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### Rings and Ideals

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Date: 11/13/98 at 10:19:26
From: Rachael Johnsvinski
Subject: Modern algebra

Dr. Math:

Can you show me how to solve the following exercises?

1. For every commutative ring R, prove that R[x]/(x) = R.

2. Prove that (0) is a maximal ideal in a commutative ring R if and
only if R is a field.

3. If k is a field and p(x) is an element of K[x], p(x) does not equal
zero, prove that the ideal (p(x)) is a prime ideal in k[x] if and
only if p(x) is an irreducible polynomial.
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Date: 11/13/98 at 11:27:54
From: Doctor Rob
Subject: Re: Modern algebra

1. You could construct a map from R[x] onto R whose kernel is (x).
One such map is the map F(f(x)) = f(0).

2. Suppose (0) is a maximal ideal in R. Then let a be any element of
R not in the ideal. You want to prove it has an inverse. Note that (a)
is the whole ring, because (0) < (a), and (0) is maximal. Thus 1 is in
(a), so ... .

Suppose R is a field. Let M be a maximal ideal. Clearly (0) <= M. Let
a be an element of M other than 0. Then a has an inverse b in R,
and since a is in M, b*a = 1 is also in M, ... .

3. Suppose p(x) is not irreducible. Then there exist q(x) and r(x) in
K[x] such that p(x) = q(x)*r(x), and both q and r have positive degree.
Then q(x) is not in (p(x)), and likewise r(x) is not in (p(x)) (why?),
but their product is, so (p(x)) is not a prime ideal.

Suppose that p(x) is irreducible, q(x) and r(x) are in K[x], and
q(x)*r(x) is in (p(x)). Then q(x)*r(x) = p(x)*s(x), for some s(x) in
K[x]. Now p(x) divides the righthand side, so it divides the lefthand
side. Since p(x) is irreducible, it must divide either q(x) or r(x),
so one of them must be in (p(x)). Thus (p(x)) is a prime ideal.

If you need justification of the next-to-last sentence in the preceding
paragraph, think about GCD(p(x),q(x)) and GCD(p(x),r(x)). Their product
is p(x), and p(x) is irreducible, so one of them must be 1, so the
other ... .

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
College Modern Algebra

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