3
$\begingroup$

This question is in the interest of answering one part of this question, but I think it is distinct enough to warrant a separate question.

Let $X$ be a regular 2-dimensional Noetherian scheme, for example an arithmetic surface $\pi:X \to S$.

Suppose we have line bundles $\mathcal{L}_1, \mathcal{L}_2$ and a quasi-coherent $\mathcal{O}_X$-module $\mathcal{F}$ satisying $$\mathcal{L}_1 \subset \mathcal{F} \subset \mathcal{L}_2.$$

Question: When can we conclude that $\mathcal{F}$ is locally free?

As submodules of a locally free sheaf on a smooth curve are locally free, this would hold if $X$ were a smooth curve. We do know at least that it is torsion-free.

Really the most important thing I need to know is when $\mathcal{F}$ will preserve exactness of sequences $$0 \to \mathcal{L}' \to \mathcal{L} \to \overline{\mathcal{L}} \to 0$$ $$0 \dashrightarrow \mathcal{L}'\otimes\mathcal{F} \to \mathcal{L}\otimes\mathcal{F} \to \overline{\mathcal{L}}\otimes\mathcal{F} \to 0$$

for line bundles $\mathcal{L}', \mathcal{L}$.

So flatness would suffice. But by this and this I believe flat is equivalent to projective in this situation.

In lieu of flatness, a characterization of the $\mathcal{L}$ for which $\mathcal{T}or_1(\mathcal{L}, \mathcal{F})$ vanishes could potentially satisfy my needs, but knowing $\mathcal{F}$ is locally free would be preferable.

Any help would be appreciated.

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ The question is too vague. And as @user1092847 explained, typically such $F$ is not locally free (essentially, the condition that $F$ contains a locally free subsheaf is non-restrictive). $\endgroup$
    – Sasha
    Commented Mar 16, 2022 at 19:44
  • 2
    $\begingroup$ Here's a geometric condition. Like others, I'm not sure what you want, so I don't know if it's good for you. Note that $\mathcal{L_2}/\mathcal{L_1}$ is supported on some divisor $D$. Also, $\overline{\mathcal{L}}$ is supported on another divisor $\widetilde{D}$. It seems to me you're okay if $D$ and $\widetilde{D}$ intersect transversally in their smooth loci. $\endgroup$
    – sdr
    Commented Mar 16, 2022 at 19:52
  • 2
    $\begingroup$ "Almost never" would be a reasonable answer. For any fixed $\mathcal L_1 \subset \mathcal L_2$, there are only finitely many line bundles between them, but usually infinitely many non-line bundles. $\endgroup$
    – Will Sawin
    Commented Mar 17, 2022 at 0:13
  • 1
    $\begingroup$ There's no reference, but it's easy: Fix a meromorphic section $s$ of $\mathcal L_1$ and look at its divisor as a section of $\mathcal L_1, \mathcal F, \mathcal L_2$. If $\mathcal F$ is a line bundle, the divisor of $\mathcal F$ is sandwiched between the divisor of $\mathcal L_1$ and $\mathcal L_2$, but these differ on only finitely many irreducible components, so there are only finitely many possibilities. $\endgroup$
    – Will Sawin
    Commented Mar 17, 2022 at 11:32
  • 1
    $\begingroup$ For the other way, we are lookiong for submodules of $\mathcal L_2/\mathcal L_1$, which is a sheaf supported on a curve. As long as this curve is nonempty, it is easy to see there are infinitely many submodules - e.g. the submodule of sections vanishing on some finite set of points. Only finitely many can have inverse image in $\mathcal L_2$ a line bundle. $\endgroup$
    – Will Sawin
    Commented Mar 17, 2022 at 11:33

1 Answer 1

Reset to default
1
$\begingroup$

No. Locally, $\mathcal L_2$ is isomorphic to the coordinate ring $R$ and $\mathcal F$ is an ideal $I \subseteq R$ which contains a principal ideal. So, e.g $R = k[x,y]$ and $I = (x,y)$ containing $(x)$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks, but this does not answer the question I asked. $\endgroup$
    – PrimeRibeyeDeal
    Commented Mar 16, 2022 at 19:27
  • $\begingroup$ @PrimeRibeyeDeal I think you should specify what kind of conditions would you expect to impose. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Mar 16, 2022 at 19:29
  • 3
    $\begingroup$ @PrimeRibeyeDeal That's fair, I misread. However, the point is that any ideal contains a principal ideal-- so being sandwiched between two line bundles tells you nothing more than $\mathcal F \subseteq \mathcal L_2$. Basically, you are asking when an ideal is (locally) principal. $M$ being an ideal ideal forces the depth to be $\geq 1$, but you need to be depth $2$ to be a free module/flat. $\endgroup$
    – user1092847
    Commented Mar 16, 2022 at 20:10
  • $\begingroup$ @user1092847 Thank you. It looks I may need to find a different approach to solve my problem. $\endgroup$
    – PrimeRibeyeDeal
    Commented Mar 16, 2022 at 20:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .