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Do you know a proof for the following inequality?

Suppose that $(R,m)$ is a Noetherian local ring, $q$ is an $m$-primary ideal and $M$ is a finitely generated $R$-module. Then $$ l(q^nM/q^{n+1}M) \leq l(M/qM) \cdot \mu(q^n), $$ where $\mu(q^n)$ denote the smallest number of generators of $q^n$.


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First line in the proof of Prop 11.1.10 (pg. 217?). By the way, what happened to your previous comments? It didn't prove the original inequality but it's something worth to notice too. – mr.bigproblem Feb 24 '13 at 2:06
Here is a link to the textbook of Huneke and Swanson mentioned above (from Swanson's website): – mbrown Feb 24 '13 at 17:16
up vote 1 down vote accepted

Let $n \ge 1$; for $n=0$, this is trivial. Say $x_1, \dots, x_m$ are generators of $q^n$. $(x_1+q^{n+1})M/q^{n+1}M \oplus \dots \oplus (x_m+q^{n+1})M/q^{n+1}M$ maps onto $q^nM/q^{n+1}M$ in an obvious way (just sum the components), and $M/qM^{\oplus m}$ maps onto $(x_1+q^{n+1})M/q^{n+1}M \oplus \dots \oplus (x_m+q^{n+1})M/q^{n+1}M$ in an evident way (multiply the $i$th component by $x_i$; one can check this is well-defined). Thus, $q^nM/q^{n+1}M$ is a quotient of $M/qM^{\oplus m}$.

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Verified! Thanks for the answer! I'll edit the question, getting rid of the conditions $\mu(q) = \text{dim } R$ and $M$ is $R$-finite. – mr.bigproblem Feb 24 '13 at 19:52
We should really require that $M$ is fg, so that everything has finite length; otherwise, one could end up with $\infty \cdot 0$ on the right side. – mbrown Feb 26 '13 at 2:26

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