Prove the theorem in the lecture:Euclidean Domains and UFD's Let F be a field, and let...

Prove the theorem in the lecture:Euclidean Domains and UFD's

Let F be a field, and let p(x) in F[x]. Prove that (p(x)) is a maximal ideal in F[x] if and only if p(x) is irreducible over F.

Solutions

Expert Solution

Denote I=<p(x)>.Then we have I is principal, I = <p(x)> for some p(x) ∈ F[x].
1. (⇐) Assume p(x) is irreducible and let I1 be an ideal of F[x] containing
I. Then I1 = <f(x)> for some f(x) ∈ F[x]. Since p(x) ∈ I1 we have p(x) =
f(x)q(x) for some q(x) ∈ F[x]. Since p(x) is irreducible this means that either
f(x) has degree zero (i.e. is a non-zero element of F) or q(x) has degree zero.
If f(x) has degree zero then f(x) is a unit in F[x] and I1 = F[x]. If q(x) has
degree zero then p(x) = af(x) for some nonzero a ∈ F, and f(x) = a^(-1)p(x);
then f(x) ∈ I and I1 = I. Thus either I1 = I or I1 = F[x], so I is a maximal
ideal of F[x].
(⇒): Suppose I = <p(x)> is a maximal ideal of F[x]. Then p(x) =/= 0. If
p(x) = g(x)h(x) is a proper factorization of p(x) then g(x) and h(x) both
have degree at least 1 and <g(x)> and <h(x)> are proper ideals of F[x] properly
containing I. This contradicts the maximality of I, so we conclude that p(x)
is irreducible. This completes the proof.

Colby Messinger answered 1 year ago

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