The intersection of distinct maximal ideals is not prime.

Let $P,Q$ be distinct maximal ideals of a ring $R$. Prove that $P\cap Q$ is not prime.

I am not sure how to prove this. The only facts that I can think of applying are the definitions, $R/M$ is a field iff $M$ is max, and $R/P$ is an ID iff $P$ is prime.

Any suggestions would be appreciated.

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Good idea with the R/P prime approach. Check the ring $R/P\cap Q$ and see if you have any ideals that multiply to zero. – rschwieb Apr 30 '12 at 13:39

Hint $\rm\: R/(I\cap J)\cong R/I \times R/J\:$ has obvious zero divisors $\rm (\bar i,0)(0,\bar j)=(0,0)\:$ if $\rm\:I\not\subset J, J\not\subset I$

Note that the other proofs reduce to this by factoring out $\rm\:P\cap Q.$ This ubiquitous modular reduction technique is better learned sooner than later (hence its emphasis in said link by the master algebraist Irving Kaplansky very early in his textbook).

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Thanks to Thomas, this is what I have: Since P and Q are both maximal, neither is contained in the other. Let $p\in P\backslash Q$ and $q\in Q\backslash P$. Then $pq\in P\cap Q$. But neither $p\in P\cap Q$ nor $q\in P\cap Q$. So $P\cap Q$ cannot be a prime ideal of R.

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Let $p\in P\setminus Q$ and $q\in Q\setminus P$. (Why can you find such $p,q$?)

Then what can you say about $pq$?

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