# Ideals and polynomial rings

• Apr 21st 2012, 07:20 PM
glebovg
Ideals and polynomial rings
Can anyone explain ideals and polynomial rings i.e. definitions, examples, the most important theorems, etc.?
• Apr 23rd 2012, 12:49 PM
Deveno
Re: Ideals and polynomial rings
well, that's kind of a vague question. the notion of an ideal is specific to rings. basically in the ring of integers, the integer zero has the following special properties:

0+0 = 0
k0 = 0

so, one wonders if, for a general ring R, we can find a subset I that has the properties:

I + I = I
rI = I, for all

if so, such a subset I could act as "a generalization of 0", and we could call two elements of R equivalent if they differed by an element of I (just as two elements of R are equal if they differ by 0).

and that is almost what we do. the thing is, we would like such equivalence classes to also form a ring, so we need I to be an additive subgroup of R, as well (and not just closed under addition). such a subset is called an ideal of the ring I, and is a generalization of the notion of the subgroup (kZ,+) of (Z,+) which is called the ideal of Z generated by k. "modding this out" gives us the ring of "integers mod k". in full generality we have:

DEFINITION: let R be a (commutative) ring. if I is a subset of R such that:

1. (I,+) is a subgroup of (R,+)
2. if a is in I, and r is in R, then ra is in I

I is called a (two-sided) ideal of R.

this gives us the following important theorem:

THEOREM: if R is a (commutative) ring, and I is a (two-sided) ideal of R,then if x + I = {x + a: a in I} the set R/I = {x + I : x in R} forms a ring, with the following definitions of addition and multiplication:

(x + I) + (y + I) = (x+y) + I
(x + I)(y + I) = xy + I

Proof: first, we need to show our definitions are well-defined. x + I is a SET, and isn't uniquely determined by x (which is just a single element). so we need to show that if x' + I = x + I, and y' + I = y + I, then: (x+y) + I = (x'+y') + I and xy + I = x'y' + I.

note that x + I = x' + I, means that (x - x') + I = 0 + I = I, so that x - x' must be an element of I. so what we need to show is that if x - x' is in I, and y - y' is in I, then so are:

(x + y) - (x' + y') and xy - x'y'.

note that (x + y) - (x' + y') = (x - x') + (y - y'), which must be in I, since x - x' and y - y' are in I, and (I,+) is an additive subgroup of (R,+). this shows addition in R/I is well-defined. next, note that xy - x'y' = xy - x'y + x'y - x'y' = y(x - x') + x'(y - y'). now x - x' is in I, so by property 2 of an ideal y(x - x') is in I, and similarly so is x'(y - y'), and so the sum is also in I. this shows xy - x'y' is also in I, so multiplication in R/I is also well-defined.

the rest of the ring axioms for R/I are now easy to verify, for example:

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

of special note is the fact that the 0-element of R/I is I itself, since I = 0 + I. so we truly have "replaced 0 with I".

we also have the following theorem:

THEOREM: if φ:R→S is a ring homomorphism (φ(r+r') = φ(r) + φ(r') and φ(rr') = φ(r)φ(r')), then the set ker(φ) = {r in R: φ(r) = 0} is an ideal of R.

Proof: from group theory, we know that ker(φ) is an additive subgroup of (R,+) (if r,r' are in ker(φ), then φ(r - r') = φ(r) - φ(r') = 0 - 0 = 0, so r - r' is in ker(φ), thus ker(φ) is a subgroup by the one-step subgroup test). so suppose a is in ker(φ), and suppose r is ANY element of R. then φ(ra) = φ(r)φ(a) = φ(r)0 = 0, which shows that ra is also in ker(φ). thus ker(φ) is an ideal of R.

this two facts lead to the following theorem, the fundamental isomorphism theorem:

THEOREM: if φ:R→S is a ring homomorphism, then R/ker(φ) and φ(R) are isomorphic rings.

Proof: the isomorphism we have in mind is r + ker(φ) → φ(r). again, we need to show this map is well-defined: that is, if r + ker(φ) = r' + ker(φ), then φ(r) = φ(r'). so suppose r + ker(φ) = r' + ker(φ). then r - r' is in ker(φ), hence φ(r - r') = 0. but since φ is a ring homomorphism, φ(r - r') = φ(r) - φ(r'), so φ(r) - φ(r') = 0, thus φ(r) = φ(r').

so we have a well-defined map. it now remains to be shown this is an isomorphism (of rings). so we need to show that (r + ker(φ)) + (r' + ker(φ)) maps to φ(r) + φ(r'), and that (r + ker(φ))(r' + ker(φ)) maps to φ(r)φ(r'), that is, that we have a ring homomorphism.

by definition: (r + ker(φ)) + (r' + ker(φ)) = (r+r') + ker(φ), which maps to φ(r+r') = φ(r) + φ(r'), since φ is a ring homomorphism. also by definition:
(r + ker(φ))(r' + ker(φ)) = rr' + ker(φ), which maps to φ(rr') = φ(r)φ(r'), since φ is a ring homomorphism. this shows the map r + ker(φ) → φ(r) is a ring homomorphism.

it is clearly surjective, since if s is in φ(R), then s = φ(r) for some r in R, so that r + ker(φ) maps to φ(r) = s. to prove it is an injective mapping, it suffices to show it has 0 kernel.

so suppose r + ker(φ) maps to the 0 element of S. then φ(r) = 0, that is, r is in ker(φ). but then r + ker(φ) = ker(φ), which IS the 0-element of R/ker(φ).

*****
now, with polynomials we have the following application of this theorem:

suppose that R[x] is a polynomial ring and S is another ring with a in S. then the map f(x)→f(a) (the "evaluation map" at a) is a homomorphism from R[x] to S. what is the kernel of this homomorphism? it is all polynomials f in R[x] such that f(a) = 0. here is an example. let R be the field of real numbers, and let S be the field of complex numbers, with a = i, a square root of -1.

it turns out that the kernel of this homomorphism is all polynomials of the form g(x)(x2 + 1), so we have R[x]/<x2+1> , which consists of cosets of the form:

a + bx + <x2 + 1>, is isomorphic to the set of images a+bi, which is all of S = C.