2
$\begingroup$

Let $A$ be a Noetherian ring and $R$ is a standard graded ring over $A.$ Let $M$ be a finitely generated graded $R$-module and $I$ be a graded ideal of $R.$ Then $x_1,\ldots,x_r\in I$ is called $M$-filter regular sequence with respect to $I$ if $x_i\notin P$ for all $P\in Ass(M/(x_1,\ldots,x_{i-1})M)\setminus V(I)$ for all $i=1,\ldots,r.$

Q: What can we say about maximal length of $M$-filter regular sequence with respect to $I.$

$\endgroup$
0

1 Answer 1

1
$\begingroup$

There is no maximal length.

See "Some results on associated primes of local cohomology modules", J. Asadollahi and P. Schenzel, Japan J. Math. 29 (2003), 285--296.

Proposition 2.2 in this paper establishes that for every $n \in \textbf{N}$, there is an $I$-filter regular $M$-sequence of length $n$.

(This fact is used in H. Dao and P. H. Quy's recent preprint arXiv:1602.00421, in the proof of their Main Theorem.)

Note that if $I = R$, an $R$-filter regular $M$-sequence is just a weak $M$-sequence. The condition that $(x_1, \ldots, x_n)M \neq M$, which distinguishes $M$-sequences from weak $M$-sequences, is what puts an upper bound on the length of such a sequence. For filter regular sequences there is no such condition and hence no upper bound.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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