I know that the problem of torsion in tensor products, even of torsion free modules, is a very delicate thing. Unfortunately i don't have a deeper insight into this subject, so i don't know how things behave in more complicated situations, like i ran into:

Given a regular local ring $A$ of dimension $\leq 2$ and an $A$-algebra $R$, which is free as an $A$-module. Then we have the canonical R-bimodule $\omega_R=Hom_A(R,A)$.

If $M$ is a left $R$-module, which is finitely generated and torsion free as an $A$-module, is $\omega_R\otimes_R M$ torsion free as an $A$-module?

For example if $R=M_n(A)$, then $\omega_R=R$ and $\omega_R\otimes_R M=R\otimes_R M\cong M$ as an $A$-module, so in this case it is torsion free. Since it is true for matrix algebras, it is true for all Azumaya algebras.

What about other examples, e.g. when $R$ is a maximal order in an division ring? Are there extra conditions for $M$ and $R$ such this can be true?


Instead of updating my previous answer, I've decided to add a new answer in order to keep it short(ish).

In the comments following his original question, TonyS added the extra assumption that $R$ is finitely generated over $A$. This is a strong condition, since it makes $R$ very close to being commutative. Moreover $R$ is known to be an integral domain, and a maximal order in its division ring of fractions $B$.

Under these assumptions the $A$-torsion-free module $M$ is actually $R$-torsion-free. To see this, let $F$ be the field of fractions of $A$; then since $R$ is finitely generated over $A$, the central localisation $F \otimes_A R$ is an integral domain which is finite dimensional over the field $F$, and is therefore a division ring; thus $F \otimes_A R = B$. Now $F \otimes_A M = F \otimes_A (R \otimes_R M) = (F\otimes_A R) \otimes_R M = B \otimes_R M$ so the kernels of the localisation maps $M \to F \otimes_A M$ and $M \to B \otimes_R M$ coincide. Thus $M$ is $A$-torsion-free if and only if $M$ is $R$-torsion-free.

Now let $N$ be a finitely generated $R$-bimodule which is torsion-free on both sides (for example $N$ could be $\omega_R$). Then $N \otimes_R M$ is a finitely generated $R$-module and therefore a finitely generated $A$-module. To study the torsion $T$ in this module, we study its support $Supp(T)$ in $Spec(A)$, or equivalently, the primes above the annihilator $Ann_A(M)$. Clearly $0$ is not in this support because $T$ is by definition a torsion $A$-module.

I claim that there are no primes in $Supp(T)$ of height $1$. Suppose for a contradiction that $P \in Supp(T)$ has height $1$; then localising $A$ and $R$ at $P$ produces a new maximal order $R_P$ which is free and finitely generated as an $A_P$-module. But $A$ is a commutative regular local ring, hence a UFD by Auslander-Buchsbaum, so $A_P$ is a discrete valuation ring. Since $R_P$ is finitely generated over $A_P$, it must be semilocal; since $R_P$ is also a maximal order, under these conditions it is known that $R_P$ is actually a right and left principal ideal domain: see Proposition 2.9 and Theorem 2.8 of the book "Ordres Maximaux au Sens de K.Asano" by Guy Maury and Jacques Raynaud.

Therefore the module $N_P$ is actually free over $R_P$ and hence $N_P\otimes_{R_P} M_P \cong M_P$ has no torsion. But this module is just $(N \otimes_R M)_P$ and by the exactness of localisation, $T_P$ is a torsion submodule of $(N \otimes_R M)_P$ and is therefore zero: thus $P \notin Supp(T)$, proving the claim.

Now $A$ was assumed to be of dimension at most $2$, so we see that $Supp(T) \subseteq \{ \mathfrak{m} \}$ where $\mathfrak{m}$ is the maximal ideal of $A$. This is the best possible result, because $N \otimes_R M$ can easily have $\mathfrak{m}$-torsion, as the following (commutative!) example shows.

Let $R = A$ and $N = M = \mathfrak{m}$. Pick a regular sequence $x,y$ in $\mathfrak{m}$, so that $0 \to A \stackrel{\alpha}{\to} A^2 \to \mathfrak{m} \to 0$ is a projective resolution of $\mathfrak{m}$, where $\alpha(a) = (ay, -ax)$. Then it's easy to see that $\mathfrak{m} \otimes_A \mathfrak{m} \cong \mathfrak{m}^2 / \alpha(\mathfrak{m})$. Now the image of the element $(y,-x) \in \mathfrak{m}^2$ in $\mathfrak{m}^2 / \alpha(\mathfrak{m})$ is non-zero and is killed by $\mathfrak{m}$, so the $\mathfrak{m}$-torsion submodule of $\mathfrak{m} \otimes_A \mathfrak{m}$ is non-zero.

So to show that $\omega_R \otimes_R M$ has no torsion, one would have to show that $\omega_R$ doesn't "look like" $\mathfrak{m}$ (or perhaps a finite direct sum of copies of $\mathfrak{m}$) as an $A$-module. One way to ensure this is to perhaps try to show that $\omega_R$ is reflexive as an $R$-module, since this would help you to show that there are no essential extensions $E$ of $\omega_R$ such that $E/\omega_R$ is $\mathfrak{m}$-torsion.

