## Torsion line bundles with non-vanishing cohomology on smooth ACM surfaces

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I am looking for an example of a smooth surface $X$ with a fixed very ample $\mathcal O_X(1)$ such that $H^1(\mathcal O(k))=0$ for all $k$ (such thing is called an ACM surface, I think) and a globally generated line bundle $L$ such that $L$ is torsion in $Pic(X)$ and $H^1(L) \neq 0$.

Does such surface exist? How can I construct one if it does exist? What if one ask for even nicer surface, such as arithmetically Gorenstein? If not, then I am willing to drop smooth or globally generated, but would like to keep the torsion condition.

More motivations(thanks Andrew): Such a line bundle would give a cyclic cover of $X$ which is not ACM, which would be of interest to me. I suppose one can think of this as a special counter example to a weaker (CM) version of purity of branch locus.

To the best of my knowledge this is not a homework question (: But I do not know much geometry, so may be some one can tell me where to find an answer. Thanks.

EDIT: Removed the global generation condition, by Dmitri's answer. I realized I did not really need it that much.

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Some (hopefully) positive criticism: You should add (edit) some discussion to motivate your question and show some thoughts/progress you have about it, even if it is very little.  This serves many purposes: 1) it gets people interested, 2) it shows you care about the problem (which I believe you do), and 3) it makes it easier for others to start working on it.  – Andrew Critch Dec 6 2009 at 0:27
Thanks! See above.  – Hailong Dao Dec 6 2009 at 0:43
I wonder what is so interesting about ACM surfaces (I'm sure there is something if you're asking)? – Ilya Nikokoshev Dec 6 2009 at 0:56
In general, ACM varieties have all the intermediate cohomolgy vanish. They correspond to Cohen-Macaulay rings, which have maximal depth, hence the name. From both algebraic (maximal depth) and geometric (no cohomology) I think they are nice. There are no wiki  entry for them though (:  – Hailong Dao Dec 6 2009 at 1:10
What is the ground field that you consider in this question? If this is over $C$ and the surface is smooth then a globally generated torsion line bundle is trivial.  – Dmitri Dec 9 2009 at 0:33
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## 1 Answer

Let us show that a globaly generated torsion line bundle $L$ on a (compact) complex surface is trivial. Ideed, a globally generated line bundle has at least one section, say $s$. Let us take it. If $s$ has no zeros, then $L$ is trivial. But if $s$ vanishes somewhere then any positive power $L^n$ has a section $s^n$ that vanishes at the same points. So any power of $L$ is not trivial, i.e. $L$ is not a torsion bundle, contradiction.

Notice that we did not use the fact that the surface is smooth. And we also did not use the fact that we work with a surface...

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Thanks! What about without the global generation condition? I realized I did not need it that much. – Hailong Dao Dec 9 2009 at 15:58
Dear Hailong, it will be very helpfull if you specify what you mean by $H^1(\mathcal O(k))=0$. For example is $k$ an integer number? In this case, can it be $0$? If it $k=0$, does this mean for you that $O(0)$ is just the structure sheaf of $X$? In this case it follows that your surface has $H^1(X)=0$. Also do I understand correctly that for you $mathcal O(1))$ is ANY ample line bundle? – Dmitri Dec 9 2009 at 17:00

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Dear Dmitri, I did mean all $k$(here I fixed a very ample line bundle $mathcal O_X(1)$). This is quite restrictive, I am aware of that. – Hailong Dao Dec 9 2009 at 18:06
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