## In a commutative ring, all prime ideals are maximal

Let $R$ be a finite commutative ring with $1 \neq 0$. Prove that every prime ideal of $R$ is maximal.

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.

• Eri  On February 12, 2011 at 1:16 pm

“r-s” should be “r=s” (before the word “hence”)

• nbloomf  On February 12, 2011 at 1:30 pm

Fixed. Thanks!

• maestra  On July 16, 2011 at 8:04 pm

ese R es real o es culaquier conjunto

• nbloomf  On July 18, 2011 at 12:59 pm

If I understand correctly, the question is whether $R$ is the ring of real numbers or an arbitrary ring. Here $R$ is an arbitrary (commutative) ring.