Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
    
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? –  TonyS Mar 20 '11 at 15:09
1  
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. –  Konstantin Ardakov Mar 20 '11 at 15:20
    
Ahh, very good. Thanks for your help. This are two conditions for $\omega_R$ i can work with. –  TonyS Mar 20 '11 at 15:41
add comment

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.

share|improve this answer
    
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 $R$. –  TonyS Mar 18 '11 at 20:02
    
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. –  Konstantin Ardakov Mar 18 '11 at 23:16
    
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. –  TonyS Mar 19 '11 at 11:01
    
The problem is that the trace pairing might be degenerate. For example this happens if $B$ is a (commutative) purely inseparable field extension of $F$. –  Konstantin Ardakov Mar 20 '11 at 13:16
    
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. –  TonyS Mar 20 '11 at 15:12
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.