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Math 312 – Applied Modern Algebra
Sets:
Let X and Y be sets
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Pidgeon-Hole Principle:
If f: X → Y is a map of sets, such that #Y < ∞ (i
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Y is a finite set) and #X > #Y, then
f(x) = f(y) for at least one pair of non-equal elements x, y ∈ X
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(2) It has an identity: ∃ e ∈ G such that for all a ∈ G, we have a∗e = e∗a = a
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We may write (G, ∗) to describe the group, or just G if there is no ambiguity about what ∗ is
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The set of invertible n × n matrices is denoted GLn(R), and is called the general linear
group
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In
particular:
(1) It is closed under multiplication (that is, g ∗ h ∈ H for every g, h ∈ H)
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(3) For every h ∈ H, we also have h−1 ∈ H
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All other
subgroups are called proper
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Notice that gH = hH if h−1 g ∈ H
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With this multiplication, G/H is a group,
called the quotient group of G by H
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Then a divides b if there is an integer c such that ac = b
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Let a and b be integers, with a ≠ 0
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(2) An integer d is called the greatest common divisor (gcd) of a and b if it is
maximal in the set of all common divisors of a and b
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(4) The integers a and b are coprime if (a, b) = 1
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The gcd of a and b is divisible by every common divisor of a and b
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The least common multiple (lcm) of a and b is the minimal integer m in the set of all
common multiples
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Let a and b be nonzero integers
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Apply the division theorem
recursively as follows:
b = aq1 + r1
a = r1 q 2 + r2

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Let d = (a, b), and take r and s such that ar + bs = d
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All integer solutions to the equation ax + by = 0 have the form
b
a
x = d k, y = − d k for some integer k
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Every other solution to ax + by = c has the form
b
a
x = rw + d k, y = sw – d k
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For such a and b, write a ≡ b (mod m)
For b any integer, write r = b mod m for the remainder of the division algorithm
b = qm + r
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Properties of Congruence Relations:
(1) If k is an integer, then a ≡ b (mod m) implies ka ≡ kb (mod m)
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(3) If a ≡ b (mod m) and a’ ≡ b’ (mod m), then aa’ ≡ bb’ (mod m)
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(5) If a ≡ b (mod m) and b ≡ c (mod m), then a ≡ c (mod m)
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(7) If a ≡ b (mod m), then ae ≡ be (mod m) for all e ≥ 0
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In the special case c = 1, the integer x is the multiplicative inverse
to a in Z/bZ
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(1) x + a ≡ b (mod m): Solutions are all integers x such that x ≡ b − a (mod m)
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(2) ax ≡ b (mod m): Has a solution if and only if (a, m)|b
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How many distinct solutions are there in Z/mZ?
There are (a, m)-many
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Fundamental Theorem of Arithmetic:
Every integer > 1factors uniquely (up to reordering) into a product of primes
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pkk and b = p11
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min(e1 ,f1 )
min(ek ,fk )
(2) (a,b) = p1

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We may also drop the subscript
m if it is not likely to cause confusion
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(2) The operation · is associative; that is, for any a, b, c ∈ R, we have
a · (b · c) = (a · b) · c
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Let (R, +, ·) be a ring
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(2) The ring R is unital (or “has a unit” or “has an identity”) if there is an element 1
such that 1 · a = a · 1 = a for every a ∈ R
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(I
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, it has a multiplicative inverse
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(5) A zero divisor in a ring R is a nonzero element a for which there is a nonzero b ∈
R such that a × b = 0
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A unit in R is an element a ∈ R for which there is an element b
∈ R such that ab = ba = 1
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We write b = a−1 for
this element of R
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The set U with
multiplication in R is a group
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It is only a group with respect to
addition
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The conjugacy class [a]m in Z/mZ is a unit if and only if (a, m) = 1
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A nonzero element a in R is a zero divisor if there is a nonzero b ∈ R such
that ab = 0
If R is a unital ring with finitely many elements, then every nonzero element is either a unit
or a zero divisor
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Then if b, c ∈ R are elements for
which ab = ac, then we must have b = c
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An element a of a ring R with identity cannot be both a zero divisor and a unit
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The ring Z/mZ is a field if and only if m is prime

