is equal to the intersection of all maximal ideals containing it; that is,
∩
√
a = m⊃a m
...
20
...
Also,
any quotient ring of a Jacobson ring is Jacobson too
...
In general, the right-hand side
√ of (15
...
1) contains the left
...
20
...
Recall the Scheinnullstellensatz, (3
...
Thus
∩
R is Jacobson if and only if p = m⊃p m for every prime p
...
Further, a Boolean
∩ ring B is Jacobson, as
every prime is maximal by (2
...
Finally, owing to (15
...
6) and the next lemma, a PID is Jacobson
...
21)
...
Assume every nonzero element lies in only finitely many mλ
...
∏
Proof: If Λ is finite, take a nonzero
xλ ∈ mλ for each λ, and set
√
√ x := ∩ xλ
...
But ⟨0⟩ = ⟨0⟩ as R is a domain
...
Thus R is not Jacobson
...
But every nonzero prime is
∩
maximal as R is 1-dimensional
...
□
Proposition (15
...
— A ring R is Jacobson if and only if, for any nonmaximal
prime p and any f ∈
/ p, the extension pRf is not maximal
...
Take a nonmaximal prime p and an f ∈
/ p
...
So
pR
is
not
maximal
by
(11
...
f
√
Conversely, let a be an ideal, f ∈
/ a
...
So there is a maximal
ideal n in (R/a)f
...
Further,
/ in R
...
8) and (12
...
Since n is maximal,
Rf /mRf is a field
...
Thus R is Jacobson
...
23)
...
We say a subset Y is locally
closed if Y is the intersection of an open set and a closed set; equivalently, Y is
open in its closure Y ; equivalently, Y is closed in an open set containing it
...
We say X is Jacobson if its set of closed points is very dense
...
(2) Every closed set F of X satisfies F ∩ X0 = F
...
Exercise (15
...
— Let R be a ring, X := Spec(R), and X0 the set of closed
points of X
...
30)
(1) R is a Jacobson ring
...
(3) If y ∈ X is a point such that {y} is locally closed, then y ∈ X0
...
25)
...
Assume that R′ = R[x] for some
x ∈ R′ and that there is y ∈ R′ with Ry′ a field
...
Further, if R is Jacobson, then R and R′ are fields
...
Then Q ⊂ Ry′ , so Ry′ = R[x]y ⊂ Q[x]y ⊂ Ry′
...
So Q[x]y is a field
...
20); whence, Q[x]y is not a field by (15
...
Thus x is algebraic over Q
...
Let a0 xn + · · · + an = 0 and b0 y m + · · · + bm = 0 be equations of minimal degree
with ai , bj ∈ R
...
Then z ̸= 0
...
Hence R[x]y ⊂ Rz [x] ⊂ Ry′
...
Therefore Rz [x] is a field too
...
So Rz [x] is integral over Rz (10
...
Hence Rz is a field by (14
...
Further, if R is Jacobson, then ⟨0⟩ is a maximal ideal by (15
...
Hence R = Rz
...
1)
...
26) (Generalized Hilbert Nullstellensatz)
...
Set m := m′ ∩R
...
Proof: First, assume R′ = R[x] for some x ∈ R′
...
Then R is Jacobson by
(15
...
Suppose (R1′ )y is a field
...
25), R1 ⊂ R1′ is a finite extension
of fields
...
To obtain (1), simply take q := m and y := 1
...
22) yields (2)
...
, xn ] with n ≥ 2
...
, xn−1 ] and
′′
m := m′ ∩ R′′
...
By induction on n, we may assume (1) and (2)
hold for R′′ /R
...
Hence, m is maximal, and R′′ /m′′
is algebraic over R/m by (1) for R′′ /R
...
27) implies
that R′ /m′ is algebraic over R/m, as desired
...
27)
...
26) may fail if R is not Jacobson, even if
R′ := R[Y ] is the polynomial ring in one variable Y over R
...
According to (3
...
Exercise (15
...
— Let P := Z[X1 ,
...
Assume
f ∈ P vanishes at every zero in K n of f1 ,
...
, an ) ∈ K n and f1 (a) = 0,
...
Prove there are g1 ,
...
Exercise (15
...
— Let R be a ring, R′ an algebra
...
Appendix: Jacobson Rings (15
...
30)
...
True or false: prove or give a counterexample to each of the following statements
...
(2) The localized ring S −1 R is Jacobson
...
−→
(4) In a filtered direct limit of rings Rλ , necessarily lim rad(Rλ ) = rad(lim Rλ )
...
31)
...
Show that P is locally free of
rank r if and only if dimR/m (P/mP ) = r for any maximal ideal m
...
Chain Conditions
In a ring, often every ideal is finitely generated; if so, we call the ring Noetherian
...
We characterize
Noetherian rings as those in which every ascending chain of ideals stabilizes, or
equivalently, in which every nonempty set of ideals has a member maximal under
inclusion
...
We define and characterize Noetherian
modules similarly, and we prove that, over a Noetherian ring, it is equivalent for
a module to be Noetherian, to be finitely generated, or to be finitely presented
...
(16
...
— We call a ring Noetherian if every ideal is finitely
generated
...
Here are two standard examples of non-Noetherian rings
...
6), and a fourth later in (18
...
First, form the polynomial ring k[X1 , X2 ,
...
It is
non-Noetherian as ⟨X1 , X2 ,
...
Second, in the polynomial ring k[X, Y ], form this subring R and its ideal a:
{
}
R := f := a + Xg | a ∈ k and g ∈ k[X, Y ] and
a := ⟨X, XY, XY 2 ,
...
, fm ∈ a
...
Then XY n+1 ∈
/ ⟨f1 ,
...
Thus R is non-Noetherian
...
2)
...
Show every set that generates M contains a finite subset that generates
...
3)
...
We say the maximal condition (maxc) is satisfied if every nonempty set of
ideals S contains ones maximal for inclusion, that is, properly contained in no
other in S
...
4)
...
Proof: Let a0 ⊂ a1 ⊂ · · · be a chain of ideals
...
Thus maxc implies acc
...
So the Axiom of Countable
Choice provides an infinite chain a0 ⫋ a1 ⫋ · · ·
...
□
Proposition (16
...
— Given a ring R, the following conditions are equivalent:
(1) R is Noetherian; (2) acc is satisfied; (3) maxc is satisfied
...
Let a0 ⊂ a1 ⊂ · · · be a chain of ideals
...
Clearly, a is an ideal
...
, xr
...
Set j := max{ji }
...
So
a ⊂ aj ⊂ aj+1 ⊂ · · · ⊂ a
...
Thus (2) holds
...
11)
97
Assume (2) holds
...
4)
...
Let a be an ideal, xλ for λ ∈ Λ generators, S the set of ideals
generated by finitely many xλ
...
, xλm
...
So by maximality, b = b + ⟨xλ ⟩
...
So b = a; whence, a is finitely generated
...
□
Example (16
...
— In the field of rational functions k(X, Y ), form this ring:
R := k[X, Y, X/Y, X/Y 2 , X/Y 3 ,
...
5)
...
Correspondingly, there is an
ascending chain of ideals that does not stabilize:
⟨X⟩ ⫋ ⟨X/Y ⟩ ⫋ ⟨X/Y 2 ⟩ ⫋ · · ·
...
7)
...
Then R/a and S −1 R are Noetherian
...
9) and by (11
...
Alternatively, any ideal b/a of R/a is, clearly, generated by the images of generators of b
...
19)(1)(b)
...
8)
...
Prove this statement or find a counterexample: if R[X] is Noetherian, then so is R
...
9)
...
Assume R′ is Noetherian, and prove R is too
...
10) (Cohen)
...
Proof: Suppose there are non-finitely-generated ideals
...
If a is finitely
generated, then all the generators lie in some aλ , so generate aλ as aλ = a, a
contradiction
...
Hence, by Zorn’s Lemma, there is
a maximal non-finitely-generated ideal p
...
Assume every prime is finitely generated
...
So p + ⟨a⟩ is finitely generated, say by x1 + w1 a,
...
Then
{x1 ,
...
Set b = Ann (p + ⟨a⟩)/p
...
So b is finitely generated,
say by y1 ,
...
Take z ∈ p
...
Then ya ∈ p
...
Hence y = b1 y1 + · · · + bm ym with bj ∈ R
...
, xn , ay1 ,
...
Thus there are
no non-finitely-generated ideals; in other words, R is Noetherian
...
11)
...
Proof: By way of contradiction, assume there is an ideal a of R[X] that is not
finitely generated
...
For each i ≥ 1, choose inductively fi ∈ a − ai−1
of least degree di , and set ai := ⟨f1 ,
...
Let ai be the leading coefficient of fi ,
and b the ideal generated by all the ai
...
, an ⟩ for
98
Chain Conditions (16
...
2)
...
By construction, di ≤ di+1 for all i
...
Then deg(f ) < dn+1 , so f ∈ an
...
□
Theorem (16
...
— Let R be a Noetherian ring, R′ a finitely
generated algebra
...
Proof: Say x1 ,
...
, Xr ] be the
polynomial ring in r variables
...
11) and induction
on r
...
Hence R′ is Noetherian by (16
...
□
(16
...
— We call a module M Noetherian if every submodule is finitely generated
...
We say the ascending chain condition (acc) is satisfied in M if every ascending
chain of submodules M0 ⊂ M1 ⊂ · · · stabilizes
...
It is simple to generalize (16
...
Lemma (16
...
— Let R be a ring, M a module
...
Proof: Given m2 ∈ M2 , there is m1 ∈ M1 with n := m2 − m1 ∈ N
...
Hence m2 ∈ M1
...
□
α
β
Exercise (16
...
— Let 0 → L −
→M −
→ N → 0 be a short exact sequence of
R-modules, and M1 , M2 two submodules of M
...
Proposition (16
...
— Let R be a ring, M a module, N a submodule
...
(2) Then M is Noetherian if and only if N and M/N are Noetherian
...
6) owing to (5
...
To prove (2), first assume M is Noetherian
...
A submodule of
M/N is finitely generated as its inverse image in M is so; thus M/N is Noetherian
...
Let P be a submodule of M
...
But P/(P ∩ N ) −→
by (4
...
2)
...
Thus M is Noetherian
...
First assume M is Noetherian
...
And any chain in M/N is the image
of a chain in M , so it too stabilizes
...
Conversely, assume N and M/N are Noetherian
...
Then
□
Mj = Mj+1 = · · · by (16
...
Thus M is Noetherian
...
24)
99
Corollary (16
...
— Modules M1 ,
...
Proof: The sequence 0 → M1 → M1 ⊕ (M2 ⊕ · · · ⊕ Mr ) → M2 ⊕ · · · ⊕ Mr → 0
is exact
...
16)(2) by induction on r
...
18)
...
, ar ideals such that each R/ai is a
Noetherian
ring
...
Theorem (16
...
— Let R be a Noetherian ring, and M a module
...
Proof: Assume (2)
...
Now, Rn is Noetherian by (16
...
13)
...
16)(2)
...
□
Exercise (16
...
— Let R be a Noetherian ring, M and N finitely generated
modules
...
Lemma (16
...
1])
...
Assume
that R is Noetherian, that R′′ /R is algebra finite, and that R′′ /R′ either is module
finite or is integral
...
Proof: Since R′′ /R is algebra finite, so is R′′ /R′
...
28)
...
, xm generate R′′ as an R-algebra, and y1 ,
...
Then there exist zij ∈ R′ and zijk ∈ R′ with
∑
∑
xi = j zij yj and yi yj = k zijk yk
...
21
...
Since R is Noetherian,
so is R0′ by the Hilbert Basis Theorem, (16
...
Any x ∈ R′′ is a polynomial in the xi with coefficients in R
...
21
...
Thus R′′ /R0′ is module
finite
...
So R′ /R0′ is
module finite by (16
...
Since R0′ /R is algebra finite, R′ /R is too
...
22) (Noether on Invariants)
...
Then the
subring of invariants R′G is also algebra finite; in other words, every invariant can
be expressed as a polynomial in a certain finite number of “fundamental” invariants
...
22), R′ is integral over R′G
...
21) yields the assertion
...
23) (Artin–Tate proof [1, Thm
...
4))
...
4), take a transcendence base x1 ,
...
Then R is
integral over k(x1 ,
...
3), p
...
So
k(x1 ,
...
21), say k(x1 ,
...
, ys ]
...
Write yi = Fi /Gi with Fi , Gi ∈ k[x1 ,
...
Let H be an
irreducible factor of G1 · · · Gs + 1
...
Say H −1 = P (y1 ,
...
Then H −1 = Q/(G1 · · · Gs )m
for some Q ∈ k[x1 ,
...
But H ∤ Gi for all i, a contradiction
...
So (10
...
100
Chain Conditions (16
...
24)
...
Assume R is Noetherian
...
3] Assume R′ is a field containing R
...
(2) [1, bot
...
77] Let K ′ ⊃ R be a field that embeds in R′
...
Show K/R is algebra finite and K ′ /K is finite
...
25)
...
Let’s
prove that p is finitely presented and that pRq is free of rank 1 over Rq for every
maximal ideal q of R, but that p is not free
...
51) do not imply that P is free
...
12) yields that R is Noetherian
...
19) yields that p is finitely presented
...
417, 421, 425] that p is maximal in R, but not principal
...
So (1 − δ)/3 ∈ Rp
...
So (1 + δ)Rp = pRp
...
Thus pRp is free of rank 1
...
Then p ∩ (R − q) ̸= ∅; so, pRq = Rq
by (11
...
Thus pRq is free of rank 1
...
Set S := R − 0
...
So S −1 p ≃ K n
...
∼ K, since S −1 p is a nonzero K-vector space
...
So
Hence S −1 p −→
p ≃ R
...
But p is not principal
...
Thus p is not free
...
26)
...
We say the ring itself is Artinian if it is an Artinian module
...
Proposition (16
...
— Let M1 ,
...
(1) Then M is Artinian if and only if minc is satisfied in M
...
(3) Then M1 ,
...
Proof: It is easy to adapt the proof of (16
...
16)(2),
and the proof of (16
...
□
Exercise (16
...
— Let k be a field, R an algebra
...
Prove that R is Noetherian and Artinian
...
29)
...
Prove that any Z-submodule N ⊂ M is either finite or all of M
...
Exercise (16
...
— Let R be an Artinian ring
...
Deduce that, in general, every prime ideal p of R is maximal
...
Associated Primes
Given a module, a prime is associated to it if the prime is equal to the annihilator
of an element
...
If the ring is
Noetherian, then the set of annihilators of elements has maximal members; we prove
the latter are prime, so associated
...
If also the module is finitely generated, then
the intersection is the set of nilpotents
...
Definition (17
...
— Let R be a ring, M a module
...
The set of
associated primes is denoted by Ass(M ) or AssR (M )
...
Warning: following a old custom, we mean by the associated primes of an
ideal a not those of a viewed as an abstract module, but rather those of R/a
...
2)
...
Then
p ∈ Ass(M ) if and only if there is an R-injection R/p ,→ M
...
Define a map R → M by x 7→ xm
...
Conversely, suppose there is an R-injection R/p ,→ M , and let m ∈ M be the
image of 1
...
□
Proposition (17
...
— Let M be a module
...
Proof: Let p ∈ AssR (M )
...
Then m/1 ∈ Mp is nonzero as no
x ∈ (R − p) satisfies xm = 0
...
Alternatively, (17
...
It induces an injection
(R/p)p ,→ Mp by (12
...
But (R/p)p = Frac(R/p) by (12
...
Thus Mp ̸= 0
...
4)
...
Then (1) Ann(m) = p and (2) Ass(R/p) = {p}
...
Let x ∈ R
...
Thus (1) holds
...
□
Proposition (17
...
— Let M be a module, N a submodule
...
Proof: Take m ∈ N
...
So Ass(N ) ⊂ Ass(M )
...
Then (17
...
Denote its image
by E
...
2)
...
Then Ann(m) = p by
(17
...
Thus p ∈ Ass(N )
...
14)
Exercise (17
...
— Given modules M1 ,
...
Prove
Ass(M ) = Ass(M1 ) ∪ · · · ∪ Ass(Mr )
...
7)
...
Find Ass(M ) and find
two submodules L, N ⊂ M with L + N = M but Ass(L) ∪ Ass(N ) ⫋ Ass(M )
...
8)
...
9)
...
Then
there is a submodule N of M with Ass(M/N ) = Ψ and Ass(N ) = Ass(M ) − Ψ
...
Given p ∈ Ass(N ), say p = Ann(m)
...
Conversely, Ass(Nλ ) ⊂ Ass(N ) for all λ by (17
...
Thus Ass(N ) = Ass(Nλ )
...
By (17
...
Take p ∈ Ass(M/N )
...
2)
...
5) and (17
...
Now, N ′ ⫌ N and
N is maximal with Ass(N ) ⊂ Ass(M ) − Ψ
...
Thus p ∈ Ψ
...
10)
...
If p ∩ S = ∅ and p ∈ Ass(M ), then S −1 p ∈ Ass(S −1 M ); the
converse holds if p is finitely generated
...
Then (17
...
It
induces an injection S −1 (R/p) ,→ S −1 M by (12
...
But S −1 (R/p) = S −1 R/S −1 p
by (12
...
Assume p ∩ S = ∅ also
...
19)(3)(b)
...
2)
...
Conversely, assume S −1 p ∈ Ass(S −1 M )
...
Say p = ⟨x∏
1 ,
...
Fix i
...
So there is
si ∈ S with si xi m = 0
...
Then xi ∈ Ann(sm)
...
Take b ∈ Ann(sm)
...
So b/1 ∈ S −1 p
...
19)(1)(a)
and (11
...
Thus p ⊃ Ann(sm)
...
Thus p ∈ Ass(M )
...
20)(2), as S −1 p is prime
...
11)
...
Prove every associated prime of R is minimal
...
12)
...
Suppose p is
maximal in the set of annihilators of nonzero elements m of M
...
Proof: Say p := Ann(m) with m ̸= 0
...
Now, take b, c ∈ R
with bc ∈ p, but c ∈
/ p
...
Plainly, p ⊂ Ann(cm)
...
But b ∈ Ann(cm), so b ∈ p
...
□
Proposition (17
...
— Let R be a Noetherian ring, M a module
...
Proof: Obviously, if M = 0, then Ass(M ) = ∅
...
Let S be the set of annihilators of nonzero elements of M
...
5)
...
12), p ∈ Ass(M )
...
Associated Primes (17
...
14)
...
We say x is a
zerodivisor on M if there is a nonzero m ∈ M with xm = 0; otherwise, we say x
is a nonzerodivisor
...
div(M )
...
15)
...
Then
∪
z
...
Proof: Given x ∈ z
...
Then
x ∈ Ann(m)
...
5); hence, p ∈ Ass(M ) by (17
...
∪
Thus z
...
The opposite inclusion results from the definitions
...
16)
...
Show that, if x ∈
/ p for any p ∈ Ass(M/N ), then xM ∩ N = xN
...
17)
...
Then
∪
Supp(M ) = q∈Ass(M ) V(q) ⊃ Ass(M )
...
Then Rp is Noetherian by (16
...
So Mp ̸= 0
if and only if AssRp (Mp ) ̸= ∅ by (17
...
But R is Noetherian; so AssRp (Mp ) ̸= ∅
if and only if there is q ∈ Ass(M ) with q∩(R−p) = ∅, or q ⊂ p, owing to (11
...
10)
...
□
Theorem (17
...
— Let R be a Noetherian ring, M a module, p ∈ Supp(M )
...
Proof: By (17
...
Also, q ∈ Supp(M ); so q = p if p is minimal
...
19)
...
Then
∩
nil(M ) = p∈Ass(M ) p
...
29)
...
17)
...
□
Lemma (17
...
— Let R be a Noetherian ring, M a finitely generated module
...
, n
...
, pn } ⊂ Supp(M )
...
20
...
19) and (16
...
Suppose M/N ̸= 0
...
13), the
quotient M/N contains a submodule N ′ /N isomorphic to R/p for some prime p
...
Hence N = M
...
The first inclusion of (17
...
1) follows by induction from (17
...
4)(2)
...
23)
...
27)(1) yields (17
...
1)
...
21)
...
Then the set Ass(M ) is finite
...
20)
...
27)
Exercise (17
...
— Let R be a Noetherian ring, a an ideal
...
Prove such primes are finite in number
...
23)
...
20)
...
Exercise (17
...
— Take R := Z and M := Z/⟨12⟩ in (17
...
Find all three
acceptable chains, and show that, in each case, {pi } = Ass(M )
...
25)
...
Then
∩
Ass(Hom(M, N )) = Supp(M ) Ass(N )
...
Then (17
...
Set k(p) := Frac(R/p)
...
23)
...
19) as R is Noetherian; hence,
Hom(M, N )p = HomRp (Mp , Np )
(17
...
1)
by (12
...
Therefore, by exactness, localizing yields an injection
φ : k(p) ,→ HomRp (Mp , Np )
...
For any m ∈ Mp with φ(1)(m) ̸= 0, the map k(p) → Np given by x 7→ φ(x)(m)
is nonzero, so an injection
...
22)
...
2), we
have pRp ∈ Ass(Np )
...
10)
...
Then Mp ̸= 0
...
Take any nonzero R-map
Mp /pMp → k(p), precede it by the canonical map Mp → Mp /pMp , and follow
it by an R-injection k(p) ,→ Np ; the latter exists by (17
...
10) since
p ∈ Ass(N )
...
(
But pRp is maximal,
so
is
the
entire
annihilator
...
p
(
)
Hence p ∈ Ass Hom(M, N ) by (17
...
1) and (17
...
□
Exercise (17
...
— Let R be a Noetherian ring, a an ideal, and M a finitely
generated module
...
div(M ); that is, there is a nonzerodivisor x on M in a;
(5) a ̸⊂ p for any p ∈ Ass(M )
...
27)
...
Then p ∈ Ass(M/xM ) if
and only if p ∈ Ass(M/yM )
...
Apply the functor Hom(R/p, •) to that sequence, and get the following one:
µy
0 → Hom(R/p, K) → Hom(R/p, M/xM ) −−→ Hom(R/p, M/xM )
...
18)
...
Thus
∼ Hom(R/p, M/xM )
...
27)
105
Form the following commutative diagram with exact rows:
µx
0−
→ M −−→ M −
→ M/xM −
→0
µy y
µy y
µy y
µx
0−
→ M −−→ M −
→ M/xM −
→0
µx
The Snake Lemma (5
...
Therefore,
Ker(µy ) = 0
...
(17
...
1)
Finally, p ∈ Supp(R/p) by (13
...
Thus (17
...
□
18
...
A submodule is called primary if the quotient module has
only one associated prime
...
A primary decomposition is a
representation of a submodule as a finite intersection of primary submodules
...
We
consider several illustrative examples in a polynomial ring
...
The celebrated Lasker–Noether
Theorem asserts the existence of an irredundant primary decomposition
...
The Second Uniqueness Theorem asserts
the uniqueness of the primary components whose primes are minimal among these
associated primes; the other primary components may vary
...
1)
...
If Ass(M/Q)
consists of a single prime p, we say Q is primary or p-primary in M
...
2)
...
4)(2)
...
3)
...
If Q is p-primary, then p = nil(M/Q)
...
19)
...
4)
...
Set p := nil(M/Q)
...
(2) p = z
...
(3) Given x ∈ R and m ∈ M with xm ∈ Q but m ∈
/ Q, necessarily x ∈ p
...
19), and z
...
15)
...
div(M/Q)
...
Conversely, if x ∈ q ∈ Ass(M/Q), but x ∈
/ q′ ∈ Ass(M/Q), then x ∈
/ p, but
x ∈ z
...
Thus (1) and (2) are equivalent
...
div(M/Q)
...
Thus (2) and (3) are equivalent
...
5)
...
Set
√
p := q
...
Proof: Clearly q = Ann(R/q)
...
So the assertions result directly
from (18
...
3)
...
6)
...
Show every positive power pn is p-primary
...
106
Primary Decomposition (18
...
7)
...
Let a be
the ideal ⟨X 2 , XY ⟩
...
Show a satisfies this
condition: ab ∈ a implies a2 ∈ a or b2 ∈ a
...
8)
...
Show that φ−1 q ⊂ R is φ−1 p-primary
...
Proposition (18
...
— Let R be a Noetherian ring, M a finitely generated module, Q a submodule
...
If p is maximal, then Q is p-primary
...
19), if p is maximal, then p = q for
any q ∈ Ass(M/Q), or {p} = Ass(M/Q), as desired
...
10)
...
Set p := q
...
□
Proof: Since p = nil(R/q), the assertion is a special case of (18
...
Corollary (18
...
— Let R be a Noetherian ring, m a maximal ideal
...
√
Proof: The condition mn ⊂ q ⊂ m just means that m := q by (3
...
So the
assertion results from (18
...
10)
...
12)
...
Let
Q1 and Q2 be p-primary submodules; set Q := Q1 ∩ Q2
...
Proof: Form the canonical map M → M/Q1 ⊕ M/Q2
...
Hence (17
...
5) yield
∅=
̸ Ass(M/Q) ⊂ Ass(M/Q1 ) ∪ Ass(M/Q2 )
...
□
(18
...
— Let R be a ring, M a module, and N a
submodule
...
We call the decomposition irredundant or minimal if these conditions hold:
∩
∩
(1) N ̸= j̸=i Qj , or equivalently, j̸=i Qj ̸⊂ Qi for i = 1,
...
(2) Say Qi is pi -primary for i = 1,
...
Then p1 ,
...
If so, then we call Qi the pi -primary component of the decomposition
...
12), any primary decomposition can be
made irredundant by intersecting all the primary submodules with the same prime
and then discarding those of them that are not needed
...
14)
...
Set
a := ⟨X 2 , XY ⟩
...
(18
...
1)
Here ⟨X , XY, Y ⟩ and ⟨X , Y ⟩ contain ⟨X, Y ⟩ ; so they are ⟨X, Y ⟩-primary by
(18
...
Thus (18
...
1) gives infinitely many primary decompositions of a
...
Note: the ⟨X, Y ⟩-primary component is not unique!
Plainly, a ⊂ ⟨X⟩ and a ⊂⟨X 2 , XY, Y n ⟩ ⊂ ⟨X 2 , Y ⟩
...
19)
take F ∈ ⟨X⟩ ∩ ⟨X 2 , Y ⟩
...
Then
X(G − AX) = BY
...
Say B = B ′ X
...
Example (18
...
— Let k be a field, R := k[X, Y ] the polynomial ring, a ∈ k
...
Define an automorphism α of R by X 7→ X and Y 7→ aX + Y
...
So (18
...
Moreover, if a ̸= b, then ⟨X 2 , aX +Y, bX +Y ⟩ = ⟨X, Y ⟩
...
12)
...
16)
...
Set
R := P/⟨XZ − Y 2 ⟩
...
Set p := ⟨x, y⟩
...
Let’s show that p2 = ⟨x⟩ ∩ ⟨x2 , y, z⟩ is an
irredundant primary decomposition
...
Conversely, given f ∈ ⟨x⟩ ∩ ⟨x2 , y, z⟩, represent f by GX with G ∈ P
...
So (G − AX)X = B ′ Y + C ′ Z with B ′ , C ′ ∈ P
...
Then
A′′ X = −B ′′ XY − C ′′ XZ + B ′ Y + C ′ Z = (B ′ − B ′′ X)Y + (C ′ − C ′′ X)Z;
whence, A′′ = 0
...
Thus p2 =⟨x⟩ ∩ ⟨x2 , y, z⟩
...
8)
...
Further, ⟨X, Y 2 ⟩ is ⟨X, Y ⟩-primary, as
under the map φ : P → k[Y, Z] with φ(X) = 0, clearly ⟨X, Y 2 ⟩ = φ−1 ⟨Y 2 ⟩ and
⟨X, Y ⟩ = φ−1 ⟨Y ⟩; moreover, ⟨Y 2 ⟩ is ⟨Y ⟩-primary by (18
...
6)
...
So ⟨x2 , y, z⟩ is
⟨x, y, z⟩-primary by (18
...
Thus p2 = ⟨x⟩ ∩ ⟨x2 , y, z⟩ is a primary decomposition
...
Moreover, ⟨x⟩ is the p-primary component of p2
...
17)
...
Set a := ⟨XY, X − Y Z⟩, set q1 := ⟨X, Z⟩ and set q2 := ⟨Y 2 , X − Y Z⟩
...
Exercise (18
...
— Let R := R′ × R′′ be a product of two domains
...
Lemma (18
...
— Let R be a ring, M a module, N = Q1 ∩ · · · ∩ Qr a primary
decomposition in M
...
, r
...
, pr }
...
19
...
, pr are distinct, then the decomposition is irredundant;
the converse holds if R is Noetherian
...
So
∪ map is injective: M/N ,→
(17
...
6) yield Ass(M/N ) ⊆ Ass(M/Qi )
...
19
...
If N = Q2 ∩ · · · ∩ Qr , then Ass(M/N ) ⊆ {p2 ,
...
Thus if equality holds
in (18
...
1) and if p1 ,
...
∩
Conversely, assume N = Q1 ∩ · · · ∩ Qr is irredundant
...
Primary Decomposition (18
...
Consider these two canonical injections:
Pi /N ,→ M/Qi
and Pi /N ,→ M/N
...
Then Ass(Pi /N ) ̸= ∅ by (17
...
So the first injection
yields Ass(Pi /N ) = {pi } by (17
...
Thus
□
Ass(M/N ) ⊇ {p1 ,
...
19
...
Theorem (18
...
— Let R be a Noetherian ring, and M a
module
...
, r
...
, pr are uniquely determined; in fact,
they are just the distinct associated primes of M/N
...
19)
...
21) (Lasker–Noether)
...
Proof: Let M be the module, N the submodule
...
21), M/N has finitely
many distinct associated primes, say p1 ,
...
Owing to (17
...
Set
P := Qi
...
Then P/N ⊂ Qi /N
...
5)
...
Hence∩Ass(P/N ) = ∅
...
13)
...
19)
...
22)
...
Consider the following submodule of M :
∪
Γa (M ) := n≥1 { m ∈ M | an m = 0 }
...
∩
(By convention, if a ⊂ pi for all i, then a̸⊂pi Qi = M
...
(Thus Γa (M ) is the set of all m whose support lies in V(a)
...
23)
...
If S ∩ p ̸= ∅, then
S −1 Q = S −1 M and QS = M
...
S (S
Proof: Every prime of S −1 R is of the form S −1 q where q is a prime of R with
S ∩ q = ∅ by (11
...
2)
...
10)
...
20)
...
13) yields S −1 M/S −1 Q = 0
...
−1
Finally, QS = φ−1
Q) by (12
...
So if S −1 Q = S −1 M , then QS = M
...
Given m ∈ QS , there is s ∈ S with sm ∈ Q
...
Further, p = z
...
15)
...
Thus QS ⊂ Q
...
Thus QS = Q
...
24)
...
Let N = Q1 ∩ · · · ∩ Qr ⊂ M be an irredundant
primary decomposition
...
31)
i ≤ h
...
Proof: By (12
...
Further, by (18
...
Therefore,
S −1 N = S −1 Q1 ∩ · · · ∩ S −1 Qh is a primary decomposition
...
19)
...
, S −1 ph } by
an argument like that in the first part of (18
...
Further, S −1 p1 ,
...
20)(2) as the pi are distinct
...
Owing to (12
...
But QSi = Qi by (18
...
So N S = Q1 ∩ · · · ∩ Qh is a primary
decomposition
...
13)(1) and (2) hold for it, since
they hold for N = Q1 ∩ · · · ∩ Qr
...
25) (Second Uniqueness)
...
Assume R is Noetherian and M is finitely generated
...
Then, in any irredundant primary decomposition of N in M , the
p-primary component Q is uniquely determined; in fact, Q = N S where S := R − p
...
24), take S := R − p
...
□
Exercise (18
...
— Let R∩be a Noetherian ring, M a finitely generated module,
N a submodule
...
Exercise (18
...
— Let R be a Noetherian ring, p a prime
...
(1) Show p(n) is the p-primary component of pn
...
(3) Show p(n) = pn if and only if pn is p-primary
...
Exercise (18
...
— Let R be a Noetherian ring, ⟨0⟩ = q1 ∩· · ·∩qn an irredundant
√
primary decomposition
...
, n
...
Show qi = pi for all large r
...
Show that replacing qi by pi for large
r gives infinitely many distinct irredundant primary decompositions of ⟨0⟩
...
29) (Krull Intersection)
...
Set N := n≥0 an M
...
Proof: By (16
...
So the desired x ∈ a exists by
(10
...
Clearly N ⊃ aN
...
21): take
a decomposition aN = Qi with Qi pi -primary
...
If there’s a ∈ a − pi , then
there’s ni with ani M ⊂ Qi
aN ⊂ Qi , and so (18
...
If a ⊂ pi , then ∩
by (18
...
32), and so again N ⊂ Qi
...
□
Exercise (18
...
— Let R be a Noetherian ring, m ⊂ rad(R) an ideal, M a
finitely generated module, and M ′ a submodule
...
Primary Decomposition (18
...
31) (Another non-Noetherian ring)
...
∼ R
...
Then, Taylor’s Theorem yields
f (x) = f (0) + f ′ (0)x + · · · +
f (n−1) (0) n−1
(n−1)! x
where fn (x) :=
∫1
0
+ xn fn (x)
(1−t)n−1 (n)
(xt) dt
...
1),
p
...
So, if f ∈ m, then f (x) = xf1 (x)
...
But, obviously, m ⊃ ⟨x⟩
...
Therefore, mn = ⟨xn ⟩
...
Conversely, assume f (x) = xn g(x) for some g ∈ R
...
f (k) (x) = j=0 kj (n−j+1)!
Hence f (k) vanishes at 0 if n > k
...
But ⟨xn ⟩ = mn
...
There is a well-known nonzero C ∞ -function all of whose derivatives vanish at 0:
{
2
e−1/x if x ̸= 0,
h(x) :=
0
if x = 0;
∩
n
see [9, Ex
...
82]
...
Given g ∈ m, let’s show (1 + g)h ̸= 0
...
Hence 1 + g(x) ≥ ∩
1/2 if |x| < δ
...
Thus (1 + g)( mn ) ̸= 0
...
29) fails for R, and so R is non-Noetherian
...
Length
The length of a module is a generalization of the dimension of a vector space
...
Our main result is the Jordan–H¨older Theorem: any two
composition series do have the same length and even the same successive quotients;
further, their annihilators are just the primes in the support of the module, and the
module is equal to the product of its localizations at these primes
...
We
also prove the Akizuki–Hopkins Theorem: a ring is Artinian if and only if it is
Noetherian and every prime is maximal
...
(19
...
— Let R be a ring, and M a module
...
We call a chain of submodules,
M = M0 ⊃ M1 ⊃ · · · ⊃ Mm = 0
(19
...
1)
a composition series of length m if each successive quotient Mi−1 /Mi is simple
...
(19
...
2)
By convention, if M has no composition series, then ℓ(M ) := ∞
...
For example, if R is a field, then M is a vector space and ℓ(M ) = dimR (M )
...
24) are composition series, but those in (17
...
Exercise (19
...
— Let R be a ring, M a module
...
(2) M is simple if and only if M ≃ R/m for some maximal ideal m, and if so,
then m = Ann(M )
...
Theorem (19
...
— Let R be a ring, and M a module with a
composition series (19
...
1)
...
Also,
Supp(M ) = { m ∈ Spec(R) | m = Ann(Mi−1 /Mi ) for some i };
the m ∈ Supp(M ) are maximal; there is a canonical isomorphism
∏
∼
M −→
m∈Supp(M ) Mm ;
and ℓ(Mm ) is equal to the number of i with m = Ann(Mi−1 /Mi )
...
Let’s show that
ℓ(M ′ ) < ℓ(M )
...
Then Mi−1
∩ Mi = Mi′
...
112
(19
...
1)
Length (19
...
′
(19
...
2)
′
If (19
...
2) holds and if Mi ⊂ M , then Mi−1 ⊂ M
...
3
...
all i, then M ⊂ M ′ , a contradiction
...
Omit Mi whenever Mi−1 /Mi = 0
...
Therefore, ℓ(M ′ ) < m for any
choice of (19
...
1)
...
3
...
Next, given a chain N0 ⫌ · · · ⫌ Nn = 0, let’s prove n ≤ ℓ(M ) by induction on
ℓ(M )
...
Assume ℓ(M ) ≥ 1
...
If n ≥ 1, then ℓ(N1 ) < ℓ(M ) by (19
...
1); so n − 1 ≤ ℓ(N1 ) by
induction
...
If Ni−1 /Ni is not simple, then there is N ′ with Ni−1 ⫌ N ′ ⫌ Ni
...
Repeating, we can refine
the given chain into a composition series in at most ℓ(M ) − n steps
...
Then ℓ(M ) ≤ n by (19
...
2)
...
Thus n = ℓ(M ), and the first assertion is proved
...
Exactness of Localization, (12
...
(19
...
3)
Now, consider a maximal ideal m
...
4) and
(12
...
If p ̸= m, then there is s ∈ m − p; so (R/m)p = 0
...
So Mi−1 /Mi ≃ R/mi and mi is maximal by (19
...
Then Exactness of Localization yields (Mi−1 /Mi )p = (Mi−1 )p /(Mi )p
...
Thus Supp(M ) = {m1 ,
...
If we omit the duplicates from the chain (19
...
3), then we get a composition
series from the (Mi )p with Mi−1 /Mi ≃ R/p
...
Finally, consider the canonical map φ : M → m∈Supp(M ) Mm
...
43), to prove φp is for each maximal ideal p
...
11)
...
Thus φp = 1
...
4)
...
Prove the equivalence of the following three conditions:
(1) that M has finite length;
(2) that Supp(M ) consists entirely of maximal ideals;
(3) that Ass(M ) consists entirely of maximal ideals
...
Exercise (19
...
— Let R be a Noetherian ring, q a p-primary ideal
...
Show (1) all such chains have length at most
ℓ(A)−1 where A := (R/q)p and (2) all maximal chains have length exactly ℓ(A)−1
...
6)
...
114
Length (19
...
3)
...
Conversely, assume M is both Artinian and Noetherian
...
Set M0 := M
...
By the dcc, this recursion terminates
...
□
Example (19
...
— Any simple Z-module is finite owing to (19
...
Hence, a
Z-module is of finite length if and only if it is finite
...
Of course, Z is Noetherian, but not Artinian
...
Then M is an Artinian Z-module,
but not Noetherian by (16
...
Since M is infinite, ℓ(M ) = ∞
...
8)
...
Prove that R
is Artinian if and only if R is a finite-dimensional k-vector space
...
9) (Additivity of Length)
...
Then ℓ(M ) = ℓ(M ′ ) + ℓ(M/M ′ )
...
The latter
yields a pair of composition series: M/M ′ = M0 /M ′ ⊃ · · · ⊃ M ′ /M ′ = 0 and
M ′ ⊃ · · · ⊃ Mm = 0
...
Therefore, ℓ(M ) < ∞ if and only if ℓ(M/M ′ ) < ∞ and
ℓ(M ′ ) < ∞; furthermore, if so, then ℓ(M ) = ℓ(M ′ ) + ℓ(M/M ′ ), as desired
...
10)
...
Assume the map from
k to the residue field is bijective
...
Theorem (19
...
— A ring R is Artinian if and only if R
is Noetherian and dim(R) = 0
...
Proof: If dim(R) = 0, then every prime is maximal
...
4)
...
6)
...
Let m be a minimal product of maximal ideals
of R
...
Let S be the set of ideals a contained in m such that am ̸= 0
...
Then am2 = am ̸= 0; hence, am = a by minimality of
a
...
Let n be any maximal ideal
...
But nm ⊂ n
...
But a = ⟨x⟩
...
So xm = 0 for any x ∈ a
...
Hence S = ∅
...
But m2 = m
...
Say m = m1 · · · mr with mi maximal
...
Consider the chain
R =: a0 ⊃ a1 ⊃ · · · ⊃ ar = 0
...
Set Vi := ai−1 /ai
...
Suppose dim(Vi ) = ∞
...
∈ Vi ,
let Wj ⊂ Vi be the subspace spanned by xj , xj+1 ,
...
Thus dim(Vi ) < ∞
...
9)
...
6)
...
26
yields Supp(R) = Spec(R)
...
4), every prime is maximal, and there
are only finitely many primes
...
12)
...
18)
115
(1) that R is Artinian;
(2) that Spec(R) is discrete and finite;
(3) that Spec(R) is discrete
...
13)
...
Show that rad(R) is nilpotent
...
14)
...
Then M has finite length, and Ass(M ) and Supp(M ) are equal and finite
...
11) every prime is maximal, so Supp(M ) consists of maximal
ideals
...
11)
...
4) yields the assertions
...
15)
...
Proof: Simply take M := R in (19
...
6)
...
16)
...
Show that R′ has only finitely many primes p′ over p, as follows: reduce
to the case that R is a field by localizing at p and passing to the residue rings
...
17)
...
Proof: A finite product of rings is Artinian if and only if each factor is Artinian
∏
by (16
...
If R is Artinian, then ℓ(R) < ∞ by (19
...
Thus the assertion holds
...
18)
...
Prove the following four conditions are equivalent:
(1) that M has finite length;
∏
(2) that M is annihilated by some finite product of maximal ideals mi ;
(3) that every prime p containing Ann(M ) is maximal;
(4) that R/Ann(M ) is Artinian
...
