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

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.25


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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