Polarizations of K3 surfaces over finite fields - MathOverflow

Polarizations of K3 surfaces over finite fields

2012-01-03T18:03:50Z

Suppose that $X$ is a (projective) K3 surface over a field $k$. A polarization of $X$ is an element $\lambda\in Pic_X(k)$ that is represented over an algebraic closure $\overline{k}$ by an ample line bundle $L$ over $X_{\overline{k}}$. Given such a $\lambda$, we can consider its self-intersection number $(\lambda,\lambda)\in 2\mathbb{Z}$. Suppose that $k$ has finite characteristic $p>2$. Is it possible to find a finite extension $k'/k$ and a polarization $\lambda$ of $X_{k'}$ such that $(\lambda,\lambda)$ is co-prime to $p$?

This question has a negative answer for polarizations of abelian varieties (see BCnrd's answer here), so I'm not optimistic that things are any better for K3 surfaces. On the other hand, one has an affirmative answer to the analogue for polarizations of abelian varieties, if one is allowed to modify the abelian variety up to isogeny.

So here's my backup question: Is there an analogous notion of an 'isogeny' of K3 surfaces that might help here? At the least, two K3 surfaces that are `isogenous' should have isometric Neron-Severi lattices (up to tensoring with $\mathbb{Q}$) and also isomorphic $l$-adic realizations, compatible with the cycle class map from the Neron-Severi lattices.

Answer by Donu Arapura for Polarizations of K3 surfaces over finite fields

2012-01-03T20:08:28Z

To expand my comment, suppose $k=\bar k$ has characteristic $p=3$. Let $X$ be a generic complete intersection in $\mathbb{P}^4$ of multidegree $(2,3)$. Apply Deligne's version of Noether-Lefschetz (SGA 7, exp XIX) to conclude that $Pic(X)$ is generated by $\mathcal{O}_X(1)$. This implies that any polarization has degree divisible by $6$. (There goes my resolution.)

Answer by Remke Kloosterman for Polarizations of K3 surfaces over finite fields

2012-01-05T10:37:43Z

Let $S$ be a smooth projective surface and $m:=\gcd(\deg \lambda)$ where $\lambda$ runs through the ample cone. Let $m'$ be the gcd of the entries of the intersection matrix of $S$

Then $m$ equals either $m'$ or $2m'$. The fact that $m'$ divides $m$ is obvious. To prove that $m$ divides $2m'$, note that there exist an effective divisor $D$ on $m$ such that at least one of $gcd(D^2,H^2)=m$ or $gcd(D.H,H^2)=m$ holds.

In particular, there exist infinitely many positive $n$ such that $\gcd(H^2, (nH+D)^2)=\gcd(H^2,2n(H.D)+D^2)\in {m,2m}$.

For $n$ sufficiently large one has that $D+nH$ is ample.

For a $K3$ surface over a finite field it is believed that the geometric Picard number is always even. So if you want to produce a counterexample you might try to construct a complete intersection of degree 3,2 with geometric Picard number 2, such that the second generator of the Picard group has genus 1 modulo 3, and such that the intersection number of this curve with $O(1)$ is divisible by 3.