Hilbert Functions
The Hilbert Function of a graded module lists the lengths of its components
...
This series
is, under suitable hypotheses, a rational function, according to the Hilbert–Serre
Theorem, which we prove
...
We prove Samuel’s Theorem: if the ring is Noetherian, if the module
is finitely generated, and if the filtration is stable, then the Hilbert–Samuel Series
is a rational function with poles just at 0 and 1
...
In a brief appendix, we study further one notion that arose: homogeneity
...
1) (Graded rings and modules)
...
For example, a polynomial ring R with coefficient ring R0 is graded if Rn is the
R0 -submodule generated by the monomials of (total) degree n
...
Obviously, R0 is closed
∑under addition and under
multiplication, but
we
must
check
1
∈
R
...
Given
0
∑
z ∈ R, say z =
z∑
xm zn with
n with zn ∈ Rn
...
Then zn = 1 · zn =
xm zn ∈ Rm+n
...
Hence xm zn = 0 for m > 0
...
So xm z = 0 for m > 0
...
Taking z := 1 yields
xm = xm · 1 = 0 for m > 0
...
We call an R-module
⊕ M (compatibly) graded if there are additive subgroups Mn
for n ∈ Z with M =
Mn and Rm Mn ⊂ Mm+n for all m, n
...
Obviously,
Mn is an R0 -module
...
Then M (m) is another graded module;
its nth graded component M (m)n is Mm+n
...
⊕
⊕
Lemma (20
...
— Let R =
Rn be a graded ring, and M =
Mn a graded
R-module
...
∑
Proof: Say R = R0 [x1 ,
...
If xi = j xij with xij ∈ Rj , then replace the
xi by the nonzero xij
...
, ms∑
with
∑
mi ∈ Mli
...
Say fi =
fij
with fij ∈ Rj , and replace fi by fik with k := n − li or by 0 if n < li
...
□
⊕
⊕
(20
...
— Let R =
Rn be a graded ring, and M =
Mn
a graded R-module
...
Then each Mn is a finitely generated
R0 -module by (20
...
14)
...
8)
117
the Hilbert Series of M
...
7) below
...
, xr ] with xi ∈ R1 , then by (20
...
Example (20
...
— Let R := R0 [X1 ,
...
Then Rn is free over R0 on the monomials of degree n, so of rank r−1+n
...
Then ℓ(Rn ) = ℓ(R0 ) r−1 by Additivity of Length,
(19
...
Thus the Hilbert Function
is,) for n ≥ 0,
(
( a )polynomial of degree r − 1
...
Therefore, Newton’s binomial
theorem for negative exponents yields this computation for the Hilbert Series:
(
)n ∑
(−r)
/
∑
n
r
H(R, t) = n≥0 ℓ(R0 ) r−1+n
n≥0 ℓ(R0 ) n (−t) = ℓ(R0 ) (1 − t)
...
5)
...
Show ⟨X, Y 2 ⟩
and ⟨X 2 , Y 2 ⟩ have different Hilbert Series, but the same Hilbert Polynomial
...
6)
...
Let N =
Nn be a homogeneous submodule; that is, Nn = N ∩ Mn
...
Set
N ′ := { m ∈ M | there is k0 such that Rk m ⊂ N for all k ≥ k0 }
...
∩, and that N is the largest such submodule containing ⊕
(2) Let N = ∩
Qi be a decomposition with Qi pi -primary
...
Prove that N ′ = pi ̸⊃R+ Qi
...
7) (Hilbert–Serre)
...
Assume R0 is Artinian, R is a finitely generated
R0 -algebra, and M is a finitely generated R-module
...
, kr ≥ 1
...
, xr ] with xi ∈ Rki
...
Say M is
generated over R by m1 ,
...
Then R = R0
...
Hence t−l0 H(M, t) is a polynomial
...
Since x1 ∈ Rk1 , the grading on M
induces a grading on K and on L
...
As R0 is Artinian, R0 is Noetherian by the Akizuki–Hopkins Theorem, (19
...
So, since R is a finitely generated R0 -algebra, R is Noetherian by (16
...
Since
M is a finitely generated R-module, obviously so is M (−k1 )
...
16)(2)
...
, xr ]
...
Therefore, H(K, t) and H(L, t) may be
written in the desired form by induction on r
...
Therefore, Additivity of Length, (19
...
Thus the assertion holds
...
11)
Corollary (20
...
— Under the conditions of (20
...
, xr ]
with xi ∈ R1
...
Then H(M, t) can be written uniquely in the form
/
H(M, t) = e(t) tl (1 − t)d
(20
...
1)
with e(t) ∈ Z[t] and e(0), e(1) ̸= 0 and l ∈ Z and r ≥ d ≥ 0; also, there is a
polynomial h(M, n) ∈ Q[n] with degree d − 1, leading coefficient e(1)/(d − 1) ! and
ℓ(Mn ) = h(M, n)
for n ≥ deg e(t) − l
...
8
...
7)
...
Set d := r − s
...
Thus H(M, t) has the asserted form
...
∑N
∑ −d)
∑ (d−1+n) n
n
Say e(t) = i=0 ei ti
...
Hence
(d−1+n+l−i)
(d−1+n−i)
∑N
d−1
ℓ(Mn ) = i=0 ei
for n + l ≥ N
...
d−1
d−1
Therefore, ℓ(Mn ) = e(1) nd−1 /(d − 1) ! + · · · , as asserted
...
9)
...
Set R := P/⟨f ⟩
...
Exercise (20
...
— Under the conditions of (20
...
Prove deg h(R, n) > deg h(M, n); start
with the case M := R/⟨f k ⟩
...
11) (Filtrations)
...
A filtration F • M of M is an infinite descending chain of submodules:
M ⊃ · · · ⊃ F n M ⊃ F n+1 M ⊃ · · ·
...
This condition means
that there are µ and ν with M = F µ and qn F ν M = F n+ν M for n > 0
...
It is called the q-adic filtration
...
We form the product of an element in qi /qi+1 and one in qj /qj+1 by choosing
representatives, forming their product, and taking its residue in qi+j /qi+j+1
...
As each F n M is an R-module, so is the direct sum
⊕
G• M := n∈Z Gn M where Gn M := F n M/F n+1 M
...
Given m ∈ Z, let M [m] denote M with the filtration F • M reindexed by shifting
it m places to the left; that is, F n (M [m]) := F n+m M for all n
...
If the quotients M/F n M have finite length, call n 7→ ℓ(M/F n M ) the Hilbert–
Samuel Function, and call the generating function
∑
P (F • M, t) := n≥0 ℓ(M/F n M )tn
Hilbert Functions (20
...
If the function n 7→ ℓ(M/F n M ) is, for n ≫ 0,
a polynomial p(F • M, n), then call it the Hilbert–Samuel Polynomial
...
Lemma (20
...
— Let R be a Noetherian ring, q an ideal, M a finitely generated
module with a stable q-filtration
...
Proof: Since R is Noetherian, q is a finitely generated ideal, say by x1 ,
...
Then, clearly, the residues of the xi in q/q2 generate G• R as an R/q-algebra
...
Hence G• M is generated by F µ M/F µ+1 M,
...
But R
is Noetherian and M is finitely generated over R; hence, every F n M is finitely
generated over R
...
Thus G• M is a finitely generated G• R-module
...
13) (Samuel)
...
Assume ℓ(M/qM ) < ∞
...
(20
...
1)
Proof: Set a := Ann(M )
...
Then R′ /q′
is Noetherian as R is
...
So G• R′ is generated as an R′ /q′ -algebra by
finitely many elements of degree 1, and G• M is a finitely generated G• R′ -module
by (20
...
Therefore, each F n M/F n+1 M is a finitely generated R′ /q′ -module by
(20
...
12)
...
1) and (13
...
31) yield, respectively,
∩
∩
V(a + q) = V(a) V(q) = Supp(M ) V(q) = Supp(M/qM )
...
4) as ℓ(M/qM ) < ∞
...
But R′ /q′ is Noetherian
...
11)
...
14)
...
Then Additivity of Length, (19
...
So induction on n yields ℓ(M/F n+1 M ) < ∞ for every n
...
□
Corollary (20
...
— Under the conditions of (20
...
Then P (F • M, t) can be written uniquely in the form
/
(20
...
1)
P (F • M, t) = e(t) tl−1 (1 − t)d+1
with e(t) ∈ Z[t] and e(0), e(1) ̸= 0 and l ∈ Z and r ≥ d ≥ 0; also, there is a
polynomial p(F • M, n) ∈ Q[n] with degree d and leading coefficient e(1)/d ! such that
ℓ(M/F n M ) = p(F • M, n)
for n ≥ deg e(t) − l
...
14
...
18)
Finally, pq (M, n)−p(F • M, n) is a polynomial with degree at most d−1 and positive
leading coefficient; also, d and e(1) are the same for every stable q-filtration
...
13) shows that G• R′ and G• M satisfy the hypotheses
of (20
...
So (20
...
1) and (20
...
1) yield (20
...
1)
...
13
...
14
...
8)
...
Dividing into M and extracting lengths, we get
ℓ(M/F n M ) ≤ ℓ(M/qn M ) ≤ ℓ(M/F n+m M )
...
14
...
The two extremes are polynomials in n with the same degree d and the same leading
coefficient c where c := e(1)/d !
...
Thus the degree and leading coefficient are the same for every stable q-filtration
...
□
Exercise (20
...
— Let R be a Noetherian ring, q an ideal, and M a finitely
√
generated module
...
Set m := q
...
(20
...
— Let R be an arbitrary ring, q an ideal
...
Let M be a module with a q-filtration F • M
...
Lemma (20
...
— Let R be a Noetherian ring, q an ideal, M a finitely generated
•
module with a q-filtration F • M
...
Also, F M is
stable if and only if R(F • M ) is module finite over R(q) and F n M = M
...
, xr
...
These r + 1 elements generate R(q) over R
...
∪ n that F M is stable: say
•
Then F M = M
...
, F ν M over R(q)
...
Thus R(F • M ) is a finitely generated R(q)-module
...
, ms
...
Then given n, any
∑
m ∈ F n M can be written as m
=
fij mij with fij ∈ Rn−j (q)
...
Suppose F n M = M
...
But if j ≤ ν ≤ n,
then fij ∈ qn−j = qn−ν qν−j
...
Thus F • M is stable
...
20)
121
Lemma (20
...
— Let R be a Noetherian ring, M a finitely generated module, N a submodule, q an ideal, F • M a stable q-filtration
...
•
Then the F N form a stable q-filtration F N
...
17), the extended Rees Algebra R(q) is finitely generated over
R, so Noetherian by the Hilbert Basis Theorem (16
...
By (20
...
19)
...
∪ n
But F M = M , so F n N = N
...
17)
...
19)
...
29), from the
Artin–Rees Lemma, (20
...
Proposition (20
...
— Let R be a Noetherian ring, q an ideal, and
0 → M ′ → M → M ′′ → 0
an exact sequence of finitely generated modules
...
If so, then the polynomial
pq (M ′ , n) − pq (M, n) + pq (M ′′ , n)
has degree at most deg pq (M ′ , n) − 1 and has positive leading coefficient; also then
deg pq (M, n) = max{ deg pq (M ′ , n), deg pq (M ′′ , n) }
...
31) and (13
...
31) again yield
(
)∩
∩
∪
Supp(M/qM ) = Supp(M ) V(q) = Supp(M ′ ) Supp(M ′′ )
V(q)
(
) ∪(
)
∩
∩
′
′′
= Supp(M ) V(q)
Supp(M ) V(q)
∪
′
′
= Supp(M /qM ) Supp(M ′′ /qM ′′ )
...
4)
...
Then the F n M ′ form a stable q-filtration
•
′
F M by the Artin–Rees Lemma
...
So the Nine Lemma yields this exact sequence:
0 → M ′ /F n M ′ → M/qn M → M ′′ /qn M ′′ → 0
...
Then Additivity of Length and (20
...
(20
...
1)
Hence pq (M ′ , n) − pq (M, n) + pq (M ′′ , n) is equal to pq (M ′ , n) − p(F • M ′ , n)
...
14) again, the latter is a polynomial with degree at most deg pq (M ′ , n) − 1
and positive leading coefficient
...
20
...
But deg p(M•′ , n) = deg pq (M ′ , n) by (20
...
20
...
21) (Homogeneity)
...
We call the Mn the homogeneous components of M
...
Call the finitely many nonzero
mn the homogeneous components of m
...
If n is lowest, call mn the initial component of m
...
Call a map α : M ′ → M of graded modules with components Mn′ and Mn homogeneous of degree r if α(Mn′ ) ⊂ Mn+r for all n
...
Further, Coker(α) is canonically graded, and the
quotient map M → Coker(α) is homogeneous of degree 0
...
22)
...
Assume M = aM
...
⊕
⊕
Exercise (20
...
— Let R =
Rn be ⊕
a Noetherian graded ring, M =
Mn a
finitely generated graded R-module, N =
Nn a homogeneous submodule
...
Show that N ′ is the largest homogeneous submodule of M containing N and having,
for all n ≫ 0, its degree-n homogeneous component Nn′ equal to Nn
...
24)
...
Suppose Q possesses this
property: given any homogeneous x ∈ R and homogeneous m ∈ M with xm ∈ Q
but m ∈
/ Q, necessarily x ∈ p := nil(M/Q)
...
Proof: Given x∑∈ R and m ∈ M ∑
, decompose them into their homogeneous
xm ∈ Q, but m ∈
/ Q
...
Suppose
i≥r xi and m =
∑
Then mt ∈
/ Q for some t; take t minimal
...
Then m′ ∈ Q
...
Then xm′′ ∈ Q
...
But Q is homogeneous
...
But mt ∈
/ Q
...
Say xs ,
...
Set x′ := i=s xi
...
So x′k ∈ Ann(M/Q) for some
′k ′′
′′
k ≥ 1
...
Set x := x − x′
...
Suppose x ∈
/ p
...
And its initial component is xv with v > u
...
But Q is homogeneous
...
But mt ∈
/ Q
...
□
Thus x ∈ p
...
4)
...
25)
...
Prove that a and Ann(M ) and nil(M ) are homogeneous
...
26)
...
Let Q∗ ⊂ Q be the submodule generated by the homogeneous
elements of Q
...
122
Appendix: Homogeneity (20
...
27)
...
Then all the associated primes of
M/N are homogeneous, and N admits an irredundant primary decomposition in
which all the primary submodules are homogeneous
...
21)
...
Let Q∗j ⊂∩Qj be the
∩ submodule∩generated
∗
∗
∗
Trivially, Qj ⊂ Qj = N ⊂ Qj
...
26)
...
And,
owing
to (18
...
Finally, M/Q∗j is graded
by (20
...
20) and (20
...
□
⊕
(20
...
— Let R = n≥0 Rn be a graded domain, and set
K := Frac(R)
...
Clearly, n is well defined
...
Clearly, the
⊕ set of all such z, plus 0
...
Thus n≥0 Kn is a graded subring of
K
...
The n with Kn ̸= 0 form a subgroup of Z
...
Fix any nonzero x ∈ K1
...
If z ∈ Kn ,
then z/xn ∈⊕
K0
...
So (2
...
Any
w
∈
Kn can be written w = a/b with a, b ∈ ∏
R and b homogeneous:
say
∑
∑
w = (an /bn ) with an , bn ∈ R homogeneous; set b := bn and a := (an b/bn )
...
29)
...
Then R is a graded subring of K
...
28)
...
34)
...
So every y ∈ R can be written as
∑r+n
y = i=r yi , with yi homogeneous and nonzero
...
Since y is integral over R, the R-algebra R[y] is module finite by (10
...
So
(20
...
Hence by j ∈ R for all j ≥ 0
...
Hence byrj ∈ R
...
Then yrj ∈ Rz
...
Hence yr ∈ R
...
Thus yi ∈ R for all i by induction on n
...
□
Exercise (20
...
— Under the conditions of (20
...
(1) Prove that there is a homogeneous f ∈ R with Rf = Rf
...
21
...
The Dimension Theorem, which we prove, characterizes the
dimension of a nonzero finitely generated semilocal module over a Noetherian ring
in two ways
...
Second, the dimension is the smallest number
of elements in the radical that span a submodule of finite colength
...
We bound the height by the minimal number of
generators of an ideal over which the prime is minimal
...
Finally, we study regular
local rings: Noetherian local rings whose maximal ideal has the minimum number
of generators, namely, the dimension
...
1) (Dimension of a module)
...
The dimension of M , denoted dim(M ), is defined by this formula:
dim(M ) := sup{ r | there’s a chain of primes p0 ⫋ · · · ⫋ pr in Supp(M ) }
...
Then M has finitely
many minimal (associated) primes by (17
...
They are also the minimal primes
p0 ∈ Supp(M ) by (17
...
Thus (1
...
(21
...
1)
(21
...
— Let R be a ring, M a nonzero module
...
If there are only finitely many such maximal ideals, call M semilocal
...
Assume M is finitely generated
...
27)(3)
...
Assume, in addition, R is Noetherian; so M is Noetherian by (16
...
Fix an
ideal q
...
6), M/qM is Artinian if and only if ℓ(M/qM ) < ∞
...
4) and (17
...
Also, by (13
...
27)(3), and (13
...
Set q′ := Ann(M ) + q
...
11), if and only if R/q′ is Artinian
...
18) implies that R/q′ is Artinian if and only if q′ contains a product of
maximal ideals each containing q′
...
Set m := rad(M )
...
Assume, in addition, M is semilocal, so that Supp(M ) contains only finitely many
maximal ideals
...
Thus, conversely, if q′ ⊃ mn
for some n > 0, then R/q′ is Artinian
...
2
...
33) if and only if m = q , or by (13
...
In
particular, mn is a parameter ideal for any n
...
4)
125
Assume q is a parameter ideal
...
14)
...
15) since m = q′ and pq′ (M, n) = pq (M, n)
...
Denote this common degree by d(M )
...
Indeed, that order is 1 less than the order of pole at 1 of the Hilbert–
Samuel series Pq (M, t) by (20
...
In turn, the latter order is d(M )+1 by (20
...
Denote by s(M ) the smallest s such that there are x1 ,
...
, xs ⟩M ) < ∞
...
2
...
We say that x1 ,
...
2
...
Note that a
sop generates a parameter ideal
...
3)
...
Set K := Ker(M −−→ M )
...
(2) Then dim(M/xM ) ≤ dim(M ) − 1 if x ∈
/ p for any p ∈ Supp(M ) with
dim(R/p)(= dim(M )
...
Proof: For (1), set s := s(M/xM )
...
, xs ∈ rad(M/xM ) with
ℓ(M/⟨x, x1 ,
...
Now, Supp(M/xM ) = Supp(M ) ∩ V(⟨x⟩) by (13
...
However, x ∈ rad(M )
...
Therefore,
rad(M/xM ) = rad(M )
...
Thus (1) holds
...
Again,
Supp(M/xM ) = Supp(M ) ∩ V(⟨x⟩) by (13
...
So x ∈ p0 ∈ Supp(M )
...
Hence r ≤ dim(M ) − 1
...
To prove (3), note that xM := Im(µx ), and form these two exact sequences:
0 → K → M → xM → 0,
and 0 → xM → M → M/xM → 0
...
20) yields d(K) ≤ d(M ) and d(xM ) ≤ d(M )
...
20) again, both
pq (K, n) + pq (xM, n) − pq (M, n) and pq (xM, n) + pq (M/xM, n) − pq (M, n) are
of degree at most d(M ) − 1
...
Thus (3) holds
...
4) (Dimension)
...
Then
dim(M ) = d(M ) = s(M ) < ∞
...
Set m := rad(M )
...
We proceed by induction on d(M )
...
Then
ℓ(M/mn M ) stabilizes
...
Hence mn M = 0 by
Nakayama’s Lemma (10
...
Hence
ℓ(M ) < ∞
...
4)
...
By (21
...
1), dim(R/p0 ) = dim(M ) for some p0 ∈ Supp(M )
...
So p0 ∈ Ass(M ) by (17
...
Hence M has a submodule N
isomorphic to R/p0 by (17
...
Further, by (20
...
Take a chain of primes p0 ⫋ · · · ⫋ pr in Supp(N )
...
Suppose r ≥ 1
...
Further, since p0 is not maximal, for
126
Dimension (21
...
Set x := x1 xn
...
Then p1 ⫋ · · · ⫋ pr lies in Supp(N ) V(⟨x⟩)
...
31)
...
However, µx is injective on N as N ≃ R/p0 and x ∈
/ p0
...
3)(3) yields
d(N/xN ) ≤ d(N ) − 1
...
So dim(N/xN ) ≤ d(N/xN ) by the
induction hypothesis
...
Thus dim(M ) ≤ d(M )
...
Let q be a parameter ideal of M with s(M )
generators
...
But deg pq (M, n) ≤ s(M ) owing to
(20
...
Thus d(M ) ≤ s(M )
...
Set r := dim(M ), which is finite since
r ≤ d(M ) by the first step
...
If r = 0, then
M has finite length by (19
...
Suppose r ≥ 1
...
, pk be the primes of Supp(M ) with dim(R/pi ) = r
...
So m lies in no pi
...
19), there
is an x ∈ m such that x ∈
/ pi for all i
...
3)(1), (2) yield s(M ) ≤ s(M/xM ) + 1
and dim(M/xM ) + 1 ≤ r
...
Hence s(M ) ≤ r, as desired
...
5)
...
Then dim(M/xM ) ≥ dim(M ) − 1, with equality if x ∈
/p
for p ∈ Supp(M ) with dim(R/p) = dim(M ); equality holds if x ∈
/ z
...
Proof: By (21
...
So the asserted inequality
holds by (21
...
If x ∈
/ p ∈ Supp(M ) when dim(R/p) = dim(M ), then (21
...
Finally, if x ∈
/ z
...
18) and (17
...
□
Exercise (21
...
— Let A be a Noetherian local ring, N a finitely generated
module, y1 ,
...
Set Ni := N/⟨y1 ,
...
Show dim(Ni ) = r − i
...
7) (Height)
...
The height of p, denoted
ht(p), is defined by this formula:
ht(p) := sup{ r | there’s a chain of primes p0 ⫋ · · · ⫋ pr = p }
...
20)(2) yields this formula:
ht(p) = dim(Rp )
...
7
...
Corollary (21
...
— Let R be a Noetherian ring, p a prime
...
Proof: Assume p is minimal containing an ideal a generated by r elements
...
20)
...
Hence pRp = aRp by the Scheinnullstellensatz
...
2)
...
4), and dim(Rp ) = ht(p)
by (21
...
1)
...
Conversely, assume ht(p) ≤ r
...
, yr by (21
...
1) and (21
...
Say yi = xi /si with si ∈
/ p
...
, xr ⟩
...
Suppose there is a prime q with √
a ⊂ q ⊂ p
...
20)(2)
...
So qRp = pRp
...
17)
127
(11
...
Thus p is minimal containing a, which is generated by r elements
...
9)
...
, xr
...
, xr′ ⟩ and
p′ := p/⟨x1 ,
...
Assume ht(p) = r
...
Theorem (21
...
— Let R be a Noetherian ring, x ∈ R,
and p a minimal prime of ⟨x⟩
...
div(R), then ht(p) = 1
...
8), ht(p) ≤ 1
...
10), x ∈ z
...
□
Exercise (21
...
— Let R be a Noetherian ring, p a prime of height at least 2
...
Exercise (21
...
— Let R be a Noetherian ring
...
(2) R has only finitely many height-1 primes
...
Exercise (21
...
4])
...
Prove the following statements are equivalent:
(1) K = Rf for some nonzero f ∈ R
...
(3) Some nonzero f ∈ R lies in every nonzero prime
...
(5) R is semilocal of dimension 1
...
14)
...
Prove that, if R is a UFD, then every
height-1 prime is principal, and that the converse holds if R is Noetherian
...
15)
...
Prove A is a domain by showing any prime q ⫋ p is ⟨0⟩
...
Set R := P × P
...
Exercise (21
...
— Let R be a finitely generated algebra over a field
...
Let x ∈ R be neither 0 nor a unit
...
Prove that r − 1 is the length of any chain of primes in R′ of maximal length
...
17)
...
Let φ : A → B be a local homomorphism
...
Proof: Set s := dim(A)
...
4), there is a parameter ideal q generated by s
elements
...
2
...
Hence mB/qB is nilpotent
...
But (21
...
Thus
the inequality holds
...
Let p ⊃ mB be a prime with dim(B/p) = dim(B/mB)
...
Hence it suffices to show
128
Dimension (21
...
As n ⊃ p ⊃ mB and as φ is local, φ−1 (p) = m
...
11)
and induction yield a chain of primes of B descending from p and lying over any
given chain in A
...
□
Exercise (21
...
— Let R be a Noetherian ring
...
Exercise (21
...
— Let A be a Noetherian local ring of dimension r
...
Prove that
r ≤ dimk (m/m2 ),
with equality if and only if m is generated by r elements
...
20) (Regular local rings)
...
We say A is regular if its maximal ideal is generated by r elements
...
By (21
...
For example, a field is a regular local ring of dimension 0, and conversely
...
13) and (15
...
6)
...
21)
...
Then deg h(G• A, n) = r − 1
...
8), deg h(G• A, n) is equal to 1 less than the order of pole at 1
of the Hilbert series H(G• A, t)
...
2)
...
4)
...
□
Proposition (21
...
— Let A be a Noetherian local ring of dimension r, and m
its maximal ideal
...
Proof: Say G• A is a polynomial ring in s variables
...
By
(20
...
So s = r by (21
...
So A is regular by (21
...
Conversely, assume A is regular
...
, xr be a regular sop, and x′i ∈ m/m2
the residue of xi
...
, Xr ] be the polynomial
ring
...
⊕
Then φ is surjective as the x′i generate G• A
...
Let P =
Pn be
•
the grading by total degree
...
So a
⊕
inherits a grading: a =
an
...
(21
...
1)
Suppose a ̸= 0
...
Take n ≥ m
...
Since P is a domain, Pn−m −→
n−m f
...
22
...
The expression on the right is a polynomial in n of degree r − 2
...
8)
...
21)
...
26)
129
preceding paragraph that deg h(G• A, n) ≤ r − 2
...
Thus φ is injective, so bijective, as desired
...
23)
...
, xs ∈ A with s ≤ r
...
, xs ⟩ and B := A/a
...
, xr ∈ A with x1 ,
...
(2) B is regular of dimension r − s
...
24)
...
Proof: Use induction on r := dim A
...
Assume r ≥ 1
...
Then A/⟨x⟩ is regular of
dimension r − 1 by (21
...
By induction, A/⟨x⟩ is a domain
...
Thus A is a domain by (21
...
□
Lemma (21
...
— Let A be a local ring, m its maximal ideal, a a proper ideal
...
Then this sequence of k-vector spaces is exact:
0 → (m2 + a)/m2 → m/m2 → n/n2 → 0
...
□
Proposition (21
...
— Let A be a regular local ring of dimension r, and a an
ideal
...
Then a is
generated by s elements, and any such s elements form part of a regular sop
...
25) yields dim((m2 + a)/m2 ) = s
...
Let b be the ideal the
s generate
...
23)
...
24), both
A/b and B are domains of dimension r − s; whence, (15
...
□
22
...
First,
we discuss the topology of a filtration, and use Cauchy sequences to construct the
completion
...
We conclude that, if we use the adic
filtration of an ideal, then the functor of completion is exact on finitely generated
modules over a Noetherian ring
...
We end with a
useful version of the Cohen Structure Theorem for complete Noetherian local rings
...
1) (Topology and completion)
...
Then M has a topology: the open sets are the arbitrary unions
of sets of the form m + F n M for various m and n
...
′
(22
...
1)
′
The addition map M × M → M , given by (m, m ) 7→ m + m , is continuous, as
(m + F n M ) + (m′ + F n M ) ⊂ (m + m′ ) + F n M
...
(Similarly, inversion m 7→ −m is a homeomorphism; so M is a topological group
...
If the filtration on M is an
a-filtration, then scalar multiplication (x, m) 7→ xm too is continuous, because
(x + an )(m + F n M ) ⊂ xm + F n M
...
Thus any two stable a-filtrations give the same topology: the a-adic topology
...
Further, if R is semi-local, then it
is conventional to take a := rad(R)
...
Its closure N is equal to n (N + F n M ), as m ∈
means there’s n with (m + F n M ) ∩ N = ∅, or equivalently m ∩
∈
/ (N + F n M )
...
Also, M is separated — that is, Hausdorff — if and only if {0} is closed
...
So given m′ ̸= m, there’s n′ with m ∈
/ (m′ + F n M )
...
Then (m + F n M ) ∩ (m′ + F n M ) = ∅ owing to (22
...
1)
...
A sequence (mn )n≥0 in M is called Cauchy if, given n0 , there’s n1 with
mn − mn′ ∈ F n0 M,
or simply mn − mn+1 ∈ F n0 M,
for all n, n′ ≥ n1 ;
the two conditions are equivalent because F n0 M is a subgroup and
130
Completion (22
...
An m ∈ M is called a limit of (mn ) if, given n0 , there’s n1 with m−mn ∈ F n0 M
for all n ≥ n1
...
The Cauchy sequences form a module under termwise addition and scalar multiplication
...
The quotient
c and called the (separated) completion
...
κ: M → M
If M is complete, but not separated, then κ is surjective, but not bijective
...
Further, M
kc
k
b
F M := (F M ) where (F k M )b is the completion of F k M arising from the intersections F k M ∩ F n M for all n
...
b
Again, let a be an ideal
...
b
is a ring, κ : R → R
Further,
c is a linear functor from ((R-mod)) to ((R-mod))
...
, Xr ] the polynomial ring in
r variables
...
, Xr ⟩
...
, Xr ]]
...
Thus R
For another example, take a prime integer p, and set a := ⟨p⟩
...
The completion of Z is
∞
called the p-adic integers, and consists of the sums i=0 zi pi with 0 ≤ zi < p
...
Proposition (22
...
— Let R be a ring, and a an ideal
...
1) that R
a-adic topology
...
2)
...
a ⊂ rad(R
Exercise (22
...
— In the 2-adic integers, evaluate the sum 1 + 2 + 4 + 8 + · · ·
...
4)
...
Prove that
the following three conditions are equivalent:
∩
c is injective; (2)
(1) κ : M → M
an M = ⟨0⟩; (3) M is separated
...
Assume either (a) a ⊂ rad(R)
c
...
Conclude M ⊂ M
(22
...
— Let R be a ring
...
Given Qn ∏
and αnn+1∏for all n ∈ Z, use only those for n in the present context
...
Then
lim Qn = Ker θ
...
(22
...
1)
←−
←−
n+1
Plainly, lim Qn has this UMP: given maps βn : P → Qn with αn βn+1 = βn ,
←−
there’s a unique map β : P → lim Qn with πn β = βn for all n
...
8)
Further, the UMP yields the following natural R-linear isomorphism:
lim Hom(P, Qn ) = Hom(P, lim Qn )
...
)
For example, let R′ be a ring, and R := R′ [X1 ,
...
Set m := ⟨X1 ,
...
Then Rn is just the R-algebra
of polynomials of degree at most n, and the canonical map αnn+1 : Rn+1 → Rn is
just truncation
...
, Xr ]]
...
Then
n
Zn is just the ring of sums i=0 zi pi with 0 ≤ zi < p, and the canonical map
n+1
αn : Zn+1 → Zn is just truncation
...
←−
Exercise (22
...
— Let R be a ring
...
We say
the Qn satisfy the Mittag-Leffler Condition if the descending chain
Qn ⊃ αnn+1 Qn+1 ⊃ αnn+2 Qn+2 ⊃ · · · ⊃ αnm Qm ⊃ · · ·
stabilizes; that is, αnm Qm = αnm+k Qm+k for all k > 0
...
Show lim1 Qn = 0
...
Show lim1 Qn = 0
...
Set Pn := m≥n αnm Qm ,
which is the stable submodule
...
(4) Assume the Qn satisfy the Mittag-Leffler Condition
...
←−
Lemma (22
...
— For n ≥ 0, consider commutative diagrams with exact rows
′
γn+1
γn+1
γ′
γn
→0
0−
→ Q′n+1 −−−→ Qn+1 −−−→ Q′′n+1 −
′n+1
n+1
′′n+1
αn
αn y
αn
y
y
n
0 −−→ Q′n −−−−
−→ Qn −−−−−→ Q′′n −−→ 0
Then the induced sequence
γb′
γ
b
0 → lim Q′n −→ lim Qn −
→ lim Q′′n
(22
...
1)
←−
←−
←−
′
is exact; further, γ
b is surjective if the Qn satisfy the Mittag-Leffler Condition
...
5
...
13) yields the exact sequence (22
...
1)
and an injection Coker γ
b ,→ lim1 Q′n
...
Thus γ
b is surjective
...
Then lim Qn = 0 by (22
...
So Coker γ
←−
Proposition (22
...
— Let R be a ring, M a module, F • M a filtration
...
c −→
M
←−
Completion (22
...
Given a Cauchy
Proof: First, let us define a map α : M
←−
sequence (mν ), let qn be the residue of mν in M/F n M for ν ≫ 0
...
Clearly, qn is the residue of qn+1
in M/F n M
...
Define α by
αmν := (qn )
...
As to surjectivity, given (qn ) ∈ lim(M/F n M ), for each ν lift qν ∈ M/F ν M
←−
up to mν ∈ M
...
Hence (mν ) is Cauchy
...
□
Example (22
...
— Let R be a ring, M a module, F • M a filtration
...
By (22
...
←−
But κ is not surjective when M is not complete; for examples of such M , see the
end of (22
...
Thus lim is not always exact, nor lim1 always 0
...
10)
...
, mm be maximal ideals
...
b = ∏A
m := mi , and give A the m-adic topology
...
11)
...
Give N and M/N the induced filtrations:
F n N := N ∩ F n M
and
F n (M/N ) := F n M/F n N
...
(1) Prove N
n
c/N
b = M/N and G• M
c = G• M
...
Prove M
Exercise (22
...
— (1) Let R be a ring, a an ideal
...
If also ∩
R
n≥0 a = 0, show R is a domain
...
b is a local
Proposition (22
...
— Let A be a ring, m a maximal ideal
...
b m
b ⊃ m
b = A/m by (22
...
Next, rad(A)
b
Proof: First, A/
′
′
b
b
by (22
...
Finally, let m be any maximal ideal of A
...
Hence
b Thus m
b is the only maximal ideal
...
□
Exercise (22
...
— Let A be a Noetherian local ring, m the maximal ideal, M
b is a Noetherian local ring with m
b as
a finitely generated module
...
if A
Exercise (22
...
— Let A be a ring, and m1 ,
...
Set
∩
b is a semilocal ring,
m :=
mi and give A the m-adic topology
...
b 1,
...
19)
(22
...
— Let R be a ring, a an ideal, and
b the canonical map
...
Let’s show that κ(t) is a unit if and only if each tn is
...
Then each tn
Indeed, by (22
...
Hence tn is a unit if κ(t) is
...
Then there are un ∈ R with un t ≡ 1 (mod an )
...
Set u := (un ) ∈ R/an
...
Thus κ(t) is a unit
...
Then by the above, T consists of the t ∈ R whose residue
Set T := κ−1 (R
n
tn ∈ R/a is a unit for each n
...
31) and (1
...
(22
...
1)
Set S := 1 + a
...
16
...
Hence the UMP of localization (11
...
6), (11
...
22) yield:
R/an = S −1 R/an S −1 R = T −1 R/an T −1 R
...
8), equal to the completion of each of S −1 R and T −1 R in
Therefore, R
−1
their aS R-adic and aT −1 R-adic topologies
...
Then T = R − m by (22
...
1)
...
Thus R
Finally, assume R is Noetherian
...
Indeed,
say τ σ(x/s) = 0
...
So x ∈ an
...
29) or (20
...
So
x/s = 0 in S −1 R
...
Similarly, τ is injective
...
17) (Exactness of Completion)
...
an ideal
...
Set F n M ′ := M ′ ∩ an M
...
18), the
F n M ′ form an a-stable filtration
...
1)
...
7) and (22
...
□
Exercise (22
...
— Let A be a Noetherian semilocal ring
...
b
x ∈ A is a nonzerodivisor on A if and only if its image x
b∈A
Exercise (22
...
— Let p ∈ Z be prime
...
⊕
⊕
Set A := n≥1 Z/⟨p⟩ and B := n≥1 Z/⟨pn ⟩
...
b
(1) Show that the p-adic completion A is just A
...
⊕
Completion (22
...
)
is not exact at B
...
20)
...
Then the natural map is an isomorphism:
∼ M
b ⊗ M −→
c
...
17), the functor M 7→ M
generated modules, and so (8
...
□
c preserves
Exercise (22
...
— Let R be a ring, a an ideal
...
Corollary (22
...
— Let R be a Noetherian ring, a and b ideals, M a finitely
generated module
...
Proof: In general, the inclusion bM → M induces a commutative square
b ⊗ (bM ) −
b⊗M
R
→R
y
y
c
(bM )b −−−−→ M
b ⊗ M )
...
20), and the
c
...
17)
...
Thus (1) holds
...
Hence bM = bR M = b
bM
n
n b
nb
b
In (1), taking b for b and R for M yields (b ) = b R
...
But bn R′ = (bR′ )n for any R-algebra R′
...
□
b is flat
...
23)
...
Then R
b⊗b = b
b by
Proof: Let b be any ideal
...
20), and b
b = bR
b is flat by the Ideal Criterion (9
...
(22
...
Thus R
□
b is
Exercise (22
...
— Let R be a Noetherian ring, a an ideal
...
Exercise (22
...
— Let R be a Noetherian ring, and a and b ideals
...
a ⊂ rad(R), and use the a-adic topology
...
26)
...
Assume αF n M ⊂ F n N for all n
...
If the induced map G• α is injective or surjective, then so is α
b
...
31)
Its rows are exact
...
13) yields this exact sequence:
Ker Gn α → Ker αn+1 → Ker αn → Coker Gn α → Coker αn+1 → Coker αn
...
Then Ker Gn α = 0
...
So
by induction Ker αn = 0 for all n
...
7) and (22
...
Assume G• α is surjective, or Coker Gn α = 0
...
But N/F n N = 0 for n ≪ 0
...
So
α
n
0 → Ker αn → M/F n M −−→
N/F n N → 0
is exact
...
7) and (22
...
□
Lemma (22
...
— Let R be a ring, a an ideal, M a module, F • M an a-filtration
...
Assume G• M
is module finite over G• R
...
Proof: Take finitely many generators µi of G• M , and replace them by their
homogeneous components
...
Lift µi to mi ∈ F ni M
...
Set E := i R[−ni ]
...
Then F n E = E for n ≪ 0
...
Then αF n E ⊂ F n M for all n
...
So α
b is surjective by (22
...
Form the following canonical commutative diagram:
κE
b
E
−−−→ E
αy
α
by
κ
c
M −−M
→M
b is surjective by (22
...
Thus
As R is complete, κR : R → R
κM is surjective; that is, M is complete
...
4)
...
So α is surjective
...
□
Exercise (22
...
— Let R be a ring,
a an ideal, and M a module
...
Show
m1 ,
...
, m′n in M/aM generate
...
29)
...
Assume
R is complete, and M separated
...
Then
M is a Noetherian R-module, and every submodule N is complete
...
Then N is separated, and F n N = N for n ≪ 0
...
However, G• M is Noetherian
...
Thus N is
□
complete and is module finite over R by (22
...
Thus M is Noetherian
...
30)
...
If R is Noetherian, so is R
...
Then G• R is algebra finite over R/a by
b
(20
...
12)
...
29) with R
b for R and R
b for M
...
11)
...
33)
137
Example (22
...
— Let k be a Noetherian ring, P := k[X1 ,
...
, Xr ]] the formal power series ring
...
, Xr ⟩-adic topology by (22
...
Further, P is Noetherian
by the Hilbert Basis Theorem, (20
...
Thus A is Noetherian by (22
...
Assume k is a domain
...
Indeed, A is one if r = 1, because
(am X1m + · · · )(bn X1n + · · · ) = am bn X1m+n + · · ·
...
, Xi ]] [[Xi+1 ,
...
Set pi := ⟨Xi+1 ,
...
Then A/pi = k[[X1 ,
...
10)
...
So 0 = pr ⫋ · · · ⫋ p0 is a chain of primes of length r
...
Assume k is a field
...
, Xr ⟩ and with
residue field k by the above and either by (22
...
10)
...
19)
...
Theorem (22
...
— Let R be a ring, R′ an
R-algebra, b an ideal of R′ , and x1 ,
...
Let P := R[[X1 ,
...
If R′ is separated and complete in the b-adic topology,
then there is a unique R-algebra map π
b : P → R′ with π
b(Xi ) = xi for 1 ≤ i ≤ n
...
, Xn ] → R′ with π(Xi ) = xi
...
Alternatively, for each m, the map π induces a map
P/⟨X1 ,
...
, Xn ]/⟨X1 ,
...
Taking inverse limits yields π
b owing to (22
...
8)
...
33) (Cohen Structure)
...
Assume that A contains a coefficient field k; that
∼ A/m
...
, X ]]/a for some variables X and ideal a
...
, Xr ]]
...
, xn ∈ m
...
, Xn ]] → A be the
map with π(Xi ) = xi of (22
...
Then G• π is surjective
...
by (22
...
