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Title: algebra class notes
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1
ADVANCED ABSTRACT ALGEBRA
SYLLABUS
Unit I :
Groups – A counting principle – Normal subgroups and Quotient groups –
homomorphism – isomorphism – Cayley’s theorem – permutation groups
...
6-2
...
[Sections 3
...
13]
Unit III :
Rings – homomorphism – Ideals and quotient rings – Field of quotients of an integral
domain – Polynomial rings – Polynomial rings over rational field
...
4-3
...
1,4
...
1,6
...
3&6
...
Unit V:
Extension fields – roots of polynomials – more about roots
...
1,5
...
5]
Text :
Topics in Algebra (Second Edition) By I
...
Herstein – Willey Indian Edition
...
UNIT II
2
UNIT I
Groups
Definition 1
...
a, b ∈ G ⇒ a · b ∈ G (closure axiom),
2
...
there exists an element e ∈ G such that a · e = e · a = a, ∀a ∈ G
(Existence of identity),
4
...
Definition 1
...
Remark 1
...
Example 1
...
Define a · b = a + b
...
i
...
, (Z, +) is an abilian group
...
5 Let G = {1, −1}
...
Here G is an
abelian group of order 2
...
6 Let S3 = {x1, x2, x3}, consider
...
φ·ψ =
x1
x3
...
, ψ·φ =
x1
x1
Σ
...
x3 and ψ2 = ψ·ψ =
x2
x1
x3
Σ
x2
x1
G = {e, φ, ψ, φ · ψ, ψ · φ, ψ2} = S3
...
O(S3) = 3! = 6
...
7 Let S be a non-empty the set having finite number of ele- ments then
A(s), the set of all permutations of S (i
...
the set of all 1 − 1, onto functions from S
onto itself)
...
Example 1
...
G={
a
b
Σ
c d |a, b, c, d ∈ R and ad − bc ƒ= 0}
Then G is an infinite non-abelian group under matrix multiplication
...
9 If G is a group, then
1
...
every element a ∈ G has unique inverse in G,
3
...
for all a, b ∈ G, (ab)−1 = b−1 · a−1
...
10 subgroup: A non-empty subset H of a group G is said to be
subgroup of G, if under the product G, H itself form a group
...
11
1
...
(3Z, +) is a subgroup of (Z, +),
3
...
H={1,-1} is a subgroup of G={1,-1,i,-i} under usual multiplication
...
12 If H is a subgroup of G, and G is a subgroup of K then H
is a subgroup of K
...
13 A non-empty subset H of the group G is a subgroup of G iff
1
...
a ∈ H ⇒ a−1 ∈ H
Lemma 1
...
Example 1
...
Then A(s) is a group under composition
of mapping
...
Let H(x0) = {φ ∈ A(s)|φ(x0) = x0}
...
2
...
16 S = {x1, x2, x3} : A(s) = s3 [∵ H(x 1) = {e, ψ
...
ψ, ψ2}
Here, H(x1), H(x2) and H(x3) are subgroups of S3
H(x 2) = {e, φ
...
17 H(x1) ∩ H(x2) = H(x2) ∩ H(x3) = H(x3) ∩ H(x1) = {e}
...
18 Cyclic Group: Let G be any group, a ∈ G
...
a−2, a−1, a0, a1, a2
...
If (a) = G for some a ∈ G then G is said to be a cyclic group
...
19 Consider G = {1, −1, i, −i}, let a = i
...
Example 1
...
Then H is a subgroup of G
...
21 Let
...
H ={
a
b
0 d
Σ
|a, b, d ∈ Z}
...
Example 1
...
e
...
Let H =
{a + ib|a2 + b2 = 1} = {z ∈ C| |z| = 1}
...
Definition 1
...
Lemma 1
...
Definition 1
...
Ha is called a right cosets of H in G
...
26 Let G = {J12, ⊕}, H = {0, 4, 8}
...
Lemma 1
...
28 There is a 1 − 1 correspondence between any two right cosets of H in G
...
29 Lagrange’s Theorem: If G is a finite group and H is a subgroup of G,
then O(H) is the divisor of O(G), converse of the Lagrange’s theorem need not be
true
...
30
1
...
Then O(H)/O(G)
but H is not a subgroup of G
...
Let G = S3 = {e, p1, p2, p3, p4, p5}, H = {p − 1, p2}
...
Definition 1
...
It is denoted by iG(H)
...
32 iG(H) = O(G)
O(H)
Example 1
...
Then iG(H) = 4 = 12/3 = O(G)
O(H)
Definition
1
...
The order of a (period of a) is the least
positive integer m such that am = e
...
Example 1
...
a = −1 ⇒ a2 = (−1)2 = 1 ⇒ O(a) = 2
2
...
Example 1
...
Example 1
...
Then 1 ∈ Z is of infinite order
...
38 If G is
a finite group and a ∈ G, then O(a) divides O(G)
...
39 If G is finite and
a ∈ G, then aO(G) = e
...
40 Euler function φ(n): φ(1) = 1, φ(n) = number of posi- tive integers
less than n and relatively prime to n for n > 1
...
6
2
...
41 If n is a positive integer and a is relatively prime to n
((a, n) = 1), then aφ(n)≡ 1(mod n)
...
42 (i) If P is a finite prime number and a is any integer than
ap ≡ a(mod p)
(ii) If G is a finite group of prime order then G is cyclic
...
Let
HK = {hk/h ∈ H, K ∈ k}
...
43 Consider the group G = S3 = {e, φ, ψ, φ · ψ, ψ · φ, ψ2}
...
Then HK = {e · e, e(φ · ψ), φ · e, φ · (φ · ψ)} =
{e, φ · ψ, φ, ψ}
...
Because, it is not closed under
(·)
...
e
...
Since O(HK) doesnot divides O(G), by
Lagrange’s Theorem, HK need not be a subgroup of G
...
44 HK is a subgroup of G iff HK = KH
...
45 If H and K are subgroups of an abelian group G, then HK
is a subgroup of G
...
46 If H and K are finite subgroups of orders O(H) and O(K)
then O(HK) = O(H) O(K)
...
47 Supp√
ose H and K are the sub√
group of a group G and order
of H is greater than
O(G) (i
...
) O(H) >
O(G)
...
Normal Subgroups and Quotient groups:
Definition 1
...
Result 1
...
Result 1
...
Result 1
...
Theorem 1
...
Let X = Na, Y = Nb ∈ G/N
...
Under this product (·), G/N is a group which is calleda quotient group (or) a
factor group ofG/N
7
Example 1
...
Then G = {N
⊕ 0, N ⊕ 1, N ⊕ 2, N ⊕ 3}
...
54 O(G/N ) = O(G)
O(N )
Homomorphism:
¯ is said to be a
Definition 1
...
Example 1
...
Let G be any group
...
Then φ is a homo- morphism of G into
¯
...
G
2
...
Define φ : G → G by φ(x) = x ∀x ∈ G
...
Let a, b ∈ G ⇒ ab ∈ G, φ(a) = a; φ(b) = b, φ(a · b) = ab = a ·
b = φ(a) · φ(b) ∀a, b ∈ G
...
Let G be a group of all real numbers under addition and let G
zero real numbers with a product being ordinary multiplication of real
¯ = (R−{0}, ·)
...
Then
numbers
...
e
...
Let a, b ∈ G ⇒ a + b ∈ G
...
¯ = {e, φ}
...
Let G = S3 = {e, φ, ψ, φ · ψ, ψ · φ, ψ 2 } and G
i
j
i
¯
G by f (φ · ψ ) = φ
...
Clearly, f is a homomorphism
...
f (xy) = f (φi+r · ψj+s ) = φi+r = φi · φr = f (φi ψ j ) · f (φr
ψs ) = f (x) · f (y)
...
Let G be the group of non-zero real numbers under multiplication
...
e
...
Define φ : G → G
¯ by
G
...
Then φ(xy) = 1 = 1 · 1 = φ(x) · φ(y),
(b) x and y are negative ⇒ φ(x) = −1;φ(y) = −1 ⇒ xy is positive
...
Then φ(xy) = −1 = 1 · −1 = φ(x) · φ(y),
(d) x is negative and y is positive ⇒ φ(x) = −1 and φ(y) = 1 ⇒ xy
is negative
...
2
...
Let G be a group of integers under addition,(i
...
) G = (z, +) and
¯ be the group of integers under addition modulo n
...
e
...
Define φ : G → G
by φ(x)=Remainder of x on division
by n = r(mod n)
...
¯ be the
7
...
group of all real numbers under addition
...
Then φ(xy) = log10xy = log10x + log10y = φ(x) + φ(y)
...
a homomorphism
...
57 Suppose G is a group, N a normal subgroup of G; define a
mapping φ : G → G/N by φ(x) = Nx ∀x ∈ G
...
Proof: Let x, y ∈ G
...
Now, φ(xy) = Nxy = Nx · Ny =
φ(x) · φ(y)
...
Let X ∈ G/N , then X = Nx, x ∈ G
...
Therefore φ is onto
...
Remark 1
...
¯ , the kernal of φ (Kφ ), defined by
Definition 1
...
Clearly Kφ is a subset of G
...
60 1
...
Then Kφ =
{x ∈ G|φ(x) = e} = G
...
Define φ : G → G by φ(x) = x ∀x ∈ G
...
¯ by φ(a) = 2a ∀a ∈ G
...
3
...
Then Kφ = {φ ∈ G|f (φi ψi ) =
4
...
¯ by
5
...
φ(x) =
1
−1
if x is positive
if x is negative
...
¯
6
...
by φ(x) = log10x
...
61 If φ is a homomorphism of G × G
¯ , 2
...
φ(e),the unit element in G
φ(x−1) = (φ(x))−1, ∀x ∈ G
...
Let x ∈ G, φ(x) ∈
¯
...
By (1), we have φ(e) = e¯ ⇒ e¯ = φ(e) = φ(x · x−1 ) = φ(x) · φ(x−1 ) ⇒
φ(x−1) = [φ(x)]−1
Remark 1
...
¯ with kernal K, then
Lemma 1
...
Proof: Let x, y ∈ K
...
Now, φ(xy) = φ(x) · φ(y) =
e¯ · e¯ = e¯ ⇒ xy ∈ K
...
Now,φ(x−1) = [φ(x)]−1 (by (2) of Lemma 1
...
(ii)
...
Let g ∈ G and k ∈ K, φ(k) = e¯
...
Therefore
K is a normal subgroup of G
...
64 If φ is a homomorphism of G onto G
¯
inverse images of g¯ ∈ G under φ in G is group by Kx , where x is any particular
inverse image of g¯ in G
...
(Since k ∈ K, φ(k) = e¯) Now, φ(y) = φ(kx) = φ(k) · φ(x) = e¯· g¯ = g¯ ⇒ y ∈ φ−1 (g¯),
Therefore kx C{φ−1 (g¯)}
...
Suppose that
Z = G ∋: φ9 = (Z) =
g¯ = φ(x) ⇒ φ(Z) = φ(x) ⇒ φ(Z) · [φ(x)]−1 =
e¯ ⇒ φ(Z) φ(x−1 ) =
e¯ ⇒ φ(Z x−1 ) =
e¯ ⇒ Z x−1 ∈ K ⇒ Z ∈ Kx ⇒
{φ−1 (g¯)} C kx
...
¯ be a (function) homomorphism
...
65 Let φ : G → G
φ is 1 − 1
...
Let x, y ∈ G
...
61(1)] ⇒ φ(x) = φ(e) ⇒ x = e [∵ φ is 1− 1]
...
To
prove: φ is 1−1
...
10
2
...
Definition 1
...
Remark 1
...
68 Two groups G, G∗ are said to be isomorphic if there is an
isomorphism of G onto G∗
...
[(i
...
) φ : G → G∗ , φ is 1 − 1,
∼
onto and homomorphism if G = G∗ ]
...
69 Isomorphic ’∼
=’ is an equivalence relation
...
Let iG : G → G define by iG(x) = x ∀x ∈ G, is a identity function on G
...
So G ∼
= G is reflexive
...
Now, let G ∼
= G∗ and f : G → G∗ be an isomorphism
...
⇒ f −1 : G∗ → G is also 1−1 and onto
...
Let f −1(x∗) = x and f −1(y∗) = y
...
f
(xy) = f (x)f (y) = x∗y∗
f −1 (x∗ y ∗ ) = xy = f −1 (x∗ )f −1 (y ∗ )
...
∴ f −1 is 1 − 1,onto and homomorphism
...
3
...
Hence f
and g are bijections and g ◦ f : G → G∗∗ is also a bijection
...
⇒ (g ◦ f )(x) = (g ◦ f )(y)
Now let x, y ∈ G
...
f (x) = y(f is onto)
g ◦ f (x) = g[f (x)]
= g(y) = z
...
}
′ ∼′
∼
∗∗
⇒ G = G ⇒ transitive
...
¯ with kernal Kφ is an iso- morphism
Corollary 1
...
¯ is an isomorphism with kernal Kφ
...
⇒ φ(x) = e¯ = φ(e)
⇒x=e
⇒ Kφ = {e} [Since φ is 1 − 1]
...
Conversely suppose φ : G → G
To prove: φ is an isomorphism
...
Suppose
φ(x) = φ(y)
⇒ φ(x)[φ(y)]−1 = e
⇒ φ(x)[φ(y)−1] = e
⇒ φ(xy−1) = e
⇒ xy−1 = e ⇒ x = y
...
¯ Then
Theorem 1
...
with kernal K
...
Define a function σ : G → G/K by σ(y) = Ky
G|φ(x) = e¯, identity in G
¯ by ψ(Kg) = φ(g)
...
ψ is well defined: Suppose, Kg = Kg′ ⇒ g ∈ Kg′ ⇒ g = kg′, k ∈ K ⇒ φ(g) =
φ(kg′ ) = φ(k) φ(g′ ) = e¯φ(g ′ ) = φ(g′ ) ⇒ ψ(Kg) = ψ(Kg′ )
...
¯
...
ψ is onto: Let g¯ ∈ G
¯ has inverse Kg
such that φ(g) = g¯ ⇒ ψ(Kg) = g¯ (i
...
) Every element g¯ ∈ G
under ψ
...
3
...
Now, ψ(xy) = ψ(Kg
· Kg′ ) = ψ(Kgg′ ) = φ(gg′) (by defn) =φ(g) φ(g′)[∵ φ is homomorphism]=ψ(kg)
ψ(kg′) = ψ(x)ψ(y)
...
4
...
It is enough to prove that Kψ = {e} =
{x ∈ G/K|ψ(x) = e¯} = {x = Kg, g ∈ G|ψ(Kg) = e¯} = {g ∈ G|φ(g) = e¯} = {g ∈
¯ is 1 − 1, onto homomorphism
...
Thus ψ : G/K → G
∼
¯
Hence G/K = G
...
72 The abovetheoremtells that a groupcanbeexpectedto arise fromthe
homomorphic image of the general group must be expressible in the form of G/K
where K is normal in G
...
e
...
Thus there is a 1 − 1 correspondence between
homomorphic images of G and normal subgroups of G
...
73 Simple: A group G is said to be simple if it has no non- trivial
normal subgroups
...
UNIT II
Theorem 1
...
Then
there is an element a =ƒ e ∈ G such that ap = e
...
The
theorem is clearly true for a group having single element
...
case(i)If G
has no subgroup H ƒ= {e}
...
Consider an element a ∈ G, a ƒ= {e}
...
Then H is a subgroup of G and H
ƒ= {e}
...
⇒ G = (a), since G has no improper subgroup
...
Any
cyclic group is isomorphic to (Z, +) or (Zn, ⊕)
...
Claim that n is prime
...
e) n is composite
...
Now the subgroup generated by a, (i
...
(i
...
Therefore our assumption is wrong
...
Hence G is a cyclic group of prime order
...
By a corollary to Lagrange’s theorem ap = aO(G) = e and a ƒ= e
...
[∵ p/O(N ) and N is abelian and O(N ) < O(G) By
induction hypothesis] b ∈ N ⊂ G ⇒ b ∈ G
...
subcase:(b)
Suppose G has a proper subgroup N ƒ= {e} and p does not divides O(N )
...
[∵ subgroup of an
abelian group is normal]
...
[If not,(i
...
(i
...
Therefore by induction hypothesis there exists and element x ∈
G/N, x N such that xp = N
...
By (1), xp = N ⇒
(N b)p = N ⇒ N bp = N [∵ Ha = H ⇔ a ∈ H] ⇒ bp ∈ N and b ∈
/ N
⇒ (bp)O(N) = e [∵ a ∈ G, aO(G) = e] ⇒ bpO(N) = e ⇒ (bO(N))p = e
...
Therefore cp = e, c ∈ N ⊂ G
...
⇒ bO(N) = e ⇒ NbO(N) = Ne = N ⇒ NbO(N) = N ⇒ Nb = N ⇒ b ∈ N
⇒⇐ to b ∈
/ N
...
Thus there exists an element c ∈ G, c ƒ= e
such that cp = e
...
13
Theorem 1
...
Then G has a subgroup of order pα
...
Suppose
α ƒ= 0
...
Then p/O(G), and G is abelian
...
Therefore by Cauchy’s
theorem for finite abelian group there exists an element a ƒ= esatisfying ap = e
...
∵ e ∈ G and ep = e, e ∈ S
...
a ƒ= e, ap = e ⇒ ap = e ⇒a ∈ S ⇒ S ƒ= {e}
...
Now (xy)p =
mn
mn
n
m
x p · yp
= (xp )m · (y p )n = em · en = e ⇒ xy ∈ S
...
Hence, the Claim(i)
...
Therefore by Cauchy’s theorem, there exists an
element c ∈ S, c ƒ= e such that cq = e
...
Also, (pn, q) = 1
...
Now, c = c1 = cλp +µq = cλp · cµq
= (c ) · (c ) =
eλ · eµ = e · e ⇒ c = e ⇒⇐ to the fact that c ƒ= e
...
∵ S is a subgroup of G, by Lagranges
theorem O(S)/O(G)
...
Suppose, β < α
...
Since
β < α, and O(G/S) = O(G) , p/O(G/S) [∵
O(S)
pα/O(G) ⇒ O(G) = kpα
O(S) = pβ, β < α, α − β > 0, α − β = γ > 0 O(G/S) =
O(G)/O(S) = kpα/pβ = kpα−β = kpγ p/kpγ ⇒
p/O(G/S)]
∵ S is a normal subgroup of G and G is abelian, G/S is a group
...
Therefore our assumption is
wrong
...
e) β < α
...
∴ O(S) = pβ = pα
...
76 If G is an abelian group of O(G) and pα/O(G), pα+1 does not divides
O(G) then there is a unique subgroup of G of order pα
...
Now O(ST ) = O(S)O(T )/O(S ∩
T ) = pαpα/O(S ∩ T ) = p2α/O(S ∩ T )
...
