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$R$ is a commutative ring. $p(x)$ is an irreducible polynomial of $R[x]$. Is the ideal $(p(x))$ generated by $p(x)$ in $R[x]$ prime?

If not, under what conditions of $R$ is $(p(x))$ prime? How about maximal?

user31480
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2 Answers2

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@kahen gave a good example. Here is more detail, note that $R[x]$ is the ring we look at. Also when there is no implication means there is a counterexample.

In a commutative ring $R$ with 1

\begin{array}{|ccccc|} \hline R/(a) \text{ integral domain} &\iff &(a) \text{ prime ideal} & \iff & a \text{ prime}\\ &&& & \Downarrow & \\ \Uparrow&&(a) \text{ maximal among principal} & \Longleftarrow & a \text{ irreducible} &\\ && & & & \\ R/(a) \text{ is a field} & \iff &(a) \text{ maximal ideal}\\ \hline \end{array}

In an integral domain $R$

\begin{array}{|ccccc|} \hline R/(a) \text{ integral domain} &\iff &(a) \text{ prime ideal} & \iff & a \text{ prime}\\ &&& & \Downarrow & \\ \Uparrow&&(a) \text{ maximal among principal} & \iff & a \text{ irreducible} &\\ && & & & \\ R/(a) \text{ is a field} & \iff &(a) \text{ maximal ideal} \\ \hline \end{array}

In a UFD $R$

\begin{array}{|ccccc|} \hline R/(a) \text{ integral domain} &\iff &(a) \text{ prime ideal} & \iff & a \text{ prime}\\ &&& & \Updownarrow & \\ \Uparrow &&(a) \text{ maximal among principal} & \iff & a \text{ irreducible} &\\ && & & & \\ R/(a) \text{ is a field} & \iff &(a) \text{ maximal ideal} \\ \hline \end{array}

In a PID $R$

\begin{array}{|ccccc|} \hline R/(a) \text{ integral domain} &\iff &(a) \text{ prime ideal} & \iff & a \text{ prime}\\ && & & \Updownarrow & \\ \Uparrow &&(a) \text{ maximal among principal} & \iff & a \text{ irreducible} &\\ &&\Downarrow & & & \\ R/(a) \text{ is a field} & \iff &(a) \text{ maximal ideal} \\ \hline \end{array}

yberman
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mez
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    While searching for some relations among primes, irreducibles and ideals,I read your answer.It is extremely beneficial for me.This is something I was looking for.Thank you so much for such an effort.Only a little confusion is there,would you please clarify for me?When R is PID;"(a) maximal among principal" and "(a) maximal ideal" will these be iff condition?You have given onesided implication..Sorry,may be I am missing something. Thnx a lot again. – usermath Jun 13 '14 at 12:09
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    @usermath You are correct. But I am also correct, note that my diagram forms a loop, meaning all is equivalent. – mez Jun 13 '14 at 12:28
  • Thank you.It is clear now.I didnot think about the loop. – usermath Jun 13 '14 at 12:30
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    Thanks for the very useful answer. In the first case, i.e. R commutative with unity, does $a$ prime really imply $a$ irreducible? Following wiki, if $R = K\times K$ for $K$ a field, then $(1,0)$ is a prime element (because $K\times K/\langle (1,0) \rangle \simeq K$) but $(1,0) = (1,0) \cdot (1,0)$ implies that it is not irreducible. Am I wrong? – Giovanni De Gaetano Nov 30 '16 at 09:34
  • @Giovanni De Gaetano you are certainly right. In a ring that's not an integral domain, neither of the implications prime <=> irreducible holds. – Barbara Nov 30 '16 at 09:43
  • Can I steal that diagrams in your answer :D.. This is Nice... – Praphulla Koushik Apr 17 '20 at 06:39
  • https://twitter.com/PraphullaK/status/1251039045792067584/photo/1 – Praphulla Koushik Apr 17 '20 at 06:46
  • Very useful answer! – Olivier Bégassat Feb 06 '21 at 12:14
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    It seems in your last diagram (PID) everything is equivalent yet you don't indicate it with the corresponding $\iff$. I guess you are assuming $a\neq0$? – Olivier Bégassat Apr 19 '21 at 09:35
  • These graphs are insane. Thank you! – JacobsonRadical Jan 05 '22 at 10:08
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This is in general false. Take $R$ to be any commutative ring which is not an integral domain (e.g. $\mathbb Z \times \mathbb Z$). Then $R[x] / (x) \cong R$ which is not an integral domain, so $(x)$ cannot be prime, but $x$ is certainly irreducible in $R[x]$.

kahen
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    I'm slightly confused by the assertion: "$x$ is certainly irreducible in $R[x]$". Indeed, if my computations are not wrong, if we take $R = \mathbb{Z} \times \mathbb{Z}$ as suggested then $$ x = \left((1,0) + (0,1)x \right) \cdot \left( (0,1) + (1,0)x \right).$$

    But none of the two elements in the decomposition is a unit, therefore $x$ is not irreducible. Where am I wrong?

     – Giovanni De Gaetano Nov 30 '16 at 09:30
  • @Giovanni De Gaetano true. Even for less pathological rings $x$ is not irreducible. Take $R=\mathbb{Z}/6\mathbb{Z}$. Then $$x=(2+3x)(3+2x) $$ and none of the two factors can be a unit, as the constant terms of the two factors are zero divisors and hence the constant term of $p(x)(2+3x)$ also has to be a zero divisor and can never be 1. – Barbara Nov 30 '16 at 09:48
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    Right. That's a major oops. Of course the first part of the argument is correct (i.e. $(x)$ cannot be prime), so we "just" have to find a non-integral domain over which $x$ is irreducible. $R = \mathbb Z / 4\mathbb Z$ should do. – kahen Nov 30 '16 at 17:09