this may be too elementary for mathoverflow, but I'll give it a try.

rings are commutative here. it is well-known that every $0$-dimensional noetherian ring is artinian. the standard proof uses a filtration argument; then it's left to show that every finitely generated vector space is artinian (dimension!) and that extensions of artinian by artinian modules are artinian (tage the images and the preimages of the chain, finally both are stable). by a sheaf argument, it's easy to reduce to: every noetherian ring with exactly one prime ideal is artinian.

is there a proof which is somehow more direct? perhaps a clever manipulation of chains of ideals? I don't expect it, but it would be great for the students in my tutorial, which had to solve this as an exercise without knowing anything about artinian or noetherian rings going beyond the definitions.

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    $\begingroup$ I don't think it's too elementary at all. I do think that you should use correct capitalization, but that's a separate issue :) $\endgroup$ Jan 7, 2010 at 2:51

2 Answers 2


My take on the standard proof can be found on pp. 62-63 of my notes on commutative rings:


(The bit about ACC/DCC being preserved by extensions occurs on p. 57 and should probably be explicitly mentioned in the proof on p. 63.)

Altogether this takes about 1-1.5 pages. I have never seen anything substantially shorter or more direct.

By the way, that's a tough problem to ask a student to solve on his/her own!

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    $\begingroup$ this is the standard proof I've mentioned above. so basically your answer is not concerned with my question. $\endgroup$ Jan 6, 2010 at 18:31
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    $\begingroup$ I acknowledged that it was the standard proof in my response. My point was that this is the best proof that I know of, so that someone who is curious about the standard proof can look it up there. I agree that it is little or no progress towards the answer to your question, but I think a downvote is a bit harsh. $\endgroup$ Jan 6, 2010 at 18:36
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    $\begingroup$ What part of "I have never seen anything substantially shorter or more direct." is not an answer to "is there a proof which is somehow more direct?"? $\endgroup$ Jan 7, 2010 at 2:52

Sorry if this is merely a reformulation of what has already been said (and doubtless it is a "standard proof"), but perhaps a suggestive hint for students would be to show that if an ideal $P$ of a commutative noetherian ring $R$ is maximal for the property that $R/P$ is non-artinian, then $P \subset R$ must be prime.  A sort of philosophical underpinning for this hint is offered in a pretty paper by T. Y. Lam and M. L. Reyes, "A prime ideal principle in commutative algebra," J. Algebra 319 (2008), no. 7, 3006-3027.

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    $\begingroup$ An afterthought: the Lam-Reyes Principle is better than just a "philosophical underpinning" in this case.  The set of ideals of a commutative noetherian ring with the property that the corresponding factor ring is artinian is a monoidal filter and hence both an Oka family and an Ako family (in Lam and Reyes's terminology); thus, by the main result of their paper, maximal members of the complementary set of ideals must be prime. $\endgroup$
    – Greg Marks
    Jun 26, 2010 at 0:55

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