4
$\begingroup$

Hi.

Question 1: If $f:A\rightarrow B$ be a morphism of local noetherian rings with $B$ is $A$-flat. Let $M$ (resp. $N$) be a $B$ (resp. $A$-)-module of finite type (fin. generated). We assume that $depth_{A}(M)\geq 2$, $M$ is $B$-torsion free and N is $A$-torsion free. Then it is true that $N\otimes_{A}M$ is torsion free ?

I remind the motivation: Let $f:X\rightarrow S$ be a proper and flat morphism of reduced finite dimensional complex spaces with n-dimensional fibers. Let $\omega^{n}_{X/S}$ be the canonical relative sheaf which is fiber wise of $depth>1$ and torsion free on $X$, $G$ torsion free coherent sheaf on $S$.

Question: Is the coherent sheaf $f^{*}G\otimes \omega^{n}_{X/S}$ (which is generically torsion free) torsion free fiber wise or on all of X?

Remark: We can reduce this question to smooth $f$ and especially to the projection $S\times U\rightarrow S$ and replace $\omega^{n}_{X/S}$ by torsion free coherent sheaf on $S\times U$ which is of $depth>1$ fiber wise...

Thank you.

P.S: Thanks to Boyarski for his remark. The last question on flatness and torsion freeness is not deleted but in another count of kaddar with the same name "kaddar".

$\endgroup$
3
  • $\begingroup$ You make no reducedness hypotheses, even though that holds in your motivating situation. Can you say exactly what you mean by "torsion-free" in this generality (without reducedness hypotheses)? Is $B$ always torsion-free over itself? You want if $N \otimes_ A M$ is torsion-free over $B$ (as opposed to over $A$), right? And $f$ is a local map, yes? Please clarify these in the question $\endgroup$
    – Boyarsky
    Jun 27, 2010 at 16:16
  • $\begingroup$ Excuse me. We assume that $A$ and $B$ are reduced or without embedded components... $\endgroup$
    – kaddar
    Jun 28, 2010 at 7:18
  • $\begingroup$ To merge account: you can email the moderators or go to tea.mathoverflow.net $\endgroup$ Jun 28, 2010 at 13:03

1 Answer 1

3
$\begingroup$

Tensor products of non-free modules typically will not be torsion-free, even if you assume good depth conditions on the modules.

In discussing the counter-example I will assume $B=A$ and depth $A$ at least $2$. Let $(m,k)$ be the maximal ideal and residue field respectively. Let $M$ be the first syzygy of $m$ (second syzygy of $k$). Then $\text{depth}_A(M)=2$ and $M$ is certainly torsion-free (being a submodule of a free module). Let $N$ be a torsion-free f.g $A$-module.

Claim: $M\otimes_AN$ is torsion free iff $\text{pd}_AN\leq 1$.

Proof: Tensor the exact sequence $0\to M\to F \to m\to 0$ with $N$ we get an exact sequence: $$0 \to \text{Tor}_1^A(m,N)\to M\otimes N \to F\otimes N$$

$\text{Tor}_1^A(m,N) = \text{Tor}_2^A(k,N)$ is killed by $m$, so it is the torsion part of $M\otimes N$ (note the quotient embeds into the torsion -free $F\otimes N$). Thus, the tensor product is torsion-free iff $\text{Tor}_2^A(k,N)=0$, which forces $\text{pd}_AN\leq 1$.

This also shows that regarding the second question, one can not improve much from the EGA result, since $\text{pd}_A\omega_A <\infty$ forces the canonical module to be free.

PS: you should merge your accounts and link to your other questions, for the benefits of the readers.

$\endgroup$
2
  • $\begingroup$ The "EGA result" was referred to in the other question by kaddar: mathoverflow.net/questions/29588/… $\endgroup$ Jun 28, 2010 at 1:57
  • $\begingroup$ Thank you very much Dao and Boyarsky. Also, in the relative setting of proper flat and surjective morphism $f:X\rightarrow S$ of complex reduced spaces with canonical relative sheaf $\omega^{n}_{X/S}:=H^{-n}(f^{!}{\cal O}_{S})$, we have: $f^{*}G\otimes \omega^{n}_{X/S}$ is torsion free (on $X$ ) for every torsion free coherent sheaf on $S$ if and only if $\omega^{n}_{X/S}$ is $S$ -flat and, then, if and only if $f$ is a Cohen-Macaulay morphism. That was i think... P.S: I dont know how to merge my differents accounts.... – kaddar 7 mins ago – kaddar 38 mins ago $\endgroup$
    – kaddar
    Jun 28, 2010 at 8:27

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.