UNIT II
14
O(S) = pα
...
e) O(S ∩ T ) < pα
...
∵ G is abelian, ST =
TS
...
(i
...
This contradition
shows that S = T
...
For
Lemma 1
...
Then
H
(i) H is a subgroup of G
(ii) H contains K and
¯ is normal in G
¯ then H is normal in G
...
¯
Proof: Let x, y ∈ H
...
Now, φ(xy) = φ(x) · φ(y) ∈ H
¯ is a subgroup of G
¯ ]
...
(i)
[∵ H
−1
By definition, xy ∈ H
...
Therefore x ∈ H (by definition)
...
By (i) and
(ii) , H is a subgroup of G
To prove: H ⊃ K
¯ ] ⇒ φ(x) = e¯ ∈ H
¯ [∵
is a subgroup of G
¯ [∵ H
¯
Let x ∈ K ⇒ φ(x) = e¯ ∈ G
¯ ] ⇒ x ∈ H (by definition) ∴ K ⊂ H
identity is unique in G
To prove: H is normal in G
¯ is normal in G
¯
...
Therefore φ(g) ∈ G
¯
Given:H
¯
and φ(h) ∈ H
¯
(by definition) and φ is onto map
...
(i
...
Therefore H is
φ(g)φ(h)φ(g−1 ) ∈ H
¯
¯ , we have a subgroup of H of G
normal G
...
Conversely, suppose that L is a subgroup of G and L ⊃ K
...
Claim that ¯L is a subgroup of G
¯
...
ThereG
fore x¯ = φ(x), y¯ = φ(y) ⇒ x¯ · y¯ = φ(x) · φ(y), x, y ∈ L = φ(x · y), x, y ∈ L ⇒
x¯ · y¯ ∈ ¯L (by definition) and (x¯)−1 = [φ(x)]−1 , x ∈ L = φ(x−1 ), x−1 ∈ L
...
Define: T = {x ∈ G|φ(x) ∈ L¯ }
...
L = T
...
(iii)
Conversely, let t ∈ T , φ(t) ∈ ¯L ⇒ φ(t) = φ(e), e ∈ L
...
Therefore te−1 ∈ K ⊂ L ⇒ te−1 ∈ L ⇒ t ∈ Le ⇒ t
∈ L
...
¯
Theorem 1
...
¯ be a normal subgroup of G
¯ , N = {x ∈ G|φ(x) ∈ N
¯}
...
Equivalently G/N ∼
isomorphic to G
= (G/K)/(N/K)
¯
¯
Proof: Let φ : G → G
is a homomorphism
...
Now, define ψ : G → G
¯ /N
¯
θ(g¯) = N
¯
by ψ(g) = θ · φ(g)
...
e) ψ(g) = Nφ(g)
...
ψ is a homomorphism:
¯
...
Now, ψ(a ·
Suppose a, b ∈ G
N
N
¯φ(ab) = N
¯[φ(a) · φ(b)] = N
¯φ(a) · N
¯φ(b) = ψ(a) · ψ(b)
...
Let T be the kernal of ψ
...
e) T = {g ∈ G|ψ(g) = N
¯ /N
¯
...
Let n ∈ N ⇒ φ(n) ∈
N
¯}] ⇒
N
¯φ(n) = N
¯
N
∴ N ⊂ T
...
¯ · φ(t) = N
¯
⇒N
¯ ⇒ t ∈ N [by the
⇒ φ(t) ∈ N
¯]
...
(2)
definition of N
¯ /N
¯ with kernal
From (1) and (2) T = N
...
By the fundamental theorem of homomorphism, G/N = G/N
...
∵ φ : G onto G
¯ is a homomorphism with kernal K
...
Theorem 1
...
(i
...
) set if all bijection on S
...
Define Tg : S(G) → S(G) by
xTg = xg
...
Tg is onto:
For y ∈ S, choose x = yg−1
...
(i
...
) every element y
∈ G has pre-image yg−1 ∈ G under Tg
...
Suppose xTg = yTg ⇒ xg = yg ⇒ x = y (By RCL) ⇒ Tg is 1 − 1
...
Now, consider the map ψ : G → A(S),defined by ψ(g) = Tg
...
∴ Tg · Th = Tgh ⇒ ψ(gh)
= ψ(g) · ψ(h)
...
Let K be the kernal of ψ , then K = {g
∈ G|ψ(g) = e¯, e¯ is the identity in A(S)} = {g ∈ G|Tg = I, Identity function S → S}
= {g ∈ G|xTg = x,
16
2
...
∴ ψ is an isomorphism of G into S
...
64] A homomorphism φ : G → G
¯ ⇔ Kφ = {e}
...
80 If G is a group, H is a subgroup of G, and S is the set of all right cosets
of H in G, then there is a homomorphism θ of G into A(S), and the kernal of θ is the
largest normal subgroup of G,which is contained in H
...
Let
H be a subgroup of G and S = {Hg|g ∈ G}
...
Clearly, tg is well defined
...
Consider, x = Hyg−1 ∈ S
...
∴ tg is onto
...
Suppose (Hx)tg = (Hy)tg ⇒ Hxg = Hyg ⇒ Hx = Hy
...
∴ tg is 1 − 1 and onto, ⇒ t gn ∈ A(S)
...
Clearly θ is well defined
...
This is true for every Hx ∈ S
...
∴ θ is a homomorphism
...
K = {g ∈ G|θ(g) = e¯, e¯ is
identity in A(S)}={g ∈ G|tg = I, I : S → S is identity} ={g ∈ G|(Hx)tg = (Hx)I, I : S
→ S is identity} ={g ∈ G|Hxg = HxI, I : S → S is identity} ={g ∈ G|Hxg = Hx}
...
∴ kernal of a ho- momorphism
is a normal subgroup of G
...
∴ Hxb = Hx [by(1)] ∀x ∈ G
...
⇒ K ⊂ H
...
To prove: K is the largest normal subgroup of G
...
Let n ∈ N
...
∴ K is the largest normal subgroup of G contained in H
...
81 If G is a finite group and H ƒ= G is subgroup of G such thatO(G) does
not divides i(H)!
...
In
particular G cannot be simple
...
∴ O(S) = i(H), index of H
...
If O(G) does not divides i(H)! = O(A(S)) ⇒ O(G) does not divides
O(A(S))
...
Hence no subgroup is isomorphic to G
...
Hence, θ(G) cannot be
17
isomorphic to G
...
e
...
(i
...
) Here G has a non-trivial normal subgroup
kernal of θ
...
Example 1
...
Then prove
that, H contains a normal subgroup of order 3
...
∴ By above theorem, H contains non-trivial normal subgroup
N
...
e
...
∴ By Lagrange’s Theorem, O(N )/O(H)
...
If
O(N ) = 1, then N = {e} ⇒⇐ N ƒ= {e}
...
∴ O(N ) = 3
...
83 Suppose G is a group of order 99 and H is a subgroup of G
order 11
...
99
Proof: O(G) = 99; O(H) = 11; i(H) = O(G)
= 9; i(H)! = 9!
...
∴ O(N ) = 11 = O(H) ⇒ N = H
...
Permutation Group:
Suppose Sn is a finite set, having n elements, S = {x1, x2,
...
Then
the set of all 1 − 1 mapping of S onto itself, written as A(S) = Sn
...
84 Let S be a set and θ ∈ A(S)
...
[i can be +ve, −ve or zero]
Result 1
...
Proof: (i) ≡ θ reflexive: a = aθ0 = ae ⇒ a ≡ θa ∀a ∈ S
...
(ii) ≡ θ is symmetric: Suppose a ≡ θb then b = aθi for some integer i ⇒
a = bθ−i ⇒ b ≡ θa
...
(iii) ≡ θ is transitive: Suppose a ≡ θb and b ≡ θc ⇒ b = aθi and c = bθj for some
integer i and j
...
∴≡ θ istransitive
...
Let S be a set and θ ∈ A(S)
...
≡ θ is an equivalence relation which induces a decomposition of S into
disjoint subsets namely the equivalence classes
...
If S is a finite set and s ∈ S, there is a smallest positive integer l = l(s),
2
...
The orbit of s under θ then consists of an
element {s, sθ, sθ2, sθ3,
...
A cycle of θ is the ordered set of
{s, sθ, sθ2, sθ3,
...
Example 1
...
...
Σ
Σ
...
θ · ψ=
1 2 3 4
3 1 2 4
1234
1 3 2 4
...
=
Σ
1 2 3 4
2 1 3 4
Σ
1 2 3 4
2 3 1 4
Example 1
...
θ=
1 2 3 4 56
2 1 3 5 64
Solution: S = {1, 2, 3, 4, 5, 6}
Orbit of 1:
1θ0, 1θ1, 1θ2, 1θ3,
...
2θ0 = 2
2θ1 = 2· θ = 1
2θ2 = (2 · θ) · θ = 1 · θ = 2
2θ3 = (2 · θ2) · θ = 2 · θ = 1
∴ Orbit of 2 consists of the element {1, 2}
Orbit of 3:
3θ0, 3θ1, 3θ2,
...
4θ0 = 4
4θ1 = 4· θ = 5
4θ2 = (4θ) · θ = 5 · θ = 6
4θ3 = (4θ2) · θ = 6 · θ = 4
4θ4 = (4θ3) · θ = 4 · θ = 5
∴ Orbit of 4 consists of the element {4, 5, 6}
Orbit of 5:
5θ0, 5θ1, 5θ2,
...
6θ0 = 6
6θ1 = 6· θ = 4
6θ2 = (6θ) · θ = 4 · θ = 5
6θ3 = (6θ2) · θ = 5 · θ = 6
∴ Orbit of 6 consists of the element {4, 5, 6}
Cycles are the ordered set of orbits
∴ Cycles is (1 2) (3) (4 5 6)
Example 1
...
θ=
1 2 3 4 5 6 7 8 9
2 3 8 1 6 4 7 5 9
Solution: Orbit of 1: 1θ0, 1θ1, 1θ2 1θ0
=1
1θ1 = 1θ = 2
1θ2 = (1θ)θ = 2 · θ = 3
Orbit of 1 = {1, 2, 3, 8, 5, 6, 4}
Orbit of 2 = {2, 3, 8, 5, 6, 4, 1}
Orbit of 3 = {3, 8, 5, 6, 4, 1, 2}
Orbit of 4 = {4, 1, 2, 3, 8, 5, 6}
Orbit of 5 = {5, 6, 4, 1, 2, 3, 8}
Orbit of 6 = {6, 4, 1, 2, 3, 8, 5}
Orbit of 7 = {7}
Orbit of 8 = {8, 5, 6, 4, 1, 2, 3}
Orbit of 9 = {9}
Σ
2
...
C=
1 2 3 4 5 6 7 8 9
2 3 8 1 6 4 7 5 9
Σ
Theorem 1
...
Proof: Let θ be a permutation on the set S
...
Let ψ be the product of all distinct cycles of θ
...
So, θ,
ψ have the same effect on every element of S
...
Example 1
...
θ=
Σ
1 2 3 4 5 6 7 8 9
3 7 5 2 1 8 4 6 9
The cycles are of the form
θ1 = (1 θ θ2) = (1 3 5); θ2 = (2 2θ 2θ2) = (2 7 4); θ3 = (6 8); θ4 = (9)
ψ = θ1 · θ2 · θ3 · θ4 = (1 3 5) (2 7 4) (6 8) (9)
Lemma 1
...
Proof: Let θ be the permutation on S = {a1, a2,
...
By above lemma, θ can be
written as the product of its cycles
...
, am) be any cycle θ of length
m(m < n)
...
, am) = (a1, a2) (a1, a3)
...
∴
An m-cycle can be written as the product of 2-cycles
...
Since every permutation is the product
of its disjoint cycles and every cycle is a product of 2-cycles, it follows that every
permutation is a product of 2-cycles
...
92 We shall refer to 2-cycles as transpositions
...
93 A permutation θ ∈ Sn is said to be an even permutation if it can be
represented as a product of an even number of transpositions and is said to be an
odd permutation if it can be represented as a product of an odd number of
transpositions
...
94
...
1 2 3 4 5
2 1 3 5 4
= (1 2) (3) (4 5) = odd permutation
Result 1
...
Proof: Let θ ∈ Sn
Suppose θ can be written as a product of X transpositions in one way and can be
written as a product of Y transpositions in another way
...
, xn which are the elements of S
...
, xn) =
(xi − xj)
...
, n
...
, xn) by
θ : P (x1, x2,
...
i
It is clear that θ : P (x1, x2,
...
, xn)
...
Then P (x1, x2,
...
, x5)) = (x3 − x5)(x3 − x4)(x3
− x1)(x3 − x2)(x5 − x4)(x5 − x1)(x5 − x2)(x4 − x1)(x4 − x2)(x1 − x2) = −[(x1 − x2)(x1 −
x3)(x1 − x4)(x1 − x5)(x2−x3)(x2−x4)(x2−x5)(x3−x4)(x3−x5)(x4−x5)] = −P (x1, x2,
...
Suppose θ = (1, 2) ∈ S2; P (x1, x2) = (x1 − x2); θ(P (x1, x2)) = (x2 − x1) =
−(x1−x 2) = −P (x1, x2)
...
e)The effect of a transposition on P is to change the sign of
P
...
(i) Any factor of P which contains neither the suffix r nor s remains un- changed
(ii) The single factor (xr − xs) changes its sign by replacing r by s and s by
r
(iii) The remaining factor which contain either the suffix r (or) s but not both can
be grouped into the following 3 types of products
...
[(xr−1 − xr)(xr−1 − xs)]
(b) [(xr − xr+1)(xr+1 − xs)][(xr − xr+2)(xr+2 − xs)]
...
[(xr − xn)(xs − xn)] On replacing
r by s and s by r, the signs of all types of products do not change
...
UNIT II
22
P
...
e) P operated upon by a transposition becomes −P
...
(−1)]P
(x times)
= (−1)x · P
...
(−1)]P
(y times)
= (−1)y · P
...
Lemma 1
...
Proof: We know that Sn is group
...
An is a subgroup ofSn:
Let θ1, θ2 ∈ An
...
⇒ θ1 · θ2 is an even
permutation
...
∴
An is a subgroup of Sn
...
Define ψ : Sn → W by
...
Let s, t ∈ Sn
...
∴ ψ(st) = 1 = 1 · 1 = ψ(s) · ψ(t) Case(ii)
If s, t are odd, then st is even
...
∴ ψ(st) = −1 = −1 × 1 = ψ(s) · ψ(t)
Case(iv) Le t s is even and t is odd
...
e) ψ(s) = 1 and ψ(t) = −1 ⇒ st is odd
...
Clearly ψ is onto
...
kernal ψ = {s ∈ Sn|ψ(s) = identity in W } = {s
∈ Sn|ψ(s) = 1} = An
...
∴ By Lemma 1
...
By Theorem 1
...
⇒ O(W ) = O( ) ⇒ O( Sn ) = 2 [∵
An
O(G/H) = O(G) = iG(H)] ⇒ is (An) = 2
...
O(H)
n
An
2
2
23
2
...
1 If a, b ∈ G, then b is said to be a conjugate of a in G if there exists an
element c ∈ G such that b = c−1ac
...
(i
...
) a ∼ b ⇒ b is conjugate to a ⇒ b = c−1ac, c ∈ G
...
2 Conjugation is an equivalence relation on G
...
∴∼ is reflexive
...
⇒ a = c b c−1 = (c−1)−1b(c−1) =
x−1bx, x = c−1 ∈ G ⇒ b ∼ a
...
(iii)∼ is
transitive:
Suppose a ∼ b and b ∼ c
...
Now, c = y by = y (x ax)y = (y x )a(xy) = (xy) a(xy) = z az, z = xy ∈ G
⇒ a ∼ c
...
Hence, ∼ is an equivalence relation
...
3 For any a ∈ G, let C(a) = {x ∈ G|x ∼ a}, C(a) is the equivalence class
a in G, under the relation ∼
...
4 C(a) = {x ∈ G|x ∼ a} = {x ∈ G|x ∼ a} = {x ∈ G|x = y−1ay, y ∈ G} =
{y−1ay|y ∈ G}
...
Suppose the number of elements in C(a) is denoted by Ca
...
∪ C(an)
Σ
O(G) = Ca1 + Ca2 +
...
Definition 2
...
6 (i) G = {1, −1, i, −i}
...
(ii) G = {Z5, ⊕5}
...
N (φ) = {e, φ}; N (ψ) = {e, ψ, ψ 2}; N (ψ2) =
{e, ψ2, ψ}
...
UNIT II
24
Lemma 2
...
Proof: Let x, y ∈ N (a) ⇒ ax = xa and ay = ya
...
(2)
Suppose x ∈ N (a) ⇒ ax = xa ⇒ x−1a = ax−1 [By premultiply and post multiply
by x−1] ⇒ x−1 ∈ N (a)
...
Calculation for C(a):
Let G = S3 = {e, φ, ψ, φ · ψ, ψ · φ, ψ2}
...
∴ C(φ) = {φ, φ · ψ, ψ · φ}
C
= O(C(1, 2)) = 3
...
∴ C
= O(S3)
...
8 If G is a finite group, then Ca = O(G) ; In other wards, the number
O(N (a))
of elements conjugate to a in G is the index of N (a) in G
...
In this way we shall have a 1 − 1
correspondence between conjugate of a in G and the right cosets of N (a) in G
...
Then y = nx where n
∈ N (a), [∵ y ∈ N (a)· x, y = nx] ⇒ y−1 = (nx)−1 = x−1n−1; y−1ay = x−1n−1ay =
x−1n−1anx = x−1(n−1an)x = x−1ax = x−1ax
...
In other wards if x and y are in different right cosets of N
(a) in G
...
Suppose not x−1ax = y−1ay
...
∴ x−1ax ƒ= y−1ay
...
∴ The number of elements conjugate to
a in G = number of distinct right cosets of N (a) in G
...
e
...
(i
...
) Ca = O(G)
...
O(N (a))
Corollary 2
...
This is
known as class equation of G
...
10 State and Prove class equation of a group G
...
Definition 2
...
Example 2
...
(ii) G = S3 and Z(G) = Z(S3) = {e}
...
13 a ∈ Z(G) ⇔ N (a) = G; if G is finite, a ∈ Z(G) ⇔
O(N (a)) = O(G)
Proof: Suppose a ∈ Z(G) ⇒ ax = xa ∀x ∈ G ⇒ x ∈ N (a) ∀x ∈ G ⇒ G ⊆
N (a)
...
∴ N (a) = G
...
If
G is finite, and a ∈ Z ⇔ N (a) = G
⇔ O(N (a)) = O(G)
...
14 Application-1: If O(G) = pn,where p is a prime number then centre
of G, Z(G) ƒ= {e}
...
Let O(N (a))/pnα, nα ≤ n
...
By class equation, O(G) =
, where
theO(N
sum
runs
over
the
set
of
elements
a
∈
G
...
