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Is there a way to directly show maximal ideals are prime, avoiding the usual argument with quotient rings?

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up vote 8 down vote accepted

Let $M$ be a maximal ideal of the commutative ring $R$. Assume $ab \in M$, but $b \not\in M$. Then the ideal $M+(b)$ strictly contains $M$. Since $M$ is maximal, this implies that $M+(b)=R$. In particular, this implies that there exists some $m\in M$ and $r\in R$ such that $m+rb=1$. Multiplying through by $a$ gives $ma+rab=a$. Both $ma$ and $rab$ are in $M$, and so $a$ must in $M$. Thus $M$ is prime.

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wow, that was fast. Thanks, that's exactly the type of thing I was looking for. –  James Vincent Dec 17 '12 at 7:20
    
Might clarify by saying $M +(b)$ strictly contains $M$, as $b \not \in M,$ but it is clear enough from the context. –  Geoff Robinson Dec 17 '12 at 8:14
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... and of course somehow this is just the quotient argument in disguise. –  Hagen von Eitzen Dec 17 '12 at 8:50
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