Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $R$ be a complete discrete valuation ring and let $S = \operatorname{Spec}(R)$. Let $\mathcal{X}\to S$ be a complete, regular, flat, connected $S$-scheme of finite type whose fibres are smooth, projective, geometrically connected algebraic curves. Let $s \in S$ be the closed point, let $\xi \in S$ be the generic point, let $\mathcal{X}_s$ be the special fibre of $\mathcal{X}$, and let $X = \mathcal{X}_\xi$ be the generic fibre of $\mathcal{X}$.

If $\mathscr{L}$ is an invertible sheaf on $\mathcal{X}$, then there are maps $\Gamma(\mathcal{X}, \mathscr{L})\to \Gamma(X, \mathcal{L}_\xi)$ and $\Gamma(\mathcal{X}, \mathscr{L})\to \Gamma(\mathcal{X}_s, \mathscr{L}_s)$. In the right circumstances (I think when $\mathscr{L}$ is generated by global sections; maybe we need projective normality?), these two maps induce surjections $\Gamma(\mathcal{X}, \mathscr{L}) \otimes_R k(\xi) \to \Gamma(X, \mathcal{L}_\xi)$ and $\Gamma(\mathcal{X}, \mathscr{L}) \otimes_R k(s) \to \Gamma(\mathcal{X}_s, \mathscr{L}_s)$.

My question involves going in the other direction:

Suppose we are given $\mathscr{L}_s$ on $\mathcal{X}_s$ which is generated by global sections. Under what conditions, and how, can one find an invertible sheaf $\mathscr{L}$ on $\mathcal{X}$ such that $\Gamma(\mathcal{X}, \mathscr{L}) \otimes_R k(\xi) \to \Gamma(X, \mathcal{L}_\xi)$ and $\Gamma(\mathcal{X}, \mathscr{L}) \otimes_R k(s) \to \Gamma(\mathcal{X}_s, \mathscr{L}_s)$ are surjective.

ETA: Note that I am particularly interested in the "and how" part; which is to say that, given an (explicit) basis for $\Gamma(\mathcal{X}_s, \mathscr{L}_s)$, how do I go about finding an (explicit) basis for $\Gamma(\mathcal{X}, \mathscr{L})$ that satisfies the surjectivity properties above (assuming $\mathscr{L}$ exists)?

My idea is to take a basis of $\Gamma(\mathcal{X}_s, \mathscr{L}_s)$, lift it to an $R$-module $M$, and take the saturation $\operatorname{Sat}(M) = R^n \cap (M \otimes_R k(\xi))$ where $n = \dim\Gamma(\mathcal{X}_s, \mathscr{L}_s)$. Some related "sub-questions" based on this idea:

  1. Suppose $\Gamma(\mathcal{X}_s, \mathscr{L}_s)$ is generated by global sections $\{\bar{s}_0, \ldots, \bar{s}_r\}$ and let $M$ be the $R$-module generated by lifts $\{s_0, \ldots, s_r\}$ of the $\bar{s}_i$. Let $\mathscr{M} = \widetilde{M}$ be the sheaf associated to $M$. Then $M = \Gamma(\mathcal{X}, \mathscr{M}) \to \Gamma(\mathcal{X}_s, \mathscr{L}_s)$ is surjective. Under what conditions is $\mathscr{M}$ invertible?
  2. Is this $\mathscr{M}$ isomorphic to $\pi_*\mathscr{L}_s$, where $\pi\colon\mathcal{X}_s \to \mathcal{X}$ is the canonical "projection"?

N.B. The conditions I listed in the first paragraph describe the (rather restrictive) situation I'm interested in; I would certainly be interested in hearing answers valid in a more general context. Feel free to strengthen any conditions as necessary; for example, I think projective normality might play a role somewhere, but I'm not sure where exactly.

share|improve this question
add comment

2 Answers

The fact that you've assumed such niceness about the fibers and $\mathcal{X}$ should allow you to always do this regardless of whether or not $\mathcal{L}$ is globally generated or ample or anything.

The way I'm thinking of your setup is that you have some scheme $X_s$ over $R/m=k$, and $\mathcal{X}\to S$ is a deformation of it. Given $\mathcal{L}_s$ on the the special fiber a starting point is to ask whether or not this line bundle deforms with $X_s$ all the way to $X_\xi$.

Since the fibers are curves, and the obstruction to deforming a line bundle lies in $H^2(X_s, \mathcal{O})=0$ there is no obstruction and the line bundle deforms. Now you have some $\mathcal{L}_\xi$ on $X_\xi$ that when restricted to $X_s$ is $\mathcal{L}_s$.

