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Suppose $M$ is an $R$-module. Prove that $M$ has finite length if $R$ is artinian and $M$ is finitely generated.

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Dear Alex, Can you prove that $R$ itself is finite length? Regards, – Matt E Jan 9 '13 at 3:45
I am afraid that I can't. Or $R$ might not be finite length? – Alex Jan 11 '13 at 2:42
Dear Alex, First of all, since $R$ is the simplest example of a f.g. module over itself, you should focus on this case first, before considering the general problem. (In fact, as Benjamin Lim explains in his answer, the general problem follows pretty formally from this case.) As for this special case, you could begin by thinking about what finiteness statements you do know for $R$. Regards, – Matt E Jan 11 '13 at 2:50
up vote 1 down vote accepted

The following steps lead to a solution:

  1. Every finitely generated $R$ - module $M$ can be put into an ses of the form $$0 \to \ker \phi \to R^n \stackrel{\phi}{\to} M \to 0$$ where $ \phi$ is a map that sends a basis element of $R^n$ to a generator of $M$. $n$ is the number of generators of $M$.

  2. A finite direct sum of Artinian rings is Artinian, and for any ses of $R$ - modules $ 0 \to M \to N \to L \to 0$ we have $N$ Artinian iff $M$ and $L$ are.

  3. Conclude.

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so you want to prove that $M\approx R^{n}/ker(\phi)$ is both artinian and noetherian and derive that $M$ has finite length. But is it possible to prove that $R$ is noeterian itself and then derive the conclusion? – Alex Jan 8 '13 at 19:08
or my proposition is wrong because $R$ may not be noetherian itself and neither is $ker(\phi)$. – Alex Jan 8 '13 at 19:10
@Alex See the edit. – user38268 Jan 9 '13 at 2:50
Dear Benjamin, There is some issue with 2, probably what is meant is that $N$ is Artinian iff $M$ and $L$ are (although the analogous statement with "finite length" in place of Artinian is more directly relevant here). – Matt E Jan 9 '13 at 3:45
@MattE Dear MattE, I have corrected my answer. I think I've been looking at my computer screen for too long now... – user38268 Jan 9 '13 at 4:09

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