Set a := Ker(π)
...
, Xn ]]/a −→
Assume A is regular of dimension r
...
Then G•)A is a polynomial
ring in r variables over k by (21
...
And G• k[[X1 ,
...
5)
...
4) with G• A for both R and M
...
π is bijective by (22
...
Thus k[[X1 ,
...
Discrete Valuation Rings
A discrete valuation is a homomorphism from the multiplicative group of a field
to the additive group integers such that the value of a sum is at least the minimum
value of the summands
...
We characterize these rings in
various ways; notably, we prove they are the normal Noetherian local domains of
dimension 1
...
Finally, we prove Serre’s Criterion for normality of Noetherian domains
...
(23
...
— Let K be a field
...
(23
...
1)
Condition (1) just means v is a group homomorphism
...
(23
...
2)
As a convention, we define v(0) := ∞
...
Clearly, A is a subring, so a domain, and m is an ideal
...
We call A the discrete valuation ring (DVR) of v
...
Hence, Frac(A) = K
...
Indeed, if x ∈ A× , then v(x) ≥ 0 and −v(x) = v(x−1 ) ≥ 0; so v(x) = 0
...
Therefore, by
the nonunit criterion, A is a local domain, not a field, and m is its maximal ideal
...
Such a t is irreducible, as t = ab with v(a) ≥ 0 and v(b) ≥ 0 implies v(a) = 0 or
v(b) = 0 since 1 = v(a) + v(b)
...
In particular,
t1 is uniformizing parameter if and only if t1 = ut with u ∈ A× ; also, A is a UFD
...
×
(23
...
3)
Indeed, given a nonzero x ∈ a, say x = ut where u ∈ A
...
So n ≥ m
...
Then y ∈ A and x = ytm , as desired
...
Thus A is regular local of dimension 1
...
2)
...
Let
∑k be a field, t a variable,
and K := k((t)) the field of formal Laurent series x := i≥n ai ti with n ∈ Z and
with ai ∈ k and an ̸= 0
...
Clearly, v is
a discrete valuation, the formal power series ring k[[t]] is its DVR, and m := ⟨t⟩ is
its maximal ideal
...
5)
139
The preceding example can be extended to cover any DVR A that contains a
∼ A/⟨t⟩ where t is a uniformizing power
...
4), and A
of its completion A
b restricts to that on A
...
33)
...
Let p ∈ Z be prime
...
Set v(x) := n
...
We call
v the p-adic valuation of Q
...
3)
...
Assume that m
is nonzero and principal and that n≥0 mn = 0
...
Proof: Given a nonzero x ∈ A, there is an n ≥ 0 such that x ∈ mn − mn+1
...
Then x = utn , and u ∈
/ m, so u ∈ A×
...
Given
×
m
x ∈ K , write x = y/z where y = bt and z = ctk with b, c ∈ A×
...
Define v : K × → Z by v(x) := n
...
Thus v is well defined
...
To verify (23
...
1), take x = utn and
y = wth with u, w ∈ A×
...
Thus (1) holds
...
Then x + y = th (utn−h + w)
...
Thus (2) holds
...
Clearly, A is the DVR of v
...
4) (Depth)
...
, xn ∈ R
...
, xi ⟩M
...
, xn is M -regular, or is an
M -sequence, and we call n its length if Mn ̸= 0 and xi ∈
/ z
...
Call the supremum of the lengths n of the M -sequences found in an ideal a the
depth of a on M , and denote it by depth(a, M )
...
If M is semilocal, call the depth of rad(M ) on M simply the depth of M and
denote it by depth(M )
...
11)
...
When
R is semilocal, call R Cohen–Macaulay if R is a Cohen–Macaulay R-module
...
5)
...
(1) Then depth(M ) = 0 if and only if m ∈ Ass(M )
...
div(M ) and
m ∈ Ass(M/xM )
...
Proof: Consider (1)
...
div(M ) and so depth(M ) = 0
...
Then m ⊂ z
...
Since A is Noetherian,
z
...
15)
...
21)
...
19)
...
Assume depth(M ) = 1
...
So there is an x ∈ m with x ∈
/ z
...
Then m ∈ Ass(M/xM ) by (1)
...
10)
Conversely, assume there is x ∈ m with x ∈
/ z
...
Then depth(M ) ≥ 1 by
definition
...
Then given any y ∈ m with y ∈
/ z
...
27)
...
So there is no
z ∈ m such that y, z is an M -sequence
...
Thus depth(M ) = 1
...
Given any M -sequence x1 ,
...
, xi ⟩
...
5)
...
But
depth(M ) := sup{n}
...
□
Exercise (23
...
— Let R be a ring, M a module, and x, y ∈ R
...
Prove that, given any m, n ∈ M with
xm = yn, there exists p ∈ M with m = yp and n = xp
...
div(M )
...
(3) Assume that R is local, that x, y lie in its maximal ideal m, and that M is
nonzero and Noetherian
...
Prove that x, y form an M -sequence
...
7)
...
20))
...
Exercise (23
...
— Let R be a local ring, m its maximal ideal, M a Noetherian
module, x1 ,
...
, n
...
, xn form
an M -sequence, and prove xσ1 ,
...
Exercise (23
...
— Prove that a Noetherian local ring A of dimension r ≥ 1 is
regular if and only if its maximal ideal m is generated by an A-sequence
...
Theorem (23
...
— Let A be a local ring, m its
maximal ideal
...
Then these five conditions are equivalent:
(1) A is a DVR
...
(3) A is a normal domain of depth 1
...
(5) m is principal and of height at least 1
...
Then A is UFD by (23
...
33)
...
Thus (2) holds
...
1)
...
(
)
Assume (2)
...
Then A/⟨x⟩ ̸= 0, so Ass A/⟨x⟩ ̸= ∅ by
(17
...
Now, A is a local domain of dimension 1
...
But ⟨0⟩ ∈
/ Ass(A/⟨x⟩)
...
Thus (23
...
Assume (3)
...
5)(2), there are x, y ∈ m such that x is nonzero and y has
residue y ∈ A/⟨x⟩ with m = Ann(y)
...
Set z := y/x ∈ Frac(A)
...
Suppose zm ⊂ m
...
23)
...
So y = zx ∈ ⟨x⟩, a contradiction
...
Given w ∈ m, therefore w = (wz)t with wz ∈ A
...
Finally, ht(m) ≥
∩ 1 because x ∈ m and x ̸= 0
...
Assume (5)
...
The Krull Intersection Theorem (18
...
Then 1 + x ∈ A×
...
Further, A is a domain by
Discrete Valuation Rings (23
...
15)(1)
...
3)
...
11)
...
Prove A is a maximal proper subring of K
...
Exercise (23
...
— Let k be a field, P := k[X, Y ] the polynomial ring in two
variables, f ∈ P an irreducible polynomial
...
Set R := P/⟨f ⟩ and
p := ⟨X, Y ⟩/⟨f ⟩
...
(Thus Rp is a DVR
if and only if the plane curve C : f = 0 ⊂ k 2 is nonsingular at (0, 0); see (23
...
)
Exercise (23
...
— Let k be a field, A a ring intermediate between the polynomial ring and the formal power series ring in one variable: k[X] ⊂ A ⊂ k[[X]]
...
Prove that A is a DVR
...
)
Exercise (23
...
— Let L/K be an algebraic extension of fields, X1 ,
...
, Xn
...
Prove ht(p) = ht(q)
...
Prove
that f and g have no common prime factor q ∈ Q
...
15) (Serre’s Conditions)
...
We say Serre’s
Condition (Rn ) holds if, for any prime p of height m ≤ n, the localization Rp is
regular of dimension m
...
Also,
(R1 ) holds if and only if (R0 ) does and Rp is a DVR for any p of height-1
...
Note depth(Mp ) ≤ dim(Mp ) by (23
...
Hence (Sn ) holds if and only if Mp is
Cohen–Macaulay when depth(Mp ) < n
...
15); that is, M has no embedded primes
...
16)
...
Show that M is torsionfree if and only if it satisfies (S1 )
...
17)
...
Show that R is reduced if and
only if (R0 ) and (S1 ) hold
...
18)
...
Set
Φ := { p prime | ht(p) = 1 }
Σ := { p prime | depth(Rp ) = 1 }
...
Further, R = p∈Σ Rp
...
Take 0 ̸= x ∈ q
...
So q ∈ Ass(Rp /⟨x⟩) by (17
...
Hence depth(Rp ) = 1 by (23
...
Thus Φ ⊂ Σ
...
17)
...
Thus
Φ = Σ if and only if (S2 ) holds
...
3); so R ⊂ p∈Σ Rp
...
Say x = a/b with a, b ∈ R and b ̸= 0
...
But p ∈ Σ if pRp ∈ Ass(Rp /bRp ) by (23
...
10)
...
26)
...
□
142
Discrete Valuation Rings (23
...
19)
...
Then
∩
R = p∈Φ Rp where Φ := { p prime | ht(p) = 1 }
...
32)
...
10)
...
18) yields the assertion
...
20) (Serre’s Criterion)
...
Then R
is normal if and only if (R1 ) and (S2 ) hold
...
17)
...
32); whence, (R1 ) and (S2 ) hold by (23
...
Conversely, assume R satisfies (R1 ) and (S2 )
...
Then
x is integral over Rp for any prime p
...
Hence, x ∈ Rp for all p of height 1, so for all p of depth 1 as R
satisfies (S2 )
...
18)
...
□
Example (23
...
— Let k be an algebraically closed field, P := k[X, Y ] the
polynomial ring in two variables, f ∈ P irreducible
...
13)
or by (21
...
Set R := P/⟨f ⟩
...
Let p ⊂ R be a nonzero prime
...
Then 0 ⫋ ⟨f ⟩ ⫋ m is a chain of
primes of length 2, the maximum
...
Hence m = ⟨X − a, Y − b⟩ for some a, b ∈ k by (15
...
Write
f (X, Y ) = ∂f /∂X(a, b)(X − a) + ∂f /∂Y (a, b)(Y − b) + g
where g ∈ m2
...
12) applied after making the change of
variables X ′ := X − a and Y ′ := Y − b
...
Further, R satisfies (R1 ) if and only if Rp is a DVR
for every nonzero prime p
...
(Put geometrically, R is normal if and only if the plane curve C : f = 0 ⊂ k 2 is
nonsingular everywhere
...
Exercise (23
...
— Prove that a Noetherian domain R is normal if and only if,
given any prime p associated to a principal ideal, pRp is principal
...
23)
...
Assuming (S1 ) holds for R, prove Φ ⊂ Σ, and prove Φ = Σ if and only if (S2 ) holds
...
Exercise (23
...
— Let R be a Noetherian ring, and K its total quotient ring
...
Prove these three conditions are equivalent:
(1) R is normal
...
∏
(3) (R1 ) and (S1 ) hold, and R → K → p∈Φ Kp /Rp is exact
...
Appendix: Cohen–Macaulayness
Exercise (23
...
— Let R → R′ be a flat map of Noetherian rings, a ⊂ R an
ideal, M a finitely generated R-module, and x1 ,
...
Set
M ′ := M ⊗R R′
...
Show x1 ,
...
Exercise (23
...
— Let R be a Noetherian ring, a an ideal, and M a finitely
generated module with M/aM ̸= 0
...
, xr be an M -sequence in a and
p ∈ Supp(M/aM )
...
, xr /1 is an Mp -sequence in ap , and
(2) depth(a, M ) ≤ depth(ap , Mp )
...
27) (Finished Sequences)
...
We say an M -sequence in a is finished in a, if it can not be lengthened in a
...
div(M )
...
It is finite if R is Noetherian, M is finitely generated, and
M/aM ̸= 0, as then depth(a, M ) ≤ depth(Mp ) for any p ∈ Supp(M/aM ) by
(23
...
5)(3) and dim(Mp ) < ∞ by (21
...
Proposition (23
...
— Let R be a Noetherian ring, a an ideal, and M a finitely
generated module
...
Let x1 ,
...
Then m = depth(a, M )
...
, yn be a second finished M -sequence in a
...
Induct
on m
...
Then a ⊂ z
...
Hence n = 0 too
...
Set Mi := M/⟨x1 ,
...
, yj ⟩M for all i, j
...
div(Mi ) ∪ j=0 z
...
Then U is equal to the union of all associated primes of Mi for i < m and of
Nj for j < n by (17
...
And these primes are finite in number by (17
...
Suppose a ⊂ U
...
19)
...
div(Mi ) and a ⊂ p ⊂ z
...
Thus a ̸⊂ U
...
Then z ∈
/ z
...
div(Nj ) for j < n
...
div(Mm ) by finishedness
...
26)
...
Moreover, xm and z are nonzerodivisors on Mm−1
...
So q ∈ Ass(Mm−1 /zMm−1 ) by (17
...
Hence
a ⊂ z
...
, xm−1 , z⟩M )
...
, xm−1 , z is finished in a
...
, yn−1 , z is finished in a
...
By (23
...
Thus we may
assume x1 = y1 = z
...
Further, x2 ,
...
, yn are
finished M1 -sequences in a
...
Thus m = n
...
29)
...
Let x ∈ a be a nonzerodivisor on M
...
143
144
Appendix: Cohen–Macaulayness (23
...
30)
...
div(M )
...
Proposition (23
...
— Let R → R′ be a map of Noetherian rings, a ⊂ R an
ideal, and M a finitely generated R-module with M/aM ̸= 0
...
Assume R′ /R is faithfully flat
...
Proof: By (23
...
, xr in a
...
, xi ⟩M and Mi′ := M ′ /⟨x1 ,
...
By (8
...
So M ′ /aM ′ ̸= 0 by faithful flatness
...
, xr is an M ′ -sequence by (23
...
As x1 ,
...
div(Mr )
...
26)
...
20) and (8
...
So HomR′ (R′ /aR′ , Mr′ ) ̸= 0 by faithful flatness
...
div(Mr′ ) by (17
...
So
x1 ,
...
Thus (23
...
□
Exercise (23
...
— Let A be a Noetherian local ring, and M a nonzero finitely
generated module
...
(1) depth(M ) = depth(M
c is Cohen–Macaulay
...
33)
...
Show that there is p ∈ Supp(M/aM ) with
depth(a, M ) = depth(ap , Mp )
...
34)
...
div(M )
...
div(M )
...
Then there is p ∈ Ass(M ′ ) with p ⊃ a
...
Set H := Hom(A/a, M )
...
26) as a ⊂ z
...
Further, H is
finitely generated by (16
...
So H/xH ̸= 0 by Nakayama’s Lemma (10
...
Also,
µx
0 → H −−→ H → Hom(A/a, M ′ ) is exact by (5
...
So Hom(A/a, M ′ ) ̸= 0
...
31)
...
26) yields
the desired p
...
35)
...
Assume that there is
no prime p with pi−1 ⫋ p ⫋ pi for any i
...
Proof: If r = 0, then p0 ⊂ z
...
So depth(p0 , M ) = 0, as desired
...
Assume r ≥ 1
...
17);
so Mpr ̸= 0
...
11) yields Mpr /pr Mpr ̸= 0
...
26)(2)
...
Let x1 ,
...
Then as pr−1 ⊂ pr , clearly
M/pr−1 M ̸= 0
...
28)
...
Set
Ms := M/⟨x1 ,
...
Then pr−1 ⊂ z
...
Suppose pr ⊂ z
...
Then x1 ,
...
So s = depth(pr , M )
Appendix: Cohen–Macaulayness (23
...
28), as desired
...
div(Ms )
...
div(Ms )
...
, xs , x
is an M -sequence in pr
...
34), there is p ∈ Ass(Ms /xMs ) with p ⊃ pr−1
...
Hence pr−1 ⫋ p ⊂ pr
...
Hence x1 ,
...
So (23
...
Thus depth(pr , M ) ≤ r, as desired
...
36) (Unmixedness)
...
Assume M is Cohen–Macaulay
...
Proof: Given p0 ∈ Ass(M ), take any maximal chain of primes p0 ⫋ · · · ⫋ pr
...
So depth(M ) = depth(pr , M )
...
35)
...
And r ≤ dim(M ) by
(21
...
So r = dim(M )
...
Thus M has no embedded primes
...
So p0 ∈ Ass(M ) by (17
...
Thus, as above, r = dim(M ), as desired
...
37)
...
Proposition (23
...
— Let A be a Noetherian local ring, M a finitely generated
module
...
, xn be nonunits of A, and set Mi := M/⟨x1 ,
...
Assume M is Cohen–Macaulay
...
, xn is an M -sequence if and only if
it is part of a sop; if so, then Mn is Cohen–Macaulay
...
, xn is part of a sop
...
For n = 0, the
assertion is trivial
...
By induction x1 ,
...
Now, all maximal chains of primes in Supp(Mn−1 ) have
the same length by (23
...
6)
...
But Mn−1 has no embedded primes by (23
...
So
xn ∈
/ p for any p ∈ Ass(Mn−1 )
...
div(Mn−1 ) by (17
...
Thus x1 ,
...
Finally, as Mn−1 is Cohen–Macaulay, so is Mn by (23
...
Conversely, assume x1 ,
...
By (23
...
, xr
...
30) applied recursively
...
Thus x1 ,
...
□
Proposition (23
...
— Let A be a Noetherian local ring, M a finitely generated
module, p ∈ Supp(M )
...
Assume M is Cohen–Macaulay
...
Proof: Induct on s
...
Then p ⊂ z
...
So p lies in some
q ∈ Ass(M ) by (17
...
But q is minimal in Supp(M ) by (23
...
So q = p
...
Thus Mp is a Cohen–Macaulay Ap -module of dimension 0
...
Then there is x ∈ p − z
...
Set M ′ := M/xM , and set
′
s := depth(p, M ′ )
...
31)
...
29), and
M ′ is Cohen–Macaulay by (23
...
Further, Mp′ = Mp /xMp by (12
...
But
x ∈ p
...
11)
...
Hence by
induction, Mp′ is a Cohen–Macaulay Ap -module of dimension s − 1
...
div(M ), also x ∈
/ z
...
26)(1)
...
30)
...
5)
...
45)
Definition (23
...
— Let R be a Noetherian ring, and M a finitely generated
module
...
It is equivalent that Mp be a Cohen–Macaulay
Rp -module for every p ∈ Supp(M ), because if p lies in the maximal ideal m, then
Rp is the localization of Rm at the prime ideal pRm by (11
...
39)
...
Proposition (23
...
— Let R be a Noetherian ring
...
Proof: First, assume R[X] is Cohen–Macaulay
...
Then P is prime in R[X] by (2
...
Now, R[X]/⟨X⟩ = R
and P/⟨X⟩ = p owing to (1
...
29)(1)
...
22)
yields (R[X]/⟨X⟩)P = R[X]P /⟨X⟩R[X]P
...
But
R[X]P is Cohen–Macaulay by (23
...
30)
...
Conversely, assume R is Cohen–Macaulay
...
Then R[X]M = (R[X]m )M by (11
...
30)
...
Thus, to show R[X]M is Cohen–Macaulay,
replace R by Rm , and so assume R is local with maximal ideal m
...
As R/m is a field,
we may take f monic
...
Set B := R[X]/⟨f ⟩
...
25)
...
12)
...
So dim(R) ≥ dim(BM )
...
7)
...
21)
...
12)
...
25) as
BM /mBM ̸= 0
...
But depth(R) = dim(R) and dim(R) ≥ dim(BM )
...
But the opposite inequality holds by (23
...
Thus BM is Cohen–Macaulay
...
22)
...
So
R[X]M is Cohen–Macaulay by (23
...
Thus R[X] is Cohen–Macaulay
...
42)
...
Theorem (23
...
— A Cohen–Macaulay ring R is universally catenary
...
9)
...
Notice P is Cohen–Macaulay by induction on n, as P = R if n = 0, and the
induction step holds by (23
...
Now, given nested primes q ⊂ p in P , put p in
a maximal ideal m
...
20)
...
37)
...
□
Example (23
...
— Trivially, a field is Cohen–Macaulay
...
By (23
...
Thus these rings are all universally catenary by (23
...
In
particular, we recover (15
...
Appendix: Cohen–Macaulayness (23
...
45)
...
Assume M is Cohen–Macaulay of dimension n
...
Proof: Induct on n
...
20), and so M is free
...
Let t ∈ A be an element of a regular system of parameters
...
23)
...
36); so q = ⟨0⟩ as A is
a domain by (21
...
Hence t is a nonzerodivisor on M by (17
...
So M/tM is
Cohen–Macaulay of dimension n − 1 by (23
...
5)
...
Let k be the residue field of A
...
16)(1)
...
Set p := ⟨t⟩
...
10)
...
39) as depth(⟨t⟩, M ) = 1
...
16)
...
28), so free by (10
...
Set s := rank(Mp )
...
Then Mp ⊗Ap k(p) = Mp /tMp by (8
...
Moreover, Mp /tMp = (M/tM )p by (12
...
So r = s
...
Then Mp ⊗Ap K = M ⊗A K by (11
...
Hence M ⊗A K
has rank r
...
14)
...
Dedekind Domains
Dedekind domains are defined as the normal Noetherian domains of dimension 1
...
Next we prove the Main Theorem of Classical Ideal
Theory: in a Dedekind domain, every nonzero ideal factors uniquely into primes
...
We conclude that a ring of algebraic integers is a Dedekind domain and
that, if a domain is algebra finite over a field of characteristic 0, then in the fraction
field or in any algebraic extension of it, the integral closure is module finite over
the domain and is algebra finite over the field
...
1)
...
Example (24
...
—
of Dedekind domains include the integers Z, the
]
[ √Examples
Gaussian integers Z −1 , the polynomial ring k[X] in one variable over a field,
and any DVR
...
33);
and R is of dimension 1 since every nonzero prime is maximal by (2
...
On the other hand, any local Dedekind domain is a DVR by (23
...
Example (24
...
— Let d ∈ Z be a square-free integer
...
√
Then R is the integral closure of Z in Q( d) by [2, Prp
...
14), p
...
26)
...
12); so dim(R) = 1
...
12) as Z is so and as R := Z + Zη
...
Example (24
...
— Let k be an algebraically closed field, P := k[X, Y ] the
polynomial ring in two variables, f ∈ P irreducible
...
21), R is a Noetherian
domain of dimension 1, and R is Dedekind if and only if ⟨f, ∂f /∂X, ∂f /∂Y ⟩ = 1
...
5)
...
(1) Assume dim(R) = 1
...
(2) Assume dim(R) ≥ 1
...
Exercise (24
...
— Let R be a Dedekind domain, S a multiplicative subset
...
Proposition (24
...
— Let R be a Noetherian domain, not a field
...
148
Dedekind Domains (24
...
6); so Rp is a DVR by (23
...
Conversely, suppose Rp is a DVR for every nonzero prime p
...
Since R is not a field,
□
dim(R) ≥ 1; whence, dim(R) = 1 by (24
...
Thus R is Dedekind
...
8)
...
By first
reducing to the case that R is local, prove that
a ∩ (b + c) = (a ∩ b) + (a ∩ c),
a + (b ∩ c) = (a + b) ∩ (a + c)
...
9)
...
, pr } = Ass(R/a) and qi = aRpi ∩ R for each i
...
21), yields an irredundant primary
∩
decomposition a = qi
...
Then by (18
...
The qi are pairwise comaximal for the following reason
...
Now, pi := qi by (18
...
32)
...
So pi ⊂ m by (2
...
But 0 ̸= a ⊂ pi ; hence, pi is maximal since
dim(R) = 1
...
Similarly, pj = m
...
Thus the ∏
qi are
pairwise comaximal
...
14), yields a = i qi
...
The pi are minimal containing a as dim(R) = 1; so the pi lie
in Ass(R/a)
(17
...
By the above
reasoning, the qi are pairwise comaximal
∏ by ∩
∩
and so
qi = qi
...
19)
...
20), and
qi = aRpi ∩R by the Second Uniqueness Theorem, (18
...
17)(3)
...
10) (Main Theorem of Classical Ideal Theory)
...
Assume R is Dedekind
...
In fact, if vp denotes the valuation of Rp , then
∏
a=
pvp (a) where vp (a) := min{ vp (a) | a ∈ a }
...
9), write a = qi with the qi primary, their primes pi distinct and unique, and qi = aRpi ∩ R
...
7)
...
1
...
But
i
vpi (1/s) = 0
...
Hence mi := vpi (a)
...
10) as pi is maximal; so pi Rpi ∩ R = pi by (18
...
Thus qi = pm
i
...
11)
...
Proof: If R is Dedekind, every primary ideal is a power of its radical by (24
...
Conversely, given a nonzero prime p, set m := pRp
...
So m ̸= m2 by
Nakayama’s Lemma
...
Then m is the only prime containing t, as
dim(Rp ) = 1 by (24
...
So tRp is m-primary by (18
...
Set q := tRp ∩R
...
8)
...
But qRp = tRp by
/ m2
...
So Rp is a DVR by (23
...
(11
...
So tRp = mn
...
7)
...
17)
Exercise (24
...
— Prove that a semilocal Dedekind domain A is a PID
...
Exercise (24
...
— Let R be a Dedekind domain, a and b two nonzero ideals
...
Lemma (24
...
— Let L be a field, G a group, σi : G → L×
distinct homomorphisms
...
∑m
Proof: Suppose there’s an equation i=1 ai σi = 0 with nonzero ai ∈ L
...
Now, σi ̸= 0 as σ∑
Since σ1 ̸= σ2 , there’s an
i : G → L ; so m ≥ 2
...
Then i=1 ai σi (x)σi (y) = i=1 ai σi (xy) = 0 for every
y ∈ G since σi( is a homomorphism
...
Then
m
∑
i=1
bi σ i =
m
∑
1 ∑
ai σi (x)σi = 0
...
□
(24
...
— Let L/K be a finite Galois field extension
...
tr : L → K by tr(x) :=
σ∈Gal(L/K)
Clearly, tr is K-linear
...
14) applied with G := L×
...
(24
...
1)
It is nondegenerate for this reason
...
Now, given x ∈ L× , set y := z/x
...
Lemma (24
...
— Let R be a normal domain, K its fraction field, L/K a finite
Galois field extension, and x ∈ L integral over R
...
Proof: Let xn + a1 xn−1 + · · · + an = 0 be an equation of integral dependence
for x over R
...
Then
(σx)n + a1 (σx)n−1 + · · · + an = 0;
so σx is integral over R
...
Thus
tr(x) ∈ R since R is normal
...
17) (Finiteness of integral closure)
...
Then R′ is module finite over R, and is Noetherian
...
Let z1 ,
...
Using (11
...
Clearly, y1 ,
...
Let x1 ,
...
15
...
Given b ∈ R′ , write b = ci xi with ci ∈ K
...
Then
(∑
) ∑
tr(byj ) = tr
for each j
...
21)
151
∑
But byj ∈ R1′
...
16)
...
Since R is Noetherian,
R′ is module finite over R-module and Noetherian owing to (16
...
□
Corollary (24
...
— Let R be a Dedekind domain, K its fraction field, L/K a
finite separable field extension
...
Proof: First, R′ is module finite over R by (24
...
19)
...
32)
...
12),
and dim(R) = 1 as R is Dedekind
...
□
Theorem (24
...
— A ring of algebraic integers is a Dedekind domain
...
2), Z is a Dedekind domain; whence, so is its integral closure in
any field that is a finite extension of Q by (24
...
□
Theorem (24
...
— Let k be a field
of characteristic 0, and R an algebra-finite domain over k
...
Let
L/K be a finite field extension (possibly L = K), and R′ the integral closure of R
in L
...
Proof: By the Noether Normalization Lemma, (15
...
Then P is normal by Gauss’s Theorem, (10
...
12); also, L/ Frac(P ) is a finite field
extension, which is separable as k is of characteristic 0
...
17), and so R′ is plainly algebra finite over k
...
21) (Other cases)
...
18), even if L/K is inseparable, the integral
closure R′ of R in L is still Dedekind; see (26
...
However, Akizuki constructed an example of a DVR R and a finite inseparable
extension L/ Frac(R) such that the integral closure of R is a DVR, but is not
module finite over R
...
9
...
5]
...
17)
...
20), remains valid in positive characteristic, but the
proof is more involved
...
13), p
...
25
...
A fractional ideal is called invertible if its product with another fractional ideal
is equal to the given domain
...
We prove that, in a Dedekind domain, any two nonzero ordinary ideals
have an invertible fractional ideal as their quotient
...
Further, we prove a Noetherian domain
is Dedekind if and only if every torsionfree module is flat
...
Definition (25
...
— Let R be a domain, and set K := Frac(R)
...
We call M principal if there is an
x ∈ K with M = Rx
...
We call them the product of M and N and the quotient of M by N
...
2)
...
Prove that M is principal if and only if there exists some isomorphism M ≃ R
...
and φ : (M : N ) −→
Proposition (25
...
— Let R be a domain, and K := Frac(R)
...
(2) There exists an x ∈ K × with xM ⊂ R
...
(4) M is finitely generated
...
Further, all four
conditions are equivalent for every M if and only if R is Noetherian
...
Take any nonzero x ∈ b
...
Thus (2) holds
...
Write x = a/b with a, b ∈ R and a, b ̸= 0
...
Thus (3) holds
...
Assume
(4) holds
...
, yn /xn ∈ K × generate M with xi , yi ∈ R
...
Then x ̸= 0 and xM ⊂ R
...
Assume (3) holds and R is Noetherian
...
So xM is finitely
generated, say by y1 ,
...
Then y1 /x,
...
Thus (4) holds
...
If M is ordinary,
then (3) holds with x := 1, and so (4) holds
...
□
152
Fractional Ideals (25
...
4)
...
Let S be a
multiplicative subset
...
∑
Proof: Given x ∈ S −1 (M N
i ∈ M , with ni ∈ N ,
∑), write x = ( mi ni )/s with m−1
and with s ∈ S
...
Thus
S −1 (M N ) ⊂ (S −1 M )(S −1 N )
...
Set s := si and t := ti
...
Thus S −1 (M N ) ⊃ (S −1 M )(S −1 N ), so equality holds
...
Given
y ∈ S −1 N , write y = n/t with n ∈ N and t ∈ S
...
So z ∈ (S −1 M : S −1 N )
...
−1
−1
Conversely, say N is generated by n1 ,
...
Given
∏z ∈ (S M : S N ), write
zni /1 = mi /si with mi ∈ M and si ∈ S
...
Then sz · ni ∈ M
...
Hence z ∈ S −1 (M : N ), as desired
...
5)
...
We call a fractional ideal M locally
principal if, for every maximal ideal m, the localization Mm is principal over Rm
...
6)
...
Prove that
the map π : M ⊗ N → M N is an isomorphism if M is locally principal
...
7) (Invertible fractional ideals)
...
A fractional ideal M
is said to be invertible if there is some fractional ideal M −1 with M M −1 = R
...
Proposition (25
...
— Let R be a domain, M an invertible fractional ideal
...
Proof: Clearly M −1 ⊂ (R : M ) as M M −1 = R
...
Thus (R : M ) ⊂ M −1 , as desired
...
9)
...
(1) Assume N is invertible, and show that (M : N ) = M · N −1
...
Lemma (25
...
— An invertible ideal is finitely generated and nonzero
...
Say 1 =
mi ni with mi ∈ M and
ni ∈ M −1
...
Then m =
mi mni
...
So the mi generate M
...
□
Lemma (25
...
— Let A be a local domain, M a fractional ideal
...
∑
Proof: Assume M is invertible
...
×
As A is local, A − A is an ideal
...
Let m ∈ M
...
Set a := (mnj )(mj nj )−1 ∈ A
...
Thus M = Amj
...
7)
...
19)
Exercise (25
...
— Let R be a UFD
...
Theorem (25
...
— Let R be a domain, M a fractional ideal
...
Proof: Say M N = R
...
10)
...
Then (S −1 M )(S −1 N ) = S −1 R by (25
...
Let m
be a maximal ideal
...
Thus Mm is principal by (25
...
Conversely, set a := M (R : M ) ⊂ R
...
Then
(25
...
In addition, assume Mm is principal and
nonzero
...
7) and (25
...
Hence (13
...
□
Theorem (25
...
— Let R be a Dedekind domain, a, b nonzero ordinary ideals,
M := (a : b)
...
Further, vp (M ) = min{vp (xi )} if the xi generate M
...
So (25
...
Also, each Rp is a DVR by
(24
...
So xMp is principal by (23
...
3)
...
13)
...
10), yields xM = pvp (xM )
∏ Theorem
and xR = pvp (x)
...
Thus (25
...
∑n
Further, given x ∈ M , say x = i=1 ai xi with ai ∈ R
...
1
...
Thus vp (M ) = min{vp (xi )}
...
15)
...
(25
...
— Let R be an arbitrary ring
...
Up to (noncanonical) isomorphism, N is unique if it exists: if N ′ ⊗ M ≃ R, then
N = R ⊗ N ≃ (N ′ ⊗ M ) ⊗ N = N ′ ⊗ (M ⊗ N ) ≃ N ′ ⊗ R = N ′
...
17)
...
Prove that M
is finitely generated, and that, if R is local, then M is free of rank 1
...
18)
...
(2) M is finitely generated, and Mm ≃ Rm at each maximal ideal m
...
Assuming these conditions hold, show that M ⊗ Hom(M, R) = R
...
19)
...
Then the
following conditions are equivalent:
Fractional Ideals (25
...
(2) M is an invertible abstract module
...
Proof: Assume (1)
...
13)
...
6) yields
M ⊗ M −1 = M M −1 by (1)
...
Thus (2) holds
...
18); so (13
...
Finally, assume (3)
...
23), there’s an M ′ with M ⊕ M ′ ≃ R⊕Λ
...
Define φλ : M ,→ R⊕Λ → R to be the composition of the injection with the
projection φλ ∑
on the λth factor
...
∑
Fix a nonzero y ∈ M
...
Set N :=
Rqλ
...
Then
a, c ∈ M ; whence, adφλ∑
(y)φλ (ac) = bcφλ (x)
...
Hence
M · N ⊂ R
...
Thus M · N = R
...
□
Theorem (25
...
— Let R be a domain
...
(2) Every nonzero ordinary ideal a is invertible
...
(4) Every nonzero ordinary ideal a is finitely generated and flat
...
If R is Dedekind, then (25
...
Assume (2)
...
10)
...
Let
p be any nonzero prime of R
...
So by (25
...
So Rp is a DVR by (23
...
Hence R is Dedekind by (24
...
Thus (1) holds
...
By (25
...
But (2) implies that R is Noetherian by
(25
...
Thus (3) and (4) are equivalent by (16
...
51)
...
21)
...
Then R is
Dedekind if and only if every torsionfree module is flat
...
28)
...
Let M be a torsionfree module, m a maximal ideal
...
Let z ∈ Rm be nonzero, and say z = x/s
with x, s ∈ R and s ∈
/ m
...
So
µx : Mm → Mm is injective by the Exactness of Localization
...
So µx/s is injective
...
Since R is Dedekind, Rm is a DVR by (24
...
1)
...
28)
...
Thus by (13
...
Conversely, assume every torsionfree module is flat
...
But R is Noetherian
...
20)
...
22) (The Picard Group)
...
We denote the collection of
isomorphism classes of invertible modules by Pic(R)
...
17), every invertible
module is finitely generated, so isomorphic to a quotient of Rn for some integer n
...
Further, Pic(R) is, clearly, a group under tensor product
156
Fractional Ideals (25
...
We call Pic(R) the Picard Group of R
...
Set K := Frac(R)
...
Set S := R − 0, and form
the canonical map M → S −1 M
...
17) if the multiplication
map µx : M → M is injective for any x ∈ S
...
Let m be a maximal ideal
...
So Mm ≃ Rm
by (25
...
Hence µx : Mm → Mm is injective
...
43)
...
Now, S −1 M is a localization
of Mm , so is a 1-dimensional K-vector space, again as Mm ≃ Rm
...
It yields the desired embedding of M into K
...
19) implies M is also invertible as a fractional ideal
...
Sending an N to
its isomorphism class yields a map κ : F(R) → Pic(R) by (25
...
By the above,
κ is surjective
...
6)
...
We call F(R)/P(R) the Ideal Class Group of R
...
Every invertible fractional ideal is, by (25
...
3)
...
14) and (25
...
In fact, then F(R) is the free abelian group on the prime ideals
...
15), a
PID
...
11, Sects
...
424–437] for a discussion of the case in which
R is a ring of quadratic integers, including many examples where Pic(R) ̸= 0
...
Arbitrary Valuation Rings
A valuation ring is a subring of a field such that the reciprocal of any element
outside the subring lies in it
...
They
are maximal under domination of local rings; that is, one contains the other, and
the inclusion map is a local homomorphism
...
Given a 1dimensional Noetherian domain and a finite extension of its fraction field with
a proper subring containing the domain, that subring too is 1-dimensional and
Noetherian, this is the Krull–Akizuki Theorem
...
Definition (26
...
— A subring V of a field K is said to be a valuation ring
of K if, whenever z ∈ K − V , then 1/z ∈ V
...
2)
...
Then V is local, m is its maximal ideal, and K is its fraction field
...
Let’s show m is an ideal
...
Suppose ax ∈
/ m
...
So a(1/ax) ∈ V
...
×
So x ∈ V , a contradiction
...
Now, by hypothesis, either x/y ∈ V or
y/x ∈ V
...
Then 1 + (y/x) ∈ V
...
Thus
m is an ideal
...
6)
...
□
Exercise (26
...
— Let V be a domain
...
Exercise (26
...
— Let V be a valuation ring of K, and V ⊂ W ⊂ K a subring
...
Exercise (26
...
— Prove that a valuation ring V is normal
...
6)
...
Then either 1 ∈
/ aR[x] or 1 ∈
/ aR[1/x]
...
Then there are equations
1 = a0 + · · · + an xn
and 1 = b0 + · · · + bm /xm
with all
ai , bj ∈ a
...
Multiply through by 1 − b0 and an xn , getting
1 − b0 = (1 − b0 )a0 + · · · + (1 − b0 )an xn
(1 − b0 )an x = an b1 x
n
n−1
and
+ · · · + an bm x
n−m
...
Simplify, getting an equation of the form 1 = c0 + · · · + cn−1 xn−1 with ci ∈ a, which
contradicts the minimality of n
...
11)
(26
...
— Let A, B be local rings, and m, n their maximal ideals
...
Proposition (26
...
— Let K be a field, A any local subring
...
Proof: Let m be the maximal ideal of A
...
Then (A, m) ∈ S
...
Let (Rλ , nλ ) form a totally∩ordered
subset
...
Plainly N ∩ Rλ = nλ and B/N = Rλ /nλ
for all λ
...
Hence B/N is a field and is
algebraic over A/m
...
For any nonzero x ∈ K, set V ′ := V [x] and V ′′ := V [1/x]
...
6), either
1∈
/ MV ′ or 1 ∈
/ MV ′′
...
Then MV ′ is proper, so it is contained in a
′
′
maximal ideal M of V
...
Further V ′ /M′ is generated as a ring over V /M by the residue x′ of x
...
Hence (V ′ , M′ ) ∈ S, and (V ′ , M′ ) ≥ (V, M)
...
Thus V is a valuation ring of K
...
Finally, (V, M) ∈ S; so V dominates A with algebraic residue field extension
...
9)
...
Show that the valuation rings of K are the maximal elements of S
...
10)
...
Then the integral closure
R of R in K is the intersection of all valuation rings V of K containing R
...
Proof:
(∩ Every
) valuation ring V is normal by (26
...
So if V ⊃ R, then V ⊃ R
...
To prove the opposite inclusion, take any x ∈ K − R
...
If 1/y ∈ R[y], then for some n,
1/y = a0 y n + a1 y n−1 + · · · + an
with
aλ ∈ R
...
So x ∈ R, a contradiction
...
So there is a maximal ideal m of R[y] containing y
...
Its kernel is m ∩ R,
so m ∩ R is a maximal ideal of R
...
8), there is a valuation ring V that
dominates R[y]m with algebraic residue field extension; whence, if R is local, then
V also dominates R, and the residue field of R[y]m is equal to that of R
...
□
n
n+1
n
(26
...
— We call an additive abelian group Γ totally ordered if
Γ has a subset Γ+ that is closed under addition and satisfies Γ+ ⊔ {0} ⊔ −Γ+ = Γ
...
Note that either x > y or x = y or
y > x
...
Let V be a domain, and set K := Frac(V ) and Γ := K × /V ×
...
It is a homomorphism:
v(xy) = v(x) + v(y)
...
11
...
14)
159
(
)
Set Γ+ := v V − 0 − 0
...
Clearly, V is a valuation
ring if and only if −Γ+ ⊔ {0} ⊔ Γ+ = Γ, so if and only if Γ is totally ordered
...
Let’s prove that, for all x, y ∈ K × ,
v(x + y) ≥ min{v(x), v(y)} if
x ̸= −y
...
11
...
Then z := x/y ∈ V
...
Hence
v(x + y) = v(z + 1) + v(y) ≥ v(y) = min{v(x), v(y)},
Note that (26
...
1) and (26
...
2) are the same as (1) and (2) of (23
...
Conversely, start with a field K, with a totally ordered additive abelian group Γ,
and with a surjective homomorphism v : K × → Γ satisfying (26
...
2)
...
Then V is a valuation ring, and Γ = K × /V ×
...
For example, a DVR V of K is just a valuation ring with value group Z, since
any x ∈ K × has the form x = utn with u ∈ V × and n ∈ Z
...
