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?
 A: @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}
A: 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 $R[x]$.
