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In a finite commutative ring, all prime ideals are maximal

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.19


We begin with a lemma.

Lemma: If $R$ is a finite integral domain, then $R$ is a field.

Proof: Let $u \in R$ be nonzero, and define $\varphi_u : R \rightarrow R$ by $\varphi_u(r) = ur$. We claim that $\varphi_u$ is injective. To see this, suppose $\varphi_u(r) = \varphi_u(s)$; then $ur = us$, so that $u(r-s) = 0$. Since $u \neq 0$, $r-s=0$, so that $r=s$. Hence $\varphi_u$ is injective. Since $R$ is finite, $\varphi_u$ is also surjective. In particular, there exists $v \in R$ such that $\varphi_u(v) = uv = 1$, so that $u$ is a unit. Thus every nonzero element of $R$ is a unit, and $R$ is a field. $\square$

Now if $P \subseteq R$ is prime, then $R/P$ is a finite integral domain. Then $R/P$ is a field, and thus $P \subseteq R$ is maximal.


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