O(G) =
Σ
a∈Z(G)
O(G)
O(N (a)) +
Σ
=
O(N (a))=O(G)
=
ΣO(G)
a∈Z
O(G)
Σ
=
1+
a∈G
Σ
a∈
/ Z(G)
O(G)
O(N (a)) +
Σ
+
O(N (a))=ƒ O(G)
Σ
= O(Z)+
O(N (a))
O(G)
Σ
O(N (a))ƒ=O(G)
O(G)
O(N (a))ƒ=O(G)
Σ
O(G) O(N
(a))
O(N (a))
O(G)
O(N (a))
O(G)
O(N (a))
O(G) O(N
(a))
2
...
Σ
O(G) =Z +
O(G)
O(N (a))
nƒ=nα
Σ
pn
pnα
=Z+
nƒ=nα
Σ
=Z+
pn−nα
n nα
Σ
∴ pn = Z +
pk, k = n − nα > 0
...
∴ O(Z) > 1
...
Hence, Z is non-trivial and the theorem
...
15 If O(G) = p2 where p is a prime number, then G is abelian
...
Since
O(G) = p2, by previous theorem Z(G) ƒ= {e}
...
Case(i): O(Z(G)) ƒ= 1[∵ Z(G) is non-trival]
...
Case(iii) Suppose, O(Z(G)) = p
...
Suppose not,(i
...
) if a ∈ Z ⇒ O(N (a)) = O(G) (by Lemma 2
...
By class
equation,
O(G) =
Σ
O(G)
a∈Z
O(N (a))
Σ
=
O(N (a))=O(G)
O(G)
O(N (a))
Σ
=
1
a∈Z
∴ O(G) = O(Z) ⇒ O(G) = p ⇒⇐ to the fact O(G) = p2
...
Consider the subgroup N (a)
in G, then Z(G) ⊂ N (a) ⊂ G [∵ a ∈ N (a) and a ∈
/
Z]
...
(2)
By Lagrange’s theorem O(N(a))O(G)
= p2 ⇒ O(N (a))/p2
...
∴ The only possibility of
O(Z(G)) = p2 = O(G) ⇒ Z(G) = G
...
27
Theorem 2
...
Proof: We have to find an element a ƒ= e ∈ G such that it satisfier ap = e
...
The result is clearly true for a group of order
1
...
Now, we have to prove the theorem for G
...
such that p/O(W ) [pdivides the order of any non-trivial subgroup of
G]
...
By induction hypothesis, the theorem is true for W , then
there exists an element a ∈ W, a ƒ= e such that ap = e ⇒ a ƒ= e, a ∈ G,
such thatap = e (∵ a ∈ W ⊂ G)
...
Case(ii): Suppose we assume that p does not divide order of any proper subgroup
of G
...
13)
...
e
...
By assumption p does not divide O(N (a))
...
(1)
Since p/O(G) and p does not divide O(N (a)), p/ O(G)
O(N (a))
⇒
Σ
N (a)ƒ=G
O(G) O(N
(a))
Thus p/O(G) and
Σ
O(G)
O(N
(a))
N (a)ƒ=G
⇒ p/O(Z(G)) [from(1)], where Z(G) is a proper subgroup of G
...
∴ Z(G) cannot be a
proper subgroup of G
...
Thus p/O(G) and G is abelian
...
UNIT II
28
Theorem 2
...
Proof: We prove this theorem by induction on O(G)
...
If O(G) = 2, the theorem only relevant prime number is 2
...
Certainly G has a sub- group of order 2 namely itself
...
Suppose, we assume that the theorem is true for all
groups of order less than O(G)
...
Suppose assume that pm/O(G), pm+1 does not divides O(G), where p is a prime
number, m ≥ 1
...
∵ H is a proper subgroup of G, O(H) < O(G)
...
Since T is a subgroup of H and H is a subgroup
of G, T is a subgroup of G of order pm
...
[(i
...
) pm does not divide any proper subgroup of O(G)]
...
If a ∈
/ Z(G) then N (a) ƒ= G
...
e
...
∴ Byour assumption pm does not divides O(N
(a))
...
Σ
O(G)
⇒ O(G) =
O(N (a))
Σ
Σ
O(G)
O(G)
=
+
a∈Z
O(N (a))
Σ O(G)
=
a∈Z
a∈
/Z
Σ
O(G)
+
O(G)
a∈
/Z
O(N (a))
O(N (a))
[∵ if a ∈ Z ⇒ O(N (a)) = O(G)]
=
Σ
a∈Z
1+
Σ
O(G)
a∈
/Z
O(N (a))
O(G)
Σ
⇒ O(G) = O(Z) +
a∈
/
Z
O(N (a))
...
O(G)
O(N (a))
)
O(G)
O(N (a))
29
Since pm/O(Z) by Cauchy’s theorem, Z has an element b ƒ= e such thatbp = e
...
∴ B is normal in G [∵ Every subgroup of an
abelian group is normal]
...
Now, O(G
¯ ) = O(G/B) = O(G)
G
p
<
O(B) =
¯ ) [∵ pm /O(G) ⇒ O(G) = tpm ,some
O(G)
...
Also, pm does
integer t]
...
∴ By induction hypothesis (G
¯ ) has a subgroup P¯ of order pm−1
...
∴ pm−1 = O(P¯ ) = O(P ) = O (P ) ⇒O(B)
O(P ) = pmp
...
Hence the theorem
...
18 External Direct Product: Let A and B be any two
groups
...
Let x = (a1, b1)/a1 ∈ A, b1 ∈ B; y = (a2, b2)/a2 ∈ A, b2 ∈ B
...
Result 2
...
Proof: (i) (·) is closed: Let x = (a1, b1) ∈ G, y = (a2, b2) ∈ G
...
∴ (·) is closed
(ii) (·) is associative: Let x = (a1, b1) ∈ G, y = (a2, b2) ∈ G and z = (a3, b3) ∈ G
...
∴ (·) isassociative
...
Now, x·e = (a1, b1)·(e 1, e2) = (a1 · e1, b1 · e2) =
(a1, b1) = x and e · x = x ∀x ∈ G
...
(iv) Existence of inverse: let x = (a1, b1) ∈ G; x−1 = (a−1, b−1) ∈ G,where
1
1
a−1 ∈ A, b−1 ∈ B
...
∴ x · x = e
...
1
1
1
∴ G is a group
...
20 Let G1, G2, G3
...
Let G = G1 × G2 ×
G3 ×
...
gn)/g1 ∈ G1, g2 ∈ G2
...
Let x =
(g1, g2
...
g′ ) ∈ G
...
gn) ·
1 2
n
(g′ , g′
...
gng′ )
...
Gn
2
...
Consider G = A × B
¯ = {(b, f ) ∈ G|a ∈
and A¯ = {(a, f ) ∈ G|a ∈ A, f is the identity element in B}; B
¯
¯
B, e is the identity element in A}
...
Define
a map φ : A → A¯ by φ(a) = (a, f ), a ∈ A
...
∴ φ is 1 − 1
...
∴ φ is a homomorphism
...
∴ φ is onto
...
isomorphism of A onto A¯
...
e
...
Similarly Define ψ : B → B
¯
∼
¯
Then ψ is a homomorphism of B onto B
...
To
¯ are normal subgroups of G
...
Now,
1
xnx−1 = (a1, b1)(a, f)(a−1, b−1)
1
= (a1 a, b1 f )(a−1 , b−1 )
1
= (a1 aa−1 , b1 f b−1 ))
1
∈ B
...
−1
¯ is the normal
∴ A¯ is a normal subgroup of G
...
Claim that G =
¯a
¯
...
{(a, b)|a ∈ A, b ∈ B}; g = (a, b) = (a, f )(e, f ) ⇒ g = a
¯¯b, a
¯ = (a, f ) ∈ A¯; ¯b = (e, b) ∈ B
¯
¯
Uniqueness: Let g ∈ G can be written as g =
x¯y¯, x
¯ ∈ A, y¯ ∈ B, x¯ =
(x, f ), x ∈ A, f is the identity element in B and y¯ = (e, y), y ∈ B, e is
the identity element in A
...
∴ (a, b) = (x, y) ⇒ a = x and b = y
...
∴ g ∈ G can be uniquely written as g = a
¯
...
chosen G = A¯B
¯ be a normal subgroup of G and A¯ ∼
Definition 2
...
Then G is called the internal direct product of A¯ and B
¯
∈B
Definition 2
...
, Nn be normal subgroups of G
such that
(i) G = N1N2 · · · Nn
(ii) Given g ∈ G then g = m1m2· · · mn, mi ∈ Ni is a unique representation
...
, Nn
...
23 Suppose G is the internal product of N1, N2,
...
Then for
i ƒ= j, Ni ∩ Nj = {e} and if a ∈ Ni, b ∈ Nj then ab = ba
Proof: Let x ∈ Ni ∩ Nj ⇒ x ∈ Ni and x ∈ Nj
...
(1)
Here e1 = e2 =
...
= en = e ∈ Ni where ei ∈ Ni, i = 1, 2,
...
Similarly x ∈ Nj then, x = e1e2 · · · ei−1eiei+1 · · ·
ej−1xej+1 · · · en
...
= ei−1 = ei+1 =
...
= en = e
...
∴ The two composition [i
...
(1) and (2)] in this form of x must coincide and the
entry from Ni in each must be equal
...
∴ x = e
...
To prove ab = ba, ∀a ∈ Ni, b ∈ Nj , it is enough to
prove that aba−1b−1 ∈ Ni ∩ Nj
...
(i
...
) b ∈ G, a−1 ∈ Ni
...
(1)
Since b ∈ Nj, b−1 ∈ Nj
...
(i
...
) a ∈ G, b−1 ∈ Nj
...
(2)
From (1) and (2), aba−1b−1 ∈ Ni ∩ Nj = {e} ⇒ aba−1b−1 = e ⇒ ab =
ba ∀a ∈ Ni, b ∈ Nj
Remark 2
...
Theorem 2
...
Nn
...
Then G and T are isomorphic
...
Let
x ∈ G
...
Define a
map ψ : T → G by ψ(a1, a2,
...
, n
...
, an), y = (b1, b2,
...
Suppose ψ(x) = ψ(y) ⇒ (a1 · a2 · · ·
an), y = (b1 · b2 · · · bn) ⇒ a1 = b1, a2 = b2, , an =
bn (By the uniqueness of the internal direct products) ⇒ (a1, a2,
...
∴ ψ is 1−1
...
, Nn if x ∈ G, then x = (a1, a2,
...
, an ∈
Nn
...
, an) = a1 · a2 · · · an = x
...
Now, ψ(x y) =
ψ((a1, a2,
...
, bn)) = ψ(a1b1, a2b2,
...
23 aj bj = bj ai for i =ƒ ja1b 1·a2b2···anbn =
a1·a2···an ·b1·b2···bn ] = ψ(a1, a2,
...
, bn)=
ψ(x) · ψ(y)
...
∴ ψ is an isomorphism
...
2
...
UNIT III
Rings
Definition 3
...
a + b ∈ R
2
...
a + (b + c) = (a + b) + c
4
...
There is an element −a in R ∋: a + (−a) = 0
6
...
a · (b · c) = (a · b) · c
8
...
(b + c) · a = b · a + c · a
Example 3
...
Definition 3
...
If a · b = b · a ∀a, b ∈ R then we call R is a
commutative ring
...
4 (J, +, ·) is a commutative ring with unit element
Example 3
...
(2Z, ⊕, ⊙) is a commutative ring with unit element
Definition 3
...
Example 3
...
(J6 ,
⊕, ⊙) is a ring
...
Thus itsi
33
possible in a ring R, that a · b with neither a = 0 nor b = 0
...
This is an example for a ring R which is not a field
...
e
...
X=Y=
X+Y=
2
Σ
αijeij =
2
Σ
i,j=1
i,j=1
2
Σ
2
Σ
i,j=1
αijeij +
2
Σ
i,j=1
βijeij ⇔ αij = βij ∀i, j = 1, 2
...
a = e11 − e21 + e22 = 1 · e11 + 0 · e12 + (−1)e21 + 1 · e22
b = e22 + 3e12 = 0 · e11 + 3 · e12 + 0 · e21 + 1 · e22
a · b = (e11 + 0 · e12 + (−1)e21 + e22)(0 · e11 + 3 · e12 + 0 · e21 + 1 · e22)
= 0 + 3 · e12 + 0 + 0 + 0 + 0 + 0 + 0 + (−3)e22 + 0 + 0 + 0 + 0 + 0 + e22
= 3 · e12 − 3 · e22 + e22 = 3 · e12 − 2e22
∴ R is a ring
...
Example 3
...
(C, +, ·) is a field
...
9 Let Q = {α0 + α1˙i + α2˙j + α3˙k/α0 , α1 , α2 α3 ∈ R} and X = α0 + α1˙i + α2˙j +
α3˙k; Y = β0 + β1˙i + β2˙j + β3˙k
...
Define
X + Y = (α0 + α1˙i + α2˙j + α3˙k) + (β0 + β1˙i + β2˙j + β3˙k)
= (α0 + β0 ) + (α1 + β1 )i˙ + (α2 + β2 )j˙ + (α3 + β3 )˙k X · Y = (α0
+ α1˙i + α2˙j + α3˙k)(β0 + β1˙i + β2˙j + β3˙k)
= α0 β0 + α0 β1˙i + α0 β2˙j + α0 β3˙k + α1 β0˙i − α1 β1 + α1 β2˙k
˙ − α2 β2 + α2 β3˙i + α3 β0˙k
+ α1 β3 (−˙j) + α2 β0˙j + α2 β1 (−k)
+ α3 β1˙j + α3 β2 (−˙i) + α3 β3 (−1)
= (α0β0 − α1β1 − α2β2 − α3β3) + (α0β1 + α1β0 + α2β3 − α3β2)˙i
+ (α0 β2 + α2 β0 − α1 β3 + α3 β1 )j˙ + (α0 β3 + α3 β0 + α1 β2 − α2 β1 )˙k
34
2
...
e) (R, +, ·) with multiplicative unit element is called
a Ring of Real Quaternions
...
Example 3
...
Also, ¯3 is also a zero divisor
...
11 A commutation ring is an Integral Domain if it has no zero divisor
...
12 (Z, +, ·) us a commutation ring and it has no zero divisor
...
13 A ring is said to be divison ring if its non-zero elements form a
group under multiplication
...
14 (R, +, ·) is a division ring
...
15 A field is a commutative division ring
...
16 (R, +, ·), (C, +, ·), (J7, ⊕, ⊙)
Lemma 3
...
a · 0 = 0 · a = 0
2
...
−a × −b = +ab
4
...
(−1)(−1) = 1
Homomorphism
Definition 3
...
φ(a + b) = φ(a) + φ(b)
2
...
19 Let R′ and R′ be any two rings
...
Define φ : R → R by φ(a) = a, ∀a ∈ R is also′a homomorphism
...
20 If φ is a homomorphism of R into R′ then 1
...
φ(−a) = −φ(a) ∀a ∈ R
Remark 3
...
However if R′ is an integral domain (or) if R′ is an arbitrary
but φ is onto then φ(1) = 1′
...
22 Kernalofahomomorphism: If φ is a homomorphism of R into R′,
then the kernal of φ devoted by I(φ) is defined as, I(φ) = {a ∈ R|φ(a) = 0 , 0 is the
identity in R′ }
...
′
Example 3
...
Then I(φ) =
{a ∈ R|φ(a) = 0 ′} = R
...
Then I(φ) = {a ∈ R|φ(a) = 0′, 0′ is identity in
R} = {0}
Lemma 3
...
I(φ) is a subgroup of R under addition,
2
...
√
√
Example 3
...
J( 2) = {a + b 2|a, b ∈ J} which
√ is a ring√under
usual addition
and
m
ultiplication
...
Clearly, φ is a homom
orphism
...
2
...
Clearly,φ is a
homomorphism of J onto Jn
...
3
...
(i
...
) R = {f |f : [0, 1] → R}
...
Define φ : R → F by φ(f (x)) = f (1/2)
...
Definition 3
...
Definition 3
...
Lemma 3
...
UNIT II
Ideals and Quotient Rings
Definition 3
...
U is a subgroup of R with respect to addition,
2
...
Example 3
...
Example 3
...
Kernal of any homomorphism in a ring is an ideal of R
...
32 Let U be an ideal of R, Define R/U = {a + U |a ∈ R}
...
Then X + Y = (a + b) + U and
X · Y = ab + U
...
Remark 3
...
If R is commutative, then R/U is commutative
...
2
...
Lemma 3
...
Result 3
...
If U is an ideal of R and 1 ∈ U then U = R
...
If F is a field then its only ideals are {0} and F itself
...
36 Let R be a commutative ring with unit element whose only ideals are
{0} and R itself
...
37 An ideal M ƒ= R in a ring R is said to be a maximal ideal of R if
whenever U is an ideal of R such that M ⊂ U ⊂ R then either M = U (or) U =
R
...
38 If R is commutative ring and M is an ideal of R, then M
is a maximal ideal of R ⇔ R/M is a field
...
39 Let R = J and U be an ideal of R
...
U is a maximal ideal of R
⇔ n0 is prime ⇒ U = (2), (3), (5) are maximal ideal in (J, +, ·)
...
40 A ring R can be imbedded in a ring R′ if there is an iso- morphism
R into R′
...
Field of Quotients of an Integral Domain
Theorem 3
...
Specify the elements of the field F
...
Define (+) and (·) in F
...
Prove that F is a field
...
D can be imbedded in F
...
Define M = {(a, b)|a, b ∈ D, b ƒ= 0},where (a,
b) represents the quotient elements a/b
...
Claim that ∼ is an equivalence relation
∼ is reflexive:
Since ab = ba, ∀a, b ∈ D[∵ D is an integral domain and so it is a commutative ring]
⇒ (a, b) ∼ (a, b) ∀a, b ∈ M
...
∼ is symmetric:
Let (a, b) ∼ (c, d) ⇒ ad = bc ⇒ da = cb ⇒ cb = da ⇒ (c, d)∼ (a, b)
...
∼ is transitive:
Let (a, b) ∼ (c, d) and (c, d) ∼ (e, f ) ⇒ ad = bc and cf = de
...
Hence, ∼ is an equivalence relation
...
Let F = {[a, b]|(a, b) ∈ M, a, b ∈ D, b ƒ= 0} Step 2:
Define + and · in F as follows: Let [a, b], [c, d] ∈ F
...
Step 3:
2
...
To
Prove: [ad+bc, bd] = [a′d′+b′c ′, b′d′]
...
Now, [a, b] = [a′, b′] ⇒ ab′ = ba′ (1)
and [c, d] = [c′, d′] ⇒ cd′ = dc′ (2)
(ad + bc)b′d′ = adb′d′ + bcb′d′
=ab′dd′+bb′cd′
= ba′dd′ + bb′dc
= bd(a′d′ + b′c′)
∴ + is well defined
...
