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Let $A$ be a ring. How can I prove that:

$A$ is an Artinian ring $\Leftrightarrow \exists$ a faithful $A$-module which is of finite length.

I know that if a ring has a faithful $A$-module which is Noetherian, then the ring is Noetherian. Can I use this to prove the result above? I need your help to solve this problem. Thank you.

rgl4
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  • Related to http://math.stackexchange.com/questions/50178/are-artinian-modules-over-non-artinian-rings-relevant –  May 01 '13 at 19:41

1 Answers1

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It looks like you must intend this for commutative rings, because it doesn't hold for general noncommutative rings.

The answer to the question follows from the same argument here in Manny Reyes's solution.

If the faithful and finite length module is $M$, then the trick is just to embed $R$ into $M^n$. As a submodule of an Artinian module, it is also then Artinian.

Added more detail: Since $M$ has finite length, it has a finite generating set $m_1,\dots m_n$. Look at the map $R\to M^n$ given by $r\mapsto (m_1r,m_2r,m_3r,\dots,m_nr)$. Since $M$ is faithful, this map is an injection. Since $M$ is Artinian, so is $M^n$. Thus, $R$ is a submodule of an Artinian module, and hence is Artinian itself.

The converse is obvious since if $R$ is Artinian, then it itself is a faithful $R$ module of finite length.

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rschwieb
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  • I don't understand if I change Artinian for Noetherian in the solution that you say me. You can explain it, please – rgl4 May 12 '13 at 10:54
  • @rgl4 I added to the solution to address your comment. – rschwieb May 12 '13 at 11:01
  • Yes, I have seen the solution but the implication right has not proved not? – rgl4 May 12 '13 at 11:09
  • @rgl4 I don't understand. I think (but I can't remember, since you edited it) that your last comment asked for this implication. Can you be more clear? – rschwieb May 12 '13 at 11:30
  • @rgl4 It's not enough for the module to be Artinian, it has to be finite length. This is because an Artinian module need not have a finite generating set, but one of finite length (which is Noetherian) must have a finite generating set. – rschwieb May 12 '13 at 11:31
  • To see if I understand the solution: The implication $\Rightarrow$ is obvious and it doesn't more explications. And the implication $\Leftarrow$ will be:

    I call $M$ an $A$-module finitelly generated and faithful.

    So $M$ is finitelly generated then exists a finite set of generators $m_1 , ... , m_n$.

    If I consider the homomorphism $f: A \rightarrow M^n$ given by $a \mapsto (m_1 a, ... , m_n a)$ how $M$ is faithful, then $f$ is injective.

    (Why I can say that $M$ is artinian?) If $M$ is artinian then $M^n$ is artinian.

    Therefore, $A$ is a submodule of an Artinian module, so $A$ is Artinian.

     – rgl4 May 14 '13 at 15:55
  • I could say that $M$ is Artinian because $M$ has finite length, not? – rgl4 May 14 '13 at 16:05
  • @rgl4 Correct, a module has finite length iff it is Artinian and Noetherian. A submodule of an Artinian module is Artinian. – rschwieb May 14 '13 at 17:01