Every Prime Ideal of a Finite Commutative Ring is Maximal

Prime Ideal Problems and Solution in Ring Theory in Mathematics

Problem 723

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

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Proof.

We give two proofs. The first proof uses a result of a previous problem. The second proof is self-contained.

Proof 1.

Let $I$ be a prime ideal of the ring $R$. Then the quotient ring $R/I$ is an integral domain since $I$ is a prime ideal. Since $R$ is finite, $R/I$ is also finite.

By Problem Finite Integral Domain is a Field, any finite integral domain is a field. This yield that $R/I$ is a field, and hence $I$ is a maximal ideal.

Proof 2.

In this proof, we prove the problem from scratch.

Let $I$ be a prime ideal of the ring $R$. Then the quotient ring $R/I$ is an integral domain since $I$ is a prime ideal. Since $R$ is finite, $R/I$ is also finite.
We claim that $R/I$ is a field. For any nonzero element $a\in R/I$, define the map
\[f_a: R/I \to R/I\] by sending $x\in R/I$ to $ax \in R/I$.

We show that the map $f_a$ is injective.
If $ax=ay$ for $x, y \in R/I$, then we have $a(x-y)=0$, and we have $x-y=0$ as $R/I$ is an integral domain and $a\neq 0$. Thus $x=y$ and the map $f_a$ is injective.
Since $R/I$ is a finite set, the map $f_a$ is surjective as well. Hence there exists $b \in R/I$ such that $f_a(b)=1$, that is, $ab=1$. Thus $a$ is a unit in $R/I$.
Since $a$ is an arbitrary nonzero element of $R/I$, we conclude that $R/I$ is a field.

Since the quotient ring $R/I$ is a field, the ideal $I$ is maximal.


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2 Responses

  1. Karthik saravanan.p says:

    Nice explanation

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