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 $I.$


1 Answer 1


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.


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