Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $q^n$.

Thanks!

share|improve this question
    
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): people.reed.edu/~iswanson/book/SwansonHunekeCUP06.pdf –  mbrown Feb 24 '13 at 17:16
add comment

1 Answer 1

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 $M/qM^{\oplus m}$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.