Then D is an integral domain, bd ƒ= 0
...
∴ + is closed
...
Additive identity:
[0, b] ∈ F acts as zero element for this addition
...
Additive inverse:
[−a, b] acts as a identive inverse of [a, b]
...
2
+ is commutative:
[a, b]+[c, d] = [ad+bc, bd] = [bc+ad, bd] = [cb+da, bd] = [c, d]+[a, b] [a, b]+ [c,
∀ d] F
...
∴ (F, +) is an abelian group
...
To Prove [a, b] [c, d]
· = [a , b ] [c , d ]·
′ ′
′ ′
′ ′
(i
...
) [ac, bd] = [a c , b d ]
...
Now,
(ac)(b′d′) = acb′d′
= ba′cd′ [∵ [a, b] = [a′, b′]]
= ba′dc′ [∵ ab′ = ba′]
= (bd)(a′c′) [∵ [c, d] = [c′, d′], cd′ = dc′]
...
· is closed:
Let [a, b], [c, d] ∈ F [b, d ∈ D, b ƒ= 0, d ƒ= 0 ∴ bd ƒ= 0]
...
∴ · is closed
...
Existence of Multiplicative Identity:
Let [a, a] ∈ F, a ƒ= 0 be the multiplicative identity
...
(i
...
) [a, b] · [a, a] = [a, a] · [a, b] = [a, b]
...
Then [b, a] ∈ F, a ƒ= 0 is the multiplicative inverse
...
· is commutative:
Let [a, b], [c, d] ∈ F
...
∴ · is
commutative
...
· is distributive over addition:
[a, b] · ([c, d] + [e, f ]) = [a, b] · [cf + de, df]
= [a(cf + de), b(df )]
= [(acf + ade), bdf ]
= [(ac)f + (ae)d, (bd)f ]
= [(ac)(bf ) + (ae)(bd), (bd)(bf )]
= [ac, bd] + [ae, bf ]
= [a, b] · [c, d] + [a, b] · [e, f ]
40
2
...
Hence F is a field
...
(i
...
) We shall find an isomor- phism
of D → F
...
Denote [ax, x] by [a, 1]
...
φ is 1 − 1:
Suppose φ(a) = φ(b), a, b ∈ D
...
∴ φ is 1 − 1
...
∴ φ is an isomorphism of D into F
...
Hence D can be imbedded into F
...
Note: Usually, the above field F is called field of quotients of D
...
42 Let F be a field
...
+ anx } where n ∈ 2Z ∪ {0} and the
n
coefficient a0,+a1, a2,
...
(i
...
) F (x) = {a0 + a1x + a2x +
...
, an ∈ F, n ∈ Z ∪ {0}}
...
43 If p(x) = a0 + a1x + a2x2 +
...
+ bnxn are in F [x]
...
p(x) = q(x) ⇔ ai = bi ∀i ≥ 0,
2
...
+ ctxt where ai + bi = ci∀i,
3
...
+ ctxt where ct = atb0 + at−1b1 +
at−2b2 +
...
Note 3
...
41
Remark 3
...
Definition 3
...
+a 1 x+a 0 ƒ= 0 in F [x] and
an ƒ= 0 ((i
...
) ai = 0 ∀i ≥ 0), then degree of f (x), denoted by deg(f (x)) is n
...
e
...
th
Remark 3
...
We say a
polynomial is constant if its degree is zero
...
48 Iff(x) andg(x) arenon-zeroelementsofF [x]
...
+ am xm , am =ƒ
0 [ai = 0, ∀i > m] in
F [x]
...
+ bnxn, bn ƒ= 0 [bj = 0∀j ≥ n]in F (x)
...
By definition, f (x)g(x) = c0 + c1x + c2x2 +
...
+ a0bt
...
Now, cm+n = am+nb0 + am+n−1b1 + am+n−2b2 +
...
(3)[∵ am ƒ= 0 and bm
0 and ambm ∈ F ]
...
Then one of aj or
bi−j is zero, so that ajbi−j = 0 ⇒ ci = aib0+ai−1b1+
...
For every i > m + n, ci = 0
...
∴ deg(f (x) · g(x)) = m + n =
deg(f (x)) +deg(g(x))
...
Corollary 3
...
By the above lemma, deg(f (x) · g(x)) =
deg(f (x)) + deg(g(x)) ≥ degf ((x))[∵ deg(g(x)) ≥ 0]
...
Proof: Clearly F [x] is a common ring with unit element
...
e)product of any
two non- zero elements in F [x] is again a non-zero el- ement in F [x]
...
+ amxm, am ƒ= 0 in F [x] and g(x) = b0 + b1x + + bnxn, bn ƒ= 0 in F [x]
...
(i
...
m+n
∴ f (x) · g(x) ƒ= 0 in F [x]
...
Lemma 3
...
Then there exists two polynomials t(x)and r(x) in F [x] such
that f (x) = t(x) · g(x) + r(x), where either r(x) = 0 (or) deg(r(x)) < deg(g(x))
...
+ amxm, am ƒ= 0 in F [x] and
g(x) = b0 + b1x + b2x2 +
...
UNIT II
42
deg(g(x)) = n
...
For put t(x) = 0 and
r(x) = f (x), where f (x), deg(f (x)) < deg(g(x))
...
we shall prove the theorem by induction on degree
of f (x)
...
Let f (x) = a
0, g(x) = b ƒ= 0, a, b ∈ F [x]
...
Now, a = (ab−1)b ƒ= 0 ⇒ f (x) = t(x) · g(x) + r(x) , where r(x) = 0
...
Assume that the result is true ∀ polynomial of degree
< m
...
(1)
n
= a0 + a1x +
...
+ bnxn)
= a0 + a1x +
...
− amxm
n
n
⇒ deg(f1 (x)) ≤ m − 1 < m ⇒ deg(f1 (x)) < m
...
From (1),
f (x) = f1 (x) + am b−1 xm−n
g(x)
n
= t1(x)g(x) + r(x) + amb−1xm−n
g(x)
n
−1 m−n
= (t1(x) + amb xn )g(x) + r(x),
where r(x) = 0 (or) deg r(x) < deg g(x)
...
(2)
If r(x) = 0 and r1(x) = 0 ⇒ t(x) = t1(x)
...
Then (2) ⇒ deg([t1(x) − t(x)]g(x)) = deg(r(x) − r1(x))
...
∴ r1(x) − r(x) = 0 ⇒ r(x) = r1(x)
...
Theorem 3
...
Proof: F [x] is an Integral Domain with unit element
...
∴ d(f (x)) ≥ 0[∵ deg(f (x)≥0]
...
e)d(f (x)) is non negative (1)
By Corollary 3
...
(i
...
(2)
By the above lemma, given two polynomials f (x) and g(x) ƒ= 0 in F [x], then there
exists two polynomials t(x) and r(x) in F [x], ∋: f (x) = t(x)·g(x)+r(x), where r(x) = 0
(or) deg(r(x)) < deg(g(x))
...
43
Lemma 3
...
Proof: Clearly, F (x) is an integral domain with unit element
...
Suppose u = (0)
...
e
...
Then F (x)is principle
ideal ring
...
Suppose u ƒ= 0
...
Claim that U = (g(x))
...
By division algorithm, there exists t(x), r(x) ∈ F
[x] ∋: f (x) = t(x) · g(x) + r(x)
where r(x) = 0 (or) deg(r(x)) < deg(g(x))
...
∵ g(x) is a least degree polynomial in U,
deg(r(x)) cannot be less than deg(g(x))
...
∴
∴ f (x) = t(x) · g(x) ∈ U
...
∴ U = (g(x))
...
∴ U is a principle ideal in
F [x] and U is arbitrary
...
Lemma 3
...
Then F [x] is a ring with
unit element
...
Then f (x) ∈ S
...
So, S ƒ= φ
...
Then, h1 (x) = s1 (x)f (x)+t1 (x)·
g(x); h2(x) = s2(x)f (x) + t2(x)g(x), where s1(x), s2(x), t1(x), t2(x) ∈ F [x]
...
Then,
p(x) · h(x) = p(x)[s(x)f (x) + t(x)g(x)]
= (p(x)s(x))f (x) + (p(x)t(x))g(x),
where p(x)s(x) ∈ F (x) and p(x)t(x) ∈ F (x)
∴ p(x) h(x)
· S (2)
∈
From (1) and (2), S is an ideal
...
∴ S is a principle
ideal
...
m(x) = s0(x)f (x) + t0(x)g(x), where s0(x)
and t0(x) ∈ F [x]
...
44
2
...
54 A polynomial over an arbitrary ring is not a principleideal ring
...
Claim 1: The ideal (2, x) of J[x] generated by (2, x) of J[x] is not a principle ideal ring
...
∴ There exists φ(x) ∈ J[x] such that 2 = φ(x) · g(x)
...
(2)
From (1) and (2),
(1) ⇒ 2x = x · φ(x) · g(x)
(2) ⇒ 2x = 2 · φ′(x)g(x)
′
⇒ x · φ(x) · g(x) = 2 · φ (x)g(x)
′
⇒ (x · φ(x) − 2 · φ (x))g(x)
=0
′
⇒ xφ(x) = 2φ (x) [∵ g(x) ƒ= 0 and J[x] is integral domain]
⇒ coefficient of φ(x) must be an even integer
...
(3)
From (1) and (3), 2 = 2h(x) · g(x) ⇒ 1 = h(x) · g(x) ⇒ 1 ∈ (g(x)) ⇒ J[x] = (g(x)) = (2,
x) [∵ 1 ∈ U ⇒ U = R]
...
53, [d(x) = λ(x)f (x) + l1(x)g(x)] ⇒
1 = 2p(x) + xq(x), p(x), q(x) ∈ J[x]
...
; q(x) =
b0+b1x+ b2x2+
...
]+x[b0+b1x+ b2x2+
...
1 ∈
/ (2, x) which is a
⇒⇐ to (A)
...
∴ J[x] is not a principle ideal ring
Definition 3
...
Then one of a(x) or b(x) has degree
0
...
e
...
Example 3
...
Lemma 3
...
Then clearly, deg(f (x)) >
0
...
Now, when f (x) is of
degree 1, it is of the form a0 + ax, where a0, a F and a∈= 0
...
Clearly, a−1a0 + x is a monic irreducible polynomial
in F [x] and a is an element of F
...
Let us assume the theorem to be
true ∀ polynomials of degree less than that of f (x) and by induction we must
show it to be true for f(x)
...
Now, if f (x) is irreducible, then so is f1 (x) and so in this
case the theorem will follow
...
∵ F [x] is a polynomial over the field F , deg[g(x) · h(x)] = deg(g(x)) + deg(h(x))
and since each one of g(x) and h(x) is a non-zero, non-unit polynomial in F [x]
...
49]
...
∴ f (x) = g(x) · h(x) = a1a2p1(x)p2(x) · · ·
pn(x)q1(x) · · · qm(x) =product of finite no of irreducible polynomial in F [x], where
each pi(x) and qj(x) are monic irreducible in F [x]
...
Now in order
to show that this decomposition is unique, let f (x) = ap1(x)p2(x) · · · pm(x) =
aq1(x)q2(x) · · · qn(x), where each pi(x) and qj(x) are the monic irreducible
polynomials in F [x]
...
(1)
...
But this means that p1(x) must divide atleast one of
q1(x)q2(x) · · · qn(x)
...
Now, p1(x)/q1(x)and p1(x), q1(x) are irreducible polynomials in F [x]
...
⇒ q1(x) is unit times p1(x)
...
[∵ units in F [x] are constant polynomial]
⇒ q1(x) = p1(x) [∵ q1(x) and p1(x) being monic, we must have u = 1]
...
Now, we
can repeat the above argument on this relation with p2(x)
...
Now, if n > m then after m steps the LHS of (1)
will become 1 and the RHS of (1) will reduce to a product of a certain number of
q(x)′s (the excess of n over m)
...
Thus, a
⇒⇐ consequently n and m
...
Hence m = n
...
Hence the decomposition is unique except for the order in
which the factors occur
...
58 The ideal A = (p(x)) in F [x] is a maximal ideal ⇔ p(x) is irreducible
over F
...
UNIT II
Proof: Suppose A = (p(x)) is a maximal ideal F [x]
...
e
...
To Prove: p(x) is irreducible over F
...
Consider the polynomial f (x), g(x) ∈ F [x] ∋: f (x)· g(x) ∈ (p(x)) ⇒ f (x) ∈ p(x) (or)
g(x) ∈ p(x) ⇒ f (x) = t(x)p(x) (or) g(x) =
r(x)p(x) ⇒ p(x)/f (x) (or) p(x)/f (x)
...
Conversely, suppose that p(x) is a
maximal ideal in F [x]
...
Suppose
there exists an ideal N of F [x] ∋: (p(x)) ⊂ N ⊂ F [x]
...
From (1),
(p(x)) ⊂ (g(x)) ⊂ F [x] ⇒ (p(x)) ⊂ (g(x)) ⇒ (p(x)) ∈ (g(x)) ⇒ (p(x)) = t(x) · g(x), t(x) ∈
F [x] ∵ p(x) is irreducible either deg(t(x)) = 0 or deg(g(x)) = 0
...
Then g(x) is a non-zeroconstant, say g(x) = a, a = 0 in F
...
Then, N = F [x] [∵ g(x) = F [x]] (2)
...
Let t(x) = b, a non-zero element in F
...
But (p(x))⊆ N
b
∴ N = (p(x))
...
F [x]
...
59 The polynomial f (x) = a0+a1x +a2x2+
...
, an are integers is said to be primitive if the GCD of a0, a1,
...
For example, f (x) = 3 + 5x + 7x2 is primitive
...
60 Apolynomial in which the leading coefficientis 1 is called as monic
polynomial
...
Definition 3
...
For example, f (x) = 5 − 6x + 12x + x is
integer monic2 polynomial
...
62 If f (x) and g(x) are primitive polynomials, then f (x) · g(x) is a
primitive polynomial
...
Proof: Let f (x) = a0 + a1(x) +
...
To Prove: f (x) · g(x) is primitive
...
e) f
(x)· g(x)is notprimitive
...
Hence some prime number p, p divides all the coefficient of f
(x)· g(x)
...
Let aj
be the first coefficient of f(x) such that p does not divides a j
...
e)p/a0, p/a1,
...
Let bk be the first coefficient of g(x) such that does not divides bk
...
e
...
, p/bk−1] (2)
cj+k = (aj+kb0 + aj+k−1b1 + + aj+1bk−1) + ajbk
+(aj−1bk+1 +
...
(3)
By our choice of aj, p/a0, p/a1, , p/aj−1
⇒ p/a0bj+k + a1bj+k+1 +
...
(4)
By our choice of bk, p/b0, p/b1, , p/bk−1
⇒ p/aj+kb0 + aj+k+1b1 +
...
(5)
But p/cj+k ⇒ p/(cj+k) − (a0bj+k + a1bj+k−1 + + aj−1bk+1)
−(aj+kb0 + aj+k+1b1 +
...
∴ Our assumption is
wrong
...
Hence, the lemma
...
63 Content of the Polynomial Let f (x) = a0 + a1x + a2x2 +
...
The content of the polynomial is the GCD of a0, a1, a2,
...
(i
...
) c(f ) = (a0, a1, a2, an)
...
64 Let p(x) = 5+10x+25x2+30x3
...
Remark 3
...
Any polynomial with integer coefficient is said to be
integer monic if the content of f (x) = 1
...
Any polynomial p(x) with integer coefficient can be written as p(x) =
d(g(x)), where d is the content of p(x), and g(x) is primitive
...
66 p(x) = 3 + 6x + 9x2 − 12x3 = 3(1 + 2x + 3x2 − 4x3) =
c(p(x))g(x), where c(p(x)) = 3 and g(x) = 1 + 2x + 3x2 − 4x3, primitive
...
67 Gauss Lemma If the primitive polynomial f (x) can be factored as
the product of two polynomials having rational coefficient, it can be factored as the
product of two polynomials having integer coefficients
...
Let u(x) = a0 + (a1 )x + (a2 )x2 +
...
Claim that f (x) = a λ(x)·l1(x),
b
where a, b are integer and λ(x), l1(x) are primitive polynomial with integer
coefficients
...
+ an(b0b1b2 · · · bn−1)xn] = 1 [c0 + m
c1x + c2x2 +
...
(1)
where m = b0b1b2 · · · bn; c0 = a0(b1b2 · · · bn); c1 = a1(b0b2b3 · · · bn);····················· ; cn =
an(b0b1b2 ····· bn−1)
...
c0+c 1 x+c2x +
...
, cn) and λ(x) is primitive
...
UNIT II
d
λ(x),
m
where d = (c0, c1, c2,
...
Similarly v(x) = d1 l1(x), where d1 and m1 are integer and l1x is
m1
d · λ(x)l1(x) = λ(x)l
a
1
primitive
...
(2)
m m1
b
where a = dd1 and b = mm1 are integers ⇒ bf (x) = aλ(x)l1(x)
...
(4)
[∵ f (x), l1(x), λ(x) are primitive, their content is 1]
...
∴ f (x) can be factored as a product of two polynomial having two
integer coefficient
...
Hence the theorem
...
68 If an integer monic polynomial factors as the product of two nonconstant polynomials having rational coefficients then it factors as the product of
two integer monic polynomials
...
(i
...
) f (x) is a primitive
polynomial factored as the product of two polynomial having rational
coefficients
...
67 f (x) can be factored as product of two
polynomials having integer coefficients
...
Let p(x) = a0 + a1x + a2x2+
...
∵ f (x) is monic, leading coefficient of f (x) is 1
...
∴ In either case, p(x), r(x) are integer monic polynomials
...
4
...
UNITIV
Vector Spaces
Definition 4
...
α(v + w) = αv + αw
2
...
α(βv) = (αβ)v
4
...
Remark 4
...
Example 4
...
Let K be a field which contains F as a subfield
...
For (K, +) is an abelian group, for α ∈ F,
v ∈ K, αv ∈ K
...
Since 1 is the identity element in K, the Axiom 4 follows from it
...
αn)|αi ∈ F }= all order of n tuples
= F (n)
...
4 (R, +, ·) is a field
...
Example 4
...
V = {(α1, α2)|αi ∈ Q∗} = Q(2) and
V = {(α1, α2, α3)|αi ∈ Q∗ = Q(3)
...
6 Let F be a field V = F [x]= set of all of polynomial x over F = {α0 +
α1x + α2x2 +
...
Then V is a vector space over F
...
7 Subspace: Let V be a vector space over a field F and if W is a subset
of V
...
Equivalently W subspace of V whenever w1, w2 ∈ W, α, β ∈ F
implies αw1 + βw2∈ W
...
8 Let F be a field
...
Under natural operations for polynomials of addition and multiplication
...
