It is well known that for rings, Artinian implies Noetherian (the famous Hopkins–Levitzki theorem) and it is also well known that there are Artinian modules which are not Noetherian. A simple example can be found in


Since rings are always finitely generated modules over themselves (all rings considered are unital), it seemed natural to me to ask whether there are finitely generated modules, which are Artinian but not Noetherian (the example given in the reference is clearly not finitely generated). I guess that if the statement "every finitely generated artinian module is noetherian" was true, I would have seen it in any standard text book on algebra, and since I haven't, I guess it's not. But still, I can't find a counter-example for this. Perhaps I'm missing something completely trivial here. I will be happy to see an example of such module or a proof that there are no (just a reference will be much appreciated too of course).

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    $\begingroup$ Wouldn't this statement reduce to cyclic modules (one generator)? Then, at least if the ring is commutative (I can't figure out whether you are making this assumption), a cyclic module is a free module over a quotient ring, and so noetherian and artinian properties appear to be equivalent. $\endgroup$
    – t3suji
    Apr 14, 2011 at 13:32
  • $\begingroup$ I changed the tag "commutative-algebra", which I put by mistake, to "noncommutative-algebra". The answer about the commutative case was great. so was the comment, that after reading Simon's answer, I understood that it says pretty much the same thing. thanks. I'm still interested in the noncommutative case. $\endgroup$
    – KotelKanim
    Apr 14, 2011 at 13:56
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    $\begingroup$ I just followed the wikipedia link that you gave and it gives links to two articles that show that cyclic Artinian modules need not be Noetherian plms.oxfordjournals.org/content/s3-35/1/55 and jlms.oxfordjournals.org/content/55/2/231 I don't have access to either so can't say much more $\endgroup$ Apr 14, 2011 at 14:25
  • $\begingroup$ thanks a lot Simon! It settles it completely. I feel stupid for missing the link from the article on wikipedia I linked myself. I promise to do my homework batter for the next question... $\endgroup$
    – KotelKanim
    Apr 14, 2011 at 14:42

1 Answer 1


Suppose you have an Artinian but not Noetherian finitely generated $R$ module $M$. Let $0\leq M_1\leq M_2\leq \cdots \leq M_n=M$ be a finite chain of $R$-modules such that each composition factor $M_i/M_{i-1}$ is cyclic for each $i$.

Certainly each composition is Artinian since subquotients of Artinian modules are Artinian. Also one of the composition factors must be non-Noetherian since extensions of Noetherian modules by Noetherian modules are Noetherian. Thus, we may assume that $M$ is a cyclic $R$-module.

Now if $R$ is commutative, $M$ is a quotient ring $R/I$ which is Artinian as such and so Noetherian also, as you say.

If $R$ is non-commutative then I'm not sure what the answer is.

Added: It seems from the wikipedia article linked from the question that Hartley showed that there are cyclic Artinian and non-Noetherian modules over certain non-commutative rings and Cohn gave another construction nearly twenty years later. See the links I give in the comments on the question for precise references.


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