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It is well known that the associated primes of a module over a commutative ring (those primes associated to primary decomposition of the zero submodule, provided such decomposition exists) are precisely the radicals of the annihilators of elements of the module that are prime. It is easy to show that the word "radical" can be omitted if the ring is Noetherian. Apparently it can also be omitted if the module (and not the ring) is Noetherian. The only proof I know constructs a huge theory of injective modules, and I'm curious to know if there is a more elementary proof.

flag	asked Jan 31 2011 at 19:15 ashpool 1,419●3●16

Perhaps I'm misunderstanding your question, but if you're asking about a more elementary proof of the fact for modules over a Noetherian ring, then a finitely generated (left, say) Noetherian ring is a (left-)Noetherian module. – Adam Hughes Jan 31 2011 at 19:30

No, I'm asking for the proof in the case of a Noetherian module over a non-Noetherian (commutative) ring. – ashpool Jan 31 2011 at 19:52

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I'm not sure what you mean by "the elements of the module that are prime." But any commutative ring with a faithful noetherian module is itself a noetherian ring. So, by factoring out the annihilator of your module, you should be able to apply any theorems you know for modules over noetherian rings. – Manny Reyes Jan 31 2011 at 21:24

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Modulo standard results (such as can be found, for example, in Eisenbud's book "Commutative Algebra"), Manny Reyes's comment completely resolves this question. – Greg Marks Feb 1 2011 at 0:01

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I don't have a reference immediately available, but the proof is straightforward. Let $M$ be an $R$-module generated by $x_1, \dots, x_n \in M$. If $M$ is faithful, then the map $r \mapsto (rx_1, \dots, rx_n)$ is an $R$-module embedding $R \hookrightarrow M^n$. If $M$ is noetherian, then so is $R$. – Manny Reyes Feb 1 2011 at 7:47

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Primary decomposition for modules

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