12)
...
Form the k-vector space R with basis the symbols X a for a ∈ Γ
...
Then R is a k-algebra
with X0 = 1
...
Define v : (R − 0) → Γ by
(∑
)
v
ra X a := min{a | ra ̸= 0}
...
Hence R is a domain
...
Thus v(x + y) ≥ min{v(x), v(y)}
...
Clearly v is well defined, surjective, and a homomorphism
...
Thus v is a valuation with group Γ
...
Clearly, R′ is a ring,
and p is a prime of R′
...
There are many choices for Γ other than Z
...
Proposition (26
...
— Let v be a valuation of a field K, and x1 ,
...
Set m := min{v(xi )}
...
(2) If x1 + · · · + xn = 0, then m = v(xi ) = v(xj ) for some i ̸= j
...
Set z := x1 /x2
...
Also v(−z) = v(z) + v(−1) > 0
...
Hence v(z + 1) = 0
...
Therefore, v(x1 + x2 ) = v(x2 ) = m
...
For (2), reorder the xi so v(xi ) = m for i ≤ k and v(xi ) > m for i > k
...
11
...
Therefore,
v(xk+1 + · · · + xn ) > m
...
So k > 1, and v(x1 ) = v(x2 ) = m, as desired
...
17)
Exercise (26
...
— Let V be a valuation ring, such as a DVR, whose value
group Γ is Archimedean; that is, given any nonzero α, β ∈ Γ, there’s n ∈ Z such
that nα > β
...
Exercise (26
...
— Let V be a valuation ring
...
Lemma (26
...
— Let R be a 1-dimensional Noetherian domain, K its fraction
field, M a torsionfree module, and x ∈ R nonzero
...
Further,
ℓ(M/xM ) ≤ dimK (M ⊗R K) ℓ(R/xR),
(26
...
1)
with equality if M is finitely generated
...
If r = ∞, then (26
...
1) is trivial; so we
may assume r < ∞
...
Given any module N , set NK := S −1 N
...
First, assume M is finitely generated
...
, mr /sr
of MK with mi ∈ M and si ∈ S
...
, mr /1 is also a basis
...
Then
its localization αK is an K-isomorphism
...
Further, S −1 Ker(α) = Ker(αK ) = 0
...
Thus α is
injective
...
Then NK = 0, and N is finitely generated
...
But dim(R) = 1 by hypothesis
...
So ℓ(N ) < ∞ by (19
...
Similarly, Supp(R/xR) is closed and proper in Spec(R)
...
Consider the standard exact sequence:
0 → N ′ → N → N → N/xN → 0 where N ′ := Ker(µx )
...
9); it yields ℓ N ′ = ℓ(N/xN )
...
Consider this commutative
diagram with exact rows:
α
0−
→ Rr −
→M
→N
→0
−
−
µx µx
µx y
y
y
α
0−
→ Rr −
→M −
→N −
→0
Apply the snake lemma (5
...
It yields this exact sequence:
0 → N ′ → (R/xR)r → M/xM → N/xN → 0
...
But ℓ (R/xR)r = r ℓ(R/xR) also
by additivity
...
16
...
Second, assume M is arbitrary, but (26
...
1) fails
...
′
′
Now, MK ⊃ MK
; so r ≥ dimK (MK
)
...
However, together these inequalities contradict the first case with M ′ for M
...
20)
161
Theorem (26
...
— Let R be a 1-dimensional Noetherian domain, K its fraction field, K ′ a finite extension field, and R′ a proper subring of
K ′ containing R
...
Proof: Given a nonzero ideal a′ of R′ , take any nonzero x ∈ a′
...
Then
a0 ∈ a′ ∩ R
...
16) yields ℓ(R/a0 R) < ∞
...
Further, R′ ⊗R K ⊂ K ′ ;
hence, dimK (R′ ⊗R K) < ∞
...
16) yields ℓR (R′ /a0 R′ ) < ∞
...
So ℓR (a′ /a0 R′ ) < ∞
...
2)(3)
...
Thus R′ is Noetherian
...
Clearly, ℓR′′ R′′ ≤ ℓR R′′
...
So, in R′′ , every
prime is maximal by (19
...
So if a′ is prime, then a′ /a0 R′ is maximal, whence a′
maximal
...
Thus R′ is 1-dimensional
...
18)
...
Let K be its fraction field, K ′ a finite extension field, and R′
the normalization of R in K ′
...
Proof: Since R is 1-dimensional, it’s not a field
...
So R′ is not a field by (14
...
Hence, R′ is Noetherian and 1-dimensional by
(26
...
Thus R′ is Dedekind by (24
...
□
Corollary (26
...
— Let K ′ /K be a field extension, V ′ a valuation ring of K ′
not containing K
...
Then V is a DVR if V ′ is, and the converse
holds if K ′ /K is finite
...
1) that V is a valuation ring, and from (26
...
Now, a nonzero subgroup of Z is
a copy of Z
...
Conversely, assume V is a DVR, so Noetherian and 1-dimensional
...
Hence, V ′ is Noetherian by (26
...
15)(2)
...
20)
...
Prove the integral closure R of R in L is
the intersection of all DVRs V of L containing R by modifying the proof of (26
...
18) to R[y]p
...
Rings and Ideals
Exercise (1
...
— Let φ : R → R′ be a map of rings, a an ideal of R, and b an
ideal of R′
...
Prove these statements:
(1) Then aec ⊃ a and bce ⊂ b
...
c
(3) If b is an extension, then b is the largest ideal of R with extension b
...
Solution: For (1), given x ∈ a, note φ(x) = x · 1 ∈ aR′
...
Thus a ⊂ aec
...
But b is an ideal of R′
...
Thus (1) holds
...
But a ⊂ aec by (1); so
e
a ⊂ aece
...
Similarly, bcec ⊃ bc by (1) applied with a := bc
...
Thus bcec = bc
...
For (3), say b = ae
...
But aece = ae by (2)
...
Further, it’s the largest such ideal, as aec ⊃ a by (1)
...
For (4), say bc1 = bc2 for extensions bi
...
Thus (4) holds
...
7)
...
, Xn ] the
polynomial ring
...
, Xn ]
...
, xn and with
aR′ = 0
...
3); then π factors through a unique
map P/aP → R′ as aR′ = 0 by (1
...
On the other hand, φ factors through a
unique map ψ : R/a → R′ as aR′ = 0 by (1
...
, Xn ] → R′ such that π(Xi ) = xi for all i by (1
...
□
Exercise (1
...
— Let R be ring, and P := R[X1 ,
...
Let m ≤ n and a1 ,
...
Set p := ⟨X1 − a1 ,
...
Prove that
P/p = R[Xm+1 ,
...
Solution: First, assume m = n
...
, Xn−1 ] and
p′ := ⟨X1 − a1 ,
...
By induction on n, we may assume P ′ /p′ = R
...
Hence
P/p′ P = (P ′ /p′ )[Xn ] by (1
...
Thus P/p′ P = R[Xn ]
...
9)
...
Hence
p(P/p′ P ) = ⟨Xn − an ⟩(P/p′ P )
...
So P/p = R by (1
...
In general, P = (R[X1 ,
...
, Xn ]
...
, Xn ]
by (1
...
□
Exercise (1
...
— Let R be a ring
...
Prove
(a) ab = a ∩ b
and (b) R/ab = (R/a) × (R/b)
...
15)
163
(2) Let a be comaximal to both b and b′
...
(3) Let a, b be comaximal, and m, n ≥ 1
...
(4) Let a1 ,
...
Prove
(a) a1 and a2 · · · an are comaximal;
(b) a1 ∩ · · · ∩ an = a1∏
· · · an ;
∼
(c) R/(a1 · · · an ) −→
(R/ai )
...
Conversely, a + b = R
implies x+y = 1 with x ∈ a and y ∈ b
...
To prove (1)(b), form the map R → R/a × R/b that carries an element to its
pair of residues
...
So we have an injection
φ : R/ab ,→ R/a × R/b
...
Say x
¯ and y¯
are the residues of x and y
...
Then φ(x + a) = (¯
x, y¯), as desired
...
To prove (2), note that
R = (a + b)(a + b′ ) = (a2 + ba + ab′ ) + bb′ ⊆ a + bb′ ⊆ R
...
Hence, bn and am are comaximal for any m ≥ 1
...
By
hypothesis, a1 and an are comaximal
...
To prove (4)(b) and (4)(c), again proceed by induction on n
...
1
2
n
□
Exercise (1
...
— First, given a prime number p and a k ≥ 1, find the idempotents in Z/⟨pk ⟩
...
Third, find the number
∏N
of idempotents in Z/⟨n⟩ where n = i=1 pni i with pi distinct prime numbers
...
Then m(m−1) is divisible
by pk
...
Hence 0 and 1 are the only idempotents in Z/⟨pk ⟩
...
14) yields
Z/⟨12⟩ = Z/⟨3⟩ × Z/⟨4⟩
...
By the previous case, we have the following possibilities:
m ≡ 0 (mod 3)
and
m ≡ 0 (mod 4);
m ≡ 1 (mod 3)
and
m ≡ 1 (mod 4);
m ≡ 1 (mod 3)
and
m ≡ 0 (mod 4);
m ≡ 0 (mod 3)
and
m ≡ 1 (mod 4)
...
ni−1
Third, for each i, the two numbers pn1 1 · · · pi−1
and pni i have no common prime
divisor
...
So the principal ideals they generate are comaximal
...
5)
Chinese Remainder Theorem yields
Z/⟨n⟩ =
N
∏
Z/⟨pni i ⟩
...
Thus there
are 2N idempotents in Z/⟨n⟩
...
16)
...
Show a = a′ × a′′ with a′ ⊂ R′ and a′′ ⊂ R′′ ideals
...
Solution: Set a′ := {x′ | (x′ , 0) ∈ a} and a′′ := {x′′ | (0, x′′ ) ∈ a}
...
Clearly,
′
a ⊃ a′ × 0 + 0 × a′′ = a′ × a′′
...
Finally, the equation R/a = (R/a′ ) × (R/a′′ ) is now clear from the construction of
the residue class ring
...
17)
...
(See (10
...
)
(1) Set a := ⟨e⟩
...
(2) Let a be a principal idempotent ideal
...
(3) Set e′′ := e + e′ − ee′
...
(4) Let e1 ,
...
Show ⟨e1 ,
...
(5) Assume R is Boolean
...
Solution: For (1), note a2 = ⟨e2 ⟩ since a = ⟨e⟩
...
Thus (1) holds
...
Then a2 = ⟨g 2 ⟩
...
So g = xg 2 for some x
...
Then f ∈ a; so ⟨f ⟩ ⊂ a
...
So a ⊂ ⟨f ⟩
...
For (3), note ⟨e′′ ⟩ ⊂ ⟨e, e′ ⟩
...
By symmetry, e′ e′′ = e′
...
Thus
(4) holds
...
Thus (3) yields (4)
...
Thus (4) yields (5)
...
Prime Ideals
Exercise (2
...
— Let a and b be ideals, and p a prime ideal
...
Solution: Trivially, (1) implies (2)
...
Finally, assume a ̸⊂ p and b ̸⊂ p
...
Hence, since p is prime, xy ∈
/ p
...
Thus (3) implies (1)
...
4)
...
Solution: Any subring of Z/⟨pn ⟩ must contain 1, and 1 generates Z/⟨pn ⟩ as an
abelian group
...
However, Z/⟨pn ⟩ is not a
domain, because in it, p · pn−1 = 0 but neither p nor pn−1 is 0
...
22)
165
Exercise (2
...
— Let R := R′ × R′′ be a product of two rings
...
Solution: Assume R is a domain
...
Correspondingly, either R′ = 0 and R = R′′ , or R′′ = 0 and
R = R′′
...
□
Exercise (2
...
— Let R be a ring, p a prime ideal, R[X] the polynomial ring
...
Solution: Note R[X]/pR[X] = (R/p)[X] by (1
...
But R/p is a domain by
(2
...
So R[X]/pR[X]/is a domain by (2
...
Thus pR[X] is prime by (2
...
Note (pR[X] + ⟨X⟩) pR[X] is/ equal to ⟨X⟩ ⊂ (R/p)[X]
...
8)
...
9)
...
9)
...
9)
...
Then R/p is a field by (2
...
But, as just noted, R/p is
equal to R[X] (pR[X] + ⟨X⟩)
...
17)
...
11)
...
Show p is prime if and only if either p = p′ × R′′ with p′ ⊂ R′ prime or p = R′ × p′′
with p′′ ⊂ R′′ prime
...
16), (2
...
5)
...
16)
...
Prove φ is injective
...
1), 1 ̸= 0 in R
...
So Ker(φ) = 0 by (2
...
Thus φ is injective
...
10)
...
, Xn ] the polynomial ring
in n variables
...
, Xm ⟩
...
Solution: Simply combine (2
...
3), and (1
...
□
Exercise (2
...
— Let R be a domain, and x, y ∈ R
...
Show
x = uy for some unit u
...
So x = 0 if and
only if y = 0; if so, take u := 1
...
Now, x = uvx, or x(1 − uv) = 0
...
So 1 − uv = 0
...
□
Exercise (2
...
— Let B be a Boolean ring
...
Solution: Given x ∈ B/p, plainly x(x − 1) = 0
...
9)
...
Thus B/p = F2
...
So p is maximal by (2
...
□
Exercise (2
...
— Let R be a ring
...
Show that every prime p is maximal
...
But R/p is a
domain by (2
...
So y = 0 or yy n−2 = 1
...
Thus p is maximal by
□
(2
...
Exercise (2
...
— Prove the following statements, or give a counterexample
...
3)
(1)
(2)
(3)
(4)
The complement of a multiplicative subset is a prime ideal
...
Given two prime ideals, their sum is prime
...
(5) In (1
...
Solution: (1) False
...
The
complement T of S contains 3 and 5, but not 8; so T is not an ideal
...
In the ring Z, consider the prime ideals ⟨2⟩ and ⟨3⟩; their intersection
⟨2⟩ ∩ ⟨3⟩ is equal to ⟨6⟩, which is not prime
...
Since 2 · 3 − 5 = 1, we have ⟨3⟩ + ⟨5⟩ = Z
...
Let φ : Z → Q be the inclusion map
...
(5) True
...
9), the operation b′ 7→ κ−1 b′ sets up an inclusion-preserving
bijective correspondence between the ideals b′ ⊃ n′ and the ideals b ⊃ κ−1 n′
...
23)
...
, Xn ] the polynomial ring,
f ∈ P nonzero
...
(1) Let S ⊂ k have at least d + 1 elements
...
, an ∈ S with f (a1 ,
...
(2) Using the algebraic closure K of k, find a maximal ideal m of P with f ∈
/ m
...
Assume n = 1
...
8), p
...
So f (a1 ) ̸= 0 for some a1 ∈ S
...
Say f = j gj X1j with gj ∈ k[X2 ,
...
But f ̸= 0
...
By induction, gi (a2 ,
...
, an ∈ S
...
, an ) = j gj (a2 ,
...
Thus (1) holds
...
As K is infinite, (1) yields a1 ,
...
, an ) ̸= 0
...
Then Im(φ) ⊂ K is the k-subalgebra generated
by the ai
...
6), p
...
Set m := Ker(φ)
...
6
...
17), and fi ∈
/ m as φ(fi ) = fi (a1 ,
...
Thus (2) holds
...
26)
...
Solution: Say ⟨x⟩ + ⟨y⟩ = ⟨d⟩
...
The
assertion is now obvious
...
29)
...
28)
...
Assume that R has infinitely many prime elements
p, or simply that there is a p such that p ∤ a0
...
Solution: Set a := ⟨p, f ⟩
...
Set
k := R/⟨p⟩
...
6) and (2
...
Let f ′ ∈ k[X]
denote the image of f
...
Hence f ′ is not a unit by
∼ k[X]/⟨f ′ ⟩ by
(2
...
Therefore, ⟨f ′ ⟩ is proper
...
7) and (1
...
So a is proper
...
3
...
3)
...
Prove that w ∈ R× if and only if w′ ∈ (R/a)×
...
16)
167
Solution: Plainly, w ∈ R× implies w′ ∈ (R/a)× , whether a ⊂ rad(R) or not
...
As every maximal ideal of R contains rad(R), the operation
m 7→ m/a establishes a bijective correspondence between the maximal ideals of R
and those of R/a owing to (1
...
So w belongs to a maximal ideal of R if and only
if w′ belongs to one of R/a
...
31)
...
Then there is a maximal ideal m ⊂ R with a ̸⊂ m
...
So there are a ∈ a and v ∈ m with a + v = w
...
For example, take R := Z and a := ⟨2⟩ and
w := 3
...
□
Exercise (3
...
— Let A be a local ring
...
Solution: Let m be the maximal ideal
...
Say e ∈
/ m
...
6)
...
Thus e = 1
...
Alternatvely, (3
...
So
(1
...
□
Exercise (3
...
— Let A be a ring, m a maximal ideal such that 1 + m is a unit
for every m ∈ m
...
Is this assertion still true if m is not maximal?
Solution: Take y ∈ A − m
...
Hence there exist
x ∈ R and m ∈ m such that xy + m = 1, or in other words, xy = 1 − m
...
Thus A is local by (3
...
No, the assertion is not true if m is not maximal
...
□
Exercise (3
...
— Let φ : R → R′ be a map of rings, p an ideal of R
...
Solution: In (1), given q, note φ(p) ⊂ q, as always φ(φ−1 (q)) ⊂ q
...
Hence φ−1 (pR′ ) ⊂ φ−1 (q) = p
...
Thus φ−1 (pR′ ) = p
...
In (2), set S := φ(R − p)
...
So there’s a prime q of R′ containing pR′ and disjoint from S by
(3
...
So φ−1 (q) ⊃ φ−1 (pR′ ) = p and φ−1 (q)∩(R−p) = ∅
...
□
Exercise (3
...
— Use Zorn’s lemma to prove that any prime ideal p contains
a prime ideal q that is minimal containing any given subset s ⊂ p
...
Then p ∈ S,
so S ̸= ∅
...
To apply Zorn’s Lemma, we must
∩ show
that, for any decreasing chain {qλ } of prime ideals, the intersection q := qλ is a
prime ideal
...
So take x, y ∈
/ q
...
Since qλ is prime, xy ∈
/ qλ
...
Thus q is prime
...
16)
...
Show that S is saturated
multiplicative if and only if R − S is a union of primes
...
18)
Solution: First, assume S is saturated multiplicative
...
Then
xy ∈
/ S for all y ∈ R; in other words, ⟨x⟩ ∩ S = ∅
...
12) gives a prime
p ⊃ ⟨x⟩ with p ∩ S = ∅
...
Conversely, assume R − S is a union of primes p
...
Take x, y ∈ R
...
Thus S is saturated multiplicative
...
17)
...
Define its
saturation to be the subset
S := { x ∈ R | there is y ∈ R with xy ∈ S }
...
(2) Show that R − S is the union U of all the primes
∪ p with p ∩ S = ∅
...
Show R − S = W
...
Given f, g ∈ R, show S f ⊂ S g if and only if ⟨f ⟩ ⊃ ⟨g⟩
...
Trivially, if x ∈ S, then x · 1 ∈ S
...
Hence 1 ∈ S as 1 ∈ S
...
Then there are y, y ′ ∈ R with
xy, x′ y ′ ∈ S
...
So (xx′′ )(yy ′ ) ∈ S
...
Thus S
is multiplicative
...
Then there is y ∈ R with
xx′ y ∈ S
...
Thus S is saturated
...
Given x ∈ S, there is y ∈ R with xy ∈ S
...
But
T is saturated multiplicative
...
Thus T ⊃ S
...
Consider (2)
...
Further, R−U is saturated multiplicative
by (3
...
So R − U ⊃ S by (1)(c)
...
Conversely, R − S is a union
of primes p by (3
...
Plainly, p ∩ S = ∅ for all p
...
Thus (2) holds
...
Then 1 ∈
/ p + a; else, 1 = p + a with
p ∈ p and a ∈ a, and so 1 − p = a ∈ p ∩ S
...
12)
...
But also p ⊂ m
...
Conversely, take p ⊃ a
...
But p ∩ (1 + p) = ∅
...
Thus U ⊃ W
...
Thus (2) yields (3)
...
By (1), S f ⊂ S g if and only if f ∈ S g
...
By definition√of radical, hf = g
for some√h and n if and only if g ∈
⟨f ⟩
...
Thus (4) holds
...
18)
...
Show S is maximal
in the set S of multiplicative subsets T of R with 0 ∈
/ T if and only if R − S is a
minimal prime — that is, it is a prime containing no smaller prime
...
Then S is equal to its saturation
S, as S ⊂ S and S is multiplicative by (3
...
So R − S is a union of primes p by (3
...
Fix a p
...
14) yields in p a minimal prime q
...
But R − q ∈ S by (2
...
As S is maximal, S = R − q, or R − S = q
...
Conversely, assume R − S is a minimal
prime q
...
1)
...
16)
...
Now,
S ⊂ T ⊂ T
...
But q is minimal
...
But p is arbitrary, and
Solutions: (3
...
Hence q = R − T
...
Hence S = T
...
□
Exercise (3
...
— Let k be a field, S ⊂ k a subset of cardinality d at least 2
...
, Xn ] be the polynomial ring, f ∈ P nonzero
...
Proceeding by induction on n, show
there are a1 ,
...
, an ) ̸= 0
...
, Wr proper subspaces
...
Show i Wi ̸= V
...
Show W ∪
⊂ Wi for some i
...
, ar ideals with a ⊂ i ai
...
Solution: For (1), first assume n = 1
...
8), p
...
So there’s a1 ∈ S with f (a1 ) ̸= 0
...
Say f =
j gj X1 with gj ∈ k[X2 ,
...
But f ̸= 0
...
By induction, there are a2 ,
...
, an ) ̸= 0
...
, an ) = j gj (a2 ,
...
Thus (1) holds
...
Form their span V∪
⊂ V
...
Then n < ∞, and it suffices to show i Wi′ ̸= V ′
...
Form the polynomial ring P := k[X1 ,
...
For each i,
take a linear form fi ∈ P that vanishes on Wi′
...
Then r is the
highest power of any variable in f
...
So (1) yields a1 ,
...
, an ) ̸= 0
...
, an ) ∈ V ′ − ∪ i Wi′
...
Then i Ui = W
...
Thus W ⊂ Wi
...
□
Exercise (3
...
— Let k be a field, R := k[X, Y ] the polynomial ring in two
variables, m := ⟨X, Y ⟩
...
Solution: Since R is a UFD, and∪m is maximal, so prime, any nonzero f ∈ m
has a prime factor p ∈ m
...
□
Exercise (3
...
— Find the nilpotents in Z/⟨n⟩
...
Solution: An integer m is nilpotent modulo n if and only if some power mk is
divisible by n
...
In particular, in Z/⟨12⟩, the nilpotents are 0 and 6
...
24)
...
(1) Assume every ideal not contained in
nil(R) contains a nonzero idempotent
...
(2) Assume R
is Boolean
...
Solution: or (1), recall (3
...
1), that nil(R) ⊂ rad(R)
...
Assume rad(R′ ) ̸= ⟨0⟩
...
Then e(1 − e) = 0
...
2)
...
Hence rad(R′ ) = ⟨0⟩
...
9) yields (1)
...
2) that every element of R is idempotent
...
Thus (1) yields (2)
...
25)
...
Prove
√
√
φ−1 b = φ−1 b
...
32)
Solution: Below, (1) is clearly equivalent to (2); and (2), to (3); and so forth:
√
√
(1) x ∈ φ−1 b;
(2) φx ∈ b;
(3) (φx)n ∈ b for some n; (4) φ(xn√
) ∈ b for some n;
(5) xn ∈ φ−1 b for some n; (6) x ∈ φ−1 b
...
38)
...
Show that
rad(R[X]) = nil(R[X]) = nil(R)R[X]
...
22
...
Next, recall
that nil(R[X]) ⊃ nil(R)R[X] by (3
...
Finally, given f := a0 + · · · + an X n in
rad(R[X]), note that 1 + Xf is a unit by (3
...
So a0 ,
...
36)(2)
...
Thus nil(R)R[X] ⊃ rad(R[X]), as desired
...
26)
...
Assume ⟨e⟩ = ⟨e′ ⟩
...
Solution: By hypothesis, en ∈ ⟨e′ ⟩ for some n ≥ 1
...
So
e = xe′ for some x
...
By symmetry, e′ = e′ e
...
□
Exercise (3
...
— Let R be a ring, a1 , a2 comaximal ideals with a1 a2 ⊂ nil(R)
...
Solution: Since a1 and a2 are comaximal, there are xi ∈ ai with x1 + x2 = 1
...
Now, x1 x2 ∈ nil(R); take n ≥ 1 so that (x1 x2 )n = 0
...
Then e1 + e2 = 1 and e1 e2 = 0
...
11)
...
28)
...
Assume a ⊂ nil(R)
...
Solution: Note that Idem(κ) is injective by (3
...
1) and (3
...
As to surjectivity, given e′ ∈ Idem(R/a), take z ∈ R with residue e′
...
And ⟨z⟩⟨1 − z⟩ ⊂ a ⊂ nil(R) as κ(z − z 2 ) = 0
...
27) yields complementary idempotents e1 ∈ ⟨z⟩ and e2 ∈ ⟨1 − z⟩
...
Then κ(e1 ) = xe′
...
Similarly,
κ(e2 ) = κ(e2 )(1 − e′ )
...
But κ(e2 ) = 1 − κ(e1 )
...
But κ(e1 ) = κ(e1 )e′
...
Thus Idem(κ) is surjective
...
30)
...
Prove the following statement equivalent:
(1) R has exactly one prime p;
(2) every element of R is either nilpotent or a unit;
(3) R/ nil(R) is a field
...
Let x ∈ R be a nonunit
...
So x is nilpotent
by the Scheinnullstellensatz (3
...
Thus (2) holds
...
Then every x ∈
/ nil(R) has an inverse
...
Assume (3)
...
15)
...
29)
...
□
√
Exercise (3
...
(√—)nLet R be a ring, and a an ideal
...
Show
a ⊂ a for all large n
...
36)
171
√
Solution: Let x∑
of a
...
, xm be generators √
m
xni i ∈ a
...
Given a ∈ a, write a = i=1 yi xi with yi ∈ R
...
Hence ji ≥ ni for some i, because if ji ≤ ni − 1 for all i, then
ji ≤ (ni − 1)
...
□
Exercise (3
...
— Let R be a ring, q an ideal, p a finitely generated prime
...
√
Solution: If p = q, then p ⊃ q ⊃ pn by (3
...
Conversely, if q ⊃ pn , then
√
√
clearly q ⊃ p
...
□
Exercise (3
...
— Let R be a ring
...
, pn
...
, xn ) ∈ Im(φ) with xi ̸= 0 but xj = 0 for j ̸= i
...
Now, R is reduced and the pi are its minimal
primes; hence, (3
...
14) yield
∩
√
⟨0⟩ = ⟨0⟩ =
pi
...
Finally,
fix i
...
Set
∏
□
a := j̸=i aj
...
So take (x1 ,
...
Exercise (3
...
— Let R be a ring, X a variable, f := a0 + a1 X + · · · + an X n
and g := b0 + b1 X + · · · + bm X m polynomials with an ̸= 0 and bm ̸= 0
...
, an ⟩ = R
...
, an are nilpotent
...
, an are nilpotent
...
(4) Then f g is primitive if and only if f and g are primitive
...
, an are nilpotent, so is f owing to (3
...
Conversely,
say ai ∈
/ nil(R)
...
29) yields a prime p ⊂ R with
ai ∈
/ p
...
But pR[X] is prime by (2
...
So plainly f ∈
/ nil(R[X])
...
Then (an X n )k = 0
...
31)
...
, an−1 are nilpotent by induction on n
...
For (2), suppose a0 is a unit and a1 ,
...
Then a1 X +· · ·+an X n
is nilpotent by (1), so belongs to rad(R) by (3
...
1)
...
2)
...
Then a0 b0 = 1
...
Further, given a prime p ⊂ R, let κp : R[X] → (R/p)[X] be the canonical map
...
But R/p is a domain by (2
...
So deg κp (f ) = 0 owing to
(2
...
1)
...
, an ∈ p
...
Thus a1 ,
...
2)
...
Set
c
:=
m−r
i
n
j+k=i aj bk
...
But f g = 1
...
Taking i := m + n yields
an bm = 0
...
Multiplying
m+1
by arn yields ar+1
b0 = 0
...
So
n bm−r = 0 by induction
...
So an X ∈ rad(R[X]) by (3
...
1)
...
So f − an X n is a
unit by (3
...
So a1 ,
...
Thus (2) holds
...
Say ar bm ̸= 0, but ar+i bm = 0 for all i > 0
...
39)
and set h := ar+i g
...
Also h = 0 or deg(h) < m
...
In particular, ar+i bm−i = 0
...
Also f g = 0
yields ar bm + ar+1 bm−1 + · · · = 0
...
Thus (3) holds
...
Then h ∈ R[X] is primitive if and only if κm (h) ̸= 0 for all m, owing to (2
...
But
R/m is a field by (2
...
So (R/m)[X] is a domain by (2
...
Hence κp (f g) = 0 if
and only if κp (f ) = 0 or κp (g) = 0
...
□
Exercise (3
...
— Generalize (3
...
, Xr ]
...
∑
Solution: Let f, g ∈ P
...
, ir ) and
i1
(i)
ir
X := X1 · · · Xr
...
Set (0) := (0,
...
Then (1)–(4) generalize as follows:
(1′ ) Then f is nilpotent if and only if a(i) is nilpotent for all (i)
...
(3′ ) Assume f is a zerodivisor
...
(4′ ) Then f g is primitive if and only if f and g are primitive
...
, Xr ], and say f = fi X1i with fi ∈ R′
...
36)(1); hence by induction on r, so
are all a(i)
...
31)
...
∑
In (2′ ), if a(0) is a unit and a(i) is nilpotent for (i) ̸= (0), then (i)̸=(0) a(i) X (i)
is nilpotent by (1), so belongs to rad(R) by (3
...
1)
...
2)
...
Then f0 is a unit, and fi is nilpotent for i > 0
by (3
...
So a(0) is a unit, and a(i) is nilpotent if i1 = 0 and (i) ̸= (0), by
induction on r
...
Thus (2′ ) holds
...
Take d larger than any exponent of
i−1
any Xi found in f or g
...
r−1
Then φ(f )φ(g) = 0
...
So φ carries distinct monomials
in f to distinct monomials in φ(f ), and the same for g
...
So φ(g) ̸= 0
...
So (3
...
Hence ca(i) = 0 for all a(i)
...
Thus (3′ ) holds
...
36)(4) with X replaced by X1 ,
...
□
Exercise (3
...
— Let R be a ring, a an ideal,
∑ X a nvariable, R[[X]] the formal
power series
ring,
M
⊂
R[[X]]
an
ideal,
and
f
:=
an X ∈ R[[X]]
...
Prove the following statements:
(1) If f is nilpotent, then an is nilpotent for all n
...
(2) Then f ∈ rad(R[[X]]) if and only if a0 ∈ rad(R)
...
Then X and m generate M
...
Then X ∈ M and m is maximal
...
The converse may fail
...
Set g := i≥n ai X i
...
So g is nilpotent by (3
...
Then am
n = 0
...
The converse is false
...
] for variables Xn
...
16)
173
∑
R := P/⟨X22 , X33 ,
...
Then ann = 0, but
an X n is
not nilpotent
...
For (2), given g =
bn X n ∈ rad(R[[X]]), note that 1 + f g is a unit if and only
if 1 + a0 b0 is a unit by (3
...
Thus (3
...
For (3), note M contains X and m, so the ideal they generate
...
So if f ∈ M, then a0 ∈ M ∩ R = m
...
For (4), note that X ∈ rad(R[[X]]) by (2)
...
So
P/n = R/m by (3
...
Thus (2
...
∑
In (5), plainly aR[[X]] ⊂ A
...
Say
m
b1 ,
...
Then an = i=1 cni bi for some cni ∈ R
...
n≥0 i=1
i=1
n≥0
For a counterexample, take a0 , a1 ,
...
∑mTake R := Z[a1 , a2 ,
...
Given g ∈ aR[[X]], say g =
i=1 bi gi with bi ∈ a and
n
gi =
b
X
...
∑Then i=1 bi bin ∈ ⟨a1 ,
...
, ap−1 ⟩
...
Thus f ∈
/ aR[[X], but f ∈ A
...
Modules
Exercise (4
...
— Let R be a ring, M a module
...
Show that ρ is an isomorphism, and describe its inverse
...
Set H := Hom(R, M )
...
It is easy to check
that αρ = 1H and ρα = 1M
...
2)
...
12)
...
Let M be the
submodule of Frac(R) generated by 1, x−1 , x−2 ,
...
Prove that x−1 ∈ R, and conclude that M = R
...
, mk
...
Set n := max{ni }
...
, x−n generate M
...
Thus
x−1 = an + · · · + a1 xn−1 + a0 xn ∈ R
...
∈ R; so M ⊂ R
...
Thus M = R
...
13)
...
Solution:
Say eλ for λ ∈ Λ form a free basis, and m1 ,
...
Then
∑
mi =
xij eλj for some xij
...
Plainly, they are finite
in number, and generate
...
□
174
Solutions: (4
...
16)
...
Endow
∏
⊕
∏
Rλ and
Rλ with componentwise addition and ⊕
multiplication
...
Solution: Consider the vector
∏ (1) whose every component is 1
...
On the other hand, no restricted vector (xλ )
can be a multiplicative identity in
Rλ ; indeed, because Λ is infinite, xµ must be
zero for some µ
...
□
Exercise (4
...
— Let R be a ring, M a module, and M ′ , M ′′ submodules
...
Solution: Assume M = M ′ ⊕ M ′′
...
15); further, M ′ is the set of (m′ , 0), and M ′ is that
of (0, m′′ )
...
Conversely, consider the map M ′ ⊕ M ′′ → M given by (m′ , m′′ ) 7→ m′ + m′′
...
It is injective if M ′ ∩ M ′′ = 0; indeed, if m′ + m′′ = 0,
then m′ = −m′′ ∈ M ′ ∩ M ′′ = 0, and so (m′ , m′′ ) = 0 as desired
...
18)
...
Consider a diagram
α
β
→
→
L−
M−
N
←
−
←
−
ρ
σ
where α, β, ρ, and σ are homomorphisms
...
Solution: If M = L ⊕ N and α = ιL , β = πN , σιN , ρ = πL , then the definitions
immediately yield αρ + σβ = 1 and βα = 0, βσ = 1, ρσ = 0, ρα = 1
...
Consider
the maps φ : M → L ⊕ N and θ : L ⊕ N → M given by φm := (ρm, βm) and
θ(l, n) := αl + σn
...
Lastly, β = πN φ and ρ = πL φ by definition of φ, and α = θιL and σ = θιN by
definition of θ
...
19)
...
Prove that the injections ικ : Mκ →
Mλ induce an injection
⊕
⊕
Hom(L, Mλ ) ,→ Hom(L,
Mλ ),
and that it is an isomorphism if L is finitely generated
...
Then
(∑
)
(
) ⊕
ιλ αλ (l) = αλ (l) ∈
Mλ
...
Thus the ικ induce an injection
...
⊕
, lk
...
Then each α(li ) lies in a finite direct subsum of
Mλ
...
Set
ακ :=
Hom(L, Mλ ),
∑πκ α for all κ ∈ Λ
...
So (ακ ) lies in
and
ικ ακ = α
...
□
Solutions: (5
...
) —
(⊕(4
...
⊕ Let a be an ideal,
∏ Λ a nonempty
∏
Prove a
Mλ =
aMλ
...
(⊕
)
⊕
Solution:
Mλ ⊂
aMλ because a · (mλ ) = (amλ )
...
(∏
) ∏
Second, a
Mλ ⊂( aM)λ as a(mλ ) = (amλ )
...
, fn
...
Indeed, take (m′λ ) ∈ aMλ
...
each λ, there is nλ such that m′λ = j=1
∑n
Write aλj = i=1 xλji fi with the xλji scalars
...
□
j=1 i=1
i=1
j=1
5
...
5)
...
Set
M := M ′ ⊕ M ′′
...
2)(1) and (5
...
4), prove M/N = M ′ /N ⊕ M ′′
...
2)(1) and (5
...
So by (5
...
Thus (5
...
□
Exercise (5
...
— Let 0 → M ′ → M → M ′′ → 0 be a short exact sequence
...
Solution: Let m′′1 ,
...
Let m ∈ M ,
and write its image in M ′′ as a linear combination of the images of the m′′i
...
Set m′ := m − m′′
...
Let m′1 ,
...
Then m′ is a
linear combination of the m′j
...
Thus the m′i and m′′j together generate M
...
11)
...
Let N be
a submodule of M containing M ′ , and set N ′′ := N ∩ M ′′
...
Solution: Form the sequence 0 → M ′ → N → πM ′′ N → 0
...
9)
as (πM ′ |N ) ◦ ιM ′ = 1M ′
...
□
Exercise (5
...
— Criticize the following misstatement of (5
...
Solution: We have α : M ′ → M , and ιM ′ : M ′ → M ′ ⊕ M ′′ , but (5
...
Let’s construct a counterexample (due to B
...
For each integer
⊕ n ≥ 2, let
Mn be the direct sum of countably many copies of Z/⟨n⟩
...
First, let us check these two statements:
176
Solutions: (5
...
(2) For any finite subgroup G ⊂ M , we have M/G ≃ M
...
4), p
...
13
...
200]
...
Then M ′ ≃ M
...
11) and (1) yield
M/G ≃ (B/G) ⊕ M ′ ≃ M
...
Take one of the
Z/⟨p2 ⟩ components of M , and let M ′ ⊂ Z/⟨p2 ⟩ be the cyclic subgroup of order p
...
Finally, take M ′′ := M/M ′
...
□
Exercise (5
...
— Referring to (4
...
Moreover, β is surjective because κ and λ are
...
15) (Five Lemma)
...
Via a chase, prove these two statements:
(1) If γ3 and γ1 are surjective and if γ0 is injective, then γ2 is surjective
...
Solution: Let’s prove (1)
...
Since γ1 is surjective, there is
m1 ∈ M1 such that γ1 (m1 ) = β2 (n2 )
...
Since γ0 is injective, α1 (m1 ) = 0
...
So β2 (γ2 (m2 ) − n2 ) = γ1 α2 (m2 ) − β2 (n2 ) = 0
...
Since γ3 is surjective,
there is m3 ∈ M3 with γ3 (m3 ) = n3
...
Hence γ2 (m2 − α3 (m3 )) = n2
...
The proof of (2) is similar
...
24)
177
Exercise (5
...
— Consider this commutative diagram:
0
0
0
y
y
y
′
0 −→ L
−−→
y
L
−−→
y
L′′ −→ 0
y
′
0−
→M
→M
→
−
−
y
y
′′
M
→0
−
y
′
0−
→N
−→
y
N′′ −
→0
y
0
N
−→
y
0
0
Assume all the columns are exact and the middle row is exact
...
Solution: The first row is exact if the third is owing to the Snake Lemma
(5
...
The converse is proved similarly
...
17)
...
Given n ∈ N and m′′ ∈ M ′′ with α′′ (m′′ ) = γ ′ (n),
show that there is m ∈ M such that α(m) = n and γ(m) = m′′
...
Then
(Solution:) Since γ ′′is surjective,
′′
′
γ n−α(m1 ) = 0 as α (m ) = γ (n) and as the right-hand square is commutative
...
Since
α′ is surjective, there is m′ ∈ M ′ with α′ (m′ ) = n′
...
Then
γ(m) = m′′ as γβ = 0
...
Thus m works
...
22)
...
Solution: Given β : M →
→ N and α : R⊕Λ → N , use the UMP of (4
...
(The Axiom of Choice permits a
simultaneous choice of all mλ if Λ is infinite
...
Thus R⊕Λ is
projective
...
24)
...
Prove Hom(P, N ) is finitely generated, and is finitely presented if N is
...
10)
...
Since P is projective, the sequence
0 → K → R⊕m → P → 0
splits by (5
...
Hence Hom(P, N ) ⊕ Hom(K, N ) = Hom(R⊕m , N ) by (4
...
2)
...
15
...
3)
...
Now, Hom(P, N ) is a
178
Solutions: (5
...
9)
...
Suppose now there is a finite presentation F2 → F1 → N → 0
...
22) and
(5
...
But the Hom(R⊕m , Fi ) are free of finite rank by (4
...
1) and (4
...
2)
...
As above, Hom(K, N ) is finitely generated
...
Thus (5
...
□
Exercise (5
...
— Let R be a ring, and 0 → L → Rn → M → 0 an exact
sequence
...
Solution: Assume M is finitely presented; say Rl → Rm → M → 0 is a finite
presentation
...
Then L′ ⊕ Rn ≃ L ⊕ Rm by Schanuel’s
Lemma (5
...
Hence L is a quotient of Rl ⊕ Rn
...
Conversely, assume L is generated by ℓ elements
...
10)(1)
...
The latter is, plainly, exact
...
□
Exercise (5
...
— Let R be a ring, X1 , X2 ,
...
Set
P := R[X1 , X2 ,
...
Is M finitely presented? Explain
...
26), the ideal ⟨X1 , X2 ,
...
, fn ∈ P , so also by X1 ,
...
□
β
α
Exercise (5
...
