# Find an element $p$ which is an irreducible element but the ideal $\langle p\rangle$ is not a maximal ideal? [duplicate]

Does there exist an element $p$ in a ring $R$ such that $p$ is an irreducible element but the ideal $\langle p\rangle$ is not a maximal ideal?
I could only find that $R$ is not a PID but I could not find any counterexample to the problem.Please help.
## marked as duplicate by rschwieb abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 7 '16 at 12:07
Take a prime number and consider it as an element in $\mathbb Z[x]$. $(p)$ is not maximal, since $\mathbb Z[x]/(p) \cong \mathbb F_p[x]$ is not a field.