Is there a way to directly show maximal ideals are prime, avoiding the usual argument with quotient rings?
Maximal ideals are prime without quotient argument
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abstract-algebra
1 Answers
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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.
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5... and of course somehow this is just the quotient argument in disguise. – 2012-12-17