— Let 0 → L −
→M −
→ N → 0 be a short exact sequence with
M finitely generated and N finitely presented
...
Solution: Let R be the ground ring
...
They
yield a surjection µ : Rm →
→ M by (4
...
As in (5
...
26), K is finitely generated
...
□
Exercise (5
...
— Let R be a ring, and a1 ,
...
Set M := (R/⟨a1 ⟩) ⊕ · · · ⊕ (R/⟨am ⟩)
...
α
Solution: Form the presentation Rm −
→ Rm → M → 0 where α has matrix
a1
0
...
0
am
Set s := m − r
...
Hence ai1 · · · ais ∈ ⟨a1 · · · as ⟩ for all
1 ≤ i1 < · · · < is ≤ m
...
□
Solutions: (6
...
37)
...
36), assume a1 is a nonunit
...
(2) Let n be the largest integer such that Fn (M ) = ⟨0⟩; set k := m − n
...
Show (a) that ai ̸= 0 for i < k and ai = 0 for i ≥ k, and (b) that
M determines each ai up to unit multiple
...
36)
...
35)
...
36)
...
Thus (1) holds
...
Hence a1 · · · ak−1 ̸= 0 and
a1 · · · ak = 0 by (5
...
But R is a domain
...
, ai ̸= 0 for i < k and
ak = 0
...
Hence ai = 0 for i ≥ k
...
For (2)(b), given b1 ,
...
Given i, (5
...
But R is a domain
...
12)
yields a unit ui such that a1 · · · ai = ui b1 · · · bi
...
If i < k, then b1 · · · bi−1 ̸= 0; whence, ui−1 ai = ui bi
...
□
6
...
3)
...
2)(1) is equivalent to the
commutativity of the corresponding diagram:
(
)
HomC (B, C) −
→ HomC′ F (B), F (C)
y
y
(6
...
1)
(
)
HomC (A, C) −
→ HomC′ F (A), F (C)
(2) Given γ : C → D, show (6
...
3
...
So the
composition of the top map and the right-hand map sends β to F (β)F (α), whereas
the composition of the left-hand map with the bottom map sends β to F (βα)
...
3
...
Thus (1) holds
if and only if (6
...
1)
...
□
Exercise (6
...
— Let C and C′ be categories, F : C → C′ and F ′ : C′ → C an
∼ Hom (A, F ′ A′ ) denote the natural
adjoint pair
...
Do the following:
180
Solutions: (6
...
We call the natural transformation A 7→ ηA the unit of (F, F ′ )
...
(3) Prove the natural map ηA : A → F ′ F A is universal from A to F ′ ; that is,
given f : A → F ′ A′ , there is a unique map f ′ : F A → A′ with F ′ f ′ ◦ ηA = f
...
Prove the equation in
(2) defines a natural bijection making (F, F ′ ) an adjoint pair, whose unit is η
...
4): the “free module” functor
and the “polynomial ring” functor
...
)
Solution: For (1), form this canonical diagram, with horizontal induced maps:
(F g)∗
(F g)∗
(F ′ F g)∗
g∗
HomC′ (F A, F A) −−−−−→ HomC′ (F A, F B) ←−−−− HomC′ (F B, F B)
φA, F A y
φA, F B y
φB, F B y
HomC (A, F ′ F A) −−−−−→ HomC (A, F ′ F B) ←−−−− HomC (B, F ′ F B)
It commutes since φ is natural
...
Follow 1F B out of the upper right
corner to find φA, F B (F g) = ηB ◦ g in HomC (A, F ′ F B)
...
For (2), form this canonical commutative diagram:
f′
∗
HomC′ (F A, F A) −−−−
−→ HomC′ (F A, A′ )
φA,A′
φA, F A y
y
(F ′ f ′ )∗
HomC (A, F ′ F A) −−−−−→ HomC (A, F ′ A′ )
Follow 1F A out of the upper left-hand corner to find φA,A′ (f ′ ) = F ′ f ′ ◦ ηA
...
−1
′
′
′
Thus f is unique
...
For (4), set ψA,A′ (f ′ ) := F ′ f ′ ◦ ηA
...
Thus ψA,A′ is a bijection:
∼ Hom (A, F ′ A′ )
...
And, ψA,A′
is natural in A′ , as F ′ is a functor
...
Thus (4) holds
...
4)
...
Further, if F is the “polynomial ring” functor and if A is the set of variables
X1 ,
...
, Xn ]
...
9)
...
Their pushout is defined as the object of C universal among objects P equipped
with a pair of maps γ : M → P and δ : N → P such that γα = δβ
...
16)
181
pushout as a direct limit
...
Show that, in ((R-mod)), the pushout is equal to the
direct sum M ⊕ N modulo the image of L under the map (α, −β)
...
Define a functor λ 7→ Mλ by Mλ := L, Mµ := M ,
Mν := N , αµλ := α, and ανλ := β
...
Then writing
−→
β
α
1R
1R
N
←−− L
−−→ M
ηµ y
ην y
ηλ y
α
as
Q ←−− Q −−→ Q
L
−→ M
ηµ y
βy
ην
N −→ Q
we see that Q is equal to the pushout of α and β; here γ = ηµ and δ = ην
...
Clearly γα = δβ
...
Thus (M ⊔ N )/ ∼ is the pushout
...
Then for all ℓ ∈ L, clearly ιM α(ℓ) − ιN β(ℓ) = (α(ℓ), −β(ℓ))
...
Hence, ιM α(ℓ) and ιN β(ℓ) have the same image in the quotient
...
Given γ ′ : M → P and δ ′ : N → P , they define a unique map M ⊕N → P ,
and it factors through the quotient if and only if γ ′ α = δ ′ β
...
□
Exercise (6
...
— Let C be a category, Σ and Λ small categories
...
(2) Assume C has direct limits indexed by Σ and by Λ
...
−→
−→
−→
Solution: Consider (1)
...
So, given a functor (σ, λ) 7→ Mσ,λ , there is a commutative diagram like (6
...
1)
...
Thus the rule σ 7→ (λ 7→ Mσλ ) is a functor from Σ to CΛ
...
Factoring (σ, λ) → (τ, µ) in two ways as above, we get a commutative cube
...
This passage from CΣ×Λ to (CΛ )Σ is reversible
...
As to (2), assume C has direct limits indexed by Σ and Λ
...
13)
...
6)
...
4)
the left adjoint of the composition of the two diagonal functors
...
So this diagonal has a left adjoint, which
is necessarily lim(σ,λ)∈Σ×Λ owing to the uniqueness of adjoints
...
□
−→
Exercise (6
...
— Let λ 7→ Mλ and λ 7→ Nλ be two functors from a small
category Λ to ((R-mod)), and {θλ : Mλ → Nλ } a natural transformation
...
−→
−→
−→
Show that the analogous statement for kernels can be false by constructing a
counterexample using the following commutative diagram with exact rows:
µ2
Z −−→
µ2
y
Z−
→ Z/⟨2⟩ −
→0
µ2
µ2
y
y
µ2
Z −−→ Z −
→ Z/⟨2⟩ −
→0
Solution: By (6
...
14), direct limits
commute; thus, the asserted equation holds
...
8)
...
The latter map is
0; so its kernel is Z/⟨2⟩
...
□
−→
7
...
2)
...
Assume Mλ = M
...
Order Λ by inclusion: λ ≤ µ if Mλ ⊂ Mµ
...
−→
Solution: Let us prove that M has the UMP characterizing lim Mλ
...
Such a λ exists as Mλ = M
...
Clearly,
β : M → P is the unique set map such that β|Mλ = βλ
...
So β(m+n) = βν (m+n) = β(m)+β(n)
and β(xm) = βν (xm) = xβ(m)
...
Thus M = lim Mλ
...
3)
...
Solution: Every element m ∈ M belongs to the submodule generated by m;
hence, M is the union of all its finitely generated submodules
...
So the
□
assertion results from (7
...
Exercise (7
...
— Show that every direct sum of modules is the filtered direct
limit of its finite direct subsums
...
10)
183
Solution: Consider an element of the direct sum
...
So it lies in the corresponding finite direct subsum
...
Now, given any two finite direct
subsums, their sum is a third
...
So the assertion results from (7
...
□
Exercise (7
...
— Keep the setup of (7
...
For each n ∈ Λ, set Nn := Z/⟨n⟩; if
n = ms, define αnm : Nm → Nn by αnm (x) := xs (mod n)
...
−→
Solution: For each n ∈ Λ, set Qn := Mn /Z ⊂ Q/Z
...
5
...
Now, Qn = Q/Z and Qn , Qn′ ⊂ Qnn′
...
2) yields
Q/Z = lim Mn
...
□
−→
−→
Exercise (7
...
— Let R := lim Rλ be a filtered direct limit of rings
...
(2) Assune that each Rλ is a domain
...
(3) Assume that each Rλ is a field
...
Solution: For (1), first assume R = 0
...
Then 1 ∈ Rκ maps to 0 ∈ R
...
8)(3) with Z for R yields some transition map αλκ : Rκ → Rλ with αλκ 1 = 0
...
Thus 1 = 0 in Rλ
...
1)
...
Then 1 = 0 in Rλ
...
Thus R = 0 by (1
...
Thus (1) holds
...
8)(1) and (7
...
Then xλ yλ maps to 0 ∈ R
...
8)(3) yields a
transition map αµλ with αµλ (xλ yλ ) = 0 in Rµ
...
So either αµλ (xλ ) = 0 or αµλ (yλ ) = 0
...
Thus R is a domain
...
For (3), given x ∈ R − 0, we can lift x back to some xλ ∈ Rλ for some λ by
(7
...
Then xλ ̸= 0 as x ̸= 0
...
So there is yλ ∈ Rλ with
xλ yλ = 1
...
Then xy = 1
...
Thus (3) holds
...
10)
...
For each λ, let
Nλ ⊂ Mλ be a submodule, and let N ⊂ M be a submodule
...
Solution: First, assume Nλ = αλ−1 N for all λ
...
So αλ−1 N = (αµλ )−1 αµ−1 N
...
∪
Further, Nλ = αλ−1 N implies αλ Nλ ⊂ N
...
Finally, for any
m ∈ M , there is λ and mλ ∈ Mλ with m = αλ mλ by (7
...
But Nλ := αλ−1 N ;
hence, if m ∈ N , then mλ ∈ Nλ , so m ∈ αλ Nλ
...
Conversely, assume (b)
...
∪
Assume (a) too
...
So there
is µ and nµ ∈ Nµ with αµ nµ = αλ mλ
...
8)(2) yields ν and ανµ and ανλ with
184
Solutions: (7
...
But ανµ Nµ ⊂ Nν and (ανλ )−1 Nν = Nλ by (a)
...
Thus Nλ ⊃ αλ−1 N
...
□
Exercise (7
...
— Let R := lim Rλ be a filtered direct limit of rings, aλ ⊂ Rλ
−→
an ideal for each λ
...
Set a := lim aλ
...
If each aλ is maximal, show a is maximal
...
/
So (7
...
14) implies that
−→
−→
/
lim(Rλ aλ ) = R/a
...
9) yields the assertions
...
16)
...
Let Nλ ⊂ Mλ be a
∪
be a submodule for all λ
...
Prove lim Nλ = αλ Nλ
...
So lim Nλ exists by
−→
(7
...
Also, by (7
...
∪
−→
Hence lim Nλ ⊃ αλ Nλ
...
Then (7
...
Thus lim Nλ = αλ Nλ
...
17)
...
Prove that
−→
lim nil(Rλ ) = nil(R)
...
As usual, denote the transition
maps by αµλ : Rλ → Rµ and the insertions by αλ : Rλ → R
...
So (7
...
Now, αλ nλ ⊂ n for all λ
...
−→
Conversely, given x ∈ n, say xn = 0
...
8)(1) yields λ and xλ ∈ Rλ with
λ n
λ
αλ xλ = x
...
So (7
...
Set xµ := αµ xλ
...
So xµ ∈ nµ
...
Thus αλ nλ = n, as desired
...
18)
...
Assume
−→
each ring Rλ is local, say with maximal ideal mλ , and assume each transition map
αµλ : Rλ → Rµ is local
...
Prove that R is local with maximal ideal
−→
m and that each insertion αλ : Rλ → R is local
...
So αµ mλ ⊂ mµ
...
16)
yields m = αλ mλ
...
8)(1)
...
So xλ is invertible as Rλ is local
with maximal ideal mλ
...
Thus R is local with maximal ideal
m by (3
...
Finally, (7
...
□
Exercise (7
...
— Let Λ and Λ′ be small categories, C : Λ′ → Λ a functor
...
Assume C is cofinal; that is,
(1) given λ ∈ Λ, there is a map λ → Cλ′ for some λ′ ∈ Λ′ , and
(2) given ψ, φ : λ ⇒ Cλ′ , there is χ : λ′ → λ′1 with (Cχ)ψ = (Cχ)φ
...
Show that
limλ′ ∈Λ′ MCλ′ = limλ∈Λ Mλ ;
−→
−→
more precisely, show that the right side has the UMP characterizing the left
...
7)
185
Solution: Let P be an object of C
...
Given λ ∈ Λ, choose a map
λ → Cλ′ , and define βλ : Mλ → P to be the composition
γ
′
λ
βλ : Mλ −→ MCλ′ −−
→ P
...
Given a second choice λ → Cλ′′ , there are maps λ′′ → µ′ and λ′ → µ′ for some
′
µ ∈ Λ′ since Λ′ is filtered
...
Therefore, λ → Cλ′′ gives rise to the same βλ , as desired
...
So the βλ
induce a map β : lim Mλ → P with βαλ = βλ for every insertion αλ : Mλ → lim Mλ
...
□
Exercise (7
...
— Show that every R-module M is the filtered direct limit over
a directed set of finitely presented modules
...
20), there is a presentation R⊕Φ1 −
→ R⊕Φ2 → M → 0
...
Clearly, an inclusion Ψi ,→ Φi yields an injection R⊕Ψi ,→ R⊕Φi , which is given by
extending vectors by 0
...
2) yields lim R⊕Ψi = R⊕Φi
...
Order Λ by componentwise inclusion
...
For λ ∈ Λ, set Mλ := Coker(αλ )
...
For i = 1, 2, the projection Ci : Λ → Λi is surjective, so cofinal
...
20)
yields limλ∈Λ R⊕Ci λ = limΨ ∈Λ R⊕Ψi
...
17) yields lim Mλ = M
...
Tensor Products
Exercise (8
...
— Let R be a ring, R′ an R- algebra, and M an R′ -module
...
Define α : M → M ′ by αm := 1 ⊗ m
...
Solution: As the canonical map R′ × M → M ′ is bilinear, α is linear
...
Plainly µ is R-bilinear
...
Then ρ is a retraction of α, as ρ(α(m)) = 1 · m
...
Then (5
...
□
Exercise (8
...
— Let R be a domain, a a nonzero ideal
...
Show that a ⊗R K = K
...
It is clearly R-bilinear
...
Then α = γβ as
α(x, y) = α(xz, y/z) = α(z, xy/z) = γ(xy) = γβ(x, y)
...
So γ is unique with this property
...
(Also, as γ is unique, γ is independent of the choice of z
...
Since β is surjective, so is φ
...
17)
∑
all xi and x in R
...
Then w = a ⊗ (1/x)
...
Thus φ is injective, so bijective
...
9)
...
Show there is a canonical R-linear map τ : M ⊗R N → M ⊗R′ N
...
Show K is equal to
Ker(τ ), and τ is surjective
...
Solution: The canonical map β ′ : M × N → M ⊗R′ N is R′ -bilinear, so Rbilinear
...
3), it factors: β ′ = τ β where β : M × N → M ⊗R N is the
canonical map and τ is the desired map
...
Then τ factors through a map τ ′ : Q → M ⊗R′ N since
each generator (x′ m) ⊗ n − m ⊗ (x′ n) of K maps to 0 in M ⊗R′ N
...
8), there is an R′ -structure on M ⊗R N with y ′ (m ⊗ n) = m ⊗ (y ′ n), and
so by (8
...
Clearly, K is a submodule
for each structure, so Q is too
...
Further,
the canonical map M × N → Q is R′ -bilinear
...
Hence Ker(τ ) is
M ⊗R′ N , furnishing an inverse to τ ′
...
Finally, suppose R′ is a quotient of R
...
So each (x′ m) ⊗ n − m ⊗ (x′ n) is equal to 0 in M ⊗R N as x′ m = xm and
x′ n = xn
...
Thus τ is an isomorphism
...
12)
...
11), find the unit ηM of each adjunction
...
A map θ : F M → P corresponds to the map M → P carrying m to θ(m ⊗ 1R′ )
...
Thus ηM : M → F M is given by ηM m = m ⊗ 1R′
...
A map
µ : M → P corresponds to the map M → F ′ P carrying m to the map ν : R′ → P
defined by νx := x(µm)
...
Thus ηM : M → F ′ M is
given by (ηM m)(x) = xm
...
15)
...
Prove M ⊗ N ̸= 0
...
Then
(8
...
6)(2)
...
□
Exercise (8
...
— Let R be a ring, a and b ideals, and M a module
...
13) to show that (R/a) ⊗ M = M/aM
...
Solution: To prove (1), view R/a as the cokernel of the inclusion a → R
...
13) implies that (R/a)⊗M is the cokernel of a⊗M → R⊗M
...
6)(2)
...
The assertion follows
...
For example, take R := Z, take a := ⟨2⟩, and take M := Z/⟨2⟩; then
a ⊗ M → M is just multiplication by 2 on Z/⟨2⟩, and so aM = 0
...
Note a(R/b) = (a + b)/b by (4
...
1)
...
The latter is equal to R/(a + b) by (4
...
2)
...
26)
187
Exercise (8
...
— Show Z/⟨m⟩ ⊗Z Z/⟨n⟩ = 0 if m and n are relatively prime
...
Thus (8
...
□
Exercise (8
...
— Let F : ((R-mod)) → ((R-mod)) be a linear functor
...
Show that θ(M ) : M ⊗ F (R) → F (M )
is surjective if F preserves surjections and M is finitely generated, and that θ(M )
is an isomorphism if F preserves cokernels and M is finitely presented
...
18), since F preserves the relations there
...
18), but with Σ and Λ finite
...
24)
...
Let M [X] be
the
in X with coefficients in M , that is, expressions of the form
∑n set of polynomials
i
m
X
with
m
∈
M
...
i
i
i=0
Solution:
Plainly,∑M [X] is an R[X]-module
...
Then b is R-bilinear, so induces an R-linear map
β : M ⊗R R[X] → M [X]
...
By (8
...
Then
βt
=
mi X i
...
Given u :=
mi X ∈ M [X], set
t := mi ⊗ X i
...
Thus β is bijective, as desired
...
∑
∑
If φM,P α = 0, then α( mi X i ) = (αmi )X i = 0, because α is R[X]-linear and
α|M
∑ = 0; thus∑φM,P is iinjective
...
Then α is R[X]-linear, and φM
...
Thus φM,P is
surjective, so bijective
...
But M 7→ M ⊗R R[X] is too by (8
...
Thus M [X] = M ⊗R R[X]
...
25)
...
For each finite
subset J of Σ, let RJ′ be the tensor product of the Rσ′ for σ ∈ J
...
Solution: Let Λ be the set of subsets of Σ, partially ordered by inclusion
...
1)
...
22) implies that RJ′
is the coproduct of the family (Rσ′ )σ∈J , so that, first, for each σ ∈ J, there is a
canonical algebra map ισ : Rσ′ → RJ′ , and second, given J ⊂ K, the ισ for σ ∈ K
J
′
induce an algebra map αK
: RJ′ → RK
...
7)
...
Further, the various φJ
are compatible, so they induce a compatible map φ : lim RJ′ → R′′
...
□
188
Solutions: (9
...
26)
...
Set k := Q(ω) and K := k[ 3 2]
...
Solution: Note ω is a root of X 2 + X + 1,√which is √
irreducible √
over Q; hence,
[k : Q] = 2
...
Therefore,
∼ K
...
So X 3 −2 is irreducible over k
...
23)
...
6)(2) and (8
...
16)(1) yield
/
/
k[X] ⟨X 3 − 2⟩ ⊗k K = k[X] ⟨X 3 − 2⟩ ⊗k[X] (k[X] ⊗k K)
/
/
= k[X] ⟨X 3 − 2⟩ ⊗k[X] K[X] = K[X] ⟨X 3 − 2⟩
...
So the Chinese Remainder Theorem, (1
...
8)
...
Flatness
Exercise (9
...
— Let R be a ring, R′ an algebra, F an R-linear functor from
((R-mod)) to ((R′ -mod))
...
Prove the following equivalent:
(1) F is faithful
...
(3) F (R/m) ̸= 0 for every maximal ideal m of R
...
Solution: To prove (1) implies (2), suppose F M = 0
...
But
always 1F M = F (1M )
...
So M = 0
...
Conversely, assume (2)
...
As F is
exact, (9
...
Hence F I = 0
...
Thus α = 0
...
Thus (1) and (2) are equivalent
...
Conversely, assume (3)
...
Set a := Ker(α)
...
We get a surjection R/a →
→ R/m
and an injection R/a ,→ M
...
But F (R/m) ̸= 0 by (3)
...
So F M ̸= 0
...
Thus (1) and (2) and (3) are equivalent
...
Now, F (βα) = 0
...
Hence I ⊂ K
...
3)
yields F I = Im(F α) and F K = Ker(F β)
...
But (1) implies (2)
...
Thus (4) holds
...
Given α : M → N with F α = 0, set K := Ker(α)
...
3) yields F K = Ker(F α)
...
So (4)
implies K → M → 0 is exact
...
Thus (1) holds, as desired
...
8)
...
Solutions: (9
...
7)
...
10)
...
Show that M ⊗ N
is flat
...
10) yields (M ⊗ N ) ⊗ • = M ⊗ (N ⊗ •); in other
words, (M ⊗ N ) ⊗ • = (M ⊗ •) ◦ (N ⊗ •)
...
Hence it is exact
...
Similarly if M and N are faithfully flat, then M ⊗ N ⊗ • is faithful and exact
...
□
Exercise (9
...
— Let R be a ring, M a flat module, R′ an algebra
...
What if “flat” is replaced everywhere by “faithfully flat”?
Solution: Cancellation (8
...
But M ⊗R •
is exact, as M is flat over R
...
Similarly, if M is faithfully flat over R, then M ⊗R • is faithful too
...
□
Exercise (9
...
— Let R be a ring, R′ a flat algebra, M a flat R′ -module
...
What if “flat” is replaced everywhere by “faithfully flat”?
Solution: Cancellation (8
...
But R′ ⊗R •
and M ⊗R′ • are exact; so their composition M ⊗R • is too
...
Similarly, as the composition of two faithful functors is, plainly, faithful, the
assertion remains true if “flat” is replaced everywhere by “faithfully flat
...
13)
...
Assume that M is flat over R and faithfully flat over R′
...
Solution: Let N ′ → N be an injective map of R-modules
...
But by Cancellation (8
...
And M is faithfully flat over R′
...
4)
...
□
Exercise (9
...
— Let R be a ring, a an ideal
...
Show a = a2
...
But the image vanishes: a ⊗ r = 1 ⊗ ar = 0
...
16)
...
Thus a = a2
...
15)
...
Prove equivalent:
(1)
(2)
(3)
(4)
(5)
(6)
R′ is faithfully flat over R
...
Every ideal a of R is the contraction of its extension, or a = φ−1 (aR′ )
...
Every maximal ideal m of R extends to a proper ideal, or mR′ ̸= R′
...
190
Solutions: (9
...
In (2), set K := Ker α
...
But α ⊗ 1 has a retraction, namely m ⊗ x ⊗ y 7→ m ⊗ xy
...
Hence K ⊗R R′ = 0
...
4)
...
Assume (2)
...
But (R/a) ⊗ R′ = R′ /aR′ by
(8
...
So φ−1 (aR′ ) = a
...
Assume (3)
...
13)(2) yields (4)
...
Then every maximal ideal m of R is the contraction of some prime
q of R′
...
Thus (5) holds
...
Consider (6)
...
As R′ is
flat, the inclusion M ′ ,→ M yields an injection M ′ ⊗ R′ ,→ M ⊗ R′
...
7)
...
16)(1)
...
Then aR′ ⊂ mR′
...
Hence
R′ /aR′ ̸= 0
...
Hence M ⊗ R′ ̸= 0
...
Finally, (6) and (1) are equivalent by (9
...
□
α
Exercise (9
...
— Let R be a ring, 0 → M ′ −
→ M → M ′′ → 0 an exact sequence
′ N ⊗α
with M flat
...
Prove M ′′ is flat
...
It yields the following commutative
diagram with exact rows by hypothesis and by (8
...
So the Snake Lemma (5
...
Thus M ′′ is flat
...
18)
...
Solution: Assume R′ is faithfully flat
...
15)
...
And,
since R′ is flat, R′ /φ(R) is flat by (9
...
Conversely, assume φ is injective and R′ /φ(R) is flat
...
16)(1), and R′ is flat by (9
...
Thus R′ is
faithfully flat by (9
...
21)
...
Show
that there is a canonical map
σ : HomR (M, N ) ⊗R R′ → HomR′ (M ⊗R R′ , N ⊗R R′ )
...
Show that if M is finitely generated, then σ is injective,
and that if M is finitely presented, then σ is an isomorphism
...
20), put P := N ⊗R R′ in
the second equation in (8
...
□
Solutions: (10
...
25) (Equational Criterion for Flatness)
...
24)(4) can be reformulated as follows: Given any relation
i xi yi = 0 with
xi ∈ R and yi ∈ M , there are xij ∈ R and yj′ ∈ M such that
∑
∑
′
(9
...
1)
j xij yj = yi for all i and
i xij xi = 0 for all j
...
, em be the standard basis of Rm
...
24)(4) holds
...
Set
φ
β
k := xi ei
...
So (9
...
Let e′1 ,
...
Let (xij ) be the n × m matrix
xji e′j
...
Now, φ(k) = 0; hence, i,j xji xi ej = 0
...
25
...
∑
Conversely, given α : Rm → M and k ∈ Ker(α), write k =
xi ei
...
25
...
Let∑
φ : Rm → Rn be the map with matrix (xij ); that is, φ(ei ) =
xji e′j
...
Define β : R → M by β(ej ) := yj
...
Thus (9
...
□
Exercise (9
...
— Let R be a ring, M a module
...
Solution: For ∑
(1), assume M is flat and xm = 0
...
25) yields xj ∈ R
and mj ∈ M with
xj mj = m and xj x = 0 for all j
...
µx
Alternatively, 0 → Ann(x) → R −−→ R is always exact
...
So Im(Ann(x) ⊗ M ) is
Ker(µx )
...
Thus (1) holds
...
26), to show α : a ⊗ M → aM is injective
...
So, given
such that z = i yi x ⊗ mi
...
Then
∑
∑
z = i x ⊗ yi mi = x ⊗ i yi mi = x ⊗ m
...
Then xm = 0
...
So
∑
m = j zj nj for some zj ∈ Ann(x) and nj ∈ M
...
Thus α is injective
...
□
10
...
6)
...
Assume α is surjective
...
Solution: Let m be a maximal ideal
...
Plainly, that map can be rewritten as (R/m)m →
→ (R/m)n
...
Thus m ≥ n
...
7)
...
Assume a is finitely generated
and idempotent (or a = a2 )
...
Solution: By (10
...
So
for all x ∈ a, we have (1 − e)x = 0, or x = ex
...
Finally, e is unique by (3
...
□
192
Solutions: (10
...
8)
...
Prove the following conditions
are equivalent:
(1) R/a is projective over R
...
(3) a is finitely generated and idempotent
...
(5) a is a direct summand of R
...
Then R/a is flat by (9
...
Further, the sequence
0 → a → R → R/a → 0 splits by (5
...
So (5
...
Hence a is principal
...
If (2) holds, then (3) holds by (9
...
If (3) holds, then (4) holds
⊕ by (10
...
If
(4) holds, then (5) holds by (1
...
If (5) holds, then R ≃ a R/a, and so (1)
holds by (5
...
□
Exercise (10
...
— Prove the following conditions on a ring R are equivalent:
(1) R is absolutely flat; that is, every module is flat
...
(3) Every finitely generated ideal is idempotent
...
Solution: Assume (1)
...
Then R/a is flat by
hypotheses
...
8)
...
Conditions (2) and (3) are equivalent by (10
...
Trivially, if (3) holds, then (4) does
...
Given a finitely
generated ideal a, say a = ⟨x1 ,
...
Then each ⟨xi ⟩ is idempotent by hypothesis
...
17)(2)
...
, fn ⟩
...
17)(4), (1)
...
Assume (2)
...
Then a is a
∼ aM
direct summand of R by hypothesis
...
6)
...
16)(1); cf
...
16)(1)
...
26)
...
□
Exercise (10
...
— Let R be a ring
...
Prove R is absolutely flat
...
Prove any quotient ring R′ is absolutely flat
...
Prove every nonunit x is a zerodivisor
...
Prove R is a field
...
Hence every
principal ideal is idempotent by (1
...
Thus (10
...
For (2), let b ⊂ R′ be principal, say b = ⟨x⟩
...
Then ⟨x⟩ is
idempotent by (10
...
Hence b is also idempotent
...
9) yields (2)
...
Then ⟨x⟩ is idempotent by (10
...
So x = ax2
for some a
...
But x is a nonunit
...
Thus (3) holds
...
Since x is a nonunit, x ∈ m
...
So ax − 1 ∈
/ m
...
But x(ax − 1) = 0
...
Thus 0
is the only nonunit
...
□
Exercise (10
...
— Let R be a ring, a an ideal, and α : M → N a map of
modules
...
Show that α is surjective
...
22)
193
Solution: Since α is surjective, α(M ) + aN = N
...
Hence α(M ) = N by (10
...
Thus α is surjective
...
15)
...
Let α, β : M ⇒ N
be two maps of finitely generated modules
...
Set γ := α + β
...
Solution: As α is surjective, given n ∈ N , there is m ∈ M with α(m) = n
...
But M/N is finitely generated as M is
...
13)(1)
...
4)
...
□
Exercise (10
...
— Let A be a local ring, m the maximal ideal, M a finitely
generated A-module, and m1 ,
...
Set k := A/m and M ′ := M/mM , and
write m′i for the image of mi in M ′
...
, m′n ∈ M ′ form a basis
of the k-vector space M ′ if and only if m1 ,
...
Solution: By (10
...
5), as an k-vector space
...
Hence, a minimal generating set of M corresponds to a minimal generating set of M ′ , that is, to a basis
...
□
Exercise (10
...
— Let A be a local ring, k its residue field, M and N finitely
generated modules
...
(2) Show
that M ⊗A N ̸= 0 if M ̸= 0 and N ̸= 0
...
Then M ⊗ k = M/mM by (8
...
So (1) is nothing but a form of Nakayama’s Lemma (10
...
In (2), M ⊗ k ̸= 0 and N ⊗ k ̸= 0 by (1)
...
15)
and (8
...
But (M ⊗ k) ⊗ (N ⊗ k) = (M ⊗ N ) ⊗ (k ⊗ k) by the associative and
commutative laws, (8
...
6)
...
16)(1)
...
19)
...
Prove that M is faithfully flat over A if and only if M is flat over
A and nonzero
...
Solution: Plainly, to prove the first assertion, it suffices to show that M ⊗A • is
faithful if and only if M ̸= 0
...
4)
...
Thus M ̸= 0
...
Denote the maximal ideals of A and B by m and n
...
11)
...
So M/mM ̸= 0
...
16)(1)
...
4) implies M ⊗A • is faithful
...
□
Exercise (10
...
— Let G be a finite group of automorphisms of a ring R
...
Show that every x ∈ R is integral over RG , in
fact, over the subring R′ generated by the elementary symmetric functions in the
conjugates gx for g ∈ G
...
2)
∏
Solution: Given an x ∈ R, form F (X) := g∈G (X − gx)
...
So F (x) = 0 is a relation of integral
dependence for x over RG , in fact, over its subring R′
...
24)
...
Set R := k[X 2 ] ⊂ P
...
Solution: Write f = fe + fo , where fe and fo are the polynomials formed by
the terms of f of even and odd degrees
...
Then the matrix of µf is
( f gX 2 )
e
...
So the Cayley–
g fe
Hamilton Theorem (10
...
□
Exercise (10
...
∏— Let R1 ,
...
Show
that their product Ri is a integral over R
...
, yn ) ∈ i=1 Ri
...
28)
...
16) and induction
∏non n
...
□
Therefore, y is integral over R by (10
...
Thus i=1 Ri is integral over R
...
31)
...
Set R := Ri , set R′ := Ri′ , and set x := (x1 ,
...
Prove
(1) x is integral over R if and only if xi is integral over Ri for each i;
(2) R is integrally closed in R′ if and only if each Ri is integrally closed in Ri′
...
Say xn + a1 xn−1 + · · · + an = 0 with
aj ∈ R
...
, arj )
...
Then xni + ai1 xn−1 + · · · + ain = 0
...
Conversely, assume each xi is integral over Ri
...
Set n := max ni , set aij := 0 for j > ni , and set aj := (a1j ,
...
Then xn + a1 xn−1 + · · · + an = 0
...
Thus (1) holds
...
□
Exercise (10
...
— Let k be a field, X and Y variables
...
Prove that R is a domain, but not a field
...
Prove that k[t] is the integral closure of R in Frac(R)
...
6); however, it is not maximal by (2
...
Hence R is a domain
by (2
...
17)
...
Hence x = t2 − 1 and y = t3 − t
...
Further, t is integral over R; so k[t] is integral over R by (2)⇒(1) of (10
...
Finally, k[t] has Frac(R) as fraction field
...
So k[t] is normal by
□
(10
...
Thus k[t] is the integral closure of R in Frac(R)
...
Localization of Rings
Solutions: (11
...
2)
...
Prove S −1 R = 0
if and only if S contains a nilpotent element
...
1), S −1 R = 0 if and only if 1/1 = 0/1
...
Finally, since S is multiplicative, 0 ∈ S if and only
if S contains a nilpotent element
...
4)
...
As a starter, prove Z[2/3] = S −1 Z where S = {3i | i ≥ 0}
...
But the opposite inclusion
holds as 1/3 = 1 − (2/3)
...
Let P ⊂ Z be the set of all prime numbers that appear as factors of the denominators of elements of R in lowest terms; recall that x = r/s ∈ Q is in lowest
terms if r and s have no common prime divisor
...
Clearly,
S is equal to the set of all products of elements of P
...
Indeed, take an element x = r/ps ∈ R in
lowest terms
...
Also the Euclidean algorithm yields m, n ∈ Z
such that mp + nr = 1
...
Hence S −1 Z ⊂ R
...
Thus S −1 Z = R
...
7)
...
Consider R := R′ × R′′ and set
S := { (1, 1), (1, 0) }
...
Solution: Let’s show that the projection map π : R′ × R′′ → R′ has the UMP
of (11
...
First, note that πS = {1} ⊂ R′×
...
Then in B,
(
)
ψ(1, 0) · ψ(0, x) = ψ (1, 0) · (0, x) = ψ(0, 0) = 0 in B
...
So ψ factors uniquely through π by (1
...
□
Exercise (11
...
— Take R and S as in (11
...
On R × S, impose this relation:
(x, s) ∼ (y, t)
if
xt = ys
...
Solution: Observe that, for any z ∈ R′′ , we have
(
) (
)
(1, z), (1, 1) ∼ (1, 0), (1, 0)
...
Thus although ∼ is reflexive and symmetric, it is not transitive if R′′ ̸= 0
...
9)
...
17)
...
Show the following:
(1) U = { x/s | x ∈ S and s ∈ S }
...
−1
−1
−1
(3) S R = T R if and only if S = T
...
196
Solutions: (11
...
Then
x/s · sy/xy = 1 in S −1 R
...
Conversely, say x/s · y/t = 1 in S −1 R
with x, y ∈ R and s, t ∈ S
...
But stu ∈ S
...
Thus (1) holds
...
Then V is saturated multiplicative by (3
...
Further,
V ⊃ S by (11
...
Thus (1)(c) of (3
...
Conversely, take x ∈ V
...
So (1) yields x/1 = y/s with y ∈ S and s ∈ S
...
But S ⊃ S by (1)(a) of (3
...
17); so yt ∈ S
...
Thus x ∈ S
...
In (3), if S −1 R = T −1 R, then (2) implies S = T
...
As to (4), note that, in any ring, a product is a unit if and only if each factor is
...
−1
Thus S R and S −1 R are characterized by equivalent UMPs
...
□
Exercise (11
...
— Let R be a ring, S ⊂ T ⊂ U and W multiplicative subsets
...
(2) Given a map φ : S −1 R → W −1 R, show S ⊂ S ⊂ W and φ = φSW
...
Assume
Sλ = S
...
Order Λ by
inclusion: λ ≤ µ if Sλ ⊂ Sµ
...
−→
Solution: For (1), note φT S ⊂ φT T ⊂ (T −1 R)×
...
5) yields a unique
R-algebra map φST : S −1 R → T −1 R
...
Thus (1) holds
...
So φ−1
R)× ⊂ φ−1
R)×
...
9)
...
17)(a)
...
For (3), notice Λ is directed
...
Then ανµ αµλ = ανλ if
λ ≤ µ ≤ ν
...
7)
...
Then βµ αµλ = βλ
...
−→
Take x ∈ Ker(β)
...
8)(1)
...
So there is s ∈ S with sxλ = 0
...
Hence αµλ (xλ /sλ ) = 0
...
Thus β is injective
...
Then s ∈ Sλ for some λ, so x/s ∈ Sλ−1 R
...
Thus β is surjective, so an isomorphism
...
□
Exercise (11
...
— Let R be a ring, S0 the set of nonzerodivisors
...
(2) Show every element x/s of S0−1 R is either a zerodivisor or a unit
...
Show R = S0−1 R
...
Then φS (sx) = 0
...
But φS (s) is a unit
...
But φS is injective
...
Thus S ⊂ S0 ; that is, (1) holds
...
Then x/1 is also a
nonzerodivisor
...
Therefore, x/s is a unit
...
∼ S −1 R by (11
...
□
In (3), by hypothesis, S0 ⊂ R×
...
29)
197
Exercise (11
...
— Let R be a ring, S a multiplicative subset, a and b ideals
...
Solution: For (1), take x ∈ aS
...
If a ⊂ b, then
sx ∈ b, and so x ∈ bS
...
To show (2), proceed by double inclusion
...
16)(2)
...
16)(2)
...
So there is t ∈ S with tsx ∈ a
...
So x ∈ aS
...
To show (3), proceed by double inclusion
...
16)(2)
...
Thus (1) yields
∑(ab) ⊂ (a b )
...
Then
there are si , ti ∈ S such that si yi ∈ a and ti zi ∈ b
...
Then u ∈ S
and ux ∈ ab
...
Thus aS bS ⊂ (ab)S
...
But ((ab)S )S = (ab)S by (2)
...
□
Exercise (11
...
— Let R be a ring, S a multiplicative subset
...
Solution: Proceed by double inclusion
...
14)(1)
...
So (x/s)n = 0
...
Thus nil(R)(S −1 R) ⊂ nil(S −1 R)
...
Then (x/s)m = 0 with m ≥ 1
...
16)(1)
...
So tx ∈ nil(R)
...
So x/s ∈ nil(R)(S −1 R) by (11
...
Thus nil(R)(S −1 R) ⊃ nil(S −1 R)
...
24)
...
Show that S −1 R′ is integral over S −1 R
...
Then
(x/s)n + (an−1 /1)(1/s)(x/s)n−1 + · · · + a0 (1/s)n = 0
is an equation of integral dependence of x/s on S −1 R, as required
...
25)
...
Show that L is the fraction field of R
...
Solution: Let x ∈ L
...
Write yi = ai /a with a1 ,
...
Then
(ax)n + a1 (ax)n−1 + · · · + an−1 a0 = 0
...
Then b ∈ R and x = b/a
...
26)
...
Assume that R′ is a finitely generated R-algebra, and that L is a finite dimensional
K-vector space
...
Solution: Let x1 ,
...
∏Using (11
...
Set f :=
ai
...
Thus Rf is module finite over Rf by (10
...
□
198
Solutions: (12
...
29)
...
(1) Set T ′ := φS (T ) and assume S ⊂ T
...
(2) Set U := {st ∈ R | s ∈ S and t ∈ T }
...
Solution: A proof similar to that of (11
...
By
(11
...
Thus (1) holds
...
So (1) yields U −1 R = U −1 (S −1 R)
...
Similarly, U −1 RS −1 (T −1 R)
...
□
Exercise (11
...
— Given a domain
R and a multiplicative subset S with 0 ∈
/ S
...
Solution: Since 0 ∈
/ S, clearly Frac(S −1 R) = Frac(R) owing to (11
...
Now,
S R is integral over S −1 R by (11
...
Thus S −1 R ⊂ S −1 R
...
∏
Say ai = bi /si with bi ∈ R and si ∈ S; set s := si
...
Hence sx ∈ R
...
Thus S −1 R ⊃ S −1 R, as desired
...
Localization of Modules
Exercise (12
...
— Let R be a ring, S a multiplicative subset, and M a module
...
Solution: If M = S −1 M , then M is an S −1 R-module since S −1 M is by (12
...
Conversely, if M is an S −1 R-module, then M equipped with the identity map has
the UMP that characterizes S −1 M ; whence, M = S −1 M
...
