Proofs of Isomorphism

Two ways to prove f is Isomorphism:

1) By definition:
f is Homomorphism + f bijective (= surjective + injective)

2) f is homomorphism + f has inverse map f^{-1}

Note: The kernel of a map (homomorphism) is the Ideal of a ring.

Two ways to construct an Ideal:
1) use Kernel of the map
2) by the generators of the map.

Two ways to prove Injective:

1) By definition of Injective Map:
f(x) = f(y)
prove x= y

2) By Kernel of homomorphism:
If f is homomorphism
f: A \mapsto B
Prove Ker f = {0}

Note: Lemma:
\text{f is injective} \iff Ker f = {0}

Proof Isomorphism 4 Steps:
1. Define function f:S -> T
Dom(f) = S
2. Show f is 1 to 1(injective)
3. Show f is onto (surjective)
4. Show f(a*b) = f(a). f(b)

Example:
Let T = even Z
Prove (Z,+) and (T,+) isomorphic
Proof:
1. Define f: Z -> T by
f(a) = 2a
2. f(a)=f(b)
2a=2b
a=b
=> f injective

3. Suppose b is any even Z
then a= b/2 ∈ Z and
f(a)=f(b/2)=2(b/2)=b
=> f onto

4. f(a+b) =2(a+b) =2a+2b =f(a)+f(b)
Hence (Z,+) ≌ (T,+)

Isomorphism (≌)
1. To prove G not isomorphic to H:
=> Prove |G| ≠ |H|

2. Isomorphism class = Equivalence Class
B’cos “≌” is an Equivalence relation.

3. Properties of Isomorphism (≌):
i) match e
\phi(e) = e'
ii) match inverse
\phi (g^{-1} )= (\phi (g))^{-1}

iii) match power
\phi(g^{k})= (\phi(g))^{k}

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Artin Field Extension

Emile Artin’s very unique book “Galois Theory” (1971) on “Finite Field Extension” interpreted by Vector Space.

Let H a Field with subfield G
F is G’s subfield:
H ⊃ G ⊃ F

Example:
Let
F = Q = Rational Field
G = Q(√2) = Larger Extended Field Q with irrational root √2
H = Q(√2, √3) = Largest Extended Field Q with irrational roots (√2 & 3)

{1, √2} forms basis of Q(√2) over Q

{1, √3} basis of Q(√2, √3) over Q(√2)
[since √3 ≠ p+ q√2 , ∀p,q ∈ Q]

=> {1,√2, √3, √6} basis of Q(√2, √3) over Q
=> Q(√2, √3) is a 4-dimensional Vector Space over Q.

Isomorphism (≌)

Q(√2) Q[x] / {x² – 2}

Read as:

Q(√2) isomorphic to the quotient of the Polynomial ring Q[x] modulo the Principal Ideal {x² – 2}
Q[x] the Polynomial Ring
{x² – 2} is the Principal Ideal
Complex Number (C)
C = R[x] / {x² + 1}
R[x] the Polynomial Ring with coefficients in the Field R
{x² + 1} is the Principal Ideal
Questions:
Since R[x] / {x² + 1} is the Field C

Why below are not Fields ?
R[x] / {x³ + 1}
R[x] / {x^4 + 1}
C[x] / {x² + 1}

Hint: they are not irreducible in that particular Field, not a Principal Ideal.

Note: C[x] the Polynomial Ring with coefficients in the Field C

20130419-124638.jpg

What is Ideal ?

Anything inside x outside still comes back inside

=> Zero x Anything = Zero

=> Even x Anything = Even

Mathematically,

1. nZ is an Ideal, represented by (n)

Eg. Even subring (2Z) x anything big Ring Z = 2Z = Even

2. (football) Field F is ‘sooo BIG’ that

(inside = outside)

=> Field has NO Ideal (except trivial 0 and F)

Why was Ideal invented ? because of ‘failure” of UNIQUE Primes Factorization” for this case (example):

6 = 2 x 3
but also
6=(1+\sqrt{-5})(1-\sqrt{-5})
=> two factorizations !
=> violates the Fundamental Law of Arithmetic which says UNIQUE Prime Factorization

Unique Prime factors exist called Ideal Primes: \mbox{gcd = 2} , \mbox{ 3}, (1+\sqrt{-5}), (1-\sqrt{-5})

