## 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.