Let $R$ be a commutative ring, $I$ is an ideal of $R$, $M$ is an $R$-module. $$IM\supset I^2M\supset I^3M\supset\cdots$$ What is $\mathop {\lim }\limits_{\begin{subarray}{c} \longrightarrow \\ \end{subarray}}I^nM$ ?
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8$\begingroup$ A direct limit whose index set has a maximal element is just the object corresponding to the maximal element. In your case, just $IM$. $\endgroup$– Mariano Suárez-ÁlvarezCommented Sep 15, 2010 at 13:19
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$\begingroup$ Is that right? I also thought that, but i am not sure. $\endgroup$– minhtringuyenCommented Sep 15, 2010 at 13:34
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1$\begingroup$ If you want to be sure, just apply the definition (for any $R$ module $N$ and for any collection of homomorphisms $f_n:I_n:M\to N$ such that...&c, there exists a unique...&c...namely $f_1$). Note: the notations for direct and inverse limits refers graphically to sequences or diagrams, whose arrows point from left to right. Are you sure what you want is not the inverse (projective) limit, i.e. in this case, the intersection of the family? $\endgroup$– Pietro MajerCommented Sep 15, 2010 at 13:57
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2$\begingroup$ In that case, the answer depends strongly on your choice of $x$. For example, if $x=0 \in I$ or is nilpotent, then the direct limit is the zero module. If $x$ is idempotent in $I$, then the direct limit is $IM$. $\endgroup$– S. Carnahan ♦Commented Sep 17, 2010 at 13:10
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1$\begingroup$ @minhtringuyen the mathoverflow bot just bumped this question to the front page, even though it's been answered correctly below. I think you should accept the answer below so that this question doesn't keep coming back to the front page. $\endgroup$– David WhiteCommented Aug 2, 2011 at 13:44
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1 Answer
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I'll interpret this question by agglomerating information given in the comments: We choose an element $x \in I$, and want to know the structure of the direct limit of $\{ I^nM\}_{n \geq 0}$, when the maps $I^nM \to I^{n+1}M$ are given by multiplication by $x$.
I think the answer is that we can't say very much at all in general without somewhat more information, and the assumption that $x$ is $M$-regular is not enough. For example, if $R = \mathbf{Q}[t]$ and $M=R$, then choosing $x = t$ yields a direct limit that is the zero module, while choosing $x=1$ yields a direct limit that is $M$ (which is $IM$ in this case). For more general rings $R$, you can get intermediate results by setting $x$ to be something that has an interesting fixed subspace, e.g., an idempotent.
answered Sep 20, 2010 at 13:14