5)
...
Set T1 := φS (T ) ⊂ S −1 R
...
Solution: Let’s check that both T −1 (S −1 M ) and T1−1 (S −1 M ) have the UMP
characterizing T −1 M
...
Then the multiplication map µs : N → N is bijective for all s ∈ T by (12
...
Hence ψ factors via a unique S −1 R-linear map
ρ : S −1 M → N by (12
...
1) again
...
Hence ψ = ρ′ φT φS , and ρ′ is clearly unique, as required
...
Hence ψ = ρ′1 φT1 φS , and
ρ′1 is clearly unique, as required
...
14)
199
Exercise (12
...
— Let R be a ring, S a multiplicative subset
...
Given a module M , define a functor S → ((R-mod)) as follows: for s ∈ S, set
Ms := M ; to each x ∈ Hom(s, t), associate µx : Ms → Mt
...
by βs (m) := m/s
...
Now, given s, t ∈ S, set u := st
...
Given x, y ∈ Hom(s, t), we have
xs = t and ys = t
...
Thus S is filtered
...
So the βs induce a homomorphism β : lim Ms → S −1 M
...
Each m ∈ lim Ms lifts to an m′ ∈ Ms for some s ∈ S by (7
...
Assume
−→
βm = 0
...
But βs m′ = m′ /s
...
So µt m′ = 0 in Mst , and µt m′ 7→ m
...
Thus β is injective,
so an isomorphism
...
7)
...
Prove S −1 M = 0 if Ann(M ) ∩ S ̸= ∅
...
Solution: Say f ∈ Ann(M )∩S
...
Then f /1·m/t = f m/t = 0
...
Thus S −1 M = 0
...
mn generate M
...
Then fi ∈ Ann(M ) ∩ S, as desired
...
8)
...
(1) Set S := 1 + a
...
(2) Use (1), Nakayama’s Lemma (10
...
7), but not the determinant
trick (10
...
3): if M = aM , then sM = 0 for an s ∈ S
...
2) as follows
...
Then for x ∈ R and c ∈ a, we have
(
)(
) (
)/
1 + a/(1 + b) x/(1 + c) = 1 + (b + c + bc + ax) (1 + b)(1 + c)
...
So a/(1 + b) ∈ rad(S −1 R) by
(3
...
Thus (1) holds
...
Then S −1 M = S −1 aS −1 M by (12
...
So S −1 M = 0
by (1) and (10
...
So (12
...
Thus (2) holds
...
12)
...
Then S −1 P is a projective S −1 R-module
...
23), there is a module K such that F := K ⊕ P is free
...
10) yields that S −1 F = S −1 P ⊕ S −1 K and that S −1 F is free over S −1 R
...
23)
...
14)
...
Show S −1 (M ⊗R N ) = S −1 M ⊗R N = S −1 M ⊗S −1 R S −1 N = S −1 M ⊗R S −1 N
...
19)
Solution: By (12
...
The latter is
equal to (S −1 R ⊗R M ) ⊗R N by associativity (8
...
Again by (12
...
Thus the first equality holds
...
11), S −1 M ⊗R N = S −1 M ⊗S −1 R (S −1 R ⊗R N ), and the
latter is equal to S −1 M ⊗S −1 R S −1 N by (12
...
Thus the second equality holds
...
9), the kernel of the map S −1 M ⊗R S −1 N → S −1 M ⊗S −1 R S −1 N
is generated by elements (xm/s) ⊗ (n/1) − (m/1) ⊗ (xn/s) with m ∈ M , n ∈ N ,
x ∈ R, and s ∈ S
...
Thus the third equality holds
...
15)
...
Show
Fr (M ⊗R R′ ) = Fr (M )R′
and
Fr (S −1 M ) = Fr (M )S −1 R = S −1 Fr (M )
...
Then, by (8
...
Further, the matrix A of α induces the matrix of α ⊗ 1
...
For the last equalities, take R′ := S −1 R
...
13)
...
2)
...
18)
...
α
(1) Let M1 −
→ M2 be a map of modules, which restricts to a map N1 → N2 of
submodules
...
α
β
(2) Let 0 → M1 −
→ M2 −
→ M3 be a left exact sequence, which restricts to a left
exact sequence 0 → N1 → N2 → N3 of submodules
...
Solution: For (1), take m ∈ N1S
...
So
α(sm) ∈ N2
...
Thus (1) holds
...
Trivially, α|N1S is injective, and
βα|N1S = 0
...
Then β(sm2 ) = sβ(m2 ) = 0
...
Also α(sm1 ) = sα(m1 ) = sm2
...
Hence sm1 = n1
...
Thus (2) holds
...
19)
...
Set T S M := ⟨0⟩S
...
Prove the following:
(1) T S (M/T S M ) = 0
...
(3) Let α : M → N be a map
...
(4) Let 0 → M ′ → M → M ′′ be exact
...
(5) Let S1 ⊂ S be a multiplicative subset
...
Solution: For (1), given an element of T S (M/T S (M )), let m ∈ M represent
it
...
So there is t ∈ S with tsm = 0
...
Thus (1) holds
...
17)(3)
...
18)(1) and (2)
...
Then sm/s1 = 0
...
Thus S1−1 T S (M ) ⊂ T S (S1−1 M )
...
4)
201
For the opposite inclusion, given m/s1 ∈ T S (S1−1 M ) with m ∈ M and s1 ∈ S1 ,
take t/t1 with t ∈ S and t1 ∈ S1 and t/t1 · m/s1 = 0
...
So there
is s′ ∈ S1 with s′ tm = 0 by (12
...
But s′ t ∈ S as S1 ⊂ S
...
Thus m/s1 ∈ S1−1 T S (M )
...
□
⊕
Exercise (12
...
— Set R := Z and S = Z − ⟨0⟩
...
Show that the map σ of (12
...
Solution: Given m > 0, let en be the nth standard basis element for some
n > m
...
Hence µR : R → HomR (M, M ) is injective
...
So Hom(S −1 M, S −1 M ) = 0
...
□
13
...
2)
...
Show that p is a closed point —
that is, {p} is a closed set — if and only if p is a maximal ideal
...
Conversely, suppose p is not maximal
...
If
p ∈ V(a), then m ∈ V(a) too
...
Thus {p} is not closed
...
3)
...
Let X1 , X2 ⊂ X be
closed subsets
...
(2) There are complementary idempotents e1 , e2 ∈ R with V(⟨ei ⟩) = Xi
...
(4) There are ideals a1 , a2 ⊂ R with a1 ⊕ a2 = R and V(ai ) = Xi
...
Solution: Assume (1)
...
Then (13
...
√
√
√
√
So a1 a2 = ⟨0⟩ and a1 + a2 = R again by (13
...
Hence (3
...
Assume (2)
...
As e1 + e2 = 1 and e1 e2 = 0, plainly (3) holds
...
As the ai are comaximal, the Chinese Remainder Theorem (1
...
But a1 a2 = 0
...
17)
...
Assume (4)
...
1) yields (1) as follows:
X1 ∪ X2 = V(a1 ) ∪ V(a2 ) = V(a1 a2 ) = V(0) = X
and
X1 ∩ X2 = V(a1 ) ∩ V(a2 ) = V(a1 + a2 ) = V(R) = ∅
...
Then ⟨ei ⟩ = ai by
□
(13
...
So eni ∈ ai for some n ≥ 1
...
Thus ei ∈ ai
...
4)
...
Set φ∗ := Spec(φ)
...
(2) If every prime of R′ is an extension of a prime, then φ∗ is injective
...
8)
Solution: Note φ∗ (q) := φ−1 (q) by (13
...
2)
...
Given two primes q1 and q2 that are extensions, if qc1 = qc2 , then q1 = q2 by
(1
...
Thus (2) holds
...
Take R to be a domain
...
Then
′
R /⟨X⟩ = R by (1
...
So ⟨X⟩ is prime by (2
...
But ⟨X⟩ is not an extension,
as X ∈
/ aR′ for any proper ideal a of R
...
Hence q is generated by q ∩ R and x
...
Thus φ∗ is injective
...
5)
...
Set X := Spec(R)
and Y := Spec(S −1 R)
...
Show
(1) that S −1 X consists of the primes p of R with p ∩ S = ∅ and (2) that φ∗S is a
homeomorphism of Y onto S −1 X
...
1
...
Hence (11
...
But φ∗S is continuous by (13
...
So it remains to show
that φ∗S is closed
...
It suffices to show
∩
φ∗S (V(b)) = S −1 X V(a)
...
5
...
Then p = φ−1
S (q) and q ⊃ b by
−1
∗
−1
(13
...
So p = φ−1
(q)
⊃
φ
(b)
=:
a
...
But
p
∈
φ
X
...
∩
Conversely, given p ∈ S −1 X V(a), say p = φ∗S (q)
...
So
φ
(q)
⊃
φ
(b)
...
So
q ⊃ b by
S
S
S
S
S
(11
...
So q ∈ V(b)
...
Thus (13
...
1) holds, as
desired
...
□
Exercise (13
...
— Let θ : R → R′ be a ring map, S ⊂ R a multiplicative subset
...
Via (13
...
23), identify Spec(S −1 R) and Spec(S −1 R′ ) with their images S −1
X ⊂ X and
S −1 Y ⊂ Y
...
Solution: Given q ∈ Y , elementary set-theory shows that q ∩ θ(S) = ∅ if and
only if θ−1 (q) ∩ S = ∅
...
5)(1)
∗
and (11
...
But φ−1
S (q) =: φS (q) by (13
...
2)
...
−1
−1 −1
Finally, (S θ)φS = φS θ by (12
...
So φ−1
θ) (q) = (S −1 θ)−1 φ−1
S (S
S (q)
...
1
...
□
Exercise (13
...
— Let θ : R → R′ be a ring map, a ⊂ R an ideal
...
Let θ : R/a → R′ /b be the induced map
...
Set
∗
′
θ∗ := Spec(θ) and θ := Spec(θ)
...
1), identify Spec(R/a) and Spec(R
/b)
∗
∗−1
∗
with V(a) ⊂ X and V(b) ⊂ Y
...
Solution: Given q ∈ Y , observe that q ⊃ b if and only if θ−1 (q) ⊃ a, as follows
...
5)(1) in its notation, q ⊃ b := ae yields qc ⊃ aec ⊃ a, and qc ⊃ a yields
q ⊃ qce ⊃ ae
...
Plainly, θ(q/b) = (θ−1 q)/a
...
1
...
□
Exercise (13
...
— Let θ : R → R′ be a ring map, p ⊂ R a prime, k the residue
field of Rp
...
Show (1) that θ∗−1 (p) is canonically homeomorphic
to Spec(R′ ⊗R k) and (2) that p ∈ Im θ∗ if and only if R′ ⊗R k ̸= 0
...
13)
203
Solution: First, take S := R − p and apply (13
...
Next, take a := pRp and apply (13
...
But θ−1 (pRp ) = p by (11
...
Therefore,
(
)−1
Spec(Rp′ /pRp′ ) = θ∗ Spec(Rp′ )
(p) = θ∗−1 (p)
...
So R′ /pR′ = k ⊗R R′
...
Finally, (1) implies p ∈ Im θ∗ if and only if Spec(R′ ⊗R k) ̸= ∅
...
□
Exercise (13
...
— Let R be a ring, p a prime ideal
...
Solution: By (13
...
Given
f ∈ R − p, note that D(f ) contains every prime contained in p, or Xp ⊂ D(f )
...
1)
...
Conversely,
given a prime q ̸⊂ p, there is g ∈ q − p
...
Thus Xp = Z, as desired
...
10)
...
Show
∩
Im Spec(θ) = Im Spec(φ) Im Spec(ψ)
...
Then (8
...
11) yield
(R′ ⊗R R′′ ) ⊗R k = R′ ⊗R (R′′ ⊗R k) = (R′ ⊗R k) ⊗k (R′′ ⊗R k)
...
15)
...
8)(2) implies that p ∈ Im Spec(θ) if and only if p ∈ Im Spec(φ) and
p ∈ Im Spec(ψ), as desired
...
11)
...
For each λ, let φλ : R′ → Rλ be a ring map with
φµ = αµλ φλ for all αµλ , so that φ := αλ φλ is independent of λ
...
Solution: Given q ∈ Spec(R′ ), set k := Rp′ /qRq′
...
13) yields
R ⊗R′ k = lim(Rλ ⊗R′ k)
...
9)(1)
...
8)(2)
implies that p ∈ Im Spec(φ) if and only if p ∈ Im Spec(φλ ) for all λ, as desired
...
12)
...
Set
K := Frac(A) and R := (A/p) × K
...
Set φ∗ := Spec(φ)
...
Solution: Note p is maximal; so A/p is a field
...
16)
...
So φ∗ is
bijective
...
□
204
Solutions: (13
...
13)
...
Set
φ∗ := Spec(φ)
...
Solution: For (1), given p ∈ φ∗ (V(b)), say p = φ−1 (P) where P is a prime
of R′ with P ⊃ b
...
So p ⊃ φ−1 b, or p ∈ V(φ−1 b)
...
But V(φ−1 b) is closed
...
√
−1
Conversely, given p ∈ V(φ−1
√b), note p ⊃√ φ b
...
Hence f ∈
/ φ b
...
25)
...
So there’s a prime P ⊃ b with φ(f ) ∈
(3
...
So φ−1 P ∈ φ∗ (V(b))
...
Therefore,
φ−1 P ∈ φ∗ (V(b)) ∩ D(f )
...
So p ∈ φ∗ (V(b))
...
For (2), take b := ⟨0⟩
...
But by (13
...
So φ∗ (Spec(R′ )) = Spec(R)
if and only if
√
V(⟨0⟩) = V(Ker(φ))
...
1),
so plainly if and only if nil(R) ⊃ Ker(φ)
...
□
Exercise (13
...
— Let R be a ring, R′ a flat algebra with structure map φ
...
Solution: Owing to the definition of Spec(φ) in (13
...
15)
...
15)
...
Show that the induced map Spec(Rq′ ) → Spec(Rp ) is surjective
...
Thus φ induces a
local homomorphism Rp → Rq′
...
13), and Rp ⊗R R′ is flat over Rp by (9
...
Also Rq′ is flat over Rp′ by
(12
...
Hence Rq′ is flat over Rp by (9
...
So Rq′ is faithfully flat over Rp by
(10
...
Hence Spec(Rq′ ) → Spec(Rp ) is surjective by (9
...
□
Exercise (13
...
— Let R be a ring
...
17)
...
(2) V(⟨g⟩)
√ ⊃ V(⟨f ⟩)
...
(4) S f ⊂ S g
...
(6) f ∈ S g
...
(8) there is an R-algebra map Rf → Rg
...
Solution: First, (1) and (2) are equivalent by (13
...
Plainly, (3) and (5) are equivalent
...
17)(4)
...
Conversely, (6) implies Sf ⊂ S g ;
whence, (3
...
Finally, (8) implies (4) by (11
...
And (4)
−1
−1
implies (7) by (11
...
But S f R = Sf−1 R and S g R = Sg−1 R by (11
...
□
Solutions: (13
...
17)
...
(1) Show that D(f ) 7→ Rf is a well-defined
contravariant functor from the category of principal open sets and inclusions to ((Ralg))
...
−→
Solution: Consider (1)
...
16), if D(g) ⊂ D(f ), then there is a unique
−1
−1
R-algebra map φfg : S f R → S g R
...
Further, if D(g) = D(f ), then S f ⊂ S g and S g ⊂ S f , so
−1
S f = S g and φfg = 1
...
9)
...
For (2), notice (13
...
Further, D(f ) ∋ p if and only if f ∈
/ p by (13
...
Set S := R − p
...
16), S is saturated multiplicative
...
17)(1)(c)
...
But Rf = S f Rf by (11
...
Thus
−1
limD(f )∋p Rf = limS ⊂S S f R
...
17), S f g ∋ f, g
...
17)(1)(b), S f g is
−1
saturated multiplicative
...
17)(1)(c)
...
10)(2)
...
Thus (2) holds
...
18)
...
Let R be a ring
...
Show that X is irreducible if and only if n is prime
...
Plainly, D(f ) = ∅; see (13
...
So in
(13
...
Suppose n is not prime
...
The
above yields D(f ) ̸= ∅ and D(g) ̸= ∅ but D(f g) = ∅
...
1
...
Thus X is not irreducible
...
By (13
...
Then D(f )∩D(g) = ∅
...
1
...
Hence, the first paragraph implies f, g ∈
/ n but f g ∈ n
...
□
Exercise (13
...
— Let X be a topological space, Y an irreducible subspace
...
(2) Show that Y is contained in a maximal irreducible subspace
...
They are called its irreducible components
...
Show that its irreducible components
are the closed sets V(p) where p is a minimal prime
...
Then U ∩Y and V ∩Y
are open in Y , and nonempty
...
So (U ∩ Y ) ∩ (V ∩ Y ) ̸= ∅
...
Thus (1) holds
...
Then Y ∈ S, and
S is partially
ordered by inclusion
...
Then Y ′ is irreducible: given nonempty open sets U, V of Y ′ , there is
Yλ with U ∩ Yλ ̸= ∅ and V ∩ Yλ ̸= ∅; so (U ∩ Yλ ) ∩ (V ∩ Yλ ) ̸= ∅ as Yλ is irreducible
...
206
Solutions: (13
...
Finally, if X is Hausdorff,
then any two points have disjoint open neighborhoods; hence, every irreducible
subspace consists of a single point
...
Then Y is closed by (1); so
Y = Spec(R/a) for some ideal
√ a by (13
...
3)
...
So nil(R/a) is
prime by (13
...
Hence a is prime
...
14)
...
Then Z = V(p) ⊃ V( a) = V(a) = Y by (13
...
Further, Z is irreducible by (13
...
So Z = Y by maximality
...
Conversely, given a minimal prime q, set Z := Spec(R/q)
...
18)
...
By
the above, Y = V(p) for some prime p
...
1)
...
Thus (4) holds
...
21)
...
Show U is quasi-compact if and only if X − U = V(a) where a is finitely generated
...
∪n Assume U is quasi-compact
...
1),
∩
So U = 1 D(fλi ) for some fλi
...
, fλn ⟩)
...
, fn ⟩)
...
But
D(fi ) = Spec(Rfi ) by (13
...
So by (13
...
Thus U is quasi-compact
...
22)
...
Set X := Spec(R)
...
Show m = 0
...
But
∪
X = D(fλ )
...
So there are
∑
∑
nλ
nλ
λ1 ,
...
, xn with 1 = xi fλi i
...
□
Exercise (13
...
— Let R be a ring; set X := Spec(R)
...
(2) X is Hausdorff
...
(4) Every prime p of R is maximal
...
Prove that X is totally disconnected; namely, no two distinct
points lie in the same connected component
...
1)
...
Assume (1)
...
Then x ∈ ⟨x2 ⟩ by
(10
...
So there is y ∈ R with x = x2 y
...
Set Xi := V(ai )
...
Further, q ∈ X2 as 1 − xy ∈ q since
x(1 − xy) = 0 ∈ q, but x ∈
/ q
...
Further a1 a2 = 0 as x(1 − xy) = 0
...
3)
...
Thus (2) holds, and X is totally disconnected
...
Thus (2) implies (3)
...
2)
...
Then every prime m is both maximal and minimal
...
31)
207
local ring with mRm as its only prime by (11
...
Hence mRm = nil(Rm ) by the
Scheinnullstellensatz (3
...
But nil(Rm ) = nil(R)m by (11
...
And nil(R) = 0
...
So Rm is a field
...
48)(2)
...
□
Exercise (13
...
— Let B be a Boolean ring, and set X := Spec(B)
...
Further, show X is a compact Hausdorff space
...
)
∪
Solution: Let f ∈ B
...
However,
∩
D(f ) D(1 − f ) = ∅; indeed, if p ∈ D(f ), then f ∈
/ p, but f (1 − f ) = 0 as B is
Boolean, so 1 − f ∈ p, so p ∈
/ D(1 − f )
...
Thus D(f ) is
closed as well as open
...
Then U is quasi-compact, as U is
closed and X is quasi-compact by (13
...
So X − U = V(a) where a is finitely
generated by (13
...
Since B is Boolean, a = ⟨f ⟩ for some f ∈ B by (1
...
Thus U = D(f )
...
Then there is f ∈ p − q
...
By the above, D(f ) is both open and closed
...
By (13
...
□
Exercise (13
...
— Show every Boolean ring B is isomorphic
to the ring of continuous functions from a compact Hausdorff space X to F2 with
the discrete topology
...
closed subsets of X; in fact, X := Spec(B), and B −→
Solution: The two statements are equivalent by (1
...
Further, X := Spec(B)
is compact Hausdorff, and its open and closed subsets are precisely the D(f ) by
(13
...
Thus f 7→ D(f ) is a well defined function, and is surjective
...
1
...
To show it preserves
addition, we must show that, for any f, g ∈ B,
∪
D(f + g) = (D(f ) − D(g)) (D(g) − D(f ))
...
25
...
There are four cases
...
Second, if g ∈
/ p but f ∈ p, then again f + g ∈
/ p
...
25
...
Third, if f ∈ p and g ∈ p, then f + g ∈ p
...
The fourth does
...
Now, B/p = F2 by (2
...
So the residues of f and g are both equal to 1
...
So again f + g ∈ p
...
25
...
Thus (13
...
1) holds
...
Then
f ∈ nil(B)
...
24)
...
√
√
Alternatively, if D(f ) = D(g), then V(⟨f ⟩) = V(⟨g⟩), so ⟨f ⟩ =
⟨g⟩ by
□
(13
...
But f, g ∈ Idem(B) as B is Boolean
...
26)
...
31)
...
Prove that
Supp(M/aM ) ⊂ Supp(M ) ∩ V(a),
with equality if M is finitely generated
...
36)
Solution: First, (8
...
But Ann(R/a) = a; hence
(13
...
Thus (13
...
□
Exercise (13
...
— Let φ : R → R′ be a map of rings, M an R-module
...
32
...
Solution: Fix a prime q ⊂ R′
...
Apply, in
order, (12
...
11), and again (12
...
(13
...
2)
First, assume q ∈ Supp(M ⊗R R′ ); that is, (M ⊗R R′ )q ̸= 0
...
32
...
Thus (13
...
1) holds
...
Then p ∈ Supp(M ), or Mp ̸= 0
...
Then Mp /pMp = Mp ⊗Rp k and Rp′ /pRp′ = Rp′ ⊗A k by (8
...
Hence Cancellation (8
...
10), and (13
...
2) yield
(Mp /pMp ) ⊗k (Rq′ /pRq′ ) = (Mp ⊗Rp k) ⊗k (Rq′ ⊗Rp k)
= Mp ⊗Rp (Rq′ ⊗Rp k) = (Mp ⊗Rp Rq′ ) ⊗Rp k
(13
...
3)
′
= (M ⊗R R )q ⊗Rp k
...
Then Mp /pMp ̸= 0 by Nakayama’s Lemma
(10
...
And Rq′ /pRq′ ̸= 0 by Nakayama’s Lemma (10
...
So (Mp /pMp ) ⊗k (Rq′ /pRq′ ) ̸= 0 by (8
...
So (13
...
3) implies
(M ⊗R R′ )q ̸= 0, or q ∈ Supp(M ⊗R R′ )
...
32
...
□
Exercise (13
...
— Let R be a ring, M a module, p ∈ Supp(M )
...
Solution: Let q ∈ V(p)
...
So Mp = (Mq )p by (11
...
Now,
p ∈ Supp(M )
...
Hence Mq ̸= 0
...
□
Exercise (13
...
— Let Z be the integers, Q the rational numbers, and set
M := Q/Z
...
Solution: Let p ∈ Spec(R)
...
20)
...
4) and (12
...
If p ̸= ⟨0⟩, then
Zp ̸= Qp since pZp Z = p by (11
...
If p = ⟨0⟩, then Zp = Qp
...
Finally, suppose Supp(M ) = V(a)
...
But ⟨0⟩ is prime
...
Thus Supp(M ) is not closed
...
36)
...
We call T (M ) the torsion submodule of M , and we say M is
torsionfree if T (M ) = 0
...
Solutions: (13
...
So (12
...
(13
...
1)
Assume M is torsionfree
...
36
...
Conversely, if Mm is torsionfree for all m, then T (M )m = 0 for all m by (13
...
1)
...
35)
...
□
Exercise (13
...
— Let R be a ring, P a module, M, N submodules
...
Show M = N
...
Solution: If M ⊂ N , then (12
...
35)
...
17)(4), (5)
...
38)
...
Suppose
Mm = 0 for all maximal ideals m containing a
...
Solution: Given any maximal ideal m, note that (aM )m = am Mm by (12
...
But Mm = 0 if m ⊃ a by hypothesis
...
14)(2)
...
Thus (13
...
Alternatively, form the ring R/a and its module M/aM
...
By hypothesis, Mm = 0
...
22)
...
So M/aM = 0 by (13
...
Thus M = aM
...
39)
...
Assume p/1 ∈ Mm for every maximal ideal m
...
Solution: Set N := M + Rp
...
But
p/1 ∈ Mm
...
So N = M by (13
...
Thus p ∈ M
...
40)
...
Show a = m aRm where
m runs through the maximal ideals and the intersection takes place in Frac(R)
...
Conversely, take x ∈ aRm
...
But aRm = am by (12
...
So (13
...
□
Exercise (13
...
— Prove these three conditions on a ring R are equivalent:
(1) R is reduced
...
(3) Rm is reduced for all maximal ideals m
...
Then nil(R) = 0
...
18)
...
Trivially (2) implies (3)
...
Then nil(Rm ) = 0
...
18) and (12
...
So nil(R) = 0 by (13
...
Thus (1) holds
...
Finally, the answer is no
...
The
primes of R are p := ⟨(1, 0)⟩ and q := ⟨(0, 1)⟩ by (2
...
Further, Rq = k1 by
(11
...
Similarly Rp = k2
...
In fact, take R := R1 × R2 for any domains Ri
...
42)(2) below
...
42)
...
Prove this:
210
Solutions: (13
...
∏n
(2) Rp is a domain for any prime p and Σ is finite if and only if R = i=1 Ri
where Ri is a domain
...
, pn } = Σ
...
Suppose p, q ∈ Σ are not comaximal
...
Hence Rm contains two minimal primes, pRm and qRm ,
by (11
...
However, Rm is a domain by hypothesis, and so ⟨0⟩ is its only minimal
prime
...
So p = q
...
Consider (2)
...
Then R∏is reduced by (13
...
Assume, also, Σ is finite
...
35), and surjective by (1) and the Chinese Remainder Theorem (1
...
Thus
R is a finite product of domains
...
Let p be a prime of R
...
11)
...
3)
...
So Rp (Ri )p for some i by (3
...
∏
Thus Rp is a domain
...
11), each pi ∈ Σ has the form pi = aj where, after renumbering, ai ⟨0⟩ and
∼ R
...
aj = Rj for j ̸= i
...
44)
...
Prove elements mλ ∈ M
generate M if and only if, at every maximal ideal m, their images mλ generate Mm
...
By (13
...
But R⊕{λ} m = Rm
by (12
...
Hence (4
...
□
Exercise (13
...
— Let R be a ring, R′ a flat algebra, p′ a prime in R′ , and p
its contraction in R
...
Solution: First, Rp′ is flat over Rp by (13
...
Next, Rp′ ′ is flat over Rp′ by
(12
...
29) as R − p ⊂ R′ − p′
...
12)
...
19)
...
48)
...
(1) Assume R is absolutely flat
...
(2) Show R is absolutely flat if and only if Rm is a field for each maximal m
...
9)
...
So there is y ∈ R with x = x2 y
...
So
s(st − 1) = 0
...
But s/1 is a unit
...
So a/s = (a/s)2 · b/t
...
Thus ⟨a/s⟩ is idempotent
...
9)
...
Alternatively, given an S −1 R-module M , note M is also an R-module, so R-flat
by (1)
...
11)
...
13), and S −1 M = M by (12
...
Thus M is S −1 R-flat
...
For (2), first assume R is absolutely flat
...
So
by (10
...
Conversely, assume each Rm is a field
...
So M is R-flat by (13
...
Thus (2) holds
...
52)
...
Solutions: (14
...
51)
...
5)
...
n
Assume Pm ≃ Rm
...
, pn /fnkn of Pm over Rm
...
Assume P is finitely generated
...
24)(1) provides f ∈ R − m such that
αf : Rfn → Pf is surjective
...
5) and (12
...
Assume Pq ≃ Rqn if also q is maximal
...
4)
...
Hence αf : Rfn → Pf is bijective owing to
(13
...
□
Exercise (13
...
— Let A be a semilocal ring, P a locally free module of rank
n
...
Solution: As P is locally free, P is finitely presented by (13
...
52)
...
So P ≃ An by (13
...
□
Exercise (13
...
— Let R be a ring, M a finitely presented module, n ≥ 0
...
Solution: Assume M is locally free of rank n
...
52)
...
39)(2)
...
15)
...
37)
...
□
14
...
4)
...
Suppose R′ has just one prime p′ over p
...
Solution: Since R′ is integral over R, the localization Rp′ is integral over Rp by
(11
...
Moreover, Rp is a local ring with unique maximal ideal pRp by (11
...
Hence, every maximal ideal of Rp′ lies over pRp by (14
...
But every maximal
ideal of Rp′ is the extension of some prime q′ ⊂ R′ by (11
...
So, by hypothesis, q′ = p′
...
So Rp′ − p′ Rp′ consists of units
...
29) and (11
...
But Rp′ is integral over Rp ; so (c) holds too
...
5)
...
Suppose R′ has at least two distinct primes p′ and q′ lying over p
...
Show that, in fact, if y lies in q′ , but not in
p′ , then 1/y ∈ Rp′ ′ is not integral over Rp
...
Say
(1/y)n + a1 (1/y)n−1 + · · · + an = 0
with n ≥ 1 and ai ∈ Rp
...
However, y ∈ q′ , so y ∈ q′ Rp′
...
So q′ ∩ (R − p) ̸= ∅ by (11
...
But q′ ∩ R = p, a contradiction
...
□
212
Solutions: (14
...
6)
...
Set R′ := k[X],
and Y := X 2 , and R := k[Y ]
...
Is Rp′ ′
integral over Rp ? Explain
...
28) as R′ is generated by 1 and X as an R-module
...
Set q′ := (X + 1)R′
...
Further X + 1 lies in q′ , but
not in p′
...
5)
...
Then (X − 1)2 = Y − 1
...
Then (X − 1)2 ∈ q′
...
But p′ is maximal
...
Thus R′
has just one prime p′ over p
...
4)
...
12)
...
Prove
that z
...
∪
Solution: If p ∈ Σ, then p ⊂ z
...
10)
...
div(R) ⊃ p∈Σ p
...
If x ∈
/ p for some p ∈ Σ, then y ∈ p
...
But p∈Σ p = ⟨0⟩ by the Scheinnullstellensatz (3
...
14)
...
Thus, if x ∈
/ p∈Σ p, then x ∈
/ z
...
Thus z
...
Thus
∪
z
...
Fix p ∈ Σ
...
41)
...
20)(2)
...
But by (12
...
Thus Rp = Frac(R/p)
...
13)
...
Assume Σ is finite
...
∪
(2) z
...
∏
(3) K/pK = Frac(R/p) for each p ∈ Σ, and K = p∈Σ K/pK
...
Then (14
...
Assume (2) holds
...
div(R)
...
∪
Then q ⊂ z
...
So (2) yields q ⊂ p∈Σ p
...
So q ⊂ p for some
p ∈ Σ by Prime Avoidance (3
...
Hence q = p since p is minimal
...
Therefore, by (11
...
Fix p ∈ Σ
...
22)
...
But
clearly S −1 (R/p) ⊂ Frac(R/p)
...
3)
...
Hence (11
...
Therefore, φS (K −pK) = R−p
...
27)
...
Thus K has only
finitely many primes, the pK; each
∏ pK is minimal, and each KpK is a domain
...
42)(2) yields K = p∈Σ K/pK
...
Assume (3) holds
...
But clearly, a product of reduced ring is reduced
...
Thus (1) holds
...
14)
...
For each p ∈ Σ, set K(p) := Frac(A/p)
...
Show that P is free of rank r if and only if dimk (P ⊗A k) = r
and dimK(p) (P ⊗A K(p)) = r for each p ∈ Σ
...
17)
213
Solution: If P is free of rank r, then dim(P ⊗ k) = r and dim(P ⊗ K(p)) = r
owing to (8
...
Conversely, suppose dim(P ⊗ k) = r
...
16) implies
P is generated by r elements
...
20) yields an exact sequence
α
0→M −
→ Ar → P → 0
...
Since A is reduced, K(p) = Rp by (14
...
So K(p)
is flat by (12
...
So the induced sequence is exact:
0 → M ⊗ K(p) → K(p)r → P ⊗ K(p) → 0
...
It then follows that M ⊗A K(p) = 0
...
And φAr is injective as φA : A → K is
...
∏
By hypothesis, A is reduced and Σ is finite; so K = p∈Σ K(p) by (14
...
So
∏
M ⊗ K = (M ⊗ K(p))
...
So M ⊗ K = 0
...
φM : M → M ⊗ K is injective
...
Thus Ar −→
□
Exercise (14
...
— Let A be a reduced semilocal ring with a finite set of minimal
primes
...
For each prime q ⊂ B, set L(q) = Frac(B/q)
...
Show that P is a free A-module of rank r
...
Since Spec(B) → Spec(A) is surjective, there
is a prime q ⊂ B whose contraction is p
...
(14
...
1)
If p is minimal, take a minimal prime q′ ⊂ q
...
Replace q by q′
...
Then the contraction of q′ contains p, so is equal to p
...
Either way, dim((P ⊗A )B) ⊗B L(q)) = r by hypothesis
...
15
...
Hence dim(P ⊗A K(p)) = r
...
14)
...
Then Spec(Bm ) → Spec(Am ) is surjective by an argument like
one in the proof of (14
...
20)(2)
...
Thus P is free of rank r by (13
...
□
Exercise (14
...
— Let R be a ring, p1
...
Prove that these three conditions are equivalent:
(1) R is normal
...
(3) R is a finite product of normal domains Ri
...
Prove the Ri are equal to the R/pj in some order
...
3)
Solution: Assume (1)
...
Then Rm is a normal domain
...
41)
...
Set S := R − m,
so that Rm := S −1 R
...
29)(2)
...
Let t ∈ S0
...
Thus (11
...
3) yield S0−1 Rm ⊂ Frac(Rm )
...
Then x/1 ∈ S −1 K is integral over S −1 R by
(11
...
But S −1 R = Rm , and Rm is a normal domain
...
Hence
x ∈ R by (13
...
Thus (2) holds
...
Set Ri := R/pi and Ki := Frac(R∏
K = Ki by (14
...
i )
...
Then R ⊂
Ri ⊂
Ri′
...
29), and the second, by (10
...
27)
...
Hence R = Ri = Ri′
...
∏
Assume (3)
...
Then Rp = (Ri )p by (12
...
32)
...
So Rp = (Ri )p for some i by (3
...
Hence Rp is a normal domain
...
Finally, the last assertion results from (13
...
□
15
...
2)
...
Set f := X q Y − XY q and R := k[X, Y ] ⟨f ⟩
...
For every a ∈ k, show that R is not module finite over
P := k[y −ax]
...
1), no k-linear combination works
...
Solution: Take a = 0
...
Any algebraic relation over P satisfied
by x is given by a polynomial in k[X, Y ], which is a multiple of f
...
So x is not integral over P
...
23), R is not
module finite over P
...
Since aq = a, after the change of variable Y ′ := Y − aX,
our f still has the same form
...
□
Exercise (15
...
— Let k be a field, and X, Y, Z variables
...
Fix a, b ∈ k, and set t := x + ay + bz
and P := k[t]
...
Solution: To see x is integral, notice xz = 1, so x2 −tx+b = −axy
...
Now,
y 3 − x2 + 1 = 0, so y is integral over P [x]
...
27),
implies that y too is integral over P
...
28)
...
Now, P ⊂ k[x, y]
...
But y 3 − x2 + 1 = 0
...
Hence z is too
...
19)
215
(10
...
Moreover, z = 1/x ∈ k(x)
...
Thus
z is not integral over P if b = 0
...
8)
...
(So
K contains a copy of every finite extension field
...
, Xn ] be the
polynomial ring, and f, f1 ,
...
Assume f vanishes at every zero in K n of
f1 ,
...
, an ) ∈ K n and f1 (a) = 0,
...
Prove that there are polynomials g1 ,
...
√
Solution: Set
√ a := ⟨f1 ,
...
We have to show f ∈ a
...
So given an m, we have to show that f ∈ m
...
By the weak Nullstellensatz, L is a finite extension field of k
...
Let ai ∈ K be the image of the
variable Xi ∈ P , and set (a) := (a1 ,
...
Then f1 (a) = 0,
...
Hence f (a) = 0 by hypothesis
...
□
Exercise (15
...
— Let R be a domain of (finite) dimension r, and p a nonzero
prime
...
Solution: Every chain of primes of R/p is of the form p0 /p ⫋ · · · ⫋ ps /p where
0 ⫋ p0 ⫋ · · · ⫋ ps is a chain of primes of R
...
Thus dim(R/p) < r
...
12)
...
Prove that
dim(R) = dim(R′ )
...
Set p′−1 := 0
...
3)(4), yields a prime p′i of R′ with p′i−1 ⊂ p′i and
p′ i ∩ R = pi
...
Thus dim(R) ≤ dim(R′ )
...
Set pi := p′ i ∩ R
...
3)(2)
...
□
Exercise (15
...
— Let k be a field, R a finitely generated k-algebra, f ∈ R
nonzero
...
Prove that dim(R) = dim(Rf )
...
13), as Rf is,
by (11
...
So since R is a finitely generated k-algebra,
Rf is one too
...
Hence both
dim(R) and dim(Rf ) are equal to tr
...
13)
...
18)
...
Set p := ⟨f ⟩ and R := Pp
...
Solution: In P , the chain of primes 0 ⊂ p is of maximal length by (2
...
25) or (15
...
So ⟨0⟩ and pR are the only primes in R by (11
...
Thus
dim(R) = 1
...
Then/ Rf = K since, if a (bf n ) ∈ K with a, b ∈ P and f ∤ b,
then a/b ∈ R and so (a/b) f n ∈ Rf
...
□
Exercise (15
...
— Let R be a ring, R[X] the polynomial ring
...
(In particular, dim(R[X]) = ∞ if and only if dim(R) = ∞
...
28)
Solution: Let p0 ⫋ · · · ⫋ pn be a chain of primes in R
...
18)
...
Let p be a prime of R, and q0 ⫋ · · · ⫋ qr be a chain of primes of R[X] with
qi ∩R = p for each i
...
9) yields a chain of primes of length r in R[X]/pR[X]
...
30) and (11
...
But k(p)[X] is
a PID
...
Take any chain P0 ⫋ · · · ⫋ Pm of primes in R[X]
...
At most two Pj contract to a given pi by the above discussion
...
Thus dim(R[X]) ≤ 1 + 2 dim(R)
...
23)
...
We say a subset Y is locally
closed if Y is the intersection of an open set and a closed set; equivalently, Y is
open in its closure Y ; equivalently, Y is closed in an open set containing it
...
We say X is Jacobson if its set of closed points is very dense
...
(2) Every closed set F of X satisfies F ∩ X0 = F
...
Solution: Assume (1)
...
Then F ∩ U is locally closed, so meets X0
...
Thus F ⊂ F ∩ X0
...
Thus (2) holds
...
In (3), the map is trivially surjective
...
Then (X −U )∩X0 = (X −V )∩X0
...
So U = V
...
Assume (3)
...
In particular, if a locally closed
set Y is nonempty, then so is Y ∩ X0
...
□
Exercise (15
...
— Let R be a ring, X := Spec(R), and X0 the set of closed
points of X
...
(2) X is a Jacobson space
...
Solution: Assume (1)
...
Trivially, F ⊃ F ∩ X0
...
Then F ∩ X0 is the set
of maximal √
ideals m containing
a by (13
...
So (1)
√
implies b ⊂ a
...
Thus F ⊂ F ∩ X0
...
23) yields (2)
...
Let y ∈ X
a point
( be ∩
) such that {y} is locally closed
...
So {y} X0 ∋ y
...
Assume (3)
...
Then {p} is closed in D(f ) by (13
...
So {p} is locally closed in X
...
Thus p is maximal
...
22) yields (1)
...
31)
217
Exercise (15
...
— Let P := Z[X1 ,
...
Assume
f ∈ P vanishes at every zero in K n of f1 ,
...
, an ) ∈ K n and f1 (a) = 0,
...
Prove there are g1 ,
...
√
Solution: Set a := ⟨f1 ,
...
Suppose f ∈
/ a
...
26)(2) and (15
...
Set K := P/m
...
26)(1)
...
Let ai
be the residue of Xi , set (a) := (a1 ,
...
Then f1 (a) = 0,√
...
□
So f (a) = 0 by hypothesis
...
Thus f ∈ a
...
29)
...
Prove that if R′ is integral
over R and R is Jacobson, then R′ is Jacobson
...
Then
R′′ is Jacobson by (15
...
So R′′ has a maximal ideal m′′ that avoids f and
contains a′ ∩R′′
...
So R′ contains a prime m′ that contains
a′ and that contracts to m′′ by Going Up (14
...
