3
$\begingroup$

I would like to know what can be said about the classification of torsion-free modules.

For my purposes, we can assume that $R$ is the function ring of a smooth affine variety over a field. How does one proceed when trying to classify finitely generated torsion-free modules over $R$, or equivalently torsion-free sheaves on $\operatorname{Spec} R$? What general theorems are available, do there exist moduli spaces and what do they look like?

I have seen that for $R=k[X_1,X_2]$, there is some relation between torsion-free sheaves on $\mathbb{A}^2$ and the Hilbert scheme of points on $\mathbb{A}^2$? Unfortunately, I have not found a precise statement. Where is the relation between isomorphism classes of torsion-free sheaves and points on the Hilbert scheme of points made explicit, and proved? Are there some classification results for torsion-free sheaves that generalize to affine spaces of arbitrary dimension, or even to smooth affine schemes in general?

Finally, what can be said about automorphism groups of finitely generated torsion-free $R$-modules, automorphisms taken of course as $R$-modules? Are there any structural statements, either from the algebraic group or the discrete group point of view? Again, I would be interested both in the special case $k[X,Y]$ as well as in the case of smooth affine schemes.

[Edit: Thanks for the answer and comments I received so far. I would still like to know more about the automorphism groups of torsion-free modules. Are there any statements about this in the literature?]

$\endgroup$
  • $\begingroup$ You can think of the Hilbert scheme of $n$ points on the plane as parameterizing ideals, which are special torsion free sheaves, of colength $n$. Does that help? $\endgroup$ – Donu Arapura Jul 12 '14 at 14:00
  • $\begingroup$ @DonuArapura: it helps a bit, thanks. But how do I get to the isomorphism classification of torsion-free sheaves, and how do I get to more general affine schemes? Do I get all torsion-free sheaves from ideals? Is it really the points of the Hilbert scheme that parametrizes isomorphism classes of sheaves, or are there additional relations to be divided out on either side? Is it true for instance, that the isomorphism classes of torsion-free sheaves on a general smooth affine variety are classified by the points of the union of the Hilbert schemes of zero-dimensional subschemes? $\endgroup$ – Matthias Wendt Jul 12 '14 at 15:29
  • $\begingroup$ This post was 2 years ago. Do you know now more about this. I am myself interested. There is a paper that deals with automorphism groups of finitely generated modules : tinyurl.com/go8vn85. I don't have access to it though. Maybe it could help nevertheless. $\endgroup$ – Jose Capco Jul 13 '16 at 10:11
2
$\begingroup$

I expect that classification in general would be very difficult, but here are a few remarks about the simplest case. Let $M$ be a f.g. torsion free module over $R=k[x_1,x_2]$. Then to answer a question in your comment, although $M$ is not ideal in general, it does behave a bit like one.

Lemma. $M$ is a submodule of a free module in a canonical way.

The double dual $N=M^{**}$ is reflexive, which implies that it has depth $2$. Therefore, it follows from Auslander-Buchsbaum-Serre, that $N$ is projective and consequently free by (a very special case of) Quillen-Suslin. We has also have a canonical map $M\to N$, which is injective by torsion freeness, so the lemma follows.

This at least gives some kind of handle on $M$. Concretely, by choosing generators, you can see that $M$ is a image of a matrix.

As for moduli. If you take the sheaf associated to the quotient $N/M$ then this will be supported on a proper closed subscheme $Z$. I would guess that you can parameterize the $N$'s where $Z$ is purely zero dimensional by a Quot scheme (generalizing the punctual Hilbert scheme). But I'm not sure that you can do anything like that in general, since the usual constructions require properness of the base. (I think you can get around it in previous case).

Added I never really thought about the automorphism group of an ideal before, but I think I can work out a simple example. Let $m=(x,y)$ be the maximal ideal of the origin. This can resolved by the Koszul complex $0 \to R\to R^2\to m$. From this it follows that $Aut(m)$ is the group of matrices in $GL_2(R)$ which stabilize the line spanned by $(-y,x)^T$

$\endgroup$
  • $\begingroup$ Thanks for the details. The relation to zero-dimensional subschemes at least implies that there are quite a lot isomorphism classes of torsion-free modules over $k[x_1,x_2]$. In the case of ideals, what can be said about automorphism groups? $\endgroup$ – Matthias Wendt Jul 12 '14 at 19:40
  • $\begingroup$ The computation of the automorphism group via the Koszul complex is really helpful, thanks very much. $\endgroup$ – Matthias Wendt Jul 13 '14 at 8:15
  • $\begingroup$ Doesn't your double-dual argument show that the automorphism group maps injectively to the automorphism group of the double dual? In this case that is $R^\times$, and clearly the map is an isomorphism. $\endgroup$ – Will Sawin Jul 13 '14 at 17:16
  • $\begingroup$ @WillSawin: I think I agree with you. So in the case of ideals, and $R=k[X,Y]$, the automorphism group would be quite small. But what happens in the case when the double dual has rank two? The automorphism group would, according to your argument, be a subgroup of $GL_2(R[X,Y])$, but is it one of the obvious ones, like upper triangular matrices? $\endgroup$ – Matthias Wendt Jul 14 '14 at 9:33
  • 1
    $\begingroup$ @MatthiasWendt well, in the case you have a $Quot$ scheme to parametrize them, you need to take the action of $GL_2(R[X,Y])$ on the Quot scheme and look at the stabilizer. If $GL_2$ is infinite-dimensional and the Quot scheme is finite-dimensional then this is a finite codimension subgroup and so can't be any nice matrix group like upper triangular matrices. $\endgroup$ – Will Sawin Jul 14 '14 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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