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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.

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One might define a ring is noetherian if every ideal is finitely generated. In that case, it is trivial of course. –  Taro Mar 9 '13 at 21:03
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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.)

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Thanks. Is there anything that needs to be added to the proof? –  J. Perry Mar 9 '13 at 21:21
    
If there are a finite number of subsets, then there are no infinite sequences of proper subsets. Then characterize countably infinite sequences of subsets. –  Loki Clock Mar 9 '13 at 21:37
    
@J.Perry Yes, you could describe more explicitly why having only finitely many ideals causes all the chains to stabilize. –  rschwieb Mar 9 '13 at 21:43
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