Then m′ avoids f as m′′ does,
and m′ is maximal by Maximality, (14
...
Thus R′ is Jacobson
...
30)
...
True or false: prove or give a counterexample to each of the following statements
...
(2) The localized ring S −1 R is Jacobson
...
−→
(4) In a filtered direct limit of rings Rλ , necessarily lim rad(Rλ ) = rad(lim Rλ )
...
13); so Rf is Jacobson by (15
...
(2) False: by (15
...
(3) False: Z⟨p⟩ isn’t Jacobson by (15
...
6)
...
□
−→
Exercise (15
...
— Let R be a reduced Jacobson ring with a finite set Σ of
minimal primes, and P a finitely generated module
...
Solution: Suppose P is locally free of rank r
...
49)
...
5)
...
11)
...
22)
...
23)
...
Consider the converse
...
Then P ⊗R K is a
K-vector space, say of dimension n
...
12)
...
24), there is an h ∈ R − p with Ph free of rank n
...
20)
...
But, by hypothesis, dimR/m (P/mP ) = r
...
Given a maximal ideal m, set A := Rm
...
41)
...
20)(2)
...
Hence
(14
...
Finally, (13
...
□
218
Solutions: (16
...
Chain Conditions
Exercise (16
...
— Let M be a finitely generated module over an arbitrary ring
...
Solution:
Say M is generated by x1 ,
...
Say
∑
xi = j zj yλij
...
□
Exercise (16
...
— Let R be a ring, X a variable, R[X] the polynomial ring
...
Solution: It’s true
...
7)
...
8); so R is Noetherian
...
9)
...
Assume R′ is Noetherian, and prove R is too
...
As R′ is Noetherian, aR′ is finitely generated
...
So by (16
...
, an that generate
aR′
...
Applying ρ yields a = a1 x1 + · · · + an xn with xi := ρ(x′i ) ∈ R
...
Thus R is Noetherian
...
Then
a1 R′ ⊂ a2 R′ ⊂ · · · stabilizes as R′ is Noetherian
...
But ρ(ai R′ ) = ai ρ(R′ ) = ai
...
5), R is Noetherian
...
15)
...
Prove or give a counterexample to
this statement: if β(M1 ) = β(M2 ) and α−1 (M1 ) = α−1 (M2 ), then M1 = M2
...
(Geometrically, we can view M1 as the line determined by the origin and the point
(1, 2), and M2 as the line determined by the origin and the point (2, 1)
...
) □
Exercise (16
...
— Let R be⊕
a ring, a1 ,
...
Prove
(1)
that
R/ai is a Noetherian R-module, and (2) that,
∩
if ai = 0, then R too is a Noetherian ring
...
Since R/ai is a Noetherian ring, such an ideal is finitely generated as an (R/a⊕
i )-module, so as an Rmodule as well
...
So
R/ai is a Noetherian
R-module by (16
...
Thus (1) holds
...
So R can be identified with a submodule of the Noetherian
⊕
R-module
R/ai
...
16)(2)
...
13)
...
30)
219
Exercise (16
...
— Let R be a Noetherian ring, M and N finitely generated
modules
...
Solution: Say M is generated m elements
...
10) yields a surjection
R⊕m →
→ M
...
18)
...
15
...
3)
...
Hence Hom(R⊕m , N ) is finitely generated, so Noetherian
by (16
...
Thus Hom(M, N ) is finitely generated
...
24)
...
Assume R is Noetherian
...
3] Assume R′ is a field containing R
...
(2) [1, bot
...
77] Let K ′ ⊃ R be a field that embeds in R′
...
Show K/R is algebra finite and K ′ /K is finite
...
Now, R ⊂ K ⊂ R′
...
Thus R′ /K is (module) finite by (15
...
22), and so
K/R is algebra finite by (16
...
Conversely, say x1 ,
...
, yn
are module generators for R′ /K
...
, xm , y1 ,
...
Thus (1) holds
...
Then L is a field,
R ⊂ K ⊂ K ′ ⊂ L, and L/R is algebra finite
...
Thus (2) holds
...
28)
...
Assume that R is finite
dimensional as a k-vector space
...
Solution: View R as a vector space, and ideals as subspaces
...
Thus R is Noetherian by (16
...
□
/
Exercise (16
...
— Let p be a prime number, and set M := Z[1/p] Z
...
Deduce that M is an
Artinian Z-module, and that it is not Noetherian
...
Then
there is an m ∈ Z with nm ≡ 1 (mod pe )
...
Therefore, either N = M , or there is a
largest integer e ≥ 0 with 1/pe ∈ N
...
Let M ⊋ N1 ⊃ N2 ⊃ · · · be a descending chain
...
Then the sequence n1 ≥ n2 ≥ · · · stabilizes; say
ni = ni+1 = · · ·
...
Thus M is Artinian
...
, mr generate M , say mi = ni /pei
...
Then 1/pe generates M , a contradiction since 1/pe+1 ∈ M
...
□
Exercise (16
...
— Let R be an Artinian ring
...
Deduce that in general every prime ideal p of R is maximal
...
22)
Solution: Take any nonzero element x ∈ R, and consider the chain of ideals
⟨x⟩ ⊃ ⟨x2 ⟩ ⊃ · · ·
...
Hence xe axe+1 for some a ∈ R
...
In general, R/p is Artinian by (16
...
Now, R/p is also a domain by (2
...
Hence, by what we just proved, R/p is a field
...
17)
...
Associated Primes
Exercise (17
...
— Given modules M1 ,
...
Prove
Ass(M ) = Ass(M1 ) ∪ · · · ∪ Ass(Mr )
...
Then N, M1 ⊂ M
...
So
(17
...
So Ass(M ) = Ass(N ) ∪ Ass(M1 )
...
□
Exercise (17
...
— Take R := Z and M := Z/⟨2⟩ ⊕ Z
...
Solution: First, we have Ass(M ) = {⟨0⟩, ⟨2⟩} by (17
...
4)(2)
...
Then the canonical maps R → L and
R → N are isomorphisms
...
4)(2)
...
□
Exercise (17
...
— If a prime p is sandwiched between two primes in Ass(M ),
is p necessarily in Ass(M ) too?
Solution: No, for example, let R := k[X, Y ] be the polynomial ring over a field
...
Then Ass(M ) = Ass(R) ∪ Ass(R/⟨X, Y ⟩)
by (17
...
Further, Ass(R) = ⟨0⟩ and Ass(R/⟨X, Y ⟩) = ⟨X, Y ⟩ by (17
...
□
Exercise (17
...
— Let R be a ring, and suppose Rp is a domain for every
prime p
...
Solution: Let p ∈ Ass(R)
...
10)
...
So pRp = ⟨0⟩ by (17
...
Hence p is a minimal prime of R by
(11
...
Alternatively, say p = Ann(x) with x ∈ R
...
However,
for any y ∈ p, we have xy/1 = 0 in Rp
...
So there exists some t ∈ R − p such that ty = 0
...
14)
...
Then there is some
y ∈ p − q
...
Thus p = q; that is, p is minimal
...
16)
...
Show that, if x ∈
/ p for any p ∈ Ass(M/N ), then xM ∩ N = xN
...
Conversely, take m ∈ M with xm ∈ N
...
Then xm′ =
/ z
...
∩ 0
...
15), x ∈
So m′ = 0
...
So xm ∈ xN
...
□
Solutions: (18
...
22)
...
Prove the primes
minimal containing a are associated to a
...
Solution: Since a = Ann(R/a), the primes in question are the primes minimal
in Supp(R/a) by (13
...
So they are associated to a by (17
...
21)
...
23)
...
20)
...
Solution: If the chain is acceptable, then M1 ̸= 0 as M1 /0 ≃ R/p1 , and M1 is
a prime ideal as M1 = Ann(M/M1 ) = p2
...
Finally, Ass(M ) = 0 by (17
...
Further, as just observed, given any acceptable
chain, p2 = M1 ̸= 0
...
□
Exercise (17
...
— Take R := Z and M := Z/⟨12⟩ in (17
...
Find all three
acceptable chains, and show that, in each case, {pi } = Ass(M )
...
Here ⟨a1 ⟩/⟨12⟩ ≃ Z/⟨p1 ⟩ with p1 prime
...
Hence the possibilities are
p1 = 2, a1 = 6 and p1 = 3, a1 = 4
...
So
a2 p2 = a1
...
In each case, a2 is
prime; hence, n = 3, and these three chains are the only possibilities
...
In each case, {pi } = {⟨2⟩, ⟨3⟩}
...
20
...
For
any M , if 0 ⊂ M1 ⊂ · · · ⊂ M is an acceptable chain, then (17
...
4)(2)
yield Ass(M ) ⊃ Ass(M1 ) = {p1 }
...
Thus Ass(M ) = {⟨2⟩, ⟨3⟩}
...
26)
...
Show that the following conditions are equivalent:
(1) V(a) ∩ Ass(M ) = ∅;
(2) Hom(N, M ) = 0 for all finitely generated modules N with Supp(N ) ⊂ V(a);
(3) Hom(N, M ) = 0 for some finitely generated module N with Supp(N ) = V(a);
(4) a ̸⊂ z
...
Solution: Assume (1)
...
Hence Ass(Hom(N, M )) = ∅ by (17
...
So Hom(N, M ) = 0
by (17
...
Thus (2) holds
...
Assume (3)
...
13)
...
25)
...
∪ Clearly (1) and (5) are equivalent
...
div(M ) = p∈Ass(M ) p by (17
...
So (4) implies (5)
...
21); so (3
...
□
18
...
22)
Exercise (18
...
— Let R be a ring, and p = ⟨p⟩ a principal prime generated by
a nonzerodivisor p
...
Solution: Let’s proceed by induction
...
Consider the map R → pn /pn+1 given by x 7→ xpn
...
But Ass(R/p) = {p}
is p as p is a nonzerodivisor
...
4)(2)
...
5) yields Ass(R/pn ) = {p} for every n ≥ 1, as desired
...
5)
...
Then
pn ⊂ q
...
Say x = ypm for some y and m ≥ 0
...
Take m maximal
...
So (18
...
Hence y = zp for some z
...
Thus q = pn
...
7)
...
Let a be
the ideal ⟨X 2 , XY ⟩
...
Show a satisfies this
condition: ab ∈ a implies a2 ∈ a or b2 ∈ a
...
10)
...
√ So a = ⟨X⟩
by (3
...
On the other hand, XY ∈ a, but X ∈
/ a and Y ∈
/ a; thus a is not
primary by (18
...
If ab ∈ a, then X | a or X | b, so a2 ∈ a or b2 ∈ a
...
8)
...
Show that φ−1 q ⊂ R is φ−1 p-primary
...
Solution: Let xy ∈ φ−1 q, but x ∈
/ φ−1 q
...
n
So φ(y) ∈ q for some n ≥ 1 by (18
...
Hence, y n ∈ φ−1 q
...
5)
...
25)
...
□
Exercise (18
...
— Let k be a field, R := k[X, Y, Z] be the polynomial ring
...
Show that
a = q1 ∩q2 holds and that this expression is an irredundant primary decomposition
...
Hence a ⊂ q1 ∩q2
...
Then F ∈ q2 , so F = GY 2 + H(X − Y Z) with G, H ∈ R
...
Then
F = (AY + B)XY + (H − BY )(X − ZY ) ∈ a
...
Thus a = q1 ∩ q2 holds
...
10)
...
8), let’s show q2 is ⟨X, Y ⟩primary
...
Clearly, q2 = φ−1 ⟨Y 2 ⟩
and ⟨X, Y ⟩ = φ−1 ⟨Y ⟩; also, ⟨Y 2 ⟩ is ⟨Y ⟩-primary by (18
...
Thus a = q1 ∩ q2 is a
primary decomposition
...
□
Exercise (18
...
— Let R := R′ × R′′ be a product of two domains
...
Solution: Set p′ := ⟨0⟩ × R′′ and p′′ := R′′ × ⟨0⟩
...
11), so primary by (17
...
Clearly ⟨0⟩ = p′ ∩ p′′
...
□
Solutions: (18
...
22)
...
Consider the following submodule of M :
∪
Γa (M ) := n≥1 {m ∈ M | an m = 0 for some n ≥ 1}
...
(2) Show Γa (M ) is the set of all m ∈ M such that m/1 ∈ Mp vanishes for every
prime p with a ̸⊂ p
...
)
Solution: For (1), given m ∈ Γa (M ), say an m = 0
...
Then an m = 0 ∈ Qi
...
4)
...
∩
Conversely, given m ∈ a̸⊂pi Qi , take any j with a ⊂ pj
...
3)
...
∩Set n := max{nj }
...
Hence an m ∈ Qi = 0
...
For (2), given m ∈ Γa (M ), say an m = 0
...
Then an m = 0 and an ∈
/ p
...
Conversely, given an m ∈ M such that ∩
m/1 ∈ Mp vanishes for every prime p
with a ̸⊂ p, consider a decomposition 0 = Qi with Qi pi -primary; one exists by
(18
...
By (1), it suffices to show m ∈ Qi if a ̸⊂ pi
...
So
there’s an a ∈ R − pi with am = 0 ∈ Qi
...
4) yields m ∈ Qi , as desired
...
26)
...
Prove N = p∈Ass(M/N ) φ−1
p (Np )
...
21) yields an irredundant primary decomposition N = 1 Qi
...
20)
...
Then {p∩i }r1 = Ass(M/N∩) (by
∩r (18
...
Thus 1 φ−1
pi (Npi ) =
1 Qi = N
...
27)
...
Its nth symbolic
power p(n) is defined as the saturation (pn )S where S := R − p
...
(2) Show p(m+n) is the p-primary component of p(n) p(m)
...
(4) Given a p-primary ideal q, show q ⊃ p(n) for all large n
...
But V(pn ) Supp(R/pn ) by (13
...
Hence p is minimal in Ass(R/pn ) by (17
...
3)
...
25) yields (1)
...
17)(3) yields (p(m) p(n) )S = p(m+n)
...
25) yields (2)
...
Conversely, if pn is p-primary, then
n
p = p(n) because primary ideals are saturated by (18
...
Thus (3) holds
...
5)
...
33)
...
But qS = q by (18
...
Thus (4) holds
...
28)
...
Set pi := qi for i = 1,
...
(r)
(1) Suppose pi is minimal for some i
...
(r)
(2) Suppose pi is not minimal for some i
...
224
Solutions: (19
...
Then A is Noetherian√by (16
...
Suppose pi is minimal
...
So m = ⟨0⟩ by the
Scheinnullstellensatz (3
...
So mr = 0 for all large r by (3
...
So p(r) = qi by
Lemma (18
...
25)
...
Suppose pi is not minimal
...
Then mr = 0
by Nakayama’s Lemma (10
...
Hence m is minimal
...
Thus by (11
...
∩
(r)
(r)
However, qi ⊃ pi for all large r by (18
...
Hence ⟨0⟩ = pi ∩ j̸=i qj
...
27)(1)
...
These decompositions are irredundant owing to two applications of (18
...
A
first yields {pi } = Ass(R) as ⟨0⟩q1 ∩ · · · ∩ qn is irredundant
...
□
Exercise (18
...
— Let R be a Noetherian ring, m ⊂ rad(R) an ideal, M a
finitely generated module, and M ′ a submodule
...
∩
Solution: Set N := n≥0 mn (M/M ′ )
...
29), there is x ∈ m such
that (1 + x)N = 0
...
2), 1 + x∩is a unit since/ m ⊂ rad(R)
...
But
mn (M/M ′ )(mn M + M ′ )/M ′
...
□
19
...
2)
...
Prove these statements:
(1) If M is simple, then any nonzero element m ∈ M generates M
...
(3) If M has finite length, then M is finitely generated
...
So it is equal to M , because
M is simple
...
Assume M is simple
...
So M ≃ R/m for m := Ann(M )
by (4
...
Since M is simple, m is maximal owing to the bijective correspondence
of (1
...
By the same token, if, conversely, M ≃ R/m with m maximal, then M is
simple
...
Assume ℓ(M ) < ∞
...
If m = 0, then M = 0
...
Then M1 has a composition series
of length m − 1
...
Further, M/M1 is simple, so finitely generated by (1)
...
16)(1)
...
□
Exercise (19
...
— Let R be a Noetherian ring, M a finitely generated module
...
Prove that, if the conditions hold, then Ass(M ) and Supp(M ) are equal and finite
...
13)
225
Solution: Assume (1)
...
3) yields (2)
...
Then (17
...
2)(2) yield (1)
...
3) yields (3)
...
Then (17
...
17) imply that Ass(M ) and Supp(M )
are equal and consist entirely of maximal ideals
...
However,
Ass(M ) is finite by (17
...
Thus the last assertion holds
...
5)
...
Consider
chains of primary ideals from q to p
...
Solution: There is a natural bijective correspondence between the p-primary
ideals containing q and the (p/q)-primary ideals of R/q, owing to (18
...
In turn,
there is one between the latter ideals and the ideals of A primary for its maximal
ideal m, owing to (18
...
23) with M := A
...
5)
...
Hence every ideal of A is m-primary
by (18
...
Further, m is the only prime of A; so ℓ(A) is finite by (19
...
Hence (19
...
□
Exercise (19
...
— Let k be a field, R an algebra-finite extension
...
Solution: As k is Noetherian by (16
...
12)
...
Then ℓ(R) < ∞ by (19
...
So R
has a composition series
...
2)(2)
...
Hence
these fields are finite extension fields of k by the Zariski Nullstellensatz
...
The converse holds by (16
...
□
Exercise (19
...
— Let k be a field, A a local k-algebra
...
Given an A-module M , prove ℓ(M ) = dimk (M )
...
If M ≃ k, then
ℓ(M ) = 1 and dimk (M ) = 1
...
Then M has a submodule
M ′ with M/M ′ ≃ k
...
9), yields ℓ(M ′ ) = ℓ(M ) − 1
and dimk (M ′ ) = dimk (M ) − 1
...
Thus ℓ(M ) = dimk (M )
...
Hence dimk (M ) = ∞
...
12)
...
Solution: Condition (1) holds, by (19
...
But a prime p is a maximal ideal if and
only if {p} is closed in Spec(R) by (13
...
It follows that (1) and (2) are equivalent
...
Conversely, (3) implies (2), since Spec(R) is quasi□
compact by (13
...
Thus all three conditions are equivalent
...
13)
...
Show that rad(R) is nilpotent
...
6)
Solution: Set m := rad(R)
...
So
m = mr+1 for some r
...
11)
...
Thus Nakayama’s Lemma (10
...
Alternatively, R is Noetherian and dim R = 0 by (19
...
So rad(R) is finitely
generated and rad(R) = nil(R)
...
32) implies mr = 0 for some r
...
16)
...
Show that R′ has only finitely many primes p′ over p, as follows: reduce
to the case that R is a field by localizing at p and passing to the residue rings
...
Hence, by (11
...
Note also that Rp′ is module-finite over
Rp
...
Similarly, we may replace R and R′ by R/p and
R′ /pR′ , and thus assume that R is a field
...
First, R′ is now Artinian by (19
...
28); hence, R′ has only finitely many primes by (19
...
Alternatively, every
prime is now minimal by incomparability (14
...
Further, R′ is Noetherian by
(16
...
22)
...
18)
...
Prove the following four conditions are equivalent:
(1)
(2)
(3)
(4)
that
that
that
that
M has finite length;
∏
M is annihilated by some finite product of maximal ideals mi ;
every prime p containing Ann(M ) is maximal;
R/Ann(M ) is Artinian
...
Let M = M0 ⊃ · · · ⊃ Mm = 0 be a composition series;
set mi := Ann(Mi−1 /Mi )
...
2)(2)
...
Hence mi · · · m1 M0 ⊂ Mi
...
∏
Assume (2)
...
Then p ⊃ mi
...
2)
...
Thus (3) holds
...
Then dim(R/Ann(M )) = 0
...
7), any quotient of R is
Noetherian
...
11) yields (4)
...
14) yields (1), because M is a finitely generated module
over R/ Ann(M )
...
Hilbert Functions
Exercise (20
...
— Let k be a field, k[X, Y ] the polynomial ring
...
Solution: Set m := ⟨X, Y ⟩ and a := ⟨X, Y 2 ⟩ and b := ⟨X 2 , Y 2 ⟩
...
So ℓ(a1 ) = 1, and ℓ(an ) = ℓ(mn ) for all n ≥ 2
...
Thus the three ideals have the
same Hilbert Polynomial, namely h(n) = n + 1, but different Hilbert Series
...
10)
227
⊕
⊕
Exercise (20
...
⊕
— Let R =
Rn be a graded ring, M =
Mn a graded Rmodule
...
Assume R0 is Artinian, R is a finitely generated R0 -algebra, and M is a finitely
generated R-module
...
(1) Prove that N ′ is a homogeneous submodule of M with the same Hilbert
′
Polynomial as N
N
...
Set R+ := n>0 Rn
...
∑
Solution: Given m =
mi ∈ N ′ , say Rk m ⊂ N
...
Hence mi ∈ N ′
...
By (19
...
12), R is Noetherian
...
19)
...
, nr be homogeneous generators of N ′ with ni ∈ Nki ; set k ′ := max{ki }
...
Given ℓ ≥ k + k , take n ∈ Nℓ , and′ write
n=
yi ni with yi ∈ Rℓ−ki
...
So n ∈ Nℓ
...
Thus N and N ′ have the same Hilbert polynomial
...
Then there is k0 with
ℓ(Nk′′ ) = ℓ(Nk ) for all k ≥ k0
...
So, if n ∈ N ′′ , then
Rk n ⊂ N for all k ≥ k0
...
Thus (1) holds
...
By (18
...
pi ̸⊃R+
′
But clearly ΓR+ (M/N ) = N /N
...
Exercise (20
...
— Let k be a field, P := k[X, Y, Z] the polynomial ring in three
variables, f ∈ P a homogeneous polynomial of degree d ≥ 1
...
Find
the coefficients of the Hilbert Polynomial h(R, n) explicitly in terms of d
...
Hence, Additivity of Length, (19
...
But
P (−d)n = P (n − d), so h(P (−d), n) = h(P, n − d)
...
4) yields
(
) (2−d+n)
h(R, n) = 2+n
−
= dn − (d − 3)d/2
...
10)
...
8), assume there is a homogeneous nonzerodivisor f ∈ R with Mf = 0
...
Solution: Suppose M := R/⟨f k ⟩
...
Form the exact sequence
µ
0 → R(−c) −
→ R → M → 0 where µ is multiplication by f k
...
9) yields h(M, n) = h(R, n) − h(R, n − c)
...
by (20
...
Thus deg h(R, n) > deg h(M, n)
...
7)
...
′
Then generators
mi ∈ Mci for 1 ≤ i ≤ r yield a surjection
→ M
...
But
deg
h(M
(−c
),
n)
=
deg
h(M
, n)
...
23)
Hence deg h(M ′ , n) ≥ deg h(M, n)
...
Thus deg h(R, n) > deg h(M, n)
...
15)
...
Assume ℓ(M/qM ) < ∞
...
Show
deg pm (M, n) = deg pq (M, n)
...
32)
...
Dividing into M and extracting lengths yields
ℓ(M/mn M ) ≤ ℓ(M/qn M ) ≤ ℓ(M/mmn M )
...
The two extremes are polynomials in n with the same degree, say d, (but not the
same leading coefficient)
...
□
Exercise (20
...
— Derive the Krull Intersection Theorem, (18
...
18)
...
29), we must prove that N = aN
...
But N ∩ an M = N for all n ≥ 0
...
□
20
...
22)
...
Assume M = aM
...
⊕
Solution: Suppose M ̸= 0; say Mn0 ̸= 0
...
Thus M = 0
...
23)
...
Set
N ′ := { m ∈ M | Rn m ∈ N for all n ≫ 0 }
...
Solution: Given m, m′ ∈ N ′ , say Rn m, Rn m′ ∈ N for n ≫ 0
...
Then Rn (m + m′ ), Rn xm ∈ N for n ≫ 0
...
Trivially
N ⊂ N ′
...
Then Rn mi ∈ N for n ≫ 0
as N is homogeneous
...
Since R is Noetherian and M is finitely generated, N ′ is finitely generated, say
by g, g ′ ,
...
Then there is n0 with Rn g, Rn g ′ ,
...
Replace g, g ′ ,
...
Say g, g ′ ,
...
, d(r) with d ≥ d′ ≥ · · · ≥ d(r)
...
Given m ∈ Nn′ with n ≥ n1 , say m = xg + x′ g ′ + · · · with x ∈ Rn−d and
′
x ∈ Rn−d′ and so on
...
Hence m ∈ Nn
...
9)
229
′
′
Nn′ ⊂ Nn
...
Thus Nn = Nn for n ≥ n′′1 , as desired
...
Let m ∈ N ′′
⊕homogeneous
′′
and p ≥ n2
...
So m ∈ N ′
...
□
Exercise (20
...
— Let √
R be a graded ring, a a homogeneous ideal, and M a
graded module
...
∑r+n
Solution: Take x = i≥r xi ∈ R with the xi the homogeneous components
...
Say xk ∈ a
...
But a is homogeneous
...
So xr ∈√ a
...
31)
...
Thus a is homogeneous
...
Let m ∈ M
...
If m
is homogeneous, then xi m = 0 for all i, since M is graded
...
Thus xi ∈ Ann(M ) for√all i, as desired
...
28)
...
26)
...
Let Q∗ ⊂ Q be the submodule generated by the
homogeneous elements of Q
...
Then Q∗ is primary too
...
Assume
x∈
/ nil(M/Q∗ )
...
So m′ has
′′
ℓ ′′
∗
ℓ ′′
a homogeneous component m with x m ∈
/ Q
...
Thus x ∈
/ nil(M/Q)
...
4)
...
Thus Q∗ is primary by (20
...
□
Exercise (20
...
— Under the conditions of (20
...
(1) Prove that there is a homogeneous f ∈ R with Rf = Rf
...
Solution: Let x1 ,
...
Write xi = ai /bi with ai , bi ∈ R homogeneous
...
Then f xi ∈ R
for each i
...
Thus (1) holds
...
Then (R/R)f = 0
by (12
...
So deg h(R/R, n) < deg h(R, n) by (20
...
But
h(R, n) = h(R, n) + h(R/R, n)
by (19
...
8)
...
□
21
...
6)
...
, yr a sop for N
...
, yi ⟩N
...
Solution: First, dim(N ) = r by (21
...
Then dim(Ni ) ≥ dim(Ni−1 ) − 1 for all
□
i by (21
...
18)
...
Exercise (21
...
— Let R be a Noetherian ring, and p be a prime minimal
containing x1 ,
...
Given r′ with 1 ≤ r′ ≤ r, set R′ := R/⟨x1 ,
...
, xr′ ⟩
...
Prove ht(p′ ) = r − r′
...
14)
Solution: Let x′i ∈ R′ be the residue of xi
...
9) and (2
...
So ht(p′ ) ≤ r − r′ by (21
...
On the other hand, Rp′ ′ = Rp′ by (11
...
, xr′ /1⟩ by
(12
...
5) with
Rp for both R and M
...
7
...
□
x′r′ +1 ,
...
11)
...
Prove that p is the union of height-1 primes, but not of finitely many
...
19), one would be equal to p, a contradiction
...
Thus we may assume R
is a domain
...
Then
ht(qx ) = 1 by the Krull Principal Theorem (21
...
Plainly qx = p
...
12)
...
Prove the following equivalent:
(1) R has only finitely many primes
...
(3) R is semilocal of dimension 1
...
Assume (2)
...
11), there’s no prime of height at least 2
...
So every prime is either of height 1 or of height 0
...
22)
...
Thus (3) holds
...
Again, every prime is either of height 1 or of height 0, and
the the height-0 primes are finite in number
...
Thus (1) holds
...
13) (Artin–Tate [1, Thm
...
— Let R be a Noetherian domain,
and set K := Frac(R)
...
(2) K is algebra finite over R
...
(4) R has only finitely many height-1 primes
...
Solution: By (11
...
Assume (2), and say K = R[x1 ,
...
Let f be a common denominator of the
xi
...
Let p ⊂ R be a nonzero prime
...
By the above, f m (1/z) ∈ R
for some m ≥ 1
...
So f ∈ p
...
√
Assume (3)
...
29) yields f ∈ ⟨y⟩
...
So 1/y = x/f n
...
Again assume (3)
...
Then f ∈ p
...
So p is one of finitely many primes by (17
...
Thus (4) holds
...
Take a nonzero element in each height-1 prime, and let
f be their product
...
But every nonzero prime
contains a height-1 prime owing to the Dimension Theorem (21
...
Thus (3) holds
...
12)
...
18)
231
Exercise (21
...
— Let R be a domain
...
Solution: Let p be a height-1 prime
...
Factor x
...
Then ⟨p⟩ is a prime ideal as p is a
prime element by (2
...
But ⟨p⟩ ⊂ p and ht(p) = 1
...
Conversely, assume every height-1 prime is principal and assume R is Noetherian
...
11, Sec
...
392–396])
...
By Krull’s
Principal Ideal Theorem, ht(p) = 1
...
Then x is prime by
(2
...
And p = xy for some y as p ∈ p
...
So y is a unit
...
□
Exercise (21
...
— (1) Let A be a Noetherian local ring with a principal prime
p of height at least 1
...
(2) Let k be a field, P := k[[X]] the formal power series ring in one variable
...
Prove that R is Noetherian and semilocal, and that R contains a
principal prime p of height 1, but that R is not a domain
...
Take y ∈ q
...
But
x∈
/ q since q ⫋ p
...
Thus q = qx
...
Hence Nakayama’s
Lemma (10
...
Thus ⟨0⟩ is prime, and so A is a domain
...
Repeating yields an ascending
chain of ideals ⟨a⟩ ⊂ ⟨a1 ⟩ ⊂ ⟨a2 ⟩ ⊂ · · ·
...
Then ak = bak−1 = bak x for some b
...
But
1 − bx is a unit by (3
...
So ak = 0
...
As to (2), every nonzero ideal of P is of the form ⟨X n ⟩ by (3
...
Hence P is
Noetherian
...
17)
...
11)
...
10)
...
Thus R is semilocal
...
Then p = m × P
...
Further, p contains
just one other prime 0 × P
...
Finally, R is not a domain as (1, 0) · (0, 1) = 0
...
16)
...
Assume
R is a domain of dimension r
...
Set R′ := R/⟨x⟩
...
Solution: A chain of primes in R′ of maximal length lifts to a chain of primes
pi in R of maximal length with ⟨x⟩ ⊆ p1 ⫋ · · · ⫋ pd
...
As x ̸= 0, also p1 ̸= 0
...
So Krull’s Principal Ideal Theorem,
(21
...
So 0 ⫋ p1 ⫋ · · · ⫋ pr is of maximal length in R
...
Hence d = dim R by (15
...
□
Exercise (21
...
— Let R be a Noetherian ring
...
232
Solutions: (22
...
Then
Rp → R[X]P is a flat local homomorphism by (13
...
Hence (21
...
(21
...
1)
/
Set k := Frac(R/p)
...
7) and (11
...
30)
...
Plainly, dim(Rp ) ≤ dim(R)
...
18
...
Thus dim(R[X]) ≤ dim(R) + 1
...
19)
...
19)
...
Let m be
the maximal ideal, and k := A/m the residue class field
...
Solution: By (21
...
But m is a parameter ideal, and the smallest number of generators
of m is dimk (m/m2 ) by (10
...
The assertion follows
...
23)
...
, xs ∈ A with s ≤ r
...
, xs ⟩ and B := A/a
...
, xr ∈ A with x1 ,
...
(2) B is regular of dimension r − s
...
Then x1 ,
...
So the residues of xs+1 ,
...
Hence dim(B) ≤ r − s by
(21
...
But dim(B) ≥ r − s by (21
...
So dim(B) = r − s
...
Assume (2)
...
, xr ∈ A
...
, xr
...
□
22
...
3)
...
Solution: In the 2-adic integers, 1 + 2 + 4 + 8 + · · · = 1/(1 − 2) = −1
...
4)
...
Prove that
the following three conditions are equivalent:
∩
c is injective; (2)
(1) κ : M → M
an M = ⟨0⟩; (3) M is separated
...
Assume either (a) a ⊂ rad(R)
c
...
Conclude M ⊂ M
Solution: A constant ∩
ssequence (m) has 0 as a limit if and only if m ∈ an M
for every n
...
Thus (1) and (2) are equivalent
...
1)
...
Assume R is Noetherian and M finitely generated
...
29) or (20
...
Assume (a)
...
2)
...
Finally, assume (b)
...
Let m ∈ M
...
Thus again (2) holds and (1) follows
...
11)
233
Exercise (22
...
— Let R be a ring
...
We say
the Qn satisfy the Mittag-Leffler Condition if the descending chain
Qn ⊃ αnn+1 Qn+1 ⊃ αnn+2 Qn+2 ⊃ · · · ⊃ αnm Qm ⊃ · · ·
stabilizes; that is, αnm Qm = αnm+k Qm+k for all k > 0
...
Show lim1 Qn = 0
...
Show lim1 Qn = 0
...
Set Pn := m≥n αnm Qm ,
which is the stable submodule
...
(4) Assume the Qn satisfy the Mittag-Leffler Condition
...
←−
∏
Solution: For (1), given (qn ) ∈
Qn , for each k ≥ n, set qk′ := αnk qk and
′
′
pn := qn + qn+1 + · · · + qm
...
So θ is surjective
...
∏
For (2), given (qn ) ∈ Qn , solve ∏
the equations pn −αnn+1 (pn+1 ) = qn recursively,
starting with p0 = 0, to get (pn ) ∈ Qn with θ(pn ) = qn
...
m
Qm
...
Thus (3) holds
...
13)
...
←−
←−
←−
For each n, the restriction αnn+1 Pn is surjective by (3)
...
←−
Further, for each n, there is m > n such that αnm Qm = Pn
...
So lim (Qn /Pn ) = 0 by (1)
...
←−
Exercise (22
...
— Let A be a ring, and m1 ,
...
Set
∩
b = ∏A
bm
...
Prove that A
i
Solution:
For each n > 0, the mni are pairwise∏
comaximal by (1
...
Hence
∏m
n
n
m = i=1 mi by (1
...
14)(4c)
...
So (A/mni )mi = A/m
∏mi by (11
...
1)
...
22)
...
Taking inverse limits, we obtain the assertion by (22
...
□
n
Exercise (22
...
— Let R be a ring, M a module, F • M a filtration, and N ⊂ M
a submodule
...
b ⊂M
c and M
c/N
b = (M/N ) b
...
(2) Also assume N ⊃ F M for n ≫ 0
...
18)
Solution: For (1), set P := M/N
...
So the induced sequence
b →M
c → Pb → 0
0→N
is exact by (22
...
8)
...
c/N
b = M/N
...
So plainly P = Pb
...
But n is
In particular, fix n and take N := F M
...
Thus G• M
c = G• M
...
Hence F n M
□
Exercise (22
...
— (1) Let R be a ring, a an ideal
...
If also ∩
R
n≥0 a = 0, show R is a domain
...
b be nonzero
...
Let x, y ∈ R
b and
positive integers r and s with x ∈ b
ar −b
a r+1 and y ∈ b
as −b
a s+1
...
Then x ̸= 0 and y ̸= 0
...
11)
...
Then x′ y ′ ̸= 0
...
Hence xy ̸= 0
...
∩
b by (22
...
Thus (1) holds
...
Then n≥0 mn = 0 by the
Krull Intersection Theorem (18
...
22), so
a domain
...
Thus (2) holds
...
14)
...
Prove (1) that A
c
maximal ideal, (2) that dim(M ) = dim(M ), and (3) that A is regular if and only
b is regular
...
30), and it’s local with m
b as maximal
Solution: First, A
ideal by (22
...
Thus (1) holds
...
11) and (22
...
So d(M ) = d(M
c/m
c ) by
b nM
Second, M/mn M = M
(20
...
Thus (2) holds by (21
...
b m
b 2 by (22
...
Hence m and m
b have generating sets with the
Third, m/m2 = m/
same number of elements by (10
...
Thus (3) holds
...
15)
...
, mm maximal ideals
...
Prove that A
b )
...
, m
b m are all its maximal ideals, and that m
b = rad(A
that m
b m
b m
b = A/m and A/
b i = A/mi
...
11) yields A/
/
∩
∏
∏ b i
b
b ⊂ (A/m
b i );
maximal
...
Hence A m
∩
b )
...
But m
b by (22
...
Thus m
b = rad(A
b= m
b i
...
Hence
Finally, let m′ be any maximal ideal of A
...
2)
...
So m′ = m
b i
...
, m
bm
m′ ⊃ m
b and so A
b is semilocal
...
24)
235
Exercise (22
...
— Let A be a Noetherian semilocal ring
...
b
x ∈ A is a nonzerodivisor on A if and only if its image x
b∈A
Solution: Assume x is a nonzerodivisor
...
So by Exactness of Completion (22
...
bx = µxb
...
Conversely, assume x
b is a nonzerodivisor and A is semilocal
...
b
on A
...
4), as the completion is taken with respect to the
Jacobson radical; further, µ
bx induces µx
...
□
Exercise (22
...
— Let p ∈ Z be prime
...
⊕
⊕
⊕
Set A := n≥1 Z/⟨p⟩ and B := n≥1 Z/⟨pn ⟩
...
b is just A
...
(3) Show that the natural sequence of p-adic completions
α
b b κ
b
b−
A
→B
−
→ (B/A) b
b (Thus p-adic completion is neither left exact nor right exact
...
Solution: For (1), note pA = 0
...
Hence
b
A = A
...
For (2), set Ak := α−1 (pk B)
...
So
)
(
⊕
⊕
Ak = α−1 0 ⊕ · · · ⊕ 0 ⊕ n>k ⟨pk ⟩/⟨pn ⟩ = (0 ⊕ · · · ⊕ 0 ⊕ n>k Z/⟨p⟩)
...
But by (22
...
Thus
←−
∏k
A = limk≥1 n=1 Z/⟨p⟩
...
, let πkk+1 : n=1 Mn → n=1 Mn be
∏k
∏∞
the projections
...
5) yields limk≥1 n=1 Mn = n=1 Mn
...
←−
For (3), note that, by (2) and (22
...
1), the following sequence is exact:
κ
b
b−
0→A→B
→ (B/A) b
...
Thus Im(b
But A
α) ̸= Ker(b
κ);
that is, (3) holds
...
21)
...
Show that M 7→ M
b
c
surjections, and that R ⊗ M → M is surjective if M is finitely generated
...
17) shows that M 7→ M
surjections
...
19) yields the desired surjectivity
...
24)
...
Prove that R
faithfully flat if and only if a ⊂ rad(R)
...
6)
b is flat over R by (22
...
Next, let m be a maximal ideal of
Solution: First, R
b ⊗R (R/m) = (R/m) b by (22
...
But (R/m) b = lim(R/m)/ar (R/m)
R
...
8)
...
Hence R
only if a ⊂ m
...
4)
...
25)
...
Assume
b is
...
Prove b is principal if bR
Solution: Since R is Noetherian, b is finitely generated
...
/
b (ab)b
Hence, b is principal if b/ab is a cyclic R-module by (10
...
But b/abb
b by (22
...
by (22
...
But, given
Assume bR = Rb
...
Then xb′ = yb′
...
□
Exercise (22
...
— Let R be a ring,
a an ideal, and M a module
...
Show
m1 ,
...
, m′n in M/aM generate
...
, m′n generate G• M over G• R
...
, mn
generate M over R by the proof of (22
...
Alternatively, M is finitely generated over R and complete by the statement of
c
...
b
(22
...
As M is also separated, M = M
As R is
b is surjective
...
1); so a is complete;
complete, κR : R → R
whence, κa : a → b
a is surjective too
...
Thus M/aM = M/b
aM
...
But a ⊂ rad(R) by (22
...
So by Nakayama’s Lemma
b so also over R as κR is surjective
...
13)(2), the mi generate M over R,
□
23
...
6)
...
(1) Assume that x, y form an M -sequence
...
(2) Assume that x, y form an M -sequence and that y ∈
/ z
...
Prove that
y, x form an M -sequence too
...
Assume that, given any m, n ∈ M with xm = yn, there
exists p ∈ M with m = yp and n = xp
...
Solution: Consider (1)
...
Then
yn1 = 0, but y ∈
/ z
...
Hence n1 = 0
...
So
x(m − yp) = 0
...
div(M )
...
Consider (2)
...
Next, set
M1 := M/yM
...
div(M1 )
...
Then xm = yn for some n ∈ M
...
Thus m1 = 0, as desired
...
The statement is symmetric in x and y
...
First, M/⟨x, y⟩M ̸= 0 by Nakayama’s Lemma
...
div(M )
...
Solutions: (23
...
Repeat with p
in place of m, obtaining p1 ∈ M such that p = yp1 and 0 = xp1
...
Then Rp1 ⊂ Rp2 ⊂ · · · is an ascending chain
...
Say Rpn = Rpn+1
...
Then pn = ypn+1 = yzpn
...
But y ∈ m
...
Hence pn = 0
...
Thus m = 0, as desired
...
div(M )
...
We must prove y ∈
/ z
...
Given n1 ∈ M1 with
yn1 = 0, lift n1 to n ∈ M
...
So there exists p ∈ M
with n = xp
...
Thus x, y form an M -sequence
...
8)
...
, xn ∈ m, and σ a permutation of 1,
...
