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
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001
1 comment :
Post a Comment