Maximal ideals and principal ideals domain

Question

Let $R$ be a ring, and an ideal $I = ( p(x) )$ of $R[x]$.

Then consider the statement: " $I$ is a maximal ideal if and only if $p(x)$ is irreducible "

Does that statement need the hypothesis "$R[x]$ is a principal ideals domain" to be true? Or is that statement true anyway?

@DietrichBurde In what way is the link you provide the same as OP's question? — Divide1918, Dec 05 '20 at 15:32
Sorry, I meant this duplicate. The domain needs to be a field. — Dietrich Burde, Dec 05 '20 at 15:34

score 1 · Answer 1 · answered Dec 05 '20 at 15:36

1

Tne statement is actually false even if $R$ is a PID.

E.g. consider the case $R=\Bbb Z$ and the chain of prime ideals with proper inclusions $$ (0)\subset(x)\subset(2,x). $$

answered Dec 05 '20 at 15:36

Andrea Mori

28,458

Sorry I made a mistake. The hypothesis I wanted to refer was R[x] is a PID (instead of R) – Lourdes LB Dec 05 '20 at 16:32
@LourdesLB, well, $R[x]$ is a PID only when $R$ is a field. Indeed if $r\in R$ is a non-invertible element the ideal $(r,x)$ is not principal. – Andrea Mori Dec 05 '20 at 22:46

Divide1918 · Accepted Answer · 2020-12-05T16:59:05.113

0

Suppose $p(x)$ is not the zero polynomial and is reducible, then there exists some non-constant polynomial such that $q(x)\mid p(x)\implies (p(x))\subsetneq (q(x))\ne R[x]$, so $(p(x))$ is not maximal. This proves the forward direction.

For the other direction, suppose $p(x)\subsetneq J$, some proper ideal of $R[x]$, and that $p(x)$ is irreducible. Then we must have $(p(x),q(x))\subset J$ whence $q(x)$ is coprime to $p(x)$. By Bézout's identity, $\exists a(x),b(x)\in R[x]$ such that $a(x)p(x)+b(x)q(x)=1\in J$, but then $J=R[x]$, a contradiction. So $p(x)$ is reducible in fact, which proves the statement.

edited Dec 05 '20 at 16:59

answered Dec 05 '20 at 16:42

Divide1918

2,225

But Bezout Identity is not true in any UFD. You need a PID to use it – Lourdes LB Dec 05 '20 at 16:46
Then you just answered your own question. – Divide1918 Dec 05 '20 at 16:51
I was wondering if there was an alternative proof to show the other implication without using the Bezout Identity since I cannot find a counterexample of a R[x] not PID in which p(x) is irreducible but (p(x)) is not maximal. – Lourdes LB Dec 05 '20 at 16:56
https://math.stackexchange.com/questions/832866/must-an-ideal-generated-by-an-irreducible-element-be-a-maximal-ideal – Divide1918 Dec 05 '20 at 17:03
There's a counterexample given by @DietrichBurde. – Divide1918 Dec 05 '20 at 17:04

Maximal ideals and principal ideals domain

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