why prime ideal and maximal ideal are same in finite integral domain?

Question

I want to prove that "prime ideal and maximal ideal are same in a finite integral domain"

I know that every maximal ideal is prime ideal. but i can't prove the converse.

How can I prove it? Please help me

Thank you in advance.

Answer 1

Let $A$ be a nonzero commutative ring and let $I$ be an ideal of $A$.

If $I$ is an prime ideal on $A$, then $A/I$ is an integral domain, but we have :

Lemma Any finite integral domain is a field

Proof

Consider a finite intégral domain $D$ and let $d\in D-\{0\}$. The map $D\to D,x\mapsto dx$ is injective because $dx=dy\implies d(x-y)=0\implies x=y$. Since $D$ is finite, this map is bijective. Hence there exists $d'\in D$ such that dd'=1.

End[Proof]

Hence, if $A/I$ is finite (which is in particular the case when $A$ itself is finite), then $A/I$ is a field and finally $I$ is a maximal ideal.

Remark

It wasn't necessary to make the assumption that $A$ is finite but only that $A/I$ is finite. One may think for example at the case $A=\mathbb{Z}$ and $I=n\mathbb{Z}$. In this special case, we get :

$\left(\mathbb{Z}/n\mathbb{Z},+,\times\right)$ is an integral domain iff $\left(\mathbb{Z}/n\mathbb{Z},+,\times\right)$ is a field

which can be easily proved directly.