Is the ideal generated by an irreducible polynomial prime?

$R$ 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?

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2 Answers

@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}\ 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} \end{array}

In an integral domain $R$ \begin{array}\ 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} \end{array}

In a UFD $R$

\begin{array}\ 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} \end{array}

In a PID $R$ \begin{array}\ 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} \end{array}

• 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
• @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
• 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 9:34
• @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 9:43

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 $R[x]$.

• 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 9:30
• @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 9:48
• 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