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I was trying to prove this statement as true. Actually I found an identical question here: "Is every Artinian module over an Artinian ring finitely generated?"

However, in the proof of the link above, a key step is

Claim:"If $M$ is not finitely generated one can assume that every proper submodule of $M$ is finitely generated. (In order to see this take the partial ordered set of submodules of $M$ which are not finitely generated and choose a minimal element.) "

I suppose that the statement above assumes we can have a minimal generating set for each submodule. However, I cannot prove that.

Or in another way, the claim above is used to prove that $Ann(M)$ is a prime ideal in $R$, so, is their any way to prove $Ann(M)$ is a prime ideal in $R$?

For some information which might be useful, we have: a commutative Artinian ring is Noetherian.(http://math.stanford.edu/~conrad/210BPage/handouts/Artinian)

and

Modules whose proper submodules are finitely generated is called almost finitely generated http://www.sciencedirect.com/science/article/pii/0021869383900753 In this article it seems that the claim is not true for a general ring, however, I don't know how to use $R$ being Artinian in the proof.

I do not have enough reputation to comment on the original question so have to open a new one.

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  • $\begingroup$ The intention was to use the Artinian condition and Zorn's lemma but there's actually a subtelty here. We can only apply the Artinian condition on a decreasing sequence of submodules indexed by the natural numbers, so the fact that Zorn's lemma applies (that is, every chain has a lower bound) is not obvious. $\endgroup$ – Matt Samuel Nov 21 '15 at 20:43
  • $\begingroup$ @Matt Samuel it is enough to show that every chain (which is indexed by natural numbers) has a minimal element, and that is true because $M$ is an Artinian module. Then Zorn's Lemma applies so we have a minimal element. I suppose we then should do the trick of showing that we can further construct a "more minimal" element which is still not finitely generated, and thus a contradiction, but that requires a minimal generating set. $\endgroup$ – kousaka Nov 21 '15 at 23:00
  • $\begingroup$ Not every chain can be indexed by natural numbers in such a way that preserves the order type, because not every linear order is isomorphic to $\mathbb{N}$. We could have a chain isomorphic to $\mathbb{Q}$, for example. $\endgroup$ – Matt Samuel Nov 21 '15 at 23:07
  • $\begingroup$ @honoka if you're willing to learn the Hopkins Levitzki theorem, you get the result almost immediately, and for noncommutative rings to boot. I've put that solution on the question you linked. $\endgroup$ – rschwieb Nov 21 '15 at 23:12
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    $\begingroup$ @Matt Samuel did you mean a chain indexed by elements in ℚ? but ℚ is countable, which means there is an injection from ℕ to ℚ. .. but suppose there is a chain indexed by uncountable set, such as real number, it seems that we could still use the descending chain condition $\endgroup$ – kousaka Nov 21 '15 at 23:20
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If $M$ is Artinian, then any nonempty collection $S$ of submodules of $M$ has a minimal element. For suppose $S$ has no minimal element; let $N_0\in S$ be any element. Since $N_0$ is not minimal, there is some $N_1\subset N_0$ in $S$. Since $N_1$ is not minimal, there is some $N_2\subset N_1$ in $S$. Continuing by induction, you get an infinite strictly descending sequence of submodules, contradicting the fact that $M$ is Artinian.

So now let $S$ be the collection of non-finitely generated submodules of $M$, and let $N\in S$ be a minimal element. Then $N$ is a non-finitely generated module with the property that every proper submodule is finitely generated. Also, $N$ is Artinian because $M$ is. We now replace $M$ by $N$: that is, apply the rest of the argument to the module $N$, instead of $M$.

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  • $\begingroup$ I guess you would let S be the set of submodules of M which are not finitely generated. And let N be a minimal element in S, but we need to show N is precisely M, which can be shown if there is a minimal generating set for M. However we don't have that. $\endgroup$ – kousaka Nov 21 '15 at 22:54
  • $\begingroup$ If $N$ isn't $M$, you just replace $M$ with $N$ in the rest of the argument. $\endgroup$ – Eric Wofsey Nov 21 '15 at 22:56
  • $\begingroup$ Oh I see, so we can show that $N$ is finitely generated, contradicts the defining property of element in $S$, and that leads to S is a empty set, is it? ||| Thanks a lot!!!! $\endgroup$ – kousaka Nov 21 '15 at 23:09

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