Homomorphisms:
Let R and S be two rings, and f: R → S a function
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(2) f(r · r’) = f(r) · f(r’) for any r, r’ ∈ R
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Properties of Homomorphisms:
Let f : R → S be a homomorphism
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(Or f(a) = 0
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Let f: R → S be a homomorphism
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e
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Let R be a unital ring
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If no such m exists, then the characteristic of R is zero
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Then char R = m
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That is, if and only if f is an injective
homomorphism from Z to R
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For example, if R = Q, then elements of Q are fractions of the form a/b with a, b ∈ Z
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If char R
= 0, then the canonical homomorphism is itself an injection
A homomorphism f is always onto its image
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The Chinese Remainder Theorem:
Let m1 ,
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e
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Then
there is an isomorphism
Z/m1
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x Z/mr Z
The isomorphism is defined by f([n]m1 ,
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, [n]mr )
The Chinese Remainder Theorem can be phrased in terms of a system of congruences
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mr Z → Z/m1 Z x Z/m2 Z x
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x ≡ ar (mod mr )
The following theorem is the Chinese Remainder Theorem stated in terms of a system of
congruences
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, mr be mutually coprime integers > 1, and a1 ,
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Then there
exists a solution to the system of congruences x ≡ ai (mod mi ), and if x0 is a solution, the
set of all solutions is equivalent to the set of all x such that x ≡ x0 (mod m1 · · · mr )
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If x0 is a solution, any other solution x is congruent to x0 modulo [m, n], the least
common multiple of m and n
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Polynomials:
Let R be a commutative ring with unit
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+ an x n = ∑n ak x k where x is an indeterminate and n is an
k=0
integer ≥ 0
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This is given the
structure of a ring with the following operations
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We assume m ≤ n and if m < n, augment q(x) with monomials bk x k where
k=0
bk = 0 for m + 1 < k ≤ n
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p(x) + q(x) = ∑n (ak + bk ) x k
k=0
p(x)q(x) = ∑n+m ai bk−i x k
k=0
Two polynomials are equal if and only if all of their coefficients are equal
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The degree of a polynomial p(x) is n if n is the highest power of x appearing in p(x) with
nonzero coefficient
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The leading coefficient is the
coefficient of the monomial of top degree in p(x)
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We can think of polynomials as functions from R to R, where if p(x) ∈ R[x] and α ∈ R, then
p(α) is again in R
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The map of evaluation at α is a surjective ring homomorphism from R[x] to R
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The map ψ above is a ring homomorphism from R[x] to S[x]
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Let m be an integer > 1 and γm : Z → Z/mZ the homomorphism of reduction modulo m
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If p(x) and q(x) are nonzero polynomials in R[x] and if the leading coefficient of p(x) is not
a zero divisor in R, then deg(p(x)q(x)) = deg(p(x)) + deg(q(x))
If p(x), q(x) ∈ R[x], then deg(p(x)+q(x)) ≤ max{deg(p(x)), deg(q(x))}

Division Theorem:
Let f(x), g(x) ∈ R[x] with f(x) ≠ 0 and with leading coefficient a unit in R
(commutative unital ring)
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A common divisor of f(x) and g(x) is a polynomial p(x) for
which f(x) = p(x)q1 (x) and g(x) = p(x)q 2 (x) for some q1 (x), q 2 (x)∈ F[x]
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The polynomials f(x) and g(x) are
coprime if their gcd is 1
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Two
polynomials which differ by a scalar multiple are called associates
A polynomial f(x) in F[x] of positive degree is irreducible if it has no divisors of degree > 0
and < deg f(x)
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rn−2 (x) = rn−1 (x)qn (x) + rn (x) where rn (x) is the last nonzero remainder
rn (x) is a gcd of f(x) and g(x)
If r(x)|f(x) and r(x)|g(x) and if k ∈ f(x) is nonzero then k ∈ F[x] is a unit, so kr(x)|f(x) and
kr(x)|g(x)
If r(x) and s(x) are gcds of f(x) and g(x) in F[x] then there is a scale k such that s(x) = kr(x)
The gcd of f(x) and g(x) in F[x] is the monic gcd of Euclid’s Algorithm (monic gcd = (f, g))