+ αnxn
+
...
Definition 4
...
Ker T = {u ∈ U |uT = 0, identity
element of addition in V }
...
10 T is an isomorphism iff Ker T = {0}
Definition 4
...
Lemma 4
...
α(0) = 0
for α ∈ F,
2
...
(−α)v = −αv,
for α ∈ F, v ∈ V ,
4
...
Lemma 4
...
Then V/W = {v
+ W |v ∈ V }
...
Define (i) (v1 + W ) + (v2 + W ) =
v1 + v2 + W,
(ii) (v1 + W ) = αv1 + W
...
Theorem 4
...
Then V is isomorphic to U/W
conversely if U is a vector space and W is a subspace of U
...
Definition 4
...
, Un be subspace of
V
...
, Un if every element v ∈
V can be written in the unique way as v = u1 + u2 +
...
Remark 4
...
Then V itself and subset of V
consisting of ¯0 vector only are the trivial subspace of V
...
For example let V = {(α1, α2, α3)|α1, α2, α3 ∈ F } and W = {(α1, α2, 0)|α1,
α2 ∈ F }
...
UNIT IV
50
Linear Independent and Spaces:
Definition 4
...
, vn
...
+ αnvn where αi ∈ F is a linear combination over F of v1,
v2,
...
Definition 4
...
Then the linear span of S, L(S) is the set of all linear combi- nation of finite sets
of element of S
...
e
...
, vn is an arbitrary finite subset of S and α1, α2,
...
Lemma 4
...
Lemma 4
...
S
⊂ T ⇒ L(S) ⊂ L(T ),
2
...
L(L(S)) = L(S)
...
Example 4
...
Let
S = {(1, 0, 0)}; L(S) = {(α, 0, 0)|α ∈ F } ⊂ V
...
22 V = F (3); S = {(1, 0, 0), (0, 1, 0)}
...
Example 4
...
Then
L(S) = V
...
24 V = α1v1 +
...
Let v = (a, b, c) ∈ F (3) = V
...
∴ L(S) = V
...
25 If V is a vector space and if v1, v2,
...
We say
that they are linearly dependent over F if there exist element λ1, λ2,
...
+ λnvn = 0
...
Remark 4
...
51
Example 4
...
, αn)}
...
, en} where e1 = {1, 0,
...
, 0};
...
, 1} is linearly independent
...
, λn ∈ F
...
+λnen = 0 ⇒ λ1(1, 0,
...
, 0)+
...
, 1) =
0 ⇒ (λ1, 0,
...
, 0) + (0, 0,
...
, λn) = 0 ⇒
λ1 = 0, λ2 = 0,
...
Remark 4
...
vn} is linearly independent
then none of the vector v1 , v2 ,
...
Example 4
...
Solution: Let λ1 = 1, λ2 = −1, λ3 = −2, λ4 = −4
...
∴ Given set is linearly dependent
...
30 If v1, v2,
...
+
λnvn with λi ∈ F
...
31 If v1, v2,
...
If V is a finite dimensional vector
space then it contains a finite set v1, v2,
...
Definition 4
...
Let set S consisting of
vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) is a basis of
F (3)
...
33
1
...
, vm
is span V then some subsets of v1, v2,
...
2
...
, vm is a basis of V over F if w1, w2,
...
3
...
For example, S1 =
{(1, 0, 0), (0, 1, 0), (0, 1, 1)} and S2 = {(1, 0, 0), (1, 1, 0), (1, 1, 1)} are
two basis of the vector space F(3)
...
F (n) ∼
= F (m) iff n = m
...
If V be a finite dimensional vector space over a field F then V ∼
= F (n) for a
unique integer n, infact n is the number of elements in any basis V over F
...
UNIT IV
52
Definition 4
...
For example, dim(F(3)) = 3 and dim(F(4)) = 4
...
35 Any two finite dimensional vector space over F of the same dimension
are isomorphic
...
Definition 4
...
Lemma 4
...
Hom(V, W ) be the set
of all vector space homomorphisms of V into W
...
Let S, T ∈ Hom(V, W )
...
Result 4
...
If V and W are of dimensions m and n respectively over F then
Hom(V, W ) is of dimension mn over F
...
2
...
3
...
Definition 4
...
^
Definition 4
...
41 if V is not finite dimensional Vwould be^ is usually too large and
of
...
Result 4
...
If V is a finite dimensional an v ƒ= 0 in V then there is
an element F ∈ V^ such that F (v) = 0
...
If V is a finite dimensional vector space then there is an isomorphism of V
onto V^
...
43 if W is a subspace of V then annihilator
of W, A(W ) =
^
{f ∈ V /f (W ) = 0 ∀w ∈ W }
...
44
1
...
53
2
...
∼
^
...
V^ /A(W ) = W
4
...
Linear Transformation:
We know that Hom(V, W ), the set of all vector space homomorphisms of V into W
is a vector space over the field F
...
Definition 4
...
Remark 4
...
Is the converse true?
Result 4
...
Proof: Let T1, T2 ∈ Hom(V, V )
...
We shall first prove that
Hom(V, V ) is a ring
...
Let T1, T2, T3 ∈ Hom(V, V )
...
0 : V → V defined by v0 = 0 ∀v ∈ V serve as additive identity element
...
Inverse of T1 is −T1 defined by, v(−T1) = −(vT1) ∀v ∈ V
...
Similarly v(−T1 + T1) = 0 ⇒ T1 + (−T1)
= (−T1) + T1 = 0
...
∴ + is commutative
...
Now,
(v1 + v2)(T1 · T2) = ((v1 + v2)T1) · T2
= (v1T1 + v2T1) · T2
= (v1T1)T2+ (v2T1)T2
= v1(T1 · T2) + v2(T1 · T2)
4
...
Clearly T1(T2 · T3) = (T1 · T2)T3 ∀T1, T2, T3 ∈
Hom(V, V )
...
T1 · (T2 + T3) = T1 · T2 + T1 · T3; (T1 + T2) · T3 = T1 · T3 + T2 · T3 ∀T1, T2, T3 ∈
Hom(V, V )
...
∴ (Hom(V, V ), +, ·) is a Ring
...
To Prove: α(T1 + T2) = αT1 + αT2
...
To prove:(α + β)T1 = αT1 + βT1
...
To Prove: α(βT1) = (αβ)T1
...
v(1·T1) = 1·(vT1) = vT1 ⇒ 1·T1 = T1 ∀T1 ∈ Hom(V, V )
...
Let v ∈ V,
v(α(T1T2)) = α(v(T1T2))
= α((vT1)T2)
= (vT1)(αT2)
= (vT1)(αT2)
⇒ α(T1T2) = T1(αT2)
...
∴ Hom(V, V ) is an algebra over F
...
48 For convenient we shall write Hom(V, V ) as A(V )
...
We shall denote it by AF (V )
...
49 A linear transformations on V over F is an element of AF (V )
...
e
...
Remark 4
...
Lemma 4
...
Then A is
isomorphic to a sub-algebra of A(V ) for some vector space V over F
...
It must be a vector space over F
...
Since A is a ring as
well as a vector space, we choose V = A
...
Claim Ta is a linear trans- formation on V
...
e
...
Let v1, v2 ∈ V and
α∈F
Now, (v1 + v2)Ta = v1 = v1a + v2a = v1Ta + v2Ta
...
(2)
From (1)and (2), Ta ∈ Hom(V, V ) = A(V )
...
e
...
Hence the claim
...
Let a, b ∈ A and α ∈ F
...
For v ∈ V, vTa+b = v(a + b) = va+vb = vTa+vTb ∀v ∈
V ⇒ Ta+b = Ta+Tb
...
For any v ∈ V, vTαa = v(αa) = α(va)
= α(vTa) ∀v ∈ V ⇒ Tαa = αTa
...
(4)
From (3) and (4), ψ is a homomorphism of A into A(V )
...
Let a ∈ Kerψ
56
4
...
Hence ψ is an isomorphism of A onto A(V )
...
Lemma 4
...
Proof: Given dim A = m
...
Let a ∈ A
...
, am are linearly dependent
...
, αm ∈ F , not all zero such that α0e + α1a + α2a2 +
...
e
...
+ αmam = 0 (1)
Let f (x) = α0 + α1x + α2x2 + + αmxm ∈ F [x]
...
Since a is arbitrary in A, every
element in A satisfy the polynomial of degree at most m
...
53 If V be an n dimensional vector space over a field F, given any
element T in A(V ) there exists a non-trivial polynomial q(x) of degree almost n2
such that q(T ) =0
...
But dim(A(V )) = dim(Hom(V, V )) = dim(V ) · dim(V )
= n × n = n2
...
By the
above lemma, there exists a non-trivial polynomial q(x) ∈ F [x] of degree at most n2
...
Definition 4
...
Remark 4
...
Proof: Let p(x) be the minimal polynomial for T then p(T ) = 0 and p(x) is of least
degree
...
(2)
From (1), 0 = h(T ) ⇒ p(T )q(T ) + r(T ) = 0 ⇒ r(T ) = 0[∵ p(T ) = 0]
...
∴ r(x) = 0
...
Hence the remark
...
56 Let A, B be any algebra’s over F
...
(a1 + a2)T = a1T +a 2 T,
2
...
(αa1)T = α(a1T )
...
Ker T = {a ∈ A|aT = 0, identity element in B}
...
57 An element T ∈ A(V ) is called a right invertible if there exists an
element S ∈ A(V ) such that TS=1
...
58 An element T ∈ A(V ) is called a left invertible if there exists an
element S ∈ A(V ) such that ST = 1
...
59 An element T ∈ A(V ) is said to be invertible (or) regular if it is both
right and left invertible (i
...
) there exists an element S ∈ A(V ) such that TS = ST =
1
...
Remark 4
...
Definition 4
...
Remark 4
...
Example 4
...
Let V = F [x] be the set of all
dx
d
∫
polynomials
as q(x)T = xin x
...
Let T ∈ A(V )
1 q(x)dx
...
Now,
q(x)TS = (q(x)T )S
=(
∫x
1
=
d(
)S
∫ x
q(x)dx)
dx 1
= (q(x)) · 1
= q(x)
⇒ TS = 1
∴ T is right invertible but not invertible
...
UNIT IV
58
Remark 4
...
Theorem 4
...
Proof: Let p(x) = α0 + α1x + α2x2 +
...
Assume that α0 ƒ= 0 and p(T ) = 0
...
Since p(x) is a minimal
polynomial for T
...
+α k T k = 0
...
+ α k T k )
α 0= −(α + 1α T + 2
...
− α T k−1 )
0
1
2
k
⇒ 1 =T ( 1 (−α − α −
...
+ α
2
Tk−1))
k
Let S = − 1 (α1 + α2 +
...
Clearly, S =ƒ 0 and T S = 1 similarly ST = 1
...
0T is invertible
...
To prove: α0
ƒ= 0
...
From(1),
α1T + α2T 2 +
...
+ α Tk−1)T = 0
...
Multiplying the above relation t−1,
−1
⇒ ((α T1 + α T 22+
...
+ α T k−1k= 0
...
+ αkxk−1
...
(i
...
) T satisfy the polynomial q(x) of
degree k − 1, which is a contradiction to the degree of minimal polynomial for T ,
which is k ⇒⇐ shows that α0 ƒ= 0
...
66 If V is finite dimensional over F and if T ∈ A(V ) is in- vertible then T
−1
is a polynomial expression in T over F
...
p(T ) = 0 ⇒ α0 + α1T + α2T 2 +
...
+ αkTk)
α0 = (−α1)T + (−α2)T 2 +
...
+ (− k )Tk
α
α
α
α10
α2 0
αk k−1 0
1 = ((− ) + (− )T +
...
+ (− )Tk−1)T · T −1
2
0
α0
α0
T −1 = β + β T +
...
, βk = (− αk )
...
Corollary 4
...
Proof: Let p(x) = α0 + α1x + α2x2 +
...
(i
...
) p(T ) = 0 ⇒ α0 + α1x + α2x2 +
...
Since T is singular (i
...
) T is noninvertible by Theorem 4
...
∴ α1T + α2T 2 +
...
+ α kT k−1)T = 0 (1)
Let S = α1 +α2T +
...
+αkxk−1
is of lower degree than p(x))
...
Similarly TS = 0
...
Corollary 4
...
Proof: Given T ∈ A(V ) is right invertible
...
Suppose T is not invertible
...
e
...
67, there exists S ƒ= 0 in A(V ) such that ST = TS = 0 (2)
From (1), TU = 0
⇒ S(TU ) = S · 1
⇒ (ST )U = S
⇒ 0 · U=S
by(2)
⇒S=0
⇒⇐ S ƒ= 0
This contradiction shows that T is invertible
...
UNIT IV
Theorem 4
...
Proof: Assume that T is singular
...
67 there exists S ƒ= 0 ∈
A(V ) such that ST = TS = 0
...
Let v = wSthen v ƒ= 0 in V,
vT = (wS)T = w(ST ) = w¯0 = 0 by(1) ⇒ vT = 0, v ƒ= 0
...
Conversely, suppose that there exists v ƒ= 0 in V such that vT = 0
...
Suppose not,T is invertible
...
Now, TU = 1 ⇒ v(TU ) = v · 1 (2)
v(TU ) = (vT )U = 0 · U = 0 → (3)
From (2) and (3), v = 0 ⇒⇐ to v ƒ= 0
...
Definition 4
...
71 (1) Range of T is a subspace of V
Proof: Let u, v ∈ V T, α, β ∈ F
...
∴ V T is a subspace of V
...
(2) If V T = V then T is onto
...
72 If V is finite dimensional over F, then T ∈ A(V ) is regular iff T maps
V onto V
...
To prove: T is onto
...
Now,
(vT −1)T = v(t−1T ) = v · 1 = v ⇒ v = (vT −1)T, v ∈ V
...
e
...
∴ T is onto
...
To
prove: T is regular
...
Since T is singular, by Theorem 4
...
Suppose α1v1 = 0 ⇒ α1 = 0 ⇒ v1is linearly independent
...
Since V is finite
dimensional, we can find vectors v2, v3,
...
∴ V T is generated by w1 = v1T, w2 =
v 2T,
...
Since w1 = v1T = 0, V T is spanned by v 2T, v 3T, , vnT
...
e
...
, wn ∴ dim(V T ) ≤ (n − 1) < n = dim(V ) ⇒ dim(V T ) <
dim(V ) ⇒ V T ⊂ V ⇒ V T ƒ= V ⇒ T is not onto
...
73 The above theorem can be replaced as T is regular ⇔ dim(V T ) =
dim(V ) (i
...
) V T = V
...
74 The above theorem suggest that we could use dim(V T ) not only as
a test for regularity but even as a measure of degree of singularity for a given T ∈
A(V )
...
75 Rank of T : If V is finite dimensional over F
...
The rank of T over F, it is denoted by r(T ) (i
...
) dim(V T )=Rank
of T = r(T )
...
76
1
...
Then T is regular,
2
...
Then T = 0
...
(2) Suppose r(T ) = 0 ⇒ dim(V T ) = 0 ⇒ V T = {0} ⇒ {vT |v ∈ V } =
{0} ⇒ {vT = 0, ∀v ∈ V } = 0 ⇒ T = ˙0
...
77 If V is finite dimensional vector space over F
...
r(ST ) ≤ r(T )
2
...
r(ST ) = r(TS) = r(T ) for S regular in V
...
(2) r(T ) = m = dim(V T ), where V T is a subspace of V
...
, wm} be basis
of V T ⇒ dim(V T ) = m
...
, wmS generate (V T )S ⇒ dim(V (TS)) ≤
m = r(T )
...
e
...
From (1) and (2),
r(ST ) ≤ r(T )andr(TS) ≤ r(S) ⇒ r(ST ) ≤ min{r(T ), r(S)}
...
(i)
Let r(T ) = m, V T = m
...
, wm} be a basis of V T
...
, wmS} generate (V T )S = V (TS)
...
, wmS}
is linearly independent for if α1(w1S) + α2(w2S) +
...
Then,
α1w1S + α2w2S +
...
+ αmwm)S = 0
...
Now,
(α1w1 + α2w2 +
...
+ αmwm = 0
...
UNIT IV
62
⇒ αi = 0 ∀ i [∵ {w1, w2,
...
Hence the claim
...
e
...
, wmS} is a basis of V (TS) ⇒ dim(V (TS)) = m ⇒
r(T ) ⇒ r(TS) = r(T )
...
Corollary 4
...
Proof: r(STS−1) = r((ST )(S−1)) = r(S−1(ST )) = r((S−1S)T ) = r(T )
...
79 S, T ∈ A(V ) and if S is regular then STS −1 and T have same
minimal polynomial
...
80 If T ∈ A(V ), then λ ∈ F is called characteristics roots (or Eigen
value of T) if λ − T is singular
...
81 The element λ ∈ F is a characteristics roots of T ∈ A(V )
iff for some v ƒ= 0 in V, vT =λv
...
Then λ − T is singular
...
69, there exists a v ƒ= 0 in V such that v(λ − T ) = 0 ⇒λv − vT = 0 ⇒ λv
= vT
...
By Theorem 4
...
∴ λ is the characteristic
root of T
...
82 If λ ∈ F is a characteristic root of T ∈ A(V ), then for any polynomial
q(x) ∈ F [x], q(λ) is a characteristic root of q(T )
...
+ αm
...
Then by Theorem 4
...
Now vT 2 = (vT )T = (λv)T = λ(vT ) = λ(λv) =λ2v
...
In
general, vT k = λkv for all positive integer k
...
+ α )
m
= v(α T0m ) + v(α T m−11 ) +
...
+ α v
0
m
m−1
= α (λ
0 v) + α (λ 1 v) +
...
+ α )v m
= (q(λ))v
...
∴ q(λ) is a characteristic root of q(T )
...
83 If λ ∈ F is a characteristic root of T ∈ A(V ), then λ is a root of
minimal polynomial of T
...
Proof: Let p(x) = α0xm + α1xm−1 +
...
Then p(T ) = 0 (i
...
) α 0T m + α 1Tm−1 +
...
Then by Theorem 4
...
(1)
We have to show that p(λ)v = vp(T )
...
Similarly vT 3 = λ3v
...
+ α )
m
= v(α T0m) + v(α Tm−11) +
...
+ α
0
m
m−1
= α (λ
0 v) + α (λ 1 v) +
...
+ α )v m
= (p(λ))v (3)
m
mv
mv
p(λ)is a characteristic root of p(T )
...
∴ λ is the root of the minimal polynomial of T and degree of p(x) ≤ n2 (by
Theorem 4
...
∴ T has only a finite number of
characteristic root in F
...
84 If T, S ∈ A(V ) and if S is regular, then T and STS −1 has the same
minimal polynomial
...
(1)
For let q(x) = α0+α1x1+α2x2+
...
