A sufficient condition for the ring property that every prime ideal is maximal

Let R be a commutative ring with 1 \neq 0 and suppose that for all a \in R, there exists an integer n > 1 such that a^n = a. Prove that every prime ideal of R is maximal.

Let P \subseteq R be a prime ideal. Now R/P is an integral domain. Let a + P \in R/P be nonzero; there exists n \geq 2 such that a^n = a. In particular, a^n + P = a + P, so that a(a^{n-1} - 1) + P = 0. Since a \notin P, we have a^{n-1} + P = 1+P, so that (a+P)(a^{n-2} + P) = 1. Thus R/P is a division ring, and since R is commutative, R/P is a field. Thus P \subseteq R is maximal.

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