Bezout’s Identity:
Let f(x), g(x) ∈ F[x] and let d(x) be any gcd of f and g
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e
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For polynomials p(x) ∈ F[x] of degree 2 or 3, p(x) irreducible if and only if p(x) has no
roots in F
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If f ∈ F[x] has leading coefficient a, then f factors into a
product of irreducible polynomials, or f(x) = ag(x), where g(x) is monic, and therefore
factors uniquely into irreducibles
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There is a scalar k such that q j = kpi
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Fundamental Theorem of Algebra:
In ℂ[x], every polynomial p(x) with deg ≥ 1 has a root in ℂ[x]
i
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if p(x) has deg > 1 then it is not irreducible (in ℂ[x])
p(x) = (x – a)q(x)
Every polynomial in ℂ[x] of degree ≥ 1 factors completely into linear factors
No polynomial f(x) ∈ ℝ[x] of deg > 2 is irreducible (in ℝ[x])
If α ∈ ℂ has nonzero imaginary part, then (x − α)(x − α) is irreducible in R[x]
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̅
The irreducibility tests will be on Z[x], and we first show that we are able to determine
irreducibility of f(x) ∈ Q[x] by proving an associate in Z[x] is irreducible
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Every f(x) in ℚ [x] has a unique primitive associate
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If f(x) = p(x)q(x) with p(x), q(x) ∈ ℚ [x], then there
are polynomials p1 (x), q1 (x) in Z[x] which are associates of p(x) and q(x),
respectively, and such that f(x) = p1 (x)q1 (x)
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Conversely, every polynomial in ℚ[x] factors in ℚ[x] if its associates in Z[x]
factor
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Reduction mod m
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By abuse of notation, also use γm for the induced homomorphism from Z[x] to
Z/mZ[x]
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The contrapositive (which is logically equivalent) is “If f(x) ∈ Z[x] is reducible, then for
every m ≥ 2, the polynomial γm (f) is reducible
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It is a very common mistake to
confuse these two statements, so be careful!
For all a, b ∈ Z and primes p, the polynomial x 4 + ax 2 + b2 factors modulo p
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Eisenstein’s Irreducibility Criterion:
If f(x) ∈ Z[x] is a polynomial for which there exists a prime p which divides all
coefficients of f except the leading coefficient, and p2 does not divide the constant
term, then f is irreducible
Quotient Rings:
Let F be a field and m(x) ∈ F[x]
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The elements of
F[x]/(m(x)) are congruence classes of polynomials in F[x] where f ∈ F[x] is congruent to g
∈ F[x] if and only if m|f − g
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A complete set of representatives for F[x]/(m(x)) is a set consisting of exactly one
polynomial in F[x] from each congruence class
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, md ∈ F[x] be pairwise coprime polynomials of positive degree
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Consequently, there is an induced isomorphism of rings: F[x]/( m1 · m2 · ·
· md )→ F[x]/(m1 ) × F[x]/(m2 ) × · · · × F[x]/(md )
Let m1 ,
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The homomorphism F[x] → F[x]/(m1 ) × F[x]/(m2 ) × · · · × F[x]/(md ) defined by
sending f ∈ F[x] to its congruence classes modulo mi has kernel (M(x))
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, md ∈ F[x] be pairwise coprime polynomials, and a1 ,
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There exists an f(x) ∈ F[x] such that f ≡ ai (mod mi) for each i
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Fields:
Let F and K be fields throughout this section
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Let F be a subfield of K
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(2) Let α be in K, and let φα : F[x] → K denote the map of evaluation at α
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If
φα is injective, then α is transcendental
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Such pα (x) is unique, irreducible, and if F(α)
denotes the image of φα in L, then F(α) F[x]/( pα (x))
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Cauchy-Kronecker-Steinitz Theorem:
If m(x) is an irreducible polynomial in F[x], then there exists a field extension of F
containing a root of m(x)
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Splitting Field Theorem:
A splitting field exists for any f(x) ∈ F[x] of positive degree
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If F is a field with q elements and m(x) ∈ F[x] is of degree d, then F[x]/(m(x)) has qd
elements
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(2) Every finite field has pd elements for some prime p and positive integer d
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(4) Any two fields with pd elements are isomorphic
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