# Maximal ideal/ Prime ideal [duplicate]

Show any maximal ideal is a prime ideal.

I am not sure how to approach this, i know i need to show that for any 'ab' in the maximal ideal, a is in the max ideal or b is in the max ideal.

## 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(); } ); }); }); May 22 '14 at 20:03

Let $R$ be a ring and $M$ a maximal ideal of $R$.

Let $x,y\in R$ such that $xy\in M$. We want to prove that $x\in M$ or $y\in M$.

Consider $I=\langle x,M \rangle = Rx + M$. Then $I\supseteq M$ and so $I=M$ or $I=R$, because $M$ is maximal.

If $I=M$, then $x\in M$.

If $I=R$, then $1=rx+m$ and $y=1y=rxy+my\in M$.

Let $I\subset R$ be a maximal ideal. The quotient ring $R/I$ is a field, hence an integral domain. Since we have $$R/I \mbox{ is integral domain}\Leftrightarrow I\mbox{ is prime ideal}$$ the claim follows.

Hint: these two properties of ideals translate into what two properties of quotient rings?

$\qquad{\rm an~ideal}~~M\triangleleft R~~{\rm is}~~\begin{array}{|l|}\hline \rm prime \\ \hline \rm maximal \\ \hline \end{array}\iff {\rm the~quotient}~~R/M~~{\rm is}~~\begin{array}{|l|} \hline \rm an~integral~domain \\ \hline \rm a~field \\ \hline\end{array}$

If you have a ring $R$ and a maximal ideal $M$, how is the quotient ring $R/M$