# Every maximal ideal is a prime ideal [duplicate]

The question formuled in the exam was exactly:

''Every maximal ideal is a prime ideal''

Maximal and prime ideals are defined for a commutative ring R, but the proof I have for maximal ideal $\Rightarrow$ prime ideal needs that R is a commutative and unitary ring, because it uses that $R/I$ is field iff $I$ is maximal, and that only happens if R is both commutative and unitary.

Then my question is:

Is there a proof for maximal ideal $\Rightarrow$ prime ideal for R a commutative ring (not neccesarily unitary)?

If not, there must be a counterexample with a maximal ideal which isn't prime in a R commutative (not unitary) ring, right?

I know there are more posts like this, but the difference is that those posts suppose R as a commutative and unitary ring (that's the difference with this one, just to be clear).

## marked as duplicate by MooS, Dietrich Burde 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(); } ); }); }); Sep 6 '17 at 14:01

• $R=2\mathbb{Z}, I=4\mathbb{Z}$. Here $I$ is a maximal ideal, but not a prime ideal. If the ring contains unity, then every maximal ideal is a prime ideal. – Krish Sep 6 '17 at 14:00
• "In a commutative ring $R,$ the ideal $I$ is prime if and only if the quotient ring $R/I$ is an integral domain." From this result you can deduce what you want. – Bumblebee Sep 6 '17 at 14:03
For example, in $2\mathbb Z/4\mathbb Z$, the zero ideal is obviously maximal, but the zero ideal is also obviously not prime.