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I'm reading Matsumura's "Commutative Algebra", specifically the chapter on Dimension. However I am having some trouble in the section 12.B, where it is presented the chain of equivalences for a module $M$ over a Noetherian (commutative) ring $A$:

  1. $M$ is an $A$-module of finite length;

  2. $A/\operatorname{Ann}(M)$ is an Artinian ring;

  3. $\operatorname{dim}(M) = 0$.

What I am not being able to justify is the implication (2) $\implies$ (1). I know that $M$ has finite length iff it is noetherian and artinian and also, since $A$ is noetherian, so is $M$. By definition of $\operatorname{Ann}(M)$, the submodules of $M$ seen as $A$-module are the same as the submodules of $M$ seen as an $A/ \operatorname{Ann}(M)$-module, and so $M$ is an artinian $A$-module iff it is an artinian $A/ \operatorname{Ann}(M)$-module.

I don't know if whenever $A$ is an artinian ring, any $A^n$ also is and if so, how to prove it, because I can't find a simple answer to this question, since there is no easier caracterization as for noetherian rings that I know of. If we do have this, then $M$ is easily shown to be artinian by the existence of an epimorphism $\pi : A^n \to M$.

Can anyone give me a hint or something? Any help is appreciated :).

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1 Answer 1

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Hint:

We know that a finite direct sum of artinian $R$-modules is artinian. Now each factor in $R=A^n$ is an artinian $A$-module by hypothesis, and a fortiori an artinian $R$-module.

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