Greatest Common Divisor (gcd or H.C.F.):
For n,m in Z
gcd (a,b)= ma+nb
Example: gcd(6,8) = (-1).6+(1).8=2
(m=-1, n=-1)

Dedekind’s Ideals (Ij):
6 =2×3= u.v =I1.I2.I3.I4 ;
u= (1+\sqrt{-5})
v=(1-\sqrt{-5})

Let gcd(2,u) = 2M+N.u
M,N in form of a+b\sqrt{-5}

1. Principal Ideals:
2M = (2) = multiple of 2

2. Ideals (nonPrincipal) = 2M+N.u

3. Ideal prime factors: 6=2 x 3=u.v
Let
I1= gcd(2, u)
I2=gcd(2, v)
I3=gcd(3, u)
I4=gcd(3, v)
Easy to verify (by definition):
I1.I2=(2)
I3.I4=(3)
I1.I3=(u)
I1.I4=(v)
=> Ij are prime & unique factors of 6=I1.I2.I3.I4
=> Fundamental Law of Arithmetic satisfied!

=>Ij “Ideal“-ly exist! hidden behind ‘compound’ (2,3,u,v) !

Verify : gcd(2, 1+√-5).gcd(2, 1-√-5)=(2) ?

Proof by definition:

[2m+n(1+√-5)][2m’+n'(1-√-5)]
=[2m+n+n√-5 ][2m’+n’-n’√-5]
= 4mm’+2mn’+2nm’+6nn’
= 2(2mm’+mn’+m’n+3nn’)
= 2M
= 2 multiples
= (2) = Principal Ideal

Chinese Remainder Theorem 中国剩余定理

中国剩余定理CRT (Chinese Remainder Theorem)
X ≡ 2 (mod 3)
X ≡ 3 (mod 5)
X ≡ 2 (mod 7)
Solve X?
明. 程大位 “算法统宗” (1593)
3人同行70稀
5树梅花21支
7子团圆半个月(15)
除百零五(105)便得知
Let remainders:
r_3=2, r_5=3, r_7=2
r= r_3.70+ r_5.21 + r_7.15 (mod 3.5.7)
r= 2.70 +3.21 +2.15 (mod 105)
r= 140 +63 +30 (mod 105)
r= 233 (mod 105)
r= 23 = x_{min}
or X= 23 +105Z (23 + multiples of 105)

——————————————-
CRT: Why 3:70, 5:21, 7:15
X ≡ 2 (mod 3)
X ≡ 3 (mod 5)
X ≡ 2 (mod 7)

1) Find A such that
A ≡ 1 (mod 3)
A ≡ 0 (mod 5)
A ≡ 0 (mod 7)

=> 5|A, 7|A => 35 |A

A=35, 70 …
70 ≡ 1 (mod 3)
=> 70×2 ≡ 2 (mod 3)
2) Find B s.t.:
B ≡ 0 (mod 3)
B ≡ 1 (mod 5)
B ≡ 0 (mod 7)
3|B, 7|B => 21|B
21 ≡ 1 (mod 5)
=> 21×3 ≡ 3 (mod 5)
3) Find C s.t. :
C ≡ 0 (mod 3)
C ≡ 0 (mod 5)
C ≡ 1 (mod 7)
=> 3|C, 5|C => 15|C
=> C=15≡ 1 (mod 7)
15×2 ≡ 2 (mod 7)
4)
X ≡ 70×2 +21×3+ 15×2 (mod 3x5x7)
X≡ 233 (mod 105)
X≡ 23 (mod 105)
X= 23+105Z

—————-Ring Theory ——————————-
Commutative Ring CRT by Ring/ Ideal Theory:
If R is a commutative ring & A1,…An are pairwise coprime ideals,
Prove that if
r_1, r_2, \cdots, r_n belong to R
Then there exists ‘a’ belongs to R with
a + A_j= r_j + A_j ; j ∈[1,n]
Interpretation:
Let R= Z ring
X ≡ 2 (mod 3)
X ≡ 3 (mod 5)
X ≡ 2 (mod 7)
Or
X ≡ rj (mod mj)

Ideal = (m) = mZ
A_3=(3), A_5=(5), A_7=(7)
r_3=2, r_5=3, r_7=2
There exists
a=23
a +A_3 = r_3 + A_3
23+(3)= 2+ (3)
23 + 1×3 =2 + 8×3 =26