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This question already has an answer here:

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.

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marked as duplicate by rschwieb abstract-algebra May 22 '14 at 20:03

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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 $y=1y=rxy+my\in M$.

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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.

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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} $

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

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