Are there any finitely generated artinian modules that are not notherian? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T23:30:05Z http://mathoverflow.net/feeds/question/61695 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61695/are-there-any-finitely-generated-artinian-modules-that-are-not-notherian Are there any finitely generated artinian modules that are not notherian? KotelKanim 2011-04-14T13:07:37Z 2011-07-09T21:38:43Z <p>It is well known that for <em>rings</em>, Artinian implies Noetherian (the famous Hopkins–Levitzki theorem) and it is also well known that there are Artinian <em>modules</em> which are not Noetherian. A simple example can be found in </p> <p><a href="http://en.wikipedia.org/wiki/Artinian_module#Relation_to_the_Noetherian_condition" rel="nofollow">http://en.wikipedia.org/wiki/Artinian_module#Relation_to_the_Noetherian_condition</a></p> <p>Since rings are always finitely generated modules over themselves (all rings considered are unital), it seemed natural to me to ask whether there are <em>finitely generated</em> 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).</p> http://mathoverflow.net/questions/61695/are-there-any-finitely-generated-artinian-modules-that-are-not-notherian/61700#61700 Answer by Simon Wadsley for Are there any finitely generated artinian modules that are not notherian? Simon Wadsley 2011-04-14T13:36:56Z 2011-04-14T14:29:05Z <p>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 $i$. </p> <p>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 <i>cyclic</i> $R$-module. </p> <p>Now if $R$ is commutative, $M$ is a quotient ring $R/I$ which is Artinian as such and so Noetherian also, as you say. </p> <p>If $R$ is non-commutative then I'm not sure what the answer is. <hr> 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.</p> http://mathoverflow.net/questions/61695/are-there-any-finitely-generated-artinian-modules-that-are-not-notherian/69502#69502 Answer by Victor Camillo for Are there any finitely generated artinian modules that are not notherian? Victor Camillo 2011-07-04T21:42:58Z 2011-07-09T21:38:43Z <p>Mistake----Adding a unit makes the radical have lots of submodules.</p>