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Question: Let $R$ be a commutative ring with $1$ that is not a field. Prove that $R[x]$ is not a PID.

Thoughts: I am seeing a lot of questions around SE about this, but most are showing that if $R[x]$ is a PID, then $R$ is a field. So, for this one, I am getting something straightforward, but I am now second guessing myself:

Consider $R[x]/(x)\cong R$. Since $R$ is not a field, then $(x)$ cannot be a maximal ideal, and since every prime ideal in a PID is also maximal, we have that $R[x]$ cannot be a PID.

Now, I am second guessing myself... can I reasonably say, without any issues, that $R[x]/(x)\cong R$?

EDIT: It appears my "proof" is far off, because I cannot claim that $(x)$ is a prime ideal. So, with that, how could I "repair" the proof?

User7238
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    You can consider the morphism of rings $$\varphi : R[x] \rightarrow R, \varphi(f)=f(0).$$ The morphism is surjective. What is the kernel? – Severin Schraven Dec 04 '21 at 23:47
  • How do you know $(x)$ is prime? – Zhen Lin Dec 04 '21 at 23:58
  • @ZhenLin Wait... now I am thinking my "proof" is way off. Because $(x)$ is prime iff $R[x]$ is an integral domain. So, since we are given that $R$ is not a field, then $R[x]$ must be infinite.... but I am not sure about that.... hmmmm... – User7238 Dec 05 '21 at 00:06
  • If $R$ is an arbitrary ring, then the ideal $(x)$ may not be prime, though. So you should first say that if $R$ is not a domain, then neither is $R[x]$. I wonder if $R[x]$ can be a principal ideal ring if $R$ is not a domain... I guess it is that if e.g. $R$ is a product of fields. – tomasz Dec 05 '21 at 00:14
  • @tomasz now I am quite confused and lost :) We have that $R[x]/(x)\cong R$, and as $R$ is not a field, then $(x)$ cannot be maximal. From this point, I don't believe there is anything more I can say that will lead us anywhere worthwhile. Or, am I missing something? – User7238 Dec 05 '21 at 00:28
  • Your reasoning is the following: since $(x)$ is not maximal and prime, $R[x]$ cannot be PID. But if $(x)$ is not prime, then your hypothesis is not true. – tomasz Dec 05 '21 at 01:06
  • @tomasz okay, so it doesn't seem like starting with "$R[x]/(x)=R, and as $R$ is not a field....." is the right way to go. Do you see a way that may be better? I am just not seeing anything right now. – User7238 Dec 05 '21 at 01:09
  • See the Lemma in the linked dupe for a simple proof, and see the remark there for generalizations. – Bill Dubuque Dec 05 '21 at 01:19
  • @BillDubuque Hi! I just have an issue: in this problem, we are assuming that $R$ is not a field, but I am not seeing that assumption in the lemma. I know I am missing something, but I am not sure what... – User7238 Dec 05 '21 at 01:33
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    The Lemma (constructively) shows if the domain $R$ contains a nonzero unit $,c,$ then $(c,x)$ is not principal in $R[x]$ (and if $R$ is not a domain then $R[x]$ is not a domain so not a PID). – Bill Dubuque Dec 05 '21 at 01:43
  • Ahhh, I see, that makes sense, thank you @BillDubuque! Just out of curiosity, how I started my (incorrect) proof as "$R[x]/(x)=R$, and $R$ is not a field, so $(x)$ cannot be maximal....", I am correct in saying that I can't really continue that argument in any way to get a proof, correct? I don't believe that $(x)$ not being maximal implies anything useful, correct? – User7238 Dec 05 '21 at 01:55
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    You can do it that way too, similar to what's done in this answer, i.e. $$R[x]\ {\rm a\ PID} \Rightarrow R\ {\rm a\ domain}\ \Rightarrow x\ {\rm prime}\ \Rightarrow\ x\ {\rm irred}\ \Rightarrow (x)\ {\rm max}\ \Rightarrow R\ {\rm a\ field}$$ but that's less constructive, i.e. it doesn't explicitly construct a nonprincipal ideal. – Bill Dubuque Dec 05 '21 at 02:12
  • @BillDubuque That is what I was going to try, but I wasn't sure how to get an implication from "not a field". For instance, could I say $(x)$ is not maximal, so $x$ is not irreducible, so $x$ is not prime, so $R$ is not a domain, and so $R[x]$ is not a PID...? – User7238 Dec 05 '21 at 02:20
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    It's just the negation (contrapositive) of the above chain of implications, i.e. negate each claim and reverse all the arrows. – Bill Dubuque Dec 05 '21 at 02:30
  • @BillDubuque Thank you so much for all your help. As you've always been when answering my questions, you're very helpful! I like the constructive proof rather than just the theoretical proof too! Thank you. – User7238 Dec 05 '21 at 06:17

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Yes this is always true. Consider the ring homomorphism $\phi : R[X] \longrightarrow R$ given by $ f \mapsto f(0)$. Then of course $\phi(f) = 0 \Leftrightarrow f(0) = 0 \Leftrightarrow f \in (x)$. So by the first isomorphism theorem it follows that $R[x]/(x) \cong R$.

Johannes Oertel
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