  • $\begingroup$ Thanks! If $dim(A)=1$ and $rk(R)=4$, one can even compute $\omega_R$: $\omega_R=p^{-1}R$, where $p$ generates the radical ideal, i.e. $rad(R)=pR$. I also found the notion of Gorenstein orders, see for example arxiv.org/abs/math/0401425, these are orders such that $\omega_R$ and $R$ are isomorphic as left and right $R$-modules, but not necessarily as $R$-bimodules. But would this be enough for $\omega_R\otimes_R M$ to be torsion free over $A$, if we consider $\omega_R$ just as a left $R$-module, i.e. forget the left module structure? $\endgroup$ – TonyS Mar 20 '11 at 15:09
  • 1
    $\begingroup$ Yes. If $\omega_R$ is free as a right $R$-module, then it's flat as such. Now $M$ is torsion-free so $M \hookrightarrow R^k$ as a left $R$-module for some integer $k$. Therefore by flatness, $\omega_R \otimes_R M \hookrightarrow \omega_R \otimes_R R^k \cong \omega_R^k$ as a left $R$-module. So if you also know that the left $R$-module $\omega_R$ is torsion-free, then so is $\omega_R \otimes_R M$. For this argument to work it is necessary and sufficient for $\omega_R$ to be projective as a right $R$-module and torsion-free as a left $R$-module. $\endgroup$ – user91132 Mar 20 '11 at 15:20
  • $\begingroup$ Ahh, very good. Thanks for your help. This are two conditions for $\omega_R$ i can work with. $\endgroup$ – TonyS Mar 20 '11 at 15:41

Since $A$ is regular local commutative ring, it's an integral domain. Let $S = A - 0$ so that the localisation $A_S$ is the field of fractions $F$ of $A$.

Assuming the algebra $A$ is central in $R$, $S$ is a central multiplicative subset in $R$. So we can localise at it and form the localised $F$-algebra $B := R_S$. The corresponding canonical $B$-bimodule $B^\ast = Hom_F(B,F)$ is the localisation of $\omega_R$ at $S$, so we see that

$\bullet$ if $\omega_R \otimes_R M$ is torsion-free as an $A$-module then $B^\ast \otimes_B M_S$ is non-zero.

So in the case when $A$ is already a field, your question is equivalent to asking whether $B^\ast$ is a faithful right $B$-module. This is not true in general: here is an example.

Let $B$ be the upper-triangular $2 \times 2$ matrix ring with entries in a field $F$, so $B = F e_{11} \oplus F e_{12} \oplus F e_{22}$. Then you can compute that the action of $B$ on $B^\ast = Fe_{11}^\ast \oplus Fe_{12}^\ast \oplus F e_{22}^\ast$ satisfies

$e_{22}^\ast \cdot e_{12} = e_{22}^\ast$

$e_{11}^\ast \cdot e_{11} = e_{11}^\ast$, and

$e_{12}^\ast \cdot e_{11} = e_{12}^\ast$.

So if $J$ is the maximal ideal of $B$ spanned over $F$ by $e_{11}$ and $e_{12}$ then

$B^\ast\cdot J = B^\ast$

which forces

$B^\ast \otimes_B (B / J) = 0$

and shows that $B^\ast$ is not a faithful right $B$-module.

In the positive direction, it is enough for $R$ to be a Frobenius extension of $A$ for $\omega_R$ to be isomorphic to $R$ as an $R$-bimodule; then of course $\omega_R \otimes_R M$ is just $M$ as an $R$-module and therefore $A$-torsionfree. You can find more information about Frobenius extensions in a paper by Bell and Farnsteiner: http://www.jstor.org/stable/2154275.

  • $\begingroup$ Thanks for the counterexample. Unfortunately in the case i'm most interested in, i.e. $R$ is a maximal order in a division ring, we have that $R_S$ is a division ring, so that $B^{\*}\cong B$, and so $B^{\*}\otimes_B M_S$ is always non trivial. So this criterion does not work in this case. Even worse, in that case we have that $R$ is properly contained in $\omega_R$, so $\omega_R$ can't be isomorphic to $R$. $\endgroup$ – TonyS Mar 18 '11 at 20:02
  • $\begingroup$ Is $R$ finitely generated as an $A$-module? And how do you know that $B^\ast \cong B$ as $B$-bimodules? As far as I know this isn't completely automatic. $\endgroup$ – user91132 Mar 18 '11 at 23:16
  • $\begingroup$ Yes , $R$ is finitely generated as an $A$-module and for my case it would be enough if $R$ is a free $A$-module of rank 4. And yes i was a little to quick. I thought, since $B$ is a division ring, i.e. an Azumaya algbera, that the trace pairing $tr: B\times B \rightarrow F$ identifies $B^{\*}$ and $B$, but this is just as $A$-modules, not as $B$-bimodules. Hm, i have to think about that. But i think i can prove it for maximal orders over rings of dimension one, i will update my question with the computations later. $\endgroup$ – TonyS Mar 19 '11 at 11:01
  • $\begingroup$ The problem is that the trace pairing might be degenerate. For example this happens if $B$ is a (commutative) purely inseparable field extension of $F$. $\endgroup$ – user91132 Mar 20 '11 at 13:16
  • $\begingroup$ Yes, but if we have a noncommutative division ring, e.g. a quaternionen algebra, then the trace pairing is not degenerate, so one should be save in that case. $\endgroup$ – TonyS Mar 20 '11 at 15:12

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