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?

Solution 1:

@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}

Solution 2:

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 $R[x]$.