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The classical Noether-Lefschetz theorem asserts the following: Over the complex numbers, a very general surface $S\subset \mathbb{P}^3$ has Picard number 1 (that is, $Pic(S)\simeq \mathbb Z$), provided that $\mbox{deg }S\ge 4$.

Over finite fields (or even countable fields), the corresponding statement does not make much sense, because here `very general' refers to the surface being chosen outside a countable union of closed proper subsets of the parameter space of surfaces. Still, I think makes sense to ask:

What evidence is there for the Noether-Lefschetz statement over countable fields? I.e., is there a sense in which the set of surfaces with low Picard number constitutes a 'large' proportion of the surfaces in $\mathbb{P}^3$ e.g., over finite fields with many elements?

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K3 surfaces over finite fields have (geometrically) even Picard number: not much hope for Noether-Lefschetz for quartic surfaces! –  M P Apr 26 '13 at 9:07
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(Though let me add that the Picard number over the ground field might be odd, so you could refine the question.) –  M P Apr 26 '13 at 9:09
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@dhagbert: The characteristic polynomial of Frobenius on $H^2$ has real coefficients and even degree, so an even number of real roots, which have to be $\pm 1/q$ by the Weil conjectures. It vanishes at $1/q$ because the Picard number is positive, so over $\mathbb{F}_{q^2}$ it has even Picard number. This works for any smooth surface of even degree in $\mathbb{P}^3$. –  Felipe Voloch Apr 26 '13 at 10:46
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@Voloch: You need to know the Tate conjecture in order to use this argument; even though it is known for K3 surfaces, at least for characteristics $> 3$, it is far from being known for all surfaces in $\mathbb{P}^3$. –  ulrich Apr 26 '13 at 11:19
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@J.C. Ottem: Hilbert's irreducibility theorem can be used to show that even over countable fields (which are not algebraic over a finite field) "most" surfaces of degree $d \geq 4$ have geometric Picard number equal to $1$. See the paper "Complete intersections with middle Picard number $1$ defined over $\mathbb{Q}$ by Terasoma. –  ulrich Apr 26 '13 at 11:25
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