Assume x1 ,
...
, xσn do too; first, say σ transposes i and i + 1
...
Set Mj := M/⟨x1 ,
...
Then
xi , xi+1 form an Mi−1 -sequence; so xi+1 , xi do too owing to (23
...
So
x1 ,
...
But M/⟨x1 ,
...
Hence xσ1 ,
...
In general, σ is a composition of transpositions of successive
integers; hence, the general assertion follows
...
7)
...
20))
...
Solution: An N -sequence x, y yields a commutative diagram with exact rows:
µx
0−
→ N −−→ N −
→ N/xN
µy y
µy y
µy y
µx
0−
→ N −−→ N −
→ N/xN
Applying the left-exact functor F yields this commutative diagram with exact rows:
µx
0−
→ F (N ) −−→ F (N ) −
→ F (N/xN )
µy y
µy y
µy y
µx
0−
→ F (N ) −−→ F (N ) −
→ F (N/xN )
Thus x is a nonzerodivisor on F (N )
...
µy
As N/xN −−
→ N/xN is injective and F is left exact, the right-hand vertical map
µy is injective
...
Thus x, y is an F (N )-sequence
...
9)
...
Prove
that, if A is regular, then A is Cohen–Macaulay
...
14)
Solution: Assume A is regular
...
, xr , let’s show it’s
an A-sequence
...
Then A1 is regular of dimension r − 1 by
(21
...
So x1 ̸= 0
...
24)
...
div(A)
...
, xr form a regular sop of A1 ; so we may assume
they form an A1 -sequence by induction on r
...
, xr is an A-sequence
...
, xn , then n ≤ depth(A) ≤ r
by (23
...
5)(3), and n ≥ r by (21
...
Thus then n = depth(A) = r,
and so A is regular and Cohen–Macaulay
...
11)
...
Prove A is a maximal proper subring of K
...
Solution: Let R be a ring, A ⫋ R ⊂ K
...
Say
x = utn where u ∈ A× and t is a uniformizing parameter
...
Set
y := u−1 t−n−1
...
So t−1 = xy ∈ R
...
Thus R = K, as desired
...
Hence Af = K by the above
...
But dim(A) = 1 by (23
...
Exercise (23
...
— Let k be a field, P := k[X, Y ] the polynomial ring in two
variables, f ∈ P an irreducible polynomial
...
Set R := P/⟨f ⟩ and
p := ⟨X, Y ⟩/⟨f ⟩
...
(Thus Rp is a DVR
if and only if the plane curve C : f = 0 ⊂ k 2 is nonsingular at (0, 0)
...
Then (12
...
4) yield
A/m = (R/p)p = k
and
m/m2 = p/p2
...
Now, the k-vector space m/m2 is generated by the images x
and y of X and Y in A
...
Also, g ∈ (X, Y )2 ; so
its image in m/m2 is also 0
...
Now, f cannot generate ⟨X, Y ⟩, so m ̸= 0; hence, m/m2 ̸= 0 by
Nakayama’s Lemma, (10
...
Therefore, m/m2 is 1-dimensional over k; hence, m
is principal by (10
...
Now, since f is irreducible, A is a domain
...
10)
...
Then f = g ∈ (X, Y )2
...
Hence, m/m2 is 2-dimensional
...
11)
...
13)
...
Suppose that A is local with maximal ideal ⟨X⟩
...
(Such
local rings arise as rings of power series with curious convergence conditions
...
Clearly, a is a
∩
subset of the corresponding ideal n≥0 ⟨X n ⟩ of k[[X]], and the latter ideal is clearly
zero
...
3) implies A is a DVR
...
14)
...
, Xn
variables, P and Q the polynomial rings over K and L in X1 ,
...
(1) Let q be a prime of Q, and p its contraction in P
...
Solutions: (23
...
Prove
that f and g have no common prime factor q ∈ Q
...
Furthermore, P is normal,
and Q is a domain
...
9)
...
Thus ht p ≤ ht q
...
3); whence, ht p ≥ ht q
...
Thus (1) holds
...
14), ht(p) + dim(P/p) = n and ht(q) + dim(Q/q) = n
as both P and Q are polynomial rings in n variables over a field
...
13), dim P/p = tr
...
degL Frac(Q/q), and
these two transcendence degrees are equal as Q/P is an integral extension
...
Suppose f and g have a common prime factor q ∈ Q, and set q := Qq
...
Hence Qq is a DVR by
(23
...
Thus ht(q) = 1
...
Then p contains f ; whence, p contains
some prime factor p of f
...
Hence p = P p
since ht p = 1 by (1)
...
Therefore, p | g, contrary to the
hypothesis
...
(Caution: if f := X1 and g := X2 , then f and g have
no common factor, yet there are no φ and ψ such that φf + ψg = 1
...
16)
...
Show that M is torsionfree if and only if it satisfies (S1 )
...
By (23
...
Hence z
...
15)
...
Conversely, assume M is torsionfree
...
Then p = Ann(m)
for some m ∈ M
...
So p = ⟨0⟩ is the only
associated prime
...
15)
...
17)
...
Show that R is reduced if and
only if (R0 ) and (S1 ) hold
...
Consider any irredundant primary de∩ (R0 ) and (S√
composition ⟨0⟩ = qi
...
Then pi ∈ Ass(R) by (18
...
20)
...
Hence the localization Rpi is a field by (R0 )
...
But pi Rpi ⊃ √
qi Rpi
...
Therefore, pi = qi by (18
...
So
∩
⟨0⟩ = pi = ⟨0⟩
...
Conversely, assume R is reduced
...
41)
...
Thus (R0 ) holds
...
So
p is minimal whenever p ∈ Ass(R) by (18
...
Thus R satisfies (S1 )
...
22)
...
Solution: Assume R normal
...
Then pRp ∈ Ass(Rp /⟨x/1⟩)
by (17
...
So depth(Rp ) = 1 by (23
...
But Rp is normal by (11
...
Hence
pRp is principal by (23
...
Conversely, assume that, given any prime p associated to a principal ideal, pRp is
principal
...
Then p is minimal
240
Solutions: (23
...
So p ∈ Ass(R/⟨x⟩) by (17
...
So, by hypothesis, pRp is principal
...
10)
...
/
Given any prime p with depth(Rp ) = 1, say pRp ∈ Ass(Rp ⟨x/s⟩) with x ̸= 0
by (23
...
Then ⟨x/s⟩ = ⟨x/1⟩ ⊂ Rp
...
10)
...
So dim(Rp ) = 1 by (23
...
Thus R also satisfies
(S2 )
...
20)
...
23)
...
Assuming (S1 ) holds for R, prove Φ ⊂ Σ, and prove Φ = Σ if and only if (S2 ) holds
...
(23
...
1)
Solution: Assume (S1 ) holds
...
15) and (23
...
Clearly, p is minimal containing ⟨x⟩
...
18)
...
5)(2)
...
However, as (S1 ) holds, (S2 ) holds if and only if Φ ⊃ Σ
...
Further, without assuming (S1 ), consider (23
...
1)
...
Conversely, take an x ∈ K that vanishes in p∈Σ Kp /Rp
...
Then a/1 ∈ bRp for all p ∈ Σ
...
Hence, if p ∈ Ass(Rp /bRp ), then p ∈ Σ
by (23
...
Therefore, a ∈ bR by (18
...
Thus x ∈ R
...
23
...
□
Exercise (23
...
— Let R be a Noetherian ring, and K its total quotient ring
...
Prove these three conditions are equivalent:
(1) R is normal
...
∏
(3) (R1 ) and (S1 ) hold, and R → K → p∈Φ Kp /Rp is exact
...
Then R is reduced by (14
...
So (23
...
But Rp is normal for any prime p by (14
...
Thus (2) holds by (23
...
Assume (2)
...
Thus (23
...
Assume (3)
...
Then x/1 ∈ K is integral over Rp
for any prime p
...
Hence,
x/1 ∈ Rp for all p ∈ Φ
...
But R is
reduced by (23
...
Thus (14
...
□
23
...
25)
...
, xr an M -sequence in a
...
Assume M ′ /aM ′ ̸= 0
...
, xr is an M ′ -sequence in aR′
...
, xi ⟩M and Mi′ := M ′ /⟨x1 ,
...
Then Mi′ = Mi ⊗R R′ by right exactness of tensor product (8
...
Moreover,
by hypothesis, xi+1 is a nonzerodivisor on Mi
...
Hence µxi+1 : Mi′ → Mi′ is injective by flatness
...
, xr ⟩ ⊂ a, so Mr′ ̸= 0
...
□
Solutions: (24
...
26)
...
Let x1 ,
...
Prove the following statements:
(1) x1 /1,
...
Solution: First, (13
...
So Mp ̸= 0 and p ∈ V(a)
...
11)
...
21)
...
25) yields (1)
...
Thus (23
...
Exercise (23
...
— Let R be a Noetherian ring, a an ideal, and M a finitely
generated module with M/aM ̸= 0
...
Show
depth(a, M/xM ) = depth(a, M ) − 1
...
29
...
, xr in a by (23
...
Then
x, x2 ,
...
Thus (23
...
29
...
□
Exercise (23
...
— Let A be a Noetherian local ring, M a finitely generated
module, x ∈
/ z
...
Show M is Cohen–Macaulay if and only if M/xM is so
...
29) yields depth(M/xM ) = depth(M ) − 1
...
5)
yields dim(M/xM ) = dim(M ) − 1
...
□
Exercise (23
...
— Let A be a Noetherian local ring, and M a nonzero finitely
generated module
...
(1) depth(M ) = depth(M
c is Cohen–Macaulay
...
24), and the maximal ideal
Solution: The completion A
b by (22
...
22)(2)
...
31)
of A extends to the maximal ideal of A
c
□
yields (1)
...
14)(2); so (1) implies (2)
...
33)
...
Show that there is p ∈ Supp(M/aM ) with
depth(a, M ) = depth(ap , Mp )
...
33
...
, xr in a by (23
...
26)(1) implies x1 /1,
...
Set Mr := M/⟨x1 ,
...
Then a ⊂ z
...
So a ⊂ p for some p ∈ Ass(Mr ) by (17
...
So pRp ∈ Ass(Mr )p by (17
...
So pRp ⊂ z
...
Hence x1 /1,
...
So (23
...
33
...
□
Exercise (23
...
— Prove that a Cohen–Macaulay local ring A is catenary
...
Fix a maximal chain from p up to the
maximal ideal and a maximal chain from q down to a minimal prime
...
36)
...
Thus A is catenary
...
Dedekind Domains
242
Solutions: (24
...
5)
...
(1) Assume dim(R) = 1
...
(2) Assume dim(R) ≥ 1
...
Solution: Consider (1)
...
Then there’s a chain of
primes 0 ⫋ p′ ⊂ S −1 R
...
Then p is as desired by (11
...
Conversely, suppose there’s a nonzero p with p ∩ S = ∅
...
20)(2); so dim(S −1 R) ≥ 1
...
Then 0 = p0 ⫋ · · · ⫋ pr ⊂ R is a chain
of primes by (11
...
So r ≤ 1 as dim(R) = 1
...
Consider (2)
...
Conversely, let 0 = p0 ⫋ · · · ⫋ pr ⊂ R be a chain of primes
...
Then 0 = p′0 ⫋ · · · ⫋ p′r is a chain of primes by (11
...
So if dim(Rpr ) = 1,
then r ≤ 1
...
□
Exercise (24
...
— Let R be a Dedekind domain, S a multiplicative subset
...
Solution: Suppose there’s a prime nonzero p with p ∩ S = ∅
...
So
S −1 R is a domain by (11
...
And S −1 R is normal by (11
...
Further, S −1 R
is Noetherian by (16
...
Also, dim(S −1 R) = 1 by (24
...
Thus S −1 R is
Dedekind
...
5)(1)
...
8)
...
By first
reducing to the case that R is local, prove that
a ∩ (b + c) = (a ∩ b) + (a ∩ c),
a + (b ∩ c) = (a + b) ∩ (a + c)
...
37), it suffices to establish the two equations after localizing
at each maximal ideal p
...
17)(4), (5)
...
Thus replacing R by Rp , we may assume R is a DVR
...
1), take a uniformizing parameter t
...
Then the two equations in questions are equivalent to these two:
{
}
{
}
max i, min{j, k} = min max{i, j}, max{i, k} ,
{
}
{
}
min i, max{j, k} = max min{i, j}, min{i, k}
...
□
Exercise (24
...
— Prove that a semilocal Dedekind domain A is a PID
...
Solution: Let p1 ,
...
Let’s prove they are
principal, starting with p1
...
11), p1 Ap1 ̸= p21 Ap1 ; so
p1 ̸= p21
...
The ideals p21 , p2 ,
...
Hence, by the Chinese Remainder
Theorem, (1
...
Solutions: (25
...
10), yields ⟨x⟩pn1 1 pn2 2 · · · pnr r
with ni ≥ 0
...
Further, x ∈ p1 − p21 ; so
n1 = 1
...
Similarly, all the other pi are principal
...
10), yields
∏ i let a be any nonzero ideal
...
Say
p
=
⟨x
⟩
...
□
i
i
i
i
Exercise (24
...
— Let R be a Dedekind domain, a and b two nonzero ideals
...
Solution:
To prove (1), let p1 ,
...
Then S is multiplicative
...
Then R′ is Dedekind
by (24
...
Let’s prove R′ is semilocal
...
Then q = pR′ by (11
...
So p is nonzero, whence maximal since R has dimension 1
...
Then p and the pi are pairwise comaximal
...
14), there is a u ∈ R that is congruent to 0 modulo p
and to 1 modulo each pi
...
Thus
p1 R′ ,
...
So R′ is a PID by (24
...
But by (12
...
Finally, S −1 (R/a) = R/a by (11
...
Thus (1) holds
...
12), as follows
...
10), yields a = pn1 1 · · · pnk k for distinct
maximal ideals pi
...
So, by the Chinese Remainder
Theorem, (1
...
R/a −→
1
k
Next, let’s prove each R/pni i is a Principal Ideal Ring (PIR); that is, every
ideal is principal
...
Then S −1 (R/pni i ) = Rpi /pni i Rpi , and the latter
ring is a PIR because Rpi is a DVR
...
6), as every
u ∈ S maps to a unit in R/pni i since p/pni i is the only prime in R/pni i
...
, Rk , we must prove their product is a
PIR
...
Then
⟨ b = b1 × ⟩· · · × bk where bi ⊂ Ri is
an ideal by (1
...
Say bi = ⟨ai ⟩
...
, ak )
...
Consider (2)
...
By (1), there is a y ∈ b whose residue
generates b/⟨x⟩
...
□
25
...
2)
...
Prove that M is principal if and only if there exists some isomorphism M ≃ R
...
and φ : (M : N ) −→
Solution: If M ≃ R, let x correspond to 1; then M = Rx
...
Then x ̸= 0 as M ̸= 0
...
It’s
surjective as M = Rx
...
Form the canonical M × N → M N with (x, y) 7→ xy
...
So it induces
a map π : M ⊗ N → M N , and clearly π is surjective
...
244
Solutions: (25
...
Say y ̸= 0
...
Finally, given θ : N → M , fix a nonzero n ∈ N , and set z := θ(n)/n
...
Then bcy = adn
...
Hence θ(y) = yz
...
Thus, φ is surjective, as desired
...
6)
...
Prove that
the map π : M ⊗ N → M N is an isomorphism if M is locally principal
...
43), it suffices to prove that, for each maximal ideal m, the
localization πm : (M ⊗ N )m → (M N )m is bijective
...
14), and (M N )m = Mm Nm by (25
...
By hypothesis, Mm = Rm x for some
x
...
And Rm ⊗ Nm = Nm by (8
...
Thus πm ≃ 1Nm
...
9)
...
(1) Assume N is invertible, and show that (M : N ) = M · N −1
...
Solution: For (1), note that N −1 = (R : N ) by (25
...
So M (R : N )N = M
...
Conversely, note that (M : N )N ⊂ M
...
Thus (1) holds
...
Conversely, if M N is invertible,
then R = (M N )(M N )−1 = M (N (M N )−1 ); thus, M is invertible
...
Thus (2) holds
...
12)
...
Show that a fractional ideal M is invertible
if and only if M is principal and nonzero
...
7), a nonzero principal ideal is always invertible
...
Then trivially M ̸= 0
...
Fix
a
nonzero
m
∈
M
...
But ni m ∈ R as m ∈ M and ni ∈ M −1
...
Then m = dx
...
Then
d′ := gcd{ni m′ } = gcd{ni ma/b} = a gcd{ni m}/b = ad/b
...
But d′ ∈ R
...
□
Exercise (25
...
— Show that a ring is a PID if and only if it’s a Dedekind
domain and a UFD
...
2), and is a UFD by (2
...
Conversely, let R be a Dedekind UFD
...
3) and (25
...
12)
...
Alternatively and more directly, every nonzero prime is of height 1 as dim R = 1,
so is principle by (21
...
But, by (24
...
Thus again, R is a PID
...
17)
...
Prove that M
is finitely generated, and that, if R is local, then M is free of rank 1
...
5)
245
∑
∼ R and 1 = α(
Solution: Say α : M ⊗ N −→
mi ⊗ ni ) with mi ∈ M and
ni ∈ N
...
Form this composition:
∼ M ⊗ M ⊗ N = M ⊗ N ⊗ M −→
∼ R ⊗ M = M
...
But β is an isomorphism
...
Suppose R is local
...
So u := α(mi ⊗ ni ) ∈ R× for
some i
...
Then α(m ⊗ n) = 1
...
Then ν(m) = 1; so ν is surjective
...
Then µν(m′ ) = ν(m′ )m = β(m′ ), or µν = β
...
So ν is injective
...
□
Exercise (25
...
— Show these conditions on an R-module M are equivalent:
(1) M is invertible
...
(3) M is locally free of rank 1
...
Solution: Assume (1)
...
17)
...
Let m be a maximal ideal
...
Hence Mm ≃ Rm
again by (25
...
Thus (2) holds
...
52)
...
Then (2) holds; so Mm ≃ Rm at any maximal ideal m
...
51); so HomR (M, R)m = HomRm (Mm , Rm ) by (12
...
Consider the evaluation map
ev(M, R) : M ⊗ Hom(M, R) → R
defined by
ev(M, R)(m, α) := α(m)
...
Clearly ev(Rm , Rm ) is bijective
...
43)
...
□
26
...
3)
...
Show that V is a valuation ring if and
only if, given any two ideals a and b, either a lies in b or b lies in a
...
Suppose also a ̸⊂ b; say x ∈ a,
but x ∈
/ b
...
Then x/y ∈
/ V ; else x = (x/y)y ∈ b
...
Hence
y = (y/x)x ∈ a
...
Conversely, let x, y ∈ V − {0}, and suppose x/y ∈
/ V
...
Hence ⟨y⟩ ⊂ ⟨x⟩ by hypothesis
...
Thus V is a valuation ring
...
4)
...
Prove that W is also a valuation ring of K, that its maximal ideal p lies in V , that
V /p is a valuation ring of the field W/p, and that W = Vp
...
Then 1/x ∈ V ⊂ W
...
Second, let y ∈ p
...
2) implies 1/y ∈ K − W ⊂ K − V
...
Third, x ∈ W − V implies 1/x ∈ V ; whence, V /p is a valuation ring of W/p
...
Conversely, let x ∈ W − V
...
But 1/x ∈
/p
as p is the maximal ideal of W
...
Thus W = Vp
...
20)
Exercise (26
...
— Prove that a valuation ring V is normal
...
Take x ∈ K
integral over V , say xn + a1 xn−1 + · · · + an = 0 with ai ∈ V
...
(26
...
1)
−1
If x ∈
/ V , then x
∈ m by (26
...
So (26
...
1) yields 1 ∈ m, a contradiction
...
Thus V is normal
...
9)
...
Show that the valuation rings of K are the maximal elements of S
...
Then
dominate V
...
Then 1/x ∈
/ m′ as 1 ∈
/ m′ ; so also 1/x ∈
/ m
...
Thus V is maximal
...
By (26
...
By maximality, V = V ′
...
2)
...
Take any nonzero
So x ∈ V by (26
...
Hence,
is dominated by a valuation
□
Exercise (26
...
— Let V be a valuation ring, such as a DVR, whose value
group Γ is Archimedean; that is, given any nonzero α, β ∈ Γ, there’s n ∈ Z such
that nα > β
...
Solution: Let R be a subring of K strictly containing V , and fix a ∈ R − V
...
Then α < 0
...
Then v(b/an ) > 0
...
So b = (b/an )an ∈ R
...
□
Exercise (26
...
— Let V be a valuation ring
...
Solution: To prove (1), say a = ⟨x1 ,
...
Let v be the
valuation
...
Then xi /x1 ∈ V for all i
...
Hence a = ⟨x1 ⟩
...
To prove (2), first assume V is Noetherian
...
2), and by
(1) its maximal ideal m is principal
...
10)
...
Then V is a PID by (23
...
Thus (2) holds
...
20)
...
Prove the integral closure R of R in L is
the intersection of all DVRs V of L containing R by modifying the proof of (26
...
18) to R[y]p
...
10)
...
Thus V ⊃R V ⊃ R
...
To find a DVR V of L
with V ⊃ R and x ∈
/ V , set y := 1/x
...
Multiplying by x yields x
− an x − · · · − a0 = 0
...
Thus y is a nonzero nonunit of R[y]
...
12)
...
20)
247
Ideal Theorem (21
...
Then R[y]p is Noetherian of dimension 1
...
Hence the integral closure R′ of
R[y]p in L is a Dedekind domain by (26
...
So by the Going-up Theorem (14
...
Then as R′ is Dedekind, Rq′ is a DVR of
L by (24
...
Further, y ∈ qRq′
...
□
Bibliography
[1] Artin, E
...
T
...
Math
...
Japan, 3
...
[2] Artin, M
...
[3] Atiyah, M
...
, “Introduction to Commutative Algebra,” Addison-Wesley,
1969
...
, “Commutative Algebra with a View Toward Algebraic Geomertry,” SpringerVerlag, 1999
...
, and Dieudonn´
e, J
...
[6] Grothendieck, A
...
, “El´
ements de G´
eom´
etrie Alg´
ebrique IV-1,” Publ
...
IHES, Vol
...
[7] Grothendieck, A
...
, “El´
ements de G´
eom´
etrie Alg´
ebrique IV-2,” Publ
...
IHES, Vol
...
[8] Judson, T
...
, “Undergraduate Analysis,” Springer-Verlag, 1997
...
, “Algebra” Graduate Texts in Mathematics 211, Springer-Verlag, 2002
...
, “Undergraduate Commutative Algebra,” Cambridge University Press, 1995
...
, “An Introduction to Number Theory,” MIT Press, 1978
...
10–16
1
...
2), 11, owing to (3
...
2
...
36), 171
3
...
37), 172
4
...
38), 170
5
...
39), 172
6
...
24), 169
7
...
20), 165
8
...
14), 167
9
...
29), 14
10
...
30), 170
11
...
19), 165, and (1
...
— Essentially (3
...
— Standard; see [10], Theorem 2
...
231
14
...
15), 13, and
(3
...
— Part of (13
...
— Best done by hand
17
...
1), 77, and (13
...
16), 204
18
...
1), 77, and (13
...
— Essentially (13
...
— Essentially (13
...
— Part of (13
...
13), 204,
and (13
...
— Essentially (13
...
13), 5
23
...
24), 207, and
(1
...
— About lattices, which we don’t treat
25
...
25), 80
26
...
— Rudimentary Algebraic Geometry,
sketched in place
28
...
31–35
1
...
17), 187
2
...
16)(1), 51
3
...
17), 193
4
...
6), 55
5
...
8), 188
6
...
24), 187
7
...
18), 165
8
...
10), 189
ii) Part of (9
...
— Part of (16
...
— Essentially (10
...
— Mostly in (10
...
6), 191,
and (5
...
— Immediate from (5
...
23), 27
13
...
4), 185
14
...
7), 43
15
...
8), 43
16
...
6), 38
17
...
2), 42, and (7
...
— Essentially (6
...
— Essentially (7
...
— Essentially (8
...
— Part of (7
...
9), 183
22
...
17), 184, and (7
...
— Essentially (8
...
— About Tor, which we don’t need
25
...
16)(2), 56
26
...
26), 59
27
...
9), 192
28
...
10), 192
Chapter 3, pp
...
— Part of (12
...
— Essentially (12
...
— Essentially (11
...
— Part of (11
...
— Essentially (13
...
— Essentially (3
...
— i) Part of (3
...
17), 168
8
...
9), 195
9
...
11), 196
10
...
48), 210
11
...
23), 206
12
...
19), 200
13
...
36), 208
14
...
38), 209
15
...
5), 61
16
...
15), 189
17
...
13), 189
18
...
15), 204
19
...
35), 81;
ii), iii), iv), v) essentially (13
...
30), 80;
vii) essentially (13
...
32), 208
20
...
4), 201,
21
...
5), 202;
ii) (13
...
7), 202;
iv) (13
...
— Essentially (13
...
— Essentially (13
...
— Covered in (13
...
3
...
199
25
...
10), 203
250
Disposition of the Exercises in [3]
26
...
11), 203
27
...
2
...
333, and
(7
...
12), p
...
2
...
338
28
...
— See 27
30
...
55–58
1
...
— To be done
3
...
— To be done
5
...
— Analysis, continuing Ex
...
3 in [3]
7
...
— To be done
9
...
— To be done
11
...
— To be done
13
...
27), 223
14
...
20), 109
15
...
— Covered in (18
...
— Technical conditions for primary
decomposition; solution sketched in place
18
...
— To be done
20
...
— Essentially (18
...
12),
107
22
...
20), 109
23
...
18), 103, and (17
...
23), 109, and (18
...
25), 110
Chapter 5, pp
...
— To be done
2
...
— To be done
4
...
— To be done
6
...
31), 194
7
...
— Part of (14
...
— To be done
10
...
— Essentially (14
...
— To be done
13
...
— To be done
15
...
— Essentially (15
...
— Part of (15
...
— Essentially (15
...
— Part of (15
...
— To be done
21
...
— To be done
23
...
— Essentially (15
...
29),
217
25
...
— Essentially (15
...
24),
216
27
...
9), 246
28
...
3), 245, and part of
(26
...
— Part of (26
...
— Part of (26
...
— Part of (26
...
— To be done
33
...
12), 159
34
...
— To be done
Chapter 6, pp
...
— Essentially (10
...
— Part of (16
...
— To be done
4
...
— To be done
6
...
— To be done
8
...
— To be done
10
...
— To be done
12
...
84–88
1
...
10), 97
2
...
— To be done
4
...
— Essentially (16
...
— To be done
7
...
12), 98
8
...
8), 218
9
...
— To be done
11
...
— To be done
13
...
— Essentially (15
...
— Essentially (10
...
26), 59
16
...
51), 83
17
...
— Essentially (18
...
2),
252
Disposition of the Exercises in [3]
101, and (17
...
— Follows easily from (18
...
— Essentially [5] (0,2
...
3), p
...
4
...
58
21
...
4
...
59
22
...
4
...
60
23
...
8
...
239
24
...
10
...
250
25
...
4
...
20
26
...
— Trivial K-theory
Chapter 8, pp
...
— Essentially (18
...
28),
223
2
...
12), 225
3
...
8), 225
4
...
— To be done
6
...
5), 225
Chapter 9, p
...
— To be done
2
...
— Part of (26
...
— Essentially (23
...
— Part of (25
...
— To be done
7
...
13), 243
8
...
8), 242
9
...
113–115
1
...
19), 235
2
...
9), 133
3
...
— To be done
5
...
— To be done
7
...
24), 135
8
...
— To be done
10
...
— To be done
12
...
125–126
1
...
— To be done
3
...
13), 92
4
...
— Trivial K-theory
6
...
19), 215
7
...
18), 231
Index
algebra: (1
...
5), 19
algebra map: (1
...
22), 53
extended Rees Algebra: (20
...
5), 55
finitely generated: (4
...
5), 55
group algebra: (26
...
1), 1
integral over a ring: (10
...
23), 69
module finite: (10
...
1), 1
subalgebra: (4
...
5), 18
tensor product: (8
...
1), 1; (4
...
8), 39
colimit: (6
...
1), 35
associative: (6
...
7), 39
direct limit: (6
...
6), 38
indexed by: (6
...
6), 38
dually: (5
...
1), 42
identity: (6
...
1), 35
inclusion: (6
...
7), 39
insertion: (6
...
1), 35
isomorphism: (6
...
1), 35
morphism: (6
...
1), 35
pushout: (6
...
6), 38
category: (6
...
1), 42
discrete: (6
...
6), 38
has direct limits: (6
...
1), 35
small: (6
...
13), 26
commutative: (1
...
1), 17
Cauchy sequence: (22
...
11), 5
constant term: (3
...
21), 63
formal power series: (3
...
10), 20
generators: (1
...
10), 20
homogeneous: (20
...
21), 122
homogeneous of degree n: (15
...
21), 122; (20
...
11), 5
initial component: (20
...
21), 63
integrally dependent on a ring: (10
...
6), 8
Kronecker delta function: (4
...
22), 177
limit: (22
...
4), 2
linearly independent: (4
...
1), 1
nilpotent: (3
...
28), 80
nonzerodivisor: (2
...
14), 103
p-adic integer: (22
...
6), 8
reciprocal: (1
...
26), 9
residue of: (1
...
10), 20
restricted vectors: (4
...
1), 138
unit: (1
...
1), 7
zerodivisorexPIRflat: (17
...
3), 7
discrete valuation: (23
...
3), 7
p-adic valuation: (23
...
3), 7
Trace Pairing: (24
...
15), 150
functor: (6
...
20), 52
adjoint: (6
...
4), 36
counit: (6
...
5), 37
universal: (6
...
20), 47
constant: (6
...
1), 36
covariant: (6
...
6), 38
direct system: (6
...
2), 54
faithful: (9
...
2), 35
isomorphic: (6
...
4), 36
left exact: (9
...
5), 49; (9
...
4), 36
natural transformation: (6
...
4), 36
right exact: (9
...
4), 2
associated prime: (17
...
3), 96
comaximal: (1
...
4), 3
extension: (1
...
35), 32
fractional: (25
...
7), 153
locally principal: (25
...
1), 152
product: (25
...
1), 152
generated: (1
...
17), 6
intersection: (1
...
10), 91
lie over: (14
...
13), 8
nested: (1
...
22), 14
parameter: (21
...
1), 7
height: (21
...
9), 91
minimal: (3
...
4), 2
product: (1
...
4), 3
radical: (3
...
15), 68
saturation: (11
...
4), 3
symbolic power: (18
...
1), 77
Lemma
Artin Character: (24
...
18), 121
Artin–Tate: (16
...
25),
58
Equational Criterion for Vanishing: (8
...
15), 26
Ideal Criterion for Flatness: (22
...
26), 59
Nakayama: (10
...
28), 136
Nine: (5
...
1), 88
Nonunit Criterion: (3
...
19), 13
Schanuel: (5
...
13), 25
Zorn’s: (2
...
10), 97; (17
...
2), 17
automorphism: (1
...
1), 48
bimodule homomorphism: (8
...
1), 1; (4
...
21), 122
homomorphism: (1
...
1), 1; (4
...
21), 27
local homomorphism: (10
...
8), 19
quotient map: (4
...
8), 25
section: (5
...
10), 50
matrix of cofactors: (10
...
1), 17
S-torsion: (12
...
1), 130
ascending chain condition (acc): (16
...
1), 17
Artinian: (16
...
11), 118
associated prime: (17
...
8), 49
bimodule homomorphism: (8
...
13), 98; (16
...
1), 60
closed: (4
...
4), 139
Cohen–Macaulay: (23
...
9), 20
cokernel: (4
...
1), 131
composition series: (19
...
7), 19
depth: (23
...
26),
100
dimension: (21
...
15), 21
direct sum: (4
...
15), 22
discrete: (22
...
1), 101
endomorphism: (4
...
8), 49
faithful: (4
...
23), 63; (12
...
5), 55
filtration: (20
...
11), 118
q-filtration: (20
...
11), 118
Hilbert–Samuel Polynomial: (20
...
11), 118
shifting: (20
...
11), 118
topology: (22
...
10), 20
finitely presented: (5
...
5), 55
free: (4
...
10), 20
free of rank ℓ: (4
...
10), 20
graded: (20
...
1), 116
Hilbert Function: (20
...
3), 117
Hilbert Series: (20
...
1), 116
homogeneous component: (20
...
2), 17
image: (4
...
5), 131
invertible: (25
...
2), 17
kernel: (4
...
1), 112
localization: (12
...
2), 72
localizaton at p: (12
...
49), 82
locally finitely presented: (13
...
49), 82
maximal condition (maxc): (16
...
26), 100
minimal generating set: (10
...
1), 101
modulo: (4
...
4), 139
M -sequence: (23
...
27), 143
Noetherian: (16
...
19), 27
projective (5
...
6), 19
quotient map: (4
...
2), 17
radical: (21
...
16), 120
residue: (4
...
5), 18
scalar multiplication: (4
...
2), 124
separated: (22
...
1), 131
Serre’s Condition: (23
...
1), 112
standard basis: (4
...
8), 20
support: (13
...
2), 125
tensor product, see also
torsion: (13
...
36), 81
notation
a + b: (1
...
4), 3
ab: (1
...
5), 49
aN : (4
...
4), 3
aS : (11
...
2), 17
R = R′ : (1
...
15), 22
M n : (4
...
11), 118
Pq (M, t): (20
...
27), 110
G• M : (20
...
15), 21
R ≃ R′ : (1
...
4), 2
∑
aλ : (1
...
11), 118
pq (M, n): (20
...
2), 17
((R-alg)): (6
...
1), 35
((Rings)): (6
...
1), 77
((Sets)): (6
...
1), 77
Ann(M ): (4
...
1), 17
Ass(M ): (17
...
1), 48
b/a: (1
...
9), 20
Coker(α): (4
...
3), 7
⨿
M : (6
...
7), 39
M ⊔ N : (6
...
1), 77
δµλ : (4
...
4), 139
depth(M ): (23
...
1), 124
dim(R): (15
...
6), 38
−→
d(M ): (21
...
10), 21
EndR (M ): (4
...
22), 156
F2 : (1
...
35), 32
Fq : (15
...
3), 7
Γa (M ): (18
...
11), 118
G• R: (20
...
3), 117
H(M, t): (20
...
2), 17
Idem(R): (1
...
11), 5
Im(α): (4
...
5), 131
←−
ικ : (4
...
2), 17
k{{X}}: (3
...
, an ⟩: (1
...
1), 112
S −1 R: (11
...
23), 69
L + M : (4
...
1), 116
(M : N ) : (25
...
1), 131
M
M −1 : (25
...
2), 72
Mp : (12
...
6), 19
M N : (25
...
15), 22
M ⊗R N : (8
...
2), 48
µR : (4
...
4), 18
nil(M ): (13
...
22), 14
1A : (6
...
2), 18
P(R) : (25
...
15), 22
(mλ ): (4
...
10), 20
φp : (11
...
2), 72
φf : (11
...
2), 72
φS : (11
...
2), 72
πκ : (4
...
22), 156
Q: (2
...
6), 3
R× : (1
...
12), 5
R[[X1 ,
...
10), 12
R[X1 ,
...
3), 2
R[{Xλ }λ∈Λ ]: (1
...
1), 11
rad(M ): (21
...
10), 20
R: (2
...
12), 67
Rp : (11
...
10), 20
R⊕Λ : (4
...
16), 120
(Rn ): (23
...
6), 117
R(q): (20
...
, xn ]: (4
...
16), 74
S: (3
...
2), 125
S − T : (1
...
15), 141
Spec(R): (13
...
22), 14
⊕
M : (4
...
15), 22
Supp(M ): (13
...
21), 27
△: (1
...
36), 81
tr: (24
...
deg: (15
...
19), 75
V(a): (13
...
10), 149
x/s: (11
...
23), 69
Z: (1
...
div(M ): (17
...
div(R): (2
...
1), 1
absolutely flat: (10
...
26), 100
ascending chain condition (acc): (16
...
11), 118
Boolean: (1
...
19), 9
catenary: (15
...
33), 137
Cohen–Macaulay: (23
...
40), 146
Dedekind domain: (24
...
10), 91
Discrete Valuation Ring (DVR): (23
...
3), 7
dominates: (26
...
1), 84
factor ring: (1
...
10), 12
graded: (20
...
1), 1
Ideal Class Group: (25
...
30), 64
integral domain: (2
...
30), 64
Jacobson: (15
...
1), 11
kernel: (1
...
11), 12
local: (3
...
19), 62
localization: (11
...
12), 67
localizaton at p: (11
...
1), 1
maximal condition (maxc): (16
...
18), 13
modulo: (1
...
1), 96
nonzerodivisor: (2
...
30), 64; (10
...
8),
85; (14
...
22), 142; (23
...
1), 148; (24
...
5),
157
normalization: (10
...
1), 131; (22
...
22), 156
polynomial ring: (1
...
1), 138
Principal Ideal Domain (PID): (2
...
28), 59
product ring: (1
...
5), 7; (4
...
29), 64; (14
...
18),
108; (19
...
6), 3
quotient ring: (1
...
1), 11
257
reduced: (3
...
20), 128
regular system of parameters: (21
...
6), 3
ring of fractions: (11
...
5), 11
Serre’s Conditions: (23
...
1), 77
principal open set: (13
...
20), 79
Zariski topology: (13
...
1), 1
total quotient ring: (11
...
25), 9; (10
...
6),
8; (23
...
12), 154
universally catenary: (23
...
1), 157
sequence
Cauchy: (22
...
1), 24
M -sequence: (23
...
27), 143
regular sequence: (23
...
3), 24
isomorphism of: (5
...
8), 25
submodule: (4
...
13), 107
homogeneous: (20
...
1), 106
primary: (18
...
13), 107
irredundant: (18
...
13), 107
saturated: (12
...
16), 74
subset
characteristic function: (1
...
1), 7
saturated: (3
...
17), 13
symmetric difference: (1
...
2), 125
regular (21
...
2), 48
adjoint associativity: (8
...
10), 50
cancellation law: (8
...
6), 49
unitary law: (8
...
9), 114
258
Index
Akizuki–Hopkins: (19
...
10), 140
Cayley–Hamilton: (10
...
10), 97
Cohen Structure: (22
...
2), 60
Dimension: (21
...
14), 41
Exactness of Localization: (12
...
17), 134
Exactness of Filtered Direct Limits: (7
...
17), 150
First Uniqueness: (18
...
33), 65
Generalized Hilbert Nullstellensatz:
(15
...
11), 86
for integral extensions: (14
...
3), 84
Hilbert Basis: (16
...
7), 90
Hilbert–Serre: (20
...
3), 84
Jordan–H¨
older: (19
...
29), 110; (20
...
10), 127
Lasker–Noether: (18
...
24), 58
Left Exactness of Hom: (5
...
3), 84
Main of Classical Ideal Theory: (24
...
14), 154
Maximality: (14
...
22), 99
Noether on Finiteness of Closure: (24
...
29), 14
Second Uniqueness: (18
...
20), 142
Stone’s: (13
...
27), 64
Watts: (8
...
4), 90; (16
...
2), 78
compact: (13
...
18), 79
irreducible component: (13
...
19), 205
Jacobson: (15
...
23), 93
quasi-compact: (13
...
23), 80
very dense subset: (15
...
1), 130
separated: (22
...
1), 77
totally ordered group: (26
...
14), 160
value group: (26
...
1), 35
Universal Mapping Property (UMP)
coequalizer: (6
...
9), 20
colimit: (6
...
7), 38
direct limit: (6
...
15), 22
direct sum: (4
...
32), 137
fraction field: (2
...
10), 21
inverse limit: (22
...
5), 66; (12
...
3), 2
pushout: (6
...
6), 19
residue ring: (1
...
3), 48
valuation
discrete: (26
...
11), 159
p-adic: (23
Title: A term of commutative algebra
Description: Commutative algebra is a branch of algebra that deals with commutative rings, which are algebraic structures that satisfy the commutative property of multiplication. One important term in commutative algebra is the prime ideal. A prime ideal is a special type of ideal in a commutative ring, which has the property that whenever two elements in the ring multiply to an element in the prime ideal, at least one of the elements must be in the prime ideal. In other words, a prime ideal is an ideal that is "prime" in the sense that it cannot be further factored into smaller ideals. Prime ideals play a crucial role in algebraic geometry, number theory, and other areas of mathematics. They are used to define algebraic varieties, which are geometric objects that can be studied using algebraic techniques. The theory of prime ideals also connects to other important concepts in commutative algebra, such as maximal ideals, local rings, and polynomial rings.
Description: Commutative algebra is a branch of algebra that deals with commutative rings, which are algebraic structures that satisfy the commutative property of multiplication. One important term in commutative algebra is the prime ideal. A prime ideal is a special type of ideal in a commutative ring, which has the property that whenever two elements in the ring multiply to an element in the prime ideal, at least one of the elements must be in the prime ideal. In other words, a prime ideal is an ideal that is "prime" in the sense that it cannot be further factored into smaller ideals. Prime ideals play a crucial role in algebraic geometry, number theory, and other areas of mathematics. They are used to define algebraic varieties, which are geometric objects that can be studied using algebraic techniques. The theory of prime ideals also connects to other important concepts in commutative algebra, such as maximal ideals, local rings, and polynomial rings.