Put $R=\mathbb{Z}[x_1,\dotsc,x_n]$ and $I=(x_1,\dotsc,x_n)$. Let $M$ be an $R$-module that is probably not finitely generated. Suppose that the sequence $x_1,\dotsc,x_n$ is regular on $M$; I believe that it is then also regular on $M^\wedge_I$. Here I am using the definition of regularity in which the quotient $M/IM$ is allowed to be zero. The claim is immediate from the Artin-Rees Lemma etc if $M$ is finitely generated. I think that I can prove it in general, but the argument is a bit long and fiddly. Does anyone know a slick proof and/or a reference?
$\begingroup$
$\endgroup$
1
asked Jul 28, 2022 at 14:41
-
$\begingroup$ When $n = 1$ you could argue like this: $M$ is $x = x_1$-torsion free, so completion is equal to derived completion. Then the short exact sequence $0 \to M \to M \to M/xM \to 0$ upon derived completion becomes the distinguished triangle $M^\wedge \to M^\wedge \to (M/xM)^\wedge$. But $(M/xM)^\wedge = M/xM$ so sits only in degree $0$ hence $x$ is regular on $M^\wedge$. For larger $n$ try induction on $n$ (using the same argument essentially). Cheers! $\endgroup$– JohanCommented Jul 28, 2022 at 19:12
Add a comment
|