Please give me an example of an almost clean module $M$ over a ring $S$ so that if $x$ is a $M$ regular element then $M/xM$ is not almost clean.
4
-
15$\begingroup$ At least one MO-user would appreciate if you'd recall briefly what "almost clean" means (it seems not to be a very difficult definition) and explain a bit your motivation. $\endgroup$– Xandi TuniCommented Mar 31, 2011 at 12:25
-
$\begingroup$ Also, do you need just one regular element, or you want $M/xM$ to be not almost clean for any regular element? And could you say more on why are you interested in such example? You will be much more likely to get help here if more efforts are put into the question. $\endgroup$– Hailong DaoCommented Mar 31, 2011 at 15:16
-
2$\begingroup$ Consider a prime filtration for $M$. Then $M$ is almost clean if $Ass M = Supp F$ , where $F$ is the prime filtration. You can find the definition for a prime filtration here at the begining of the preface : duepublico.uni-duisburg-essen.de/servlets/DerivateServlet/… . I only need one regular element, not for all. I need this because I want to find an example for sdepth $M/xM > sdepth M - 1$ where sdepth is the stanley depth, M is almost clean and x is M regular. $\endgroup$– AndreiCommented Mar 31, 2011 at 16:22
-
1$\begingroup$ $M$ is almost clean if we have $AssM=SuppF$ for only one prime filtration $F$, not for all. $\endgroup$– AndreiCommented Mar 31, 2011 at 16:32
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
This is only an observation, but perhaps it would be helpful. The problem is equivalent to the case $M$ being a cyclic prime module, i.e $M=S/P $ for some prime ideal $P$.
That is because if we take a prime filtration $\mathcal F$ whose quotient has only associated primes of $M$, then any regular element has to be outside all of those primes. By induction on the length of $\mathcal F$, $M/xM$ has a filtration with quotients of the forms $S/(P_i,x)$, where $P_i \in Ass(M)$. So if $M$ is not almost clean, then one of the quotients can't be almost clean.
So now the question really is: Is there a domain $R$, and an element $x\in R$ such that $R/xR$ is not almost clean? I think there are, but proving they work might involve some K-theoretic arguments (see this question by Steven Landsburg).
answered Jun 13, 2011 at 14:48
$\begingroup$
$\endgroup$
1
Hope that this can help: When a finitely generated $Z^n$-graded $R$-module $M$ is almost clean, we have $sdepth(M)=depth(M)$. For this, please have a look at the discussion after Proposition 1.3 of "How to compute the Stanley depth of a monomial ideal" by Jurgen Herzog, Marius Vladoiu and Xinxian Zheng.
-
$\begingroup$ Yes this is true, but doesn't help. I only need an example of what I said in the first post. My comments referring to sdepth were only to explain my motivation for Xandi Tuni. $\endgroup$– AndreiCommented Apr 19, 2011 at 6:21