# Maximal ideals are prime without quotient argument

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

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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.
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