# torsion freeness of tensor product continued

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".

• 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 – Boyarsky Jun 27 '10 at 16:16
• Excuse me. We assume that $A$ and $B$ are reduced or without embedded components... – kaddar Jun 28 '10 at 7:18
• To merge account: you can email the moderators or go to tea.mathoverflow.net – Hailong Dao Jun 28 '10 at 13:03

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.

• 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 – kaddar Jun 28 '10 at 8:27