## Tuesday, September 01, 2009

### Quotient Rings

The following definitions and lemmas are taken from Jean Tignol's Galois' Theory of Algebraic Equations.

In a previous blogs, I wrote about cosets and ideals. In today's blog, I will show how we can bring these ideas together to define quotient rings. Here are links to review the properties of groups, subgroups, or commutative rings.

Definition 1: A/I

A/I = { a + I such that a ∈ A}

Note: This is a set of sets. For example, if a + I is a coset, then A/I is the set of distinct cosets. For review of the a + I notation, see here.

Example 1.1: Modular sets

The sets Z/nZ are all examples of A/I.

Z/2Z = { 0+2z, 1+2z } since { 0 + 2z = 2 + 2z = 2z + 2z, 1 + 2z = 3 + 2z = ... }

Z/3Z = { 0 + 3z, 1 + 3z, 2 + 3z }

A/I becomes especially interesting when A is a Commutative Ring and I is in an Ideal. I will assume both of these properties for the rest of this article.

(a + I) + (b + I) = (a + b) + I

Definition 3: Multiplication for A/I

(a + I) * (b + I) = ab + I

Lemma 1: Addition for A/I is well defined

Proof:

(1) Let:

s + I = s' + I

t + I = t' + I

(2) Using Lemma 1, here, it follows that:

s -s' ∈ I

and

t - t' ∈ I

(3) So, there exists a,b such that a, b ∈ I and:

s = s' + a

t = t' + b

(4) s + t = (s' + a) + (t' + b) = s' + a + t' + b

(5) s + t + I = s' + t' + (a + b) + I

(6) Since a ∈ I and b ∈ I, it follows that a + b ∈ I (from Closure)

(7) Using Lemma 2, here, we then have:

(a+b) + I = I

(8) So that:

s + t + I = s' + t' + I

QED

Lemma 2: Multiplication for A/I is well defined

Proof:

(1) Assume that I is an ideal.

(2) Let:

s + I = s' + I t + I = t' + I

(3) Using Lemma 1, here, it follows that:

s -s' ∈ I

and

t - t' ∈ I

(4) There exists a,b such that a, b ∈ I and:

s = s' + a t = t' + b

(5) st = (s' + a)(t' + b) = s't' + at' + s'b + ab

(6) st + I = s't' + at' + s'b + ab + I

(7) Since a,t',s',b ∈ I, we have (see Definition 2, here, and Definition 1, here for details):

at' ∈ I

s'b ∈ I

ab ∈ I

(8) So, at' + s'b + ab + I = I [See Lemma 2, here]

(9) And:

st + I = s't' + I

QED

Lemma 3: If A is a Commutative Ring and I is an Ideal, then A/I is a ring

Proof:

(a + I) + (b + I) = (a + b) + I = (b + a) + I = (b + I) + (a + I)

[(a + I) + (b + I)] + (c + I) = (a + b) + I + c + I = (a + b + c) + I = a + I + (b + c) + I = (a + I) + [(b + I) + (c + I)]

0+I is the additive identity since 0 ∈ A and for all a, (a + I) + (0 + I) = (a + 0) + I = a + I

-a+I is the additive inverse since a ∈ A → -a ∈ A and (a + I) + (-a + I) = (a + -a) + I = 0 + I

(5) Associative Rule for Multiplication

[(a + I)*(b+I)](c + I) = (ab + I)(c + I) = (abc + I) = (a + I)(bc + I) = (a+I)[(b+I)(c+I)]

(6) Distributive Rule

(a + I)[(b + I) + (c + I) ] = [(a + I)(b+I)] + [(a+I)(c+I)] = (ab + I) + (ac + I)

QED

Definition 5: Quotient Ring

The ring A/I is called the quotient ring of A by the ideal I.

Example 5.1: Sets that form quotient rings

Z/2Z and Z/3Z are factor rings [See Example 1.1 above for details]

Reference
• Jean-Pierre Tignol, , World Scientific, 2001

## Monday, August 31, 2009

### Cosets

The following definitions and lemmas are taken from Jean Tignol's Galois' Theory of Algebraic Equations.

Definition 1: g + H

Let g + H = { g + h such that h ∈ H }.

Example 1.1:

Let G be the set { 0, 1, 2, 3 }

Let H be the set { 9, 10, 11 }

Then:

0 + H = { 9, 10, 11 }

1 + H = { 10, 11, 12 }

2 + H = { 11, 12, 13 }

3 + H = { 12, 13, 14 }

Definition 2: coset of H in G

The set g + H is a coset if and only if there exists a group G such that g ∈ G and H is a subgroup of G.

Note: Technically, definition 3 above describes a left coset. A coset can also be defined as H + g which is called a right coset. The notation of g + H defines a coset based on addition. Cosets can also be defined on multiplication and represented as gH or Hg.

Example 2.1:

Let G = the set of integers Z
Let H = 2Z = the set of even integers

1 ∈ Z
1+H = odd integers = { ... -3, -1, 1, 3, ... }

Example 2.2: Cyclic Group Z4

Z4 = { 0, 1, 2, 3}

It is a group since:

(1) Closure

Since Z4 is always modulo 4, it is clear that that operation of addition is closed (for example: 3 + 2 = 1)

(2) Associativity

For any elements a,b,c ∈ Z4, (a + b) + c = a + (b + c)

(3) Identity Element

0 is the identity element

(4) Inverse Element

Since it is modulo 4, each element has an inverse: 0+0=0, 1+3=0, 2+2=0

Let H = { 0 , 2 }

H is a subset and H is itself a group since:

(1) It has closure: 0+2=2, 0+0=0, 2+2=0

(2) It has associativity.