Now, (STS −1) 2 = (STS −1 )(STS −1 ) =
(ST )(S −1S)(TS−1) = (ST )(1)(TS−1) = STTS − 1 = ST 2S −1
...
(2)
−1
−1
−1 2
−1 m
q(STS
0
1 ) = α + α (STS2 ) + α (STS ) +
...
+
m α (ST S )
m −1
= S(α +0 α T +1 α T 2 +
2
...
q(STS ) = Sq(T )S
(3)
Let p(x) be the minimal polynomial of T over F
...
Suppose Let f (x)be the poly- nomial of T such
that deg(f (x)) < deg(p(x)) and f (STS −1 ) = 0
...
T satisfy the polynomial f (x) and
deg(f (x)) < deg(p(x)), which is contradiction to the minimality of p(x)
...
UNIT IV
Consequently, p(x) is the minimal polynomial of STS −1 also let g(x) be the minimal
polynomial of STS −1 (i
...
)Sg(T )S−1 = 0 ⇒ Sg(T )S−1 = 0 ⇒ g(T ) = 0
...
e) T
satisfies the polynomial of g(x)
...
Again h(STS −1) = Sh(T )S−1 = 0
...
e
...
Consequently, g(x) is a minimal polynomial of T also
...
Definition 4
...
(Theorem 4
...
86 If λ1, λ2,
...
, vk are characteristics vectors of T belonging λ1, λ2,
...
, vk are linearly independent over F
...
Case(ii): If k > 1, To prove: v1, v2,
...
Suppose the
characteristic vector v1, v2,
...
Then there exists
scalars α1, α2,
...
+ αkvk = 0
...
+ βjvj = 0
where β1 = β2 =
...
+ βjv j)T = 0 · T β1v1T +
β2v2T +
...
+ βj(v jT ) = 0
β1λ 1v1 + β2λ 2v2 +
...
+ (βjλ j)vj = 0
...
+ λ1βjvj = 0
(2) − (3) ⇒ (λ2 − λ1)β2v2 + (λ3 − λ1)β3v3 +
...
(4)
Now, (λj − λ1)βj ƒ= 0, i = 2, 3,
...
(i
...
)
γ2v2 +γ3v3 +
...
(5)
where γ2 = λ2 − λ1 ƒ= 0, γ3 = λ3− λ1
0,
...
, vj are linearly dependent
...
, vk ⇒⇐
...
, vk are linearly independent
...
65
Corollary 4
...
Proof: Letλ1, λ2,
...
To prove: m ≤ n
...
, vm be the characteristic vector T belonging to the characteristic
roots λ1, λ 2,
...
By Theorem 4
...
, vm are lin- early independent
...
Corollary 4
...
Proof: Let λ1, λ2,
...
Let v1, v2,
...
, λn
...
, vn are distinct for 1
≤ i, j ≤ n
...
, λn distinct characteristic root)
...
, vn are distinct
...
86, v1, v2,
...
Let v ∈ V , since dimF (V ) = n any subset of n + 1 vectors are linearly dependent
...
e
...
, vn, v are linearly dependent
...
, αn, α
not all zero such that α1v1 + α2v2 +
...
In particular α ƒ= 0,
αvv == −(α
− (α11vv11++αα22vv22++
...
++ααnnvvnn))1
α−1
⇒ v = (−(α α )v + (−α−1α )v +
...
2 + β v where
n n β = −α iα , i = 1, 2, i
...
⇒ v ∈ L(S) ⇒ {v1, v2,
...
∴ {v1, v2,
...
Canonical forms:
Triangular forms: Since the basis used at any time is completely at our choice for
a given linear transformation T
...
Such nice forms of matrices as
canonical forms
...
Definition 4
...
The definition already
defined interms of matrices as m2(T ) = Cm 1C−1 ⇒ A = CBC −1
...
Remark 4
...
(i
...
) S ∼ T ⇒ S is similar to T ⇒ T = CSC−1
...
UNIT IV
66
is reflexive, symmetric and transitive
...
The
equivalence class of an element in A(V ) under the relation similarity is called the
similarity class and is denoted by [S]
...
Instead, we try to establish some kind of land- mark in each
similarity classes of one of these to set if the other is in it, but this procedure is not
feasible
...
Definition 4
...
W is said to be invariant
under T ∈ A(V ) if WT ⊂ W
...
92 If W ⊂ V is a invariant under T then T induces a linear transformation
T¯ on V /W defined by (v + w)T¯ = v + W
...
If p1 (x) is a minimal polynomial of T¯over F and if p(x) is that for T
then p1(x)/p(x)
...
(i
...
) T : V → V is a homomorphism
...
Define v¯T¯ = (v +W )T¯ =
vT + W
...
Now,
(v¯1 + v¯2 )T¯ = ((v1 + W ) + (v2 + W ))T¯
= (v1 + v2)T +W
= v 1T + v 2T + W
= (v1T + W ) + (v2T + W )
= (v1 + W )T¯+ (v2 + W )T¯
= v¯1 T¯ + v¯2 T¯
67
(αv¯1 )T¯ = (α(v1 ) + W )T¯
= (αv1 + W )T¯
= (αv1)T + W
= α(v1T ) + W
= α(v1T +W )
= α((v1 + W )T¯)
= α(v¯1 T¯ )
∴ T¯ defines linear transformation on V¯
...
+αkxk ∈ F [x]
...
(i
...
) α0 + α1T + α2T 2 +
...
(1)
Claim: q(T ) = q(T¯ ) we prove that q(T¯) = 0
...
(2), for any k > 0
Consequently for any polynomial q(x) ∈ F [x], q(T ) = q(T¯), for
q(T¯) = α0 + α1 T¯ + α2 T¯ 2 +
...
+ αk (T¯k )
= α0 + α1T + α2T 2 +
...
∴ T satisfies
q(x) ∈ F [x]
...
e
...
Also
given that p(x) is minimal polynomial for T
...
e
...
We have to show that
p1 (x)/p(x)
...
p(x) satisfies T¯
...
54,
p1 /p(x) (here p(x) and h(x) = p(x) = p1 (x))
...
93 Triangular matrix A matrix M is called triangular if the entries
above the main diagonal are zero (or) equivalently T is a lin- ear transformation
on V over F matrix of T in the basis v1, v2,
...
UNIT IV
68
triangular if
v1T = α11v1
v2T = α21v1 + α22v2
v3T = α31v1 + α32v2 + α33v3
...
...
...
vnT = αn1v1 + αn2v2 + αnnvn
...
e
...
Theorem 4
...
Proof: We prove this theorem by induction on the dimension of V over F
...
Then every matrix representation of T ∈ A(V ) is a scalar
...
e
...
Suppose the theorem is true for all
vector spaces over F of dimension (n − 1)
...
Since the Linear Transformation T on V has all its characteristic root in F
...
Then there exists a non-zero vector v1 ∈
V such that v1T = λ1v1
...
Then,
WT = {(αv1)T |α ∈ F, v1 ∈ V }
= {α(v1T )|α ∈ F, v1 ∈ V }
= {αw1|w1 ∈ V, α ∈ F }
⇒ WT ⊂ W
...
Let V¯ = V /W then dim(V¯ ) = dim(V /W ) = dim(V ) − dim(W ) = (n − 1)
...
92, T induces the linear transformation T¯ on V¯
...
∴ All the roots of
minimal polynomial of T¯being the roots of minimal polynomialof
T must be in F
...
Since dim(V¯ ) = n − 1, then by induction hypotheses there is a basis consists
of the vector v¯2 , v¯3 ,
...
...
v¯n T¯ = αn2 v¯2 + αn3 v¯3 +
...
, vn be the elements of V mapping into v¯2 , v¯3 ,
...
(i
...
) v¯2 = v2 + W ; v¯3 = v3 + W ;
...
Then v1 , v2 ,
...
Since v¯2 T¯ = α22 (v2 + W ) = α22 v2 + W
(v2 + W ) + T¯= α22v2 + W v2T +
W = α22v2 + W
⇒ v2T − α22v2 ∈ W
⇒ v2T − α22v2 is a multiples of v1, say α21v1
⇒ v2T − α22v2 = α21v1
v2T = α21v1 + α22v2 Similarly
v3T = α31v1 + α32v2 + α33v3
...
...
, n)
(i
...
) the basis v1, v2,
...
, vn} istriangular
...
95 If V is a dimensional over F and T ∈ A(V ) has matrix m1(T ) in the
basis v1, v2,
...
In fact if S is the linear transformation
of V defined by viS = wi for i = 1, 2,
...
Remark 4
...
Proof: Let A ∈ Fn has all its characteristic roots in F
...
UNIT IV
70
transformation T on Fn whose matrix in the basis is precisely A
...
0)
v2 = (0, 1,
...
...
vn = (0, 0,
...
Hence by Theorem
4
...
However by
Theorem 4
...
97 characteristic root of triangular matrix is diagonal matrix
...
98 If V is n dimensional over F and if T ∈ A(V ) has all its characteristic
roots in F, then T satisfies a polynomial of degree n over F
...
∴ By Theorem 4
...
, vn such that
v1T = λ1v1
v2T = α21v1 + λ2v2
v3T = α31v1 + α32v2 + λ3v3
...
...
+ λii−1vi−1 + λivi for i = 1, 2, 3,
...
Equivalently,
v1T − λ1v1 = 0
(i
...
) v1(T − λ1) = 0
v2T − α21v1 + λ2v2 = 0 (i
...
)
v2(T − λ2) = α21v1
v3(T − λ3) = α31v1 +α32v2
...
...
+ αii−1vi−1 for i = 1, 2,
...
(2)
But (T − λ2)(T − λ1) = (T − λ1)(T − λ2)
v1(T − λ2)(T − λ1) = v1((T − λ1)(T − λ2)) = 0 (by (1))
Similarly v1((T − λ3)(T − λ2)(T − λ1)) = 0 Continuing
this type of computation fields,
v1((T − λi)(T − λi−1) ············· (T − λ2)(T − λ1)) = 0
v2((T − λi)(T − λi−1) ············· (T − λ2)(T − λ1)) = 0
...
...
, n
for i = n, let S = (T − λn)(T − λn−1) ················ (T − λ2)(T − λ1)
v1S = v1((T − λn)(T − λn−1) ················ (T − λ2)(T − λ1)) = 0
v2S = v2((T − λn)(T − λn−1) ················ (T − λ2)(T − λ1)) = 0
Similarly v3S = 0,
...
= vnS = 0
The matrix S satisfies v1S = 0, v2S = 0,
...
Since S anihilates a basis of V , S
must anihilates all of V
...
(T − λn)(T − λn−1) · · · (T − λ2)(T − λ1) = 0
...
Canonical Form:
The relation on A(V ) defined by similarly is an equivalence relation
...
Given
two linear transformation, by scanning the similarity class of one we could
determine whether or not they are similar
...
Instead we try to establish some kind of land mark in each similarity class, and
the way of going from any element in the class to this landmark
...
These matrices will be called canonical
forms
...
4
...
99 The trace of a matrix A is the sum of the elements on the main
diagonal of A we shall write trace of A as trA (i
...
) if A = (αij) i, j = 1, 2,
...
Then
n
Σ
trA =
αii
i=1
Lemma 4
...
Trace of λA = λ(trA)
...
tr(A + B) = trA + trB
...
tr(AB) = tr(BA)
...
n
λA = (λαij) i, j = 1, 2,
...
, n where (γij) = αij +
βij, i, j = 1, 2,
...
tr(A + B) = tr(γij)
=
n
Σ
γii
i=1
n
=
Σ
(αij + βii)
i=1
n
=
Σ
i=1
αii +
n
Σ
βii
i=1
= trA + trB
73
(3) Let A = (αij) ∈ Fn, B = (βij) ∈ Fn
...
(i)
i=1 k=1
n
Σ
Let BA = (λij), where λij =
αikβkj
k=1
tr(BA) =tr(λij)
=
=
n
Σ
i=1
n
Σ
λii
Σn
(
i=1 k=1
n Σn
Σ
from (i) tr(AB) =
(
=
i=1 k=1
n Σn
Σ
(
βikαki)
αikβki)
βkiαik)
k=1 i=1 n
= Σ (λkk)
k=1
n
=
Σ
(λii)
i=1
= tr(BA)
Definition 4
...
102 The above definition is meaningful and depends only on T
and not on any particular basis of V
...
103 If A is invertible then tr(ACA−1) = trC
...
Then, tr(A(CA−1)) = tr(A(B)) = tr(BA) =
tr(CA−1A) = tr(C(AA−1)) = tr(C)
...
104 If T ∈ A(V ) then tr(T ) is the sum of the characteristic root of T
...
UNIT IV
74
of p(x)
...
trJ=Sum of the characteristic root of T
...
(by (1))
Corollary 4
...
Remark 4
...
Proof: Given T is nilpotent ⇒ there exist k > 0 such that tk = 0
...
Now, vT 2 =
(vT )T = (λv)T = λ(V T ) = λ(λv) = λ 2v
...
Hence all the
characteristic roots of T are 0 (since λ is only characteristic root)
...
Since T is nilpotent, T i, for i ≥ 1 is nilpotent ⇒
trT i = 0∀i ≥ 1
...
107 Converse of trT i = 0, ∀i ≥ 1 then T is nilpotent
...
In general T need not be nilpotent
...
108 Let
...
(1)
trT= Sum of diagonal element= 1+1=2(1)=0[∵ char F=2]
...
But T is not nilpotent, since Tk = T ƒ= 0 ∀i
...
109 If F is a field of characteristic 0 and if T ∈ A(V ) is there exist trT i =
0 ∀i ≥ 1, then T is nilpotent
...
+ αm−1T be the minimal polynomial of T
...
+ αmT 0 = 0
⇒ tr(T m + α 1T m−1 + α 2T m−2 +
...
e
...
∴ T is singu- lar,by
Theorem 4
...
Then there exists v ƒ= 0 in V such that vT = 0 (by
75
Theorem 4
...
(i
...
) 0 is a characteristic root of T
...
Now in Kn, T can be
brought to the triangular form (since 0 is the characteristic root if T )
...
=
·
α2
∗
·
T2 =
·
·
∗
0
α3
2
· · · αn
·
·
βn
where
0Σ 0
∗T
···
···
0
0
is n − 1 × n − 1 matrix
∗
· · · αn
...
By using induction on dimension
2
2
(or repeating the argument on T2
...
= αn = 0 ⇒ T is brought to the triangular form and all its diagonal
elements are zero
...
Hence T is nilpotent
...
110 If F is of characteristic 0 and if S, T ∈ A(V ) are such that
ST − TS commutes with S then ST − TS is nilpotent
...
UNIT IV
76
Proof: Given F is of characteristic 0 Let k ≥ 1 then,
(ST − TS)k = (ST − TS)k−1(ST − TS)
= (ST − TS) k−1(ST ) − (ST − TS) k−1 (TS)
= S((ST − TS) k−1T ) − ((ST − TS) k−1 T )S
= SB − BS where B = (ST − TS)k−1T
⇒ tr((ST − TS) k ) = tr(SB − BS)
= tr(SB)− tr(BS)
= tr(BS − BS)
=0
∴ tr((ST − TS)k) = 0 k ≥ 0
...
109 ST − TS is nilpotent
...
111 Transpose:If A = (αij) ∈ Fn then the transpose of A, written as
A′, is the matrix A′ = (γij) where γij = αji ∀i and j
...
112 For all A, B ∈ Fn, 1
...
(A + B)′ = A′ + B′
3
...
Then A′ = (βij) where βij = αji
(A′)′ = (γij) where γij = βji ⇒ (A′)′ = (βji) = αij = A
...
113 (i) A is said to be symmetric matrix if A′ = A
For example,
d
o g
o n e
g e t
(ii) A said to be skew symmetric matrix if A′ = −A
For example,
...
114 AdjointOperator:LetF beafieldofcomplexnumber
...
Then A∗ = (γij where γij = αji the complex conjugate of αji so here ∗ is
usually called the hermitian adjoint on Fn, denoted by A∗ defined as A∗ = (γij)
where γij = αji
...
Remark 4
...
2
Definition 4
...
(A∗)∗ = A
4
...
(A + B)∗ = A∗ + B∗
3
...
117 Suppose F be a field of complex numbers and that adjoint
∗ on Fn is the hermitian adjoint
...
Definition 4
...
119
...
Any square matrix A can be uniquely written as a sum of a hermitian and a
skew1 hermitian1matrices
A = (A + A∗) + (A − A∗)
...
If A ƒ= 0 ∈ Fn then trace of AA∗ > 0
...
If A1, A2,
...
+ AkA∗
1
A1 = A2 =
...
= 0 then
2
k
¯
...
If λ is a scalar matrix then λ∗ = λ
Example 4
...
λ=
3i
0
Σ
0
∗
;λ =
3i
...
−3i
0
0
−3i
Σ
∗
¯
⇒ λ =λ
Result 4
...
e
...
Proof: Let A be a hermitian matrix then A = A∗ (i
...
) A¯′ = A and λ
be a characteristic root of T ∈ A(V )
...
+ x¯nxn =
Σ
n
2
ƒ= 0
|xi
|
i=1
¯) = 0 ⇒ λ = λ
¯
...
⇒ (λ − λ
Result 4
...
UNIT V
80
5
...
1 Let F be a field; a field K is said to be an extension of F
if K ⊃ F
...
Remark 5
...
Example 5
...
1
...
2
...
3
...
Remark 5
...
1
...
But a vector space
over F cannot be considered as an extension
...
If K is an extension of F, then under the ordinary field operation in K, K is a
vector space over F
...
in K relative to F
...
5 Let F be a given field and K be an extension of F
...
(i
...
) degree of K over
F=dimension of K over F = dimF (K)
...
6 [K : F ] will denote the degree of K over F
...
7 When K is finite dimensional as a vector space over F we say that
[K : F ] is finite and we call K is finite extension of F
...
8
...
If F is an arbitrary field then clearly, F is a subfield of F
...
Here the set S = {1} consisting of only the unity of F
...
S forms a basis of F over F
...
e
...
Here F is finite extension of F
...
Since the field of complex numbers C contains the field of real num- bers R,
C is an extension of R
...
Claim:
S is a bases of C over R
...
(1)
Let a + ib be an arbitrary element in C
...
(i
...
) Any element
in C can be uniquely written as a linear combination of 1 and i ⇒ L(S) = C
...
∴ C is an finite extension of R
...
Q( 2) = {a + b 2|a, b ∈ Q} is a field with respect to addition and
multiplication
...
Clearly Q is a subfield√of Q 2
...
e
...
Consider the set S = {1,
2}
...
√
...
Let a + b 2 be any element in Q( 2) and
√
√
√
a + b 2 = a · 1 + b · 2
...
(2)
√
From (1) and (2),
√ S forms a basis
√ of Q( 2) over Q
...
e
...
∴ Q( 2) is a finite extension of Q
...
Consider an indeterminate x over a field F
...
Then K is an extension of F
...
