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Let R be a commutative ring with an identity such that for all $r\in$ R, there exists some $n>1$ such that $r^n = r$. Show that any prime ideal is automatically maximal. Any hints?

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You are probably assuming the ring is commutative: if so, please edit the question and add that hypothesis. In any case, your hypothesis actually implies that the ring must be commutative, by a theorem of Jacobson, but the proof of this is quite not as simple as one might want. –  Mariano Suárez-Alvarez Oct 11 '12 at 7:42
    
Yes, you are correct, I apologize. –  Kiyoshi Oct 11 '12 at 8:04
    
Oh, there is no need to apologize! Just keeping things precise is good for the universe :-) –  Mariano Suárez-Alvarez Oct 11 '12 at 8:06

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up vote 2 down vote accepted

Hint: Reduce to the case that $R$ is an integral domain satisfying $\forall r \exists n (r^n=r)$, and show that $R$ is a field.

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Slightly more verbosely: in order to prove that a prime ideal $\mathfrak p\subseteq R$ is maximal, it is enough to show that the quotient $R/\mathfrak p$ is a field. Now the quotient is an integral domain and it has the same property that you supposed $R$ itself has. &c –  Mariano Suárez-Alvarez Oct 11 '12 at 7:41

Let $\,D\,$ be any integer domain and let $\,d\in D\,$ be s.t. $\,d^n=d\,\,,\,\,1<n\in\Bbb N\,$ , then:

$$d^n=d=\Longrightarrow d(d^{n-1}-1)=0\Longleftrightarrow d=0\,\,\,or\,\,\,d^{n-1}=1$$

so if $\,d\,$ is not zero then it must be a unit.

$$-------o----------o---------o---$$

In our case: let $\,I\leq R\,$ be a prime ideal and let $\, r\in R\setminus I\,$, then:

$$\exists\,n\in\Bbb N\,\,s.t.\,\,r^n=r\Longrightarrow \left(r+I\right)^n=r^n+I=r+I\in R/I$$

Now use the first part with $\,D:=R/I\,\,\,,\,\,d=r+I\,$ and deduce $\,R/I\,$ is actually a field...

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