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Let $R$ be a ring where for every element $a \in R$, there exists a positve integer $n_a \gt2$ such that $a^{n_a}=a$. Prove that every prime ideal in $R$ is maximal.

I think that I would want to prove this by contradiction, assuming that a prime ideal $P$ is not maximal. Then, there exists $M \subset R$ (properly). This would imply that there exists $a \in M\setminus P$. Then $a^{n_a}=a \in M\setminus P$. Then $M\setminus P$ is prime. And here I am stuck. I don't see how this helps.

Would I rather want to use the fact that then $R/P$ is an integral domain?

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    $\begingroup$ Yes, you would rather want to use the fact that $R/P$ is an integral domain. Then deduce from the given that it is indeed a field. $\endgroup$ – Daniel Fischer Mar 29 '14 at 12:24
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Suppose $\;P\le R\;$ is prime $\;\iff R/P\;$ is an integer domain, and let $\;0\neq a+P\in R/P\;$ .

We're given that

$$\exists\,\,2<n_a\in\Bbb N\;\;s.t.\;\;a^{n_a}=a\implies (a+P)^{n_a}=a^{n_a}+P=a+P\implies$$

$$a(a^{n_a-1}-1)+P=P(=\overline 0)$$

and as $\;R/P\;$ is an integer domain we're done...

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Hint $\ $ More generally, assume every element is the root of a reverse-monic polynomial $\,f\in R[x],\,$ $\,f = u x^n + x^{n+1} g(x),\,$ with $\,u\,$ a unit (invertible). In domain $\,R/P,$ every $\,x\,$ is a root of some $\,f = x^n(u+x g(x))\,$ so either $\,x^n = 0\,\Rightarrow\, x=0,\,$ or $\,-xg(x) = u\,$ unit, so $\,x\mid u\,$ unit $\,\Rightarrow\,x\,$ unit. Therefore $\,R/P$ is a field, so $\,P\,$ is maximal.

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