αn ∈ F
α0 · 1 + α1 · x +
...
= 0 = 0 + 0 · x + 0 · x2 +
...
The set S = {1, x, x2,
...
It is an infinite subset of K
which forms a basis of K over F
...
Theorem 5
...
Proof: Let [L : K] = m and let [K : F ] = n
...
, vm} be a basis of L over K
...
wn} be a
basis of K over F
...
, m, j = 1, 2,
...
To Prove:
S forms a basis of L over F
...
e
...
(i
...
) to show that every element in L can be written
5
...
Let
t ∈ L be any element
...
, vm} with coefficient in K, in particular t = k1v1, k 2v2,
...
(1)
Since [K : F ] = n and {w1, w2,
...
, wn}
with the coefficients in F
k1 = f11 w1 + f12 w2 +
...
+ f2n wn
·
·
·
km = fm1 w1 + fm2 w2 +
...
(∗)
Substitute these values of k1, k2,
...
+ f1n wn )v1 + (f21 w1 + f22 w2 +
...
+ (fm1 w1 + fm2 w2 +
...
, m;
j = 1, 2,
...
+ f1n wn v1 + f21 w1 v2 + f22 w2 v2 +
...
+ fm1 w1 vm + fm2 w2 vm +
...
+ f1n (wn v1 ) + f21 (w1 v2 ) + f22 (w2 v2 ) +
...
+ fm1 (w1 vm ) + fm2 (w2 vm ) +
...
(A)
(i
...
) t is a linear combination of {vjwj|i = 1, 2,
...
n} over F
∴ L(S) = L
...
, m, j = 1, 2,
...
Suppose, f11 (w1 v1 ) + f12 (w2 v1 ) +
...
f2n (wn v2 ) +
...
+ fmn (wn vm ) = 0
...
m, j = 1, 2,
...
Regrouping the (3) we get,
(f11 w1 + f12 w2 +
...
+ f2n wn )v2 +
...
+ fmn wn )vm = 0
...
e
...
+ kmvm = 0, ki ∈ K
...
, vm}form the basis of L over K so v1, v2,
...
= km = 0
k1 = 0 ⇒ f11 w1 + f12 w2 +
...
+ f2n wn = 0
·
·
·
km = 0 ⇒ fm1 w1 + fm2 w2 +
...
(5)
Since {w1, w2,
...
from (5) we have,
f11 = f12 =
...
= f2n = 0
·
·
·
fm1 = fm2 =
...
e
...
, m, j = 1, 2,
...
∴ S = {viwj|i = 1, 2,
...
,
n} is linearly independent
...
∴ [L : F ] = dimF (L) = mn = [L : K][K : F ]
...
∴ L is a finite extension
of F
...
10 If L is a finite extension of F and K is a subfield of L
which contains F, then [K : F ]/[L : F ]
...
Clearlyany
elementinL, linearly independentoverK, linearly independent over F
...
By
previous theorem, [L : F ] = [L : K][K : F ]
...
Definition 5
...
, αn ∈ F, not all zero such that α0a + α1a +
n
n−1
...
Remark 5
...
+ αn, αi ∈ F
...
+ αn = 0
⇒ p(a) = 0
...
e
...
(i
...
) p(a)=0
...
5
...
Let K be an extension of F and a ∈ K
...
The intersection of all subfields of K which are members of M is also a subfield
of K
...
Obviously F (a) contains both F and a because
each members of M contains both F and a
...
∴ F (a) is the smallest
subfield of K containing both F and a
...
Our assumption
of F (a), so for has been purely an external one, we now give an alternative and
more constructive description of F (a)
Suppose K is an extension field of F
...
+ αn
U={
β am + β a
0
m
|αiβj ∈ F, β0a
+
...
+ βn ƒ= 0},
n
where m and n non-negative integer
...
It can be
easily seen that
(i) α, β ∈ U ⇒ α − β ∈ U
(ii) α ∈ U, 0 ƒ= β ∈ U ⇒ α∈ U
...
Claim: U = F (a)
...
∴ U is
a subfield of K containing both F and a
...
e
...
(1) [Since F (a)
is the smallest subfield of K containing both F and a]
...
+ αn = 0
Since F (a) is a subfield of K contain both F and a, F (a) must contain all such
elements being a subfield of K
...
Theorem 5
...
[(i
...
) [F (a) : F ] is finite iff a ∈ K is algebraic over F ] Proof: Suppose F (a) is
a finite extension of F
...
To prove: a ∈ K is
algebraic over F
...
, am are all in F (a)
...
[∵ [F (a) : F
2
] = m] ∴ These (m +1) elements
85
of F (a) are linearly dependent over F
...
, αm ∈ F , not all
zero such that α0 · 1 + α1a + α2a2 +
...
(1)
Letp(x) = α0 + α1x +
...
By (1) p(a)=0 (i
...
) a satis- fies a non-zero
polynomial in F [x]
...
Conversely, suppose that a ∈ K is
algebraic overF
...
Let p(x) be a
polynomial in F [x] of smallest positive degree such that p(a)=0
...
Suppose not, p(x) is reducible over F
...
Now, 0 = p(a) = f (a)g(a) ⇒ g(a)f (a) = 0 ⇒ f (a) = 0(or) g(a) = 0
[∵ f (a), g(a) ∈ F and F is a field F is an integral domain and has no zero divisor]
...
We have
either deg(f (x)) ≥ deg(p(x)) or deg(g(x)) ≥ deg(p(x))
...
Which is a contradiction to
the minimality of degree of p(x)
...
Define a mapping ψ : F [x] → F (a) by (h(x))ψ = h(a)
...
Let h1(x) and h2(x) ∈ F [x]
...
(1)
(h1(x)h2(x))ψ = ((h1h2)x)ψ
= h1h2(a)
= h1(a)h2(a)
= (h1(x))ψ(h2(x))ψ
...
Let V = Kerψ = {h(x) ∈
F [x]|(h(x))ψ = 0} where 0 is identity element of F (a)
...
Let h(x), g(x) ∈ V , then (h(x))ψ = 0 and (g(x))ψ = 0 ⇒ h(a) = 0 and g(a) =
0
...
∴ S(a) = h(a) − g(a) = 0 ⇒ S(x) ∈ V ⇒ h(x) − g(x) ∈ V
...
Let t(x) = h(x)f (x); t(a) =
h(a)f (a) = 0 ⇒ t(x) ∈ V ⇒ h(x)f (x) ∈ V, h(x) ∈ V, f (x) ∈ F [x]
...
(4)
From (3) and (4), V is an ideal of F [x]
...
Since p(x) is irreducible, V is a maximal ideal
in F (x)
...
By the general homomorphism F [x]/V is
isomorphic to the image of F [x] we have shown that the image of F [x] under ψ is
a subfield of F (a)
...
More clearly F [x]/V is isomorphic
5
...
Let V = (p(x)) be the ideal generated by p(x)
...
In view of this isomorphism we obtained between F [x]/V and F (a) we get that,
[F [x]/V : F ] = deg(p(x))
degF (F (x)/V ) = deg(p(x))
degF (F (a)) = deg(p(x))
[F (a) : F ] = deg(p(x))
Hence [F (a) : F ] is finite
...
14 We have actually proved that more, namely that [F (a) : F ]
= degree of the minimal polynomial satisfied by a overF
...
15 Let F be a field and Let F [x] be a ring of polynomial in x over F
...
Prove that F [x]/V is n
dimensional vector space over F
...
Let V + f1 (x), f2 (x) ∈ F [x]/V
...
Also, we define scalar multiplication in F [x]/V over F
...
Then we define, a[V +fx] = V +afx
...
The residue class V is the zero
vector
...
Then,
(i) (a + b)[V + f1 (x)] = V + (a + b)f1 (x)
= V + af1(x) +bf1(x)
= [V + af1(x)] + [V + bf1(x)]
= a[V + f1 (x)] + b[V + f1 (x)] (ii) a[{V
+ f1 (x)} + {V + f2 (x)}] = a[V + f1 (x) + f2 (x)]
= V + a(f1 (x) + f2 (x))
= V + af1(x) +af2(x)
= [V + af1(x)] + [V + af2(x)]
= a[V + f1 (x)] + a[V + f2 (x)]
(iii) a[b(V + f1 (x))] = a[V + bf1 (x)]
= V + (ab)f1 (x)
= ab[V + f1 (x)] (iv)
[V + f1 (x)] = V + 1 · f1 (x)
= V + f1 (x)
Hence F [x]/V is a vector space over F
...
We claim that V +1, V +x, V +x2,
...
First we shall show that these n
elements of F [x]/V are linearly independent over F
...
+ an−1(V + xn−1) = V
V + a0 + a1x +
...
+ an−1xn−1 ∈ V
⇒ a0 + a1x +
...
+ an−1xn−1]
a0 + a1x +
...
= an−1 = 0
V +1, V +x, V +x2,
...
Let V +f (x) be any
element in F [x]/V
...
By division algorithm there exists q(x), r(x)
∈ F [x] such that f (x) = q(x)g(x)+r(x) where either r(x)=0 or deg(r(x)) < deg(g(x))
...
+ an−1xn−1
where a0, a1,
...
e
...
+ an−1(V + xn−1)
...
, V + xn−1 forms a basis of F [x]/V over F
...
e
...
Definition 5
...
Remark 5
...
1
...
e
...
2
...
Example 5
...
√
1
...
∴ x2 − 3 is a
min√imal polynomial
of
3 over Q
...
5
...
x3 − 2 is a minimal polynomial of cubic root 2 over Q
...
19 If p(x) is a minimal polynomial of a over F of degree n then
[F (a) : F ] = n
Proof: Suppose p(x) is a minimal polynomial for a over F of degree n
...
+ αn, αi ∈ F
...
+ αn = 0 ⇒ an = (−α1)an−1 +
...
e
...
, an−1 and ∴ an ∈ L(S) where S = {1, a, a ,
...
(3)
n+1
n
a = (−α1)a + (−α2)an−1 +
...
(2)
Sub (1) in (2) we get
an+1 = −[α1(−(α1an−1 + α2an−2 +
...
+ αna]
= − [(α2 − α21)an−1 + (α3 α1α2−)an−2 +
...
, an−1 over F
...
0, an+k
L(S) (i
...
) a linear combination
≥
∈
of 1, a,
, an−1
...
F (a) is the subfield of K generated by a over F
...
In order to show that F (a) is a finite extension of
F
...
Since F (a) is the field containing a, 1, a, a2,
...
Let L(S) denote the set of all linear combination of S
...
Consequently L(S) F (a)
...
It is clear that
L(S) = F (a)
...
It follows that the product of 2 elements
of L(S) is a linear combination
of 1, a, a2, , an−1 and
⊂
is therefore contained in L(S)
...
Hence L(S) is a
≥
∈
subring of F (a)
...
+ 0 · an−1
...
e
...
, a
...
(i
...
) L(S) contains the unit element
...
Hence L(S) is a ring with unit element and
is without zero divisor
...
Consider T = {β0 + β1a + + βn−1an−1/βi ∈ F }
...
Claim: T is a field
...
Let h(x) = β0 + β1a +
...
Since u ƒ= 0 and u = h(a) we have, h(a) ƒ= 0
...
e
...
e
...
Hence we
can find polynomial S(x)
89
and t(x) in F [x]
⇒ p(x)S(x) + h(x)t(x) = 1
But then 1 = p(a)S(a) + h(a)t(a)
= h(a)t(a) [∵ p(a) = 0]
= ut(a) [∵ h(a) = u]
u−1 = t(a)
Since t(a) is the value of polynomials t(x) at x = a follows that t(a) is a linear
combination of 1, a, a2,
...
Also since an+k is a linear combination of 1, a, a2,
...
It follows that t(x) is a linear combination of 1, a, a2,
...
e
...
e
...
T
is a field
...
We have identified
F (a) as the set of all expression β0 + β1a +
...
Hence L(S) is a field containing F and a and itself contained in F (a)
...
Also the set S is linearly independent
...
There exists elements α0, α1,
...
+ αn−1a
= 0 ⇒ a satisfy a polynomial, α0 + α1x +
...
This
contradiction shows that S is linearly independent
...
, an−1} is the basis of the vector space F (a) over the field
F
...
e
...
Theorem 5
...
Proof: Let K be a finite extension of the given field F
...
∴ There exists a minimal polynomial p(x) of degree n over F
satisfies a
...
Theorem 5
...
In otherwords, the elements in K which are algebraic
over F form a subfield of K
...
E = {a ∈ K|a is algebraic over F }
...
E is not
empty and∴is a subset of K
...
∴ a ∈ E
and [F (a) : F ] = m [by Theorem 5
...
Let T = F (a) the T is a subfield of K of
degree m over F
...
(i
...
) T (b) is a subfield of K and is of
degree atmost n over T
...
e
...
e
...
By Theorem 5
...
e
...
UNIT V
90
mn ⇒ [F (a, b) : F ] = [W : F ] = mn
...
But F (a, b), being a
field a, b ∈ F (a, b) ⇒ a ± b, ab, a (b ƒ= 0) ∈ F (a, b) (Since each element of F (a, b)is
algebraic over F ) ⇒ a ± b, ab, a (b ƒ= 0) ∈ E
...
(i
...
) the
elements in K which are algebraic over F form a subfield of bK
...
22 If a and b in K are algebraic over F of degree m and n
respectively then a ± b, ab, a (b
0) are algebraic of degree atmost mn
...
Since b ∈ K is
algebraic over F of degree n, [F (b) : F ] = n
...
But F (a), being a subfield of F , the minimal polynomial
over F (a) satisfies b is of degree atmost n
...
By Theorem
5
...
Consequently each element of F (a,
b) is algebraic of degree not exceeding
mn
...
Hence a±b, ab, a (b ƒ= 0)
b
b
are algebraic of degree at most mn over F
...
23 The extension K of F is called an algebraic extension of
F if every element in K is algebraic over F
...
24 If L is an algebraic extension of K and if K is an algebraic extension
of F, then L is an algebraic extension of F
...
To Prove: L is an algebraic extension of
F , it is enough to prove that u is algebraic over F
...
e
...
Since u ∈ L and L is an algebraic
extension of K, u satisfies a non trivial polynomial xn + σ1xn−1 + σ2xn−2 +
...
, σn ∈ K
...
, σn are
algebraic over F
...
20, M = F (σ1, σ2,
...
Since u satisfy the polynomial xn + σ1xn−1 +
...
, σn are in M = F (σ1, σ2,
...
∴ u is algebraic over M using
theorem 5
...
By Theorem 5
...
M (u) is a finite extension of F and u is an algebraic over F
...
25 A complex number is said to be algebraic number if it is an
algebraic over the field of rational number
...
26 Let a = 2 + 3i then (a − 2)2 = (3i)2 ⇒ a2 + 4 − 4a = −9 ⇒ a2 − 4a + 13 = 0
...
∴ a = 2 + 3i satisfies a polynomial over the field of
rational numbers
...
∴
91
Example 5
...
(b) What is degree of
2 +√ √3 over Q
...
√
solution: (a) Given
2
∈
R algebraic over
Q and the element
2∈R
2
2
satisfy the polynomial,√x − 2 = 0 over Q and x − 2 is an irreducible
...
2 is algebraic
of degree 2
√
√
2
over
Q
...
Sim
is algebraic
algebraic
over
Q
...
satisfies a
polynomial
Q
...
Since Q( 2 +
√
√
3+
3) is field,
√ √
√
√
√ √
( 2 + 3)3 = 11 2 + 9 3 ∈ Q( 2 + 3)
√ √
√ √
Also − 9( 2 + 3) ∈ Q( 2 + 3)
√
√
√ √
√
√ √
1/2[(11 2 + 9 3) − 9( 2 + 3)] = 2 ∈ Q( 2 + 3)
√ √ √ √
√ √
√
2 + 3 − 2 = 3 ∈ Q( 2 + 3) Thus 2,
√
√ √
√ √
3 ∈ Q( 2 + 3) Q( 2, 3) ∈
√ √
Q( 2 + 3)
√ √
√ √
Hence Q( 2, 3) = Q( 2 + 3)
...
Also x2 − 3 is an irreducible polynomial
2 over Q
...
UNIT V
92
√
over L satisfied by 3,
√
[L 3 : L] = 2
√
√
Now [L 3 : Q] = [L 3 : L][L : Q]
= 2 2· = 4
√ √
√ √
√
Let L( 3) = (Q √2) 3√= Q( 2, 3)
= Q( 2 + 3)
√ √
[Q( 2 + 3) : Q] = 4
√ √
⇒ 2 + 3 is of degree 4over Q
...
Let K = Q
Now, [L : K]=2 and [K : Q]=2
...
[L : Q] = [L : K][K :Q]
√
L=K 3
√ √
= (Q 2)( 3)
√ √
= Q( 2 3)
√ √
[L : Q] = [Q( 2 3) : Q]
= [L : K][K : Q] = 2 · 2 = 4
...
28 With the same notation as in above problem
...
√
Solution: Let
2 + 3 5
...
Given 2
over Q and the element
2 ∈ R satisfies the polynom
√ ial x −2 = 0 over Q2 and
x2−2 is a√n irreducible
...
(i
...
)
2 is algebraic
√ of degree 2 over Q
...
e
...
Similarly
3
5 algebraic over Q
...
The degree of
√3
3
algebraic of n over Q = deg(x − 5) = 3
...
[Q( 2 + 3 5) : Q] = degree of p(x) = 6
...
Roots of Polynomials:
Definition 5
...
Lemma 5
...
Proof: Since F ⊂ K, F [x] ⊂ K[x]; p(x) ∈ F [x] ⇒ p(x) ∈ K[x]
...
∴ there exists polynomials q(x) and r(x) in K[x] such that p(x) = (x − b)q(x) + r(x),
q(x) ∈ K[x], where either r(x) = 0 or deg(r(x)) < deg(x − b)
...
e
...
Let r(x) = r ∈ K (i
...
) r must be an element in K
...
From (1) deg(p(x)) = deg(p(x)) = deg(x −
b)q(x) + deg(p(b)) ⇒ n = 1 + m + 0 ⇒ m = n − 1 ⇒ deg(q(x)) = deg(p(x)) − 1
...
31 FactorTheorem: If a ∈ K is a root of p(x) ∈ F [x] where
F ⊂ K then in K[x], (x − a)/p(x)
...
Then by
Remainder theorem in K[x], we have p(x) = (x−a)q(x)+p(a) ⇒ p(x) = (x− a)q(x)+0 (∵
a is a root of p(x)) ⇒ p(x) = (x−a)q(x) ⇒ (x−a)/p(x) ∈ K[x]
...
32 The element a ∈ K is a root of p(x) ∈ F [x] ofmultiplicity
m
m+1
m if (x − a) /p(x)
where (x − a) /p(x)
...
33 A polynomial of degree n over a field can have at most n
roots in any extension field
...