(3) It has a identity element: 0

(4) From #1, it is clear that each element is its own inverse.

From Z4 and H, there are 2 distinct cosets:

0 + H = { 0, 2}
1 + H = { 1, 3}
2 + H = { 2, 0} = 0 + H
3 + H = { 3, 1} = 1 + H

Lemma 1:

For the coset H in G:

For all a,b ∈ G:

a + H = b + H if and only if a - b ∈ H

Proof:

(1) Assume that a + H = b + H

(2) So, for all x ∈ (a + H) → x ∈ (b + H)

(3) a ∈ a + H, b ∈ b + H [since H is a group and therefore 0 ∈ H]

(4) a ∈ b + H [follows directly from step #2]

(5) Let c = a - b

(6) b + c ∈ b + H [since b + c = a and a ∈ b + H]

(7) So c ∈ H [from step #6 and Definition 3 above]

(8) Assume that (a-b) ∈ H

(9) Assume that x ∈ a + H

(10) Let y = x - a

(11) y ∈ H [since x = a + y]

(12) y + (a - b) ∈ H since H is a group

(13) x - b ∈ H [since y + (a - b) = (x - a) + (a - b) = x - b]

(14) Then x ∈ b + H since [b + x-b = x]

(15) We can make the same argument if x ∈ b + H so this shows that a + H = b + H.

QED

Lemma 2:

For the coset H in G:

a ∈ H → a + H = H

Proof:

(1) Assume that a ∈ H

(2) Assume that x ∈ a + H

(3) Then there exists h such that x = a + h where h ∈ H [See Definition 1 above]

(4) But if a ∈ H and h ∈ H, then a + h ∈ H. [since H is a group, see Definition 2 above and the Closure property of groups]

(5) So x ∈ H

(6) Assume that x ∈ H

(7) Let b = x - a

(8) Since a ∈ H, it follows that -a ∈ H [See Definition 2 above and the Inverse property of groups]

(9) Since x ∈ H and -a ∈ H, it follows that b ∈ H [from the Closure property of groups]

(10) So then, x ∈ a + H [since b ∈ H and x = a + b from Definition 1 above]

QED

Reference
• Jean-Pierre Tignol, , World Scientific, 2001

## Sunday, August 30, 2009

### Ideals

The following definitions and lemmas are taken from Jean Tignol's Galois' Theory of Algebraic Equations.

Definition 1: stable under multiplication

a set I is stable under multiplication by the elements in A if and only if:

a ∈ A and x ∈ I ↔ ax ∈ I.

Example 1.1: set that is stable under multiplication

A = {1}

I = {1}

Example 1.2: set that is not stable under multiplication

A= {1,2}

I = {2}

2 ∈ A and 2 ∈ I but 2*2=4 is not in I.

Definition 2: An ideal I

Let A be a commutative ring.

a set I is an ideal if and only if I is a subgroup of the additive group of A which is stable under multiplication by the elements in A (see Definition 1 above).

Example 2.1: Example of a set that is an ideal: 2Z

The set 2Z is the set of even integers.

2Z = { ... -2, 0, 2, 4, 6, ... }

2Z is a subset of the set of integers Z and Z is a commutative ring and a group.

2Z is a group on the operation of addition so it is a subgroup of Z.

2Z is stable under multiplication by elements of Z since an even integer multiplied by any integer is an even integer.

Example 2.2: Example of a set that is not an ideal: 2W

Let 2W be the set of even whole numbers = { 0, 2, 4, ... }

2W is a subset of the set of integers Z which is a commutative ring and a group.

2W is not a group on the operation of addition since there is no inverse element.

2W is not stable under multiplication by the elements of Z since -1 ∈ Z and 2 ∈ 2W but -2 is not in 2W.

Lemma 1:

The set of multiples for a given polynomial is an ideal of the set of all polynomials for a given field

Proof:

(1) Let F[X] be the set of all polynomials in the field F

(2) Let (P) be the set of multiples of polynomials for a given polynomial P so that:

(P) = { PQ where Q ∈ F[X] }

(3) All fields are commutative rings so F[X] is a Commutative Ring. [See Definition 3, here for Fields]

(4) (P) is itself a group since:

If pq, pq' ∈ (P), then (pq+pq')=p(q+q') ∈ (P) since (q+q') ∈ F[X]

(pq + pq') + pq'' = p([q + q'] + q'') = p(q + [q' + q'']) = pq + (pq' + pq'')

(c) Identity Element

0 = P*0 ∈ (P) since 0 ∈ F[X]

(d) Inverse Element

For all pq, there exists -pq since -pq = p(-q) and -q ∈ F[X] if q ∈ F[X]

(5) Finally, (P) is stable under multiplication since:

(a) Assume pq, pq' ∈ (P)

(b) Then q,q' ∈ F[X]

(c) p ∈ F[X]

(d) So pqq' ∈ F[X] since F[X] is closed on multiplication.

(e) So p*(pqq') ∈ (P)

QED

Reference
• Jean-Pierre Tignol, , World Scientific, 2001