Since $\mathcal{X}$ is nice you should be able to think of $\mathcal{L}_\xi$ as a Weil divisor on $X_\xi$ and taking Zariski closure will give you a line bundle $\mathcal{L}$ on all of $\mathcal{X}$ that when restricted to the two fibers mentioned gives you the correct thing back (to do this all you needed was $\mathcal{X}$ to be locally factorial if I remember correctly).

Edit: I didn't interpret your maps properly. What we have now is a candidate with the property that $\mathcal{L}|_{X_s}\simeq \mathcal{L}_s$, which I thought would translate into your map upon taking sections, but it is still in theory possible that $\Gamma (\mathcal{X}, \mathcal{L})=0$ while $\Gamma (X_s, \mathcal{L}_s)\neq 0$ so we need to convert this to what you actually want somehow.

share|improve this answer
    
Thanks for your response, Matt. Consider the $\mathscr{M}$ that I constructed in my two "sub-questions". Is it the case that your $\mathcal{L}_\xi$ corresponds to $\mathscr{M} \otimes_R k(\xi)$? If so, then the sheaf $\mathscr{S}$ associated to the saturation $\operatorname{Sat}(M)$ of $M$ satisfies the surjectivity conditions in my "main question" and we are left with the problem of showing that it's invertible. Or am I on entirely the wrong track? –  Hamish Jul 18 '11 at 18:21
    
I probably should have made it clearer in my question statement that I'm really looking for an explicit construction of this line bundle $\mathcal{L}$ on $\mathcal{X}$ based on the data of $\mathcal{L}_s$, which we can assume is determined by $\Gamma(\mathcal{X}_s, \mathcal{L}_s)$. –  Hamish Jul 18 '11 at 18:23
    
This construction definitely gives you an invertible sheaf. –  Matt Jul 18 '11 at 20:57
    
Could you give a (sketch of a) demonstration that the construction yields an invertible sheaf? Sorry if it's obvious; I don't have much experience with this stuff and I don't really see how to proceed. –  Hamish Jul 20 '11 at 17:55
    
If you believe that there is an invertible sheaf $\mathcal{L}_\xi$ on $X_\xi$, then you will use the divisor associated to this and take the Zariski closure (in $\mathcal{X}$). You'll need regularity of $\mathcal{X}$ to prove that you actually get a divisor on $\mathcal{X}$. But now take the invertible sheaf associated to that divisor to get $\mathcal{L}$. –  Matt Jul 20 '11 at 20:12
add comment

This is not a complete answer just a proposal: under the assumption you make the special fibre $\mathcal{X}_s$ is a regular curve, thus invertible sheaves correspond to Weil divisors.

Assume that the sheaf $\mathcal{L}_s$ corresponds to a Weil prime divisor $\mathcal{P}$ on $\mathcal{X}_s$. There exists a Weil prime divisor $P$ on the generic fibre $X$ such that $\mathcal{P}$ is contained in the Zariski closure $\overline{P}$ of $P$ on $\mathcal{X}$. Since $R$ is henselian $\mathcal{P}$ is the only point of $\mathcal{X}_s$ lying in $\overline{P}$. Let $\mathcal{L}(P)$ be the $\mathcal{O}_\mathcal{X}$-ideal sheaf defining the reduced subscheme structure on $\overline{P}$. By construction this sheaf is reflexive and since $\mathcal{X}$ is regular it is thus invertible and should do the job.

If $\mathcal{L}_s$ corresponds to a Weil divisor $e_1\mathcal{P}_1+\ldots +e_r\mathcal{P}_r$ one can extend the construction linearly replacing $\mathcal{L}(P)$ by the product (or tensor product) $\mathcal{L}(P_1)^{e_1}\cdot\ldots\cdot$ $\mathcal{L}(P_r)^{e_r}$.

share|improve this answer
    
Thanks for your response, Hagen. Let $\mathcal{P}$ and $P$ be as you define them. If $\mathcal{L}_s = \mathcal{L}(\mathcal{P})$ is generated by global sections and we have an explicit basis $B$ for $\Gamma(\mathcal{X}_s, \mathcal{L}_s)$, do you know how to use $B$ to get a basis of $\Gamma(\mathcal{X}, \mathcal{L}(P))$? For computational reasons, I have a strong preference for answers in terms of the bases of the global sections on the line bundles (the line bundles are allowed to be quite nice (generated by globals, very ample, etc.)). –  Hamish Jul 18 '11 at 13:14
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.