## 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

#### 1 comment :

Anonymous said...

thank you!!
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