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Let $R$ be a Noetherian ring and $\mathfrak a$ be an ideal of $R$. Then

(i) $G_{\mathfrak a}(R)$ is Noetherian.

(ii) If $M$ is a finitely generated $R$-module and $\mathcal F=\{M_n\}$ is a stable $\mathfrak a$-filtration, then $G_{\mathcal F}(M)$ is a finitely generated $G_{\mathfrak a}(R)$-module.

But what is the converse of (ii), i.e., if $G_{\mathcal F}(M)$ is a finitely generated $G_{\mathfrak a}(R)$-module does that imply the filtration is stable?

I think if $\mathfrak a =\operatorname{Ann}(M)$ this gives a counter example to the converse.

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Please show why do you think that $G_{\mathcal F}(M)$ is a finitely generated $G_{\mathfrak a}(R)$-module when $\mathfrak a =\operatorname{Ann}(M)$. – user26857 Feb 1 '13 at 8:09
@YACP, thanks for the reply, I take $R = \mathbb Z$ and $ I = (0)$ then $G_I(\mathbb Z) = \mathbb Z$, then $G_I(M)$ is f.g. $\mathbb Z$ module since Z is Noetherian, but if I take $M = (p) \supset (p^2) \supset (p^3) \cdots $ $IM_n = 0$ for all n, so the filtration is not stable. – Ram Feb 1 '13 at 8:44
Actually your last sentence (from the original question) seemed to point out that, in general, $G_{\mathcal F}(M)$ is a finitely generated $G_{\mathfrak a}(R)$-module when $\mathfrak a =\operatorname{Ann}(M)$, and this is why I've asked for some explanation. – user26857 Feb 1 '13 at 11:12
@YACP yes, but I couldn't succeed in proving that, so I thought about this. – Ram Feb 1 '13 at 12:29
Okay. Now coming back to your example, I have to repeat the same question: why $G_I(M)$ is f.g. over $\mathbb Z$? (Take care that not all $\mathbb Z$-modules are f.g. despite the fact that $\mathbb Z$ is noetherian.) – user26857 Feb 1 '13 at 14:22

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