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

Let $X$ be a smooth projective surface over a finite field $k=\mathbb{F}_q$. Let us first review the proof of the finite generation of $Pic(X)$ (notice that the proof is valid for any smooth projective variety over $\mathbb{F}_q$, not only for surfaces):

Let $\bar{k}$ be the algebraic closure of $k$, we know that $NS(X)$ and $Pic(X)$ are defined classically on $\bar{k}$, $NS(X)$ is finitely generated over $\bar{k}$ and $Pic^0(X)$ are the rational points $P(\bar{k})$ of the $\textit{Picard variety}$ $P$. We will descend the result from $\bar{k}$ to $k$:

Let $G=Gal(\bar{k}/k)$, we can use the Hochschild-Serre spectral sequence to deduce a short exact sequence: $$0\to H^1(G,\bar{k}^{\times})\to Pic(X)\to Pic(X)^G.$$ By the Hilbert theorem 90, $Pic(X)\to Pic(X)^G$ is injective, so we need to show $Pic(X)^G$ is finitely generated. Let $\bar{X}=X\otimes\bar{k}$. Consider the following exact sequence: $$0\to Pic^0(\bar{X})^G\to Pic(\bar{X})^G\to NS(\bar{X})^G$$ The right term is contained in $NS(\bar{X})$ and thus finitely generated. The left term equals to: $$P(\bar{k})^G = P(k).$$ Since $P$ is projective and $k$ is a finite field, $P(k)$ is finite. So, we conclude $Pic(X)$ is finitely generated.

For a detailed (and generalized) discussion, please refer to Buno Kahn's article: Sur le groupe des classes d'un schéma arithémtique, Bull. SMF 134 (2006), 395-415, which can be found on his webpage.

My question is: from the proof, we don't have a clear picture of the generators of $Pic(X)$. Does anyone know how to find the generators? I know the generators are in general very hard to find, so let me ask a weaker question: what can we say about the generators? Can they be chosen to be very ample divisors, or from some very ample linear systems?

Thank you very much in advance!

share|improve this question
    
Any line bundle on a projective variety is the difference of two very ample line bundles (twist it by the powers of a very ample line bundle to make the twist very ample). –  Torsten Ekedahl May 29 '11 at 12:15
    
Yeah, thanks. So I guess it's always possible to choose a family of very ample generators...... But I think these generators can't be taken from one linear system of a very ample sheaf, so they belong to different linear systems (of very ample sheaves). Is that possible to find these very ample sheaves? –  Fei May 29 '11 at 12:56

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.