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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.

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    Finite integral domains are fields... they don't have proper ideals so the statement is trivially true. Or am I missing here something? – 2017-02-15
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    A very silly corollary to a much more general statement. "Integral domain" can be replaced with "commutative ring" or even just "ring". "Finite" can be replaced with "Artinian" or some even more general things (but that is too far afield.) – 2017-02-16

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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.