Proof about Noetherian rings

Question

I have to prove that every finite ring is Noetherian. I know examples of Noetherian rings which are not finite such as the field of complex numbers or a PIR like the integers. But anyway:

[Proof]:

I know by definition that for a ring R that satisfies the ascending chain condition (ACC), i.e. every sequence of ideals $$I_1 \subseteq I_2\subseteq I_3\subseteq ... $$ of R stablises. i.e. $\exists n_o$ such that $I_{n_0}=I_n$ for all n $\ge n_o$ then the ring R is Noetherian. So this would mean there is a finite number of ideals. Now assuming R is finite, this would mean it has a finite number of subsets and as an ideal $I$ is defined to be a $I\subseteq R $ , wouldn't this mean the ACC is always satisfied for a finite ring and thus all finite rings are Noetherian?

I don't know if any of this is right but would really appreciate any advice. Thanks for any help.

Answer 1

That's the right idea... in fact you've even proved that a finite ring is Artinian, which turns out to imply Noetherian (although that's not trivial.)

Answer 2

Suppose there is an ascending chain of $R$ ideals that is not stationary. Then, for each $n$ there is $m$, with $m> n$, so that $I_n$ is properly contained in $I_m$, which accounts for the finitude of $R$. Therefore, every ascending chain is stationary, and therefore, $R$ is Noetherian.