Direct proof of every prime ideal of finite commutative ring with unity is maximal

Question

The usual proof of "every prime ideal of a finite commutative ring with unity is maximal" is pretty straightforward.

$I$ is a prime ideal, so $R/I$ is a finite integral domain, so $R/I$ is a field, so $I$ is maximal.

But, how can we prove it without a quotient ring? or how can we prove it directly?

In order to illustrate what I mean by direct proof, let's look at the following example:

We can prove maximal ideals of a commutative ring with unity are prime as follows:

$I$ is maximal, so $R/I$ a is field, so $R/I$ is an integral domain, so $I$ is prime.

This approach is not direct, instead, you can check the direct approach here: https://math.stackexchange.com/q/68493

So, I want similar direct proof for my question.

Answer 1

let $I$ a prime ideal in a finite commutative ring with unity. We assume that exists an ideal J s.t. $I\subset J\subset R$ and let $a\in J \backslash I$. because of the limitedness of R there exists $n,m\in\mathbb N$ s.t. $$a^n=a^{n+m}\implies a^n(1-a^m)=0\in I.$$ because I is prime we have that $a^n\in I(\implies a\in I, \text{ contradiction})$ or $1-a^m\in I (\implies (a)^m+(1-a^m)=1\in J=R)$.