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This question already has an answer here:

Let $R$ be a commutative ring such that for each $a\in R $ there exists a positive integer $n>1$ (depending on $a$) such that $a^n=a$. Prove that every prime ideal is a maximal ideal.

My try: Let $P$ be a prime ideal. Let $P\subset Q \subseteq R$. Then there exists $q\in Q $ but $q\notin P$. Then $\exists n$ such that $q^n=q\implies q(q^{n-1}-1)=0\in P$ since $P$ is prime either $q\in P $ or $q^{n-1}-1\in P$.

Thus $q^{n-1}-1\in P$. How to complete the proof from here?

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marked as duplicate by user26857, rschwieb abstract-algebra Sep 29 '15 at 18:38

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$q^{n-1}-1\in P$ implies $1 \in \langle P,q \rangle \subseteq Q$ and so $Q=R$.

Alternatively, $a^n=a$ implies $a^n \equiv a \bmod P$. If $a\notin P$, then $a^{n-1}\equiv 1 \bmod P$ because $R/P$ is a domain. But then $a$ is a unit mod $P$ and so $R/P$ is a field.

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