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:

(a) Closure on addition

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

(b) Associativity on Addition

(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

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Leslie Lim said...
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