Show that any prime ideal from such a ring is maximal.

Question

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 maximal. (Atiyah and MacDonald, Introduction to Commutative Algebra, Chapter 1, Exercise 7.)

Any hints?

Answer 1

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.

Answer 2

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

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

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...