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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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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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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