Let p(x) be a polynomial of degree 1 over any F
...
Let a be a root of p(x) in some extension of F
...
In this case p(x)
94
5
...
e
...
The theorem is true when p(x) is of degree 1
...
Let us suppose
that p(x) be a polynomial of degree n over F
...
If p(x)
has no roots in K ,then the theorem is obviously true, because the number of
roots in K is zero which is definitely at most n
...
Let a be the root of multiplicity m then in K[x],
(x − a)m/p(x), m ≤ n (1)
m
⇒ deg((x −ma) ) ≤ deg(p(x)) ⇒ m ≤ n
...
We have p(x) = (x − a)mq(x) where q(x) ∈ K[x] ⇒ deg(p(x)) =
deg((x−a)m)+deg(q(x)); deg(q(x)) = deg(p(x))−deg((x−a)m) = (n−m) ≤ n (1 ≤ m ≤
n)
...
we have, (x − a)m+1 does not
divides p(x) = (x − a)m+1q(x)
...
∴ a is not a root of q(x)
...
Since Kis a field and 0 ƒ= (b − a)m ∈
K and q(b) ∈ K, we have q(b) = 0 ⇒ b is a rootof q(x) in K
...
Since deg(q(x)) = n − m < n, by our induction
hypothesis, q(x)∴has atmost n − m roots in K other than a
...
(i
...
) p(x) has atmost n roots in K
...
∴ By induction hypothesis the lemma
follows
...
34 If p(x) is a polynomial in F [x] of degree n ≥ 1 and is irreducible over
F, then there is an extension E of F such that [E : F ] = n in which p(x) has a root
...
Let V = (p(x)) be the ideal
generated by p(x) F [x]
...
Hence by Theorem
3
...
∴ F [x]/V = E(say) is a field
...
First we shall show that E can be regarded as an
extension of F
...
Let F¯ be the image of F
in E
...
(i) ψ is 1-1:
95
Let α, β ∈ F such that αψ = βψ,
V+α=V+β
(α − β) ∈ V = p(x)
(α − β) = f (x)p(x) for some f (x) ∈ F [x]
⇒ f(x)=0
⇒ (α − β) = 0
⇒α=βψ
is 1 − 1
...
Thus ψ is an isomorphism from F into E
...
Let F¯ = {α + V |α ∈ F }
...
If we identify F and F¯
under this isomorphism we can consider E to be an extension of F
...
First
we shall prove that the n elements {1+V, x+V, (x+V )2 = x2+V, (x+V )3 = x3 + V,
...
[E : F ] = n
...
Let p(x) = β0 + β1x + β2x2 +
...
, βk ∈
F
...
For convenience of notation Let us denote the element xψ = x + V
in the field E as aβk by βk +V, p(x) = (β0+V )+(β 1 +V )x+
...
We shall show
that x + V ∈ E satisfies p(x)
...
+ (βk + V )(x + V )k
= (β0 + V ) + (β1 + V )(x + V ) + (β2 + V )(x2 + V ) +
...
+ βkxk) + V
= p(x) + V
= v (∵ p(x) ∈ V )
= zero element of E
...
∴ An element x + V in the extension E satisfies the
polynomial p(x) ∈ F [x]
...
UNIT V
properties required in the conclusion of the theorem
...
35 If f (x) ∈ F [x] the there is a finite extension E of F in which f (x)
has a root
...
≤
Proof: Let p(x) be an irreducible factor of f (x)
...
∴ deg(p(x))
≤ deg(f (x))
...
Then p(a)
= 0 ⇒ f (a) = p(a)q(a) = 0 ⇒ f (a) = 0
...
Since p(x) is irreducible
over F , by the above theorem, [E : F ] = deg(p(x)) ≤ deg(f (x)) ⇒ [E : F ] ≤ deg(f
(x))
...
36 let f (x) ∈ F [x] be a polynomial of degree n greater than or equal to
q then there is an extension of E of F of degree atmost n! in which f (x) has n roots
...
Let f
(x) ∈ F [x] of degree 1
...
Now F itself is an
extension of F
...
e
...
Now, f (x) = a0x + a0 = 0 ⇒ x =
−a/a0 ∈ F, a0 ƒ= 0 is a root of f (x) = a0x + a
...
There is a
finite extension F of degree atmost 1=1!
...
Now
assume by our induction hypothesis that the theorem is true in any field for all
polynomials of degree less than
n
...
By Corollary 5
...
∴
By remainder theorem, in E0[x], f (x) can be factored as f (x) = (x − α)q(x) + (α)
where deg of q(x) = deg(f (x)) − 1 = (n − 1) < n (i
...
)deg(q(x)) < n By induction
hypothesis there is an extension E of degree atmost (n − 1)! (i
...
) [E : E0] = (n −
∴ roots
...
In E we obtain all n roots of f (x)
...
Thus E is an extension of F of degree atmost n! in which f (x)
has n roots
...
37 The above theorem asserts that the finite extension E of a given
field F in which the given polynomial of degree n over F has n roots
...
+ an, a0 ƒ= 0 ∈ F [x]
...
, αn be n roots of f (x) in E
...
31, f (x) can be factored over E as f (x) = a0(x − α1)(x − α2 · · · (x −
αn)
...
Definition 5
...
f (x)
can be factored as a product of linear factors
...
39 The above theorem guarantees the existence of splitting field
...
Remark 5
...
+ an, αni ∈ F and
n−1E = F (α1, α2,
...
, αn ∈
E
...
41 Let E1 and E0 be two splitting fields of the same polynomial f (x) in F
[x]
...
An isomorphic mapping:
Let F and F ′ be two fields and let E and E ′ be the extension fields of F and F ′
respectively
...
Let τ be an isomorphism of F
onto F ′ for convenience let us denote the image of any α ∈ F under τ by α′ (i
...
)
ατ = α′
...
42 In the fallowing result we can make use of τ to set up an
isomorphism between F [x] and F ′[t]
...
43 Let ψ be an isomorphism of a field F onto a field F ′ such that (α)τ =
α′
...
Proof: Given τ is a isomorphism of F onto F ′
...
Let us define τ ∗ = F [x] F ′[t] as follows, →
let f (x) = α0xn + α1xn−1 +
...
Define
(f (x))τ ∗ = (α0xn + α1xn−1 +
...
+ (αnτ )
= α′0tn + α′1xn−1 +
...
Let f (x) = α0xn + α1xn−1 +
...
+ βm be any two elements in F [x]
...
UNIT V
98
(f (x))τ ∗ = g(x)τ ∗
⇒ (α0xn + α1xn−1 +
...
+ βm)τ ∗
′
⇒ α′0tn + α′1xn−1 +
...
+ βm
⇒ n = m and αi′ = βi′, i = 0, 1,
...
n
⇒ n = m and αi = βi, i = 0, 1, 2
...
+ γn′ be any element of F ′ [t], γ′i ∈
F ′ since τ is onto, there exists γ0 , γ1 ,
...
, (γn )τ = γn′
...
, γn ∈ F [x] and (γ0 xn , γ1 xn−1 ,
...
∴ τ ∗ is onto
...
+ αn + β0xm + β1xm−1 +
...
+ α′n) + (β0′ xm + β1′ xm−1 +
...
+ αn)τ ∗ + (β0x + β1x +
...
Remark 5
...
1
...
2
...
In
particular f(x) is irreducible in F [x] iff f′(t) is irreducible in F ′[t]
...
45 Let τ be an isomorphism of a field F onto a field F ′ defined by (α)τ =
α′ ∀α ∈ F for an arbitrary polynomial f (x) = (α0xn + α1xn−1 +
...
Let us define f ′ (t) = α′0tn +α′1xn−1+
...
If f (x)
is irreducible in F [x], show that there is an isomorphism τ ∗∗ of F [x]/f (x)
onto F ′[t]/f ′[t] with the property that ατ ∗∗ = α′(x + f (x))τ ∗∗ = t + f ′(t)
...
Then by Lemma
5
...
Let f (x) be irreducible in F [x] then f
′
(t) will be irreducible in F ′[t]
...
Now, f (x) and f ′(t) are irreducible both V and v′ are
maximal ideal
...
Define τ ∗∗ : F [x]/V → F ′[t]/V ′
by (g(x) + V )τ ∗∗ = g(x)τ ∗ + V ′ = g′(t) + V ′
...
We have V + g(x) =
V +h(x) ⇒ g(x)− h(x) ∈ V ⇒ [g(x)− h(x)] = [k(x)f (x)] where k(x) ∈ F [x]
[g(x) − h(x)]τ ∗ = [k(x)f (x)]τ ∗
⇒ g(x)τ ∗ − h(x)τ ∗ = (k(x))τ ∗ · (f (x))τ ∗
⇒ g′(t) − h′(t) = k′(t)f ′(t)
⇒ g′(t) − h′(t) ∈ V ′
⇒ V ′ + g′(t) = V ′ + h′(t)
⇒ [V + g(x)]τ ∗∗ = [V + h(x)]τ ∗∗
∴ τ ∗∗ is well defined
...
[V + g(x)]τ ∗∗ = [V + h(x)]τ ∗∗
V ′ + g′(t) = V ′ + h′(t) g′(t) −
h′(t) ∈ V ′
g′(t) − h′(t) = k′(t)f ′(t) for some k′(t) ∈ F ′[t]
⇒ g(x)τ ∗ − h(x)τ ∗ = (k(x))τ ∗(f (x))τ ∗
(g(x) − h(x))τ ∗ = (k(x) · f (x))τ ∗
⇒ g(x) − h(x) = k(x)f (x)
⇒ g(x) − h(x) ∈ V
⇒ V + g(x) = V + h(x)
⇒ τ ∗∗ is 1 − 1
...
∴ corresponding to any polynomial g′(t) ∈
F ′[t]
we have a polynomial g(x) in F [x], V ′ + g′(t) ∈ F ′[t]/V ′ ⇒ V + g(x) ∈ F [x]/V
such that [V + g(x)]τ ∗∗ = V ′ + g′(t) ⇒ τ ∗∗ preserves addition and multiplication
...
UNIT V
100
Thus τ ∗ is an isomorphism of F [x]/V onto F ′[t]/V ′
...
34 we have
shown that F can be imbedded in field F [x]/V by identifying the element α ∈ F
with the residue class (coset) V + α in F [x]/V
...
Example 5
...
α, β be in
F [x]
...
By Lemma 5
...
∴ p(a) = 0
...
(1)
p(b) = b2 + αb + β
= (α + a)2 − α(a + α) − a(a + α)
= α2 + a2 + 2αa − αa − α2 − a2 − aα
=0
∴ b is root in K
Case(i): Suppose b = a
...
Since b = a, p(x) = x2 − 2ax + a2 = (x − a)2 = (x − a)(x − a)
...
Case (ii): Suppose b ƒ= 0
...
Remark 5
...
36
...
48 Let F be the field of rational numbers and f (x) = x3 − 2
...
Determine the degree of the splitting field of this polynomial
f (x) over F
...
Let f (x) = x3−2 ∈ F [x] =
101
Q[x]
...
f(x)=0
x − 2 =0
x3 = 2· 1
x3 = 2[cos0 + isin0]
x3 = 2(cos2kπ +isin2kπ)
3
1
1
x = 2 3 (cos2kπ + isin2kπ)3
2k
1
2kπ
x = 2 3 (cos( 3 ) +isin( π
))
3
Put k=0,1,2
...
The1 polynomial f (x) is1irreducible over
Q by Eisentien criterion
...
∴ [F (21/3 ) : F ]=3 by Theorem 5
...
Let F be the splitting field of f (x)
over F the field F of 21/3 cannot splits f (x) because as a subfield of
1
real field
it cannot contain the complex number but not real number w 1· 2 3
...
1
1
Also by Theorem 5
...
9) ⇒ [F (2 3 ) : F ]/[E : F ] ⇒ 3/6
...
∴ E is the splitting field of f (x) over F of degree6
...
UNIT V
Example 5
...
More about roots
Definition 5
...
+ αixn−i +
...
+(n − i)αixn−i+1+
...
Definition 5
...
If ma = 0 for some m > 0 andsome a ƒ= 0 ∈ F then F is
said to be of finite characteristic
...
Remark 5
...
1
...
2
...
3
...
e
...
Even when the
characteristic of F is p ƒ= 0 we can still describe the poly-nomial with zero
derivative
...
Lemma 5
...
1
...
2
...
3
...
Proof: (1) Let
f (x) = αn + αn−1x +
...
+ αn−mxm + αn−(m+1)xm+1
+
...
+ β2xm−2 + β1xm−1 + β0xm
103
Assume that n > m f
(x) + g(x)
= (αn + βm) + (αn−1 + βm−1)x +
...
+ α1xn−1 + α0xn
(f (x) + g(x))′
= (αn−1 + βm−1) + 2(αn−2 + βm−2)x +
...
+ (n − 1)α1xn−2
+ nα0xn−1
= (αn−1 + 2αn−2x +
...
+ (n − 1)α1xn−2 + nα0xn−1)
+ (βm−1 + 2βm−2x + (m − 1)β1xm−2 + mβ0xm−1) where n = m
= f′(x)+g′(x) (2)
α(f (x)) = ααn + ααn−1x +
...
+ αα0xn
(α(f (x)))′ = ααn−1 + 2ααn−2x +
...
+ nαα0xn−1
= α(αn−1 + 2αn−2x +
...
+ nα0xn−1)
= αf ′(x)
(3) To Prove this part it is enough to prove it in the highly special case,
f (x) = xi and g(x) = xj where i and j are positive
...
Then,
(f (x)g(x))′ = ixi−1xj = (i + j)xi+j−1
...
(2)
f (x)g′(x) = jxixj−1
...
Remark 5
...
For
where they relatively prime as elements in F [x], then they would be able to find
two polynomials a(x) and b(x) in F [x] such that a(x)f (x) + b(x)g(x)=1
...
55 The polynomial f (x) ∈ F [x] has a multiple root iff f (x) and
f ′(x) have a non-trivial (i
...
of positive degree) common factors
...
UNIT V
104
Proof: From the above remark, just may, we may assume without loss of
generality, the roots of f (x) are all lie in F (otherwise extend F to K, the splitting
field of F )
...
Then f (x) = (x − α)mq(x) and q(α) ƒ= 0, q(x) ∈ K[x]
...
Also f ′(α)=0 (i
...
) α is a root of f ′(x)
...
Conversely, suppose that f(x) and f′(x) have a non trivial common factor
...
Suppose not, (i
...
) f(x) has no multiple root
...
Claim: No root of f (x) is a root of
f ′(x) (i
...
f (x) and f ′(x) have no common factor)
f ′(αi)
= (α1 − α2)(α1 − α3) · · · (α1 − αn) + (α2 − α1)(α2 − α3) · · · (α2 − αn)+ (α3 − α1)(α3 −
α2) · · · (α3 − αn) + (αn − α1)(αn − α2) · · · (αn − αn−1)
= Y (αi − αj) ƒ= 0 (∵ αi ƒ= αj for i ƒ= j)
j i
This show that f′(x) =0 holds if one of the roots α1, α2,
...
However if f (x) and f ′(x) have a non trivial common factor, they have
common root, namely, any root of this common that f(x) has a multiple root
...
56 If f (x) ∈ F [x] is irreducible, then
1
...
if the characteristic of F is p = 0,ƒf (x) has a multiple root only if it is of the
form f (x) = g(xp)
...
+ αn−1x + αn, α0 ƒ= 0 be an irreducible
polynomial of degree n ≥ 1 over a field F of characteristic zero
...
+αn−1
...
∴ f ′(x) ƒ= 0, also deg(f ′(x)) < deg(f (x)
...
Suppose if possible f (x) has a multiple root (say α)
...
But f′(x)
0 and f (x) and f ′(x)
′
both being irreducible with deg f (x) < degf (x)
...
e
...
(2) In this case characteristic of F is p = 0
...
ƒ
Let f (x) = α0 + α1x +
...
Let f ′(x) = α1 +ƒ 2α2x +
...
Now,
since f (x) has a multiple root, f ′(x) = 0 (i
...
) α1 + 2α2x +
...
+ 0xn−1 ⇒ α1 = 2α2 = 3α3 =
...
e
...
Since F is of characteristic p 0, p/k or αk = 0
...
αk = 0 then p/k k = k1p
...
ƒ + βn(x ) for some
⇒ positive integer n then f (x) ∈ F [xp]
...
Corollary 5
...
n
n−1
Proof: Let f (x) = xp − x
...
(1)
Now p ∈ F , we mean 1+1+
...
Since F is of characteristic p, the order
of element of the additive group of F is p, p=1+1+
...
Hence pn = 0 ⇒ f (x) = −1
...
By Lemma 5
...
Hence f (x) has
distinct roots
...
58 The extension K of a field F is called a simple extension of F if K =
F (α) for some α ∈ K
...
59 If F is of characteristic zero and if a, b are algebraic over
F then there exists an element c ∈ F (a, b) such that F (a, b) = F (c)
...
Let f (x), g(x) be the irreducible polynomial
over F of a and b respectively and let m, n be their respective degrees
...
e
...
Clearly, every root
of f (x) is a root of f (x)g(x) and K contains the splitting field of f (x)
...
56)
...
, am in K and g(x) has n distinct
roots say b = b1, b2, , bn
in K
...
e
...
We can solve the
equation ai + λbj = a1 + λb1 = a + λb has only one solution λ in K namely,
5
...
These λ’s are finite numbers
...
F has infinite number of elements
...
e
...
Let√c = a +
b ∈ F (a, b)
...
Since a, b ∈ F (a, b), a +
b ∈ F (a, b) ⇒ c ∈
F (a, b) ⇒ F (c) = F (a, b)
...
Since b is a root of g(x), (x − b) is a factor of g(x)
...
Then h(b) = 0 ⇒ b is a root of h(x) ⇒ (x − b) is a factor of h(x)
...
e
...
If j ƒ= 1, h(bj) = f (c − rbj) ƒ= f (a) ƒ= 0 ⇒
f (c − rb j) ƒ= 0 (i
...
) (x − bj) is nota factor of h(x)
...
Thus, (x − b) is a gcd of h(x) and
g(x) over F of K
...
Since deg(x − b) = 1 we see that the gcd of g(x) and h(x) in
K[x] is exactly x − b
...
Hence b ∈ K = F (c) ⇒ b ∈ F (c)
...
(2)
From (1) and (2), ⇒ F (a, b) = F (c)
...
60 Any finite extension of a field of characteristic zero is a simple
extension
...
, αn be algebraic over F of characteristic zero
...
, αn) = K
= F (α1, α2), (α3, α4,
...
, αn) ∵ F (α1, α2) = F (γ1)
= F (γ1, α3), (α4,
...
, αn)
·
·
·
= F (γn−2), αn
= F (γn−1)
F (α1, α2,
...
Title: algebra class notes
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Description: This note is very helpful to understand this topic easily