5
$\begingroup$

The following fact is extremely well known:

Fact. Let $Y$ be a geometrically irreducible variety (not necessarily smooth or proper) over a finite field $k$. Then there is a constant $B$, depending only on cohomological invariants of the base change of $Y$ to an algebraic closure of $k$, such that: if $l$ is any finite extension of $k$ satisfying $|l|>B$, then $Y(l)$ is non-empty.

Unfortunately, I can't find any simple reference for this well-known fact. If $Y$ is projective (and possibly singular), then it follows from the Lang-Weil estimates. If $Y$ is smooth and proper over $k$, then it follows from Deligne's proof of the Weil conjectures. Given that we don't mind replacing $Y$ with an open subvariety, there's a plausible argument for reducing the general case to one of these cases.

Also, the phrase "depending only on cohomological invariants of..." is rather vague, especially when $Y$ might be non-proper and singular. So maybe it's better to stick to more concrete statements, such as that $B$ should remain constant in certain types of families.

I would like to cite this fact without having to devote half a page to deducing it from something else. So my question is simply:

What is a good reference for the fact stated above?

Improve this question
$\endgroup$
1
  • $\begingroup$ You can take $B$ to be constant in any family whatsoever; this is weaker that depending only on cohomological invariants. An elementary expostion of this is given in a blog post of Terry Tao (seach for Lang-Weil). $\endgroup$
    – naf
    Commented Oct 29, 2013 at 13:16

3 Answers 3

Reset to default
6
$\begingroup$

This is already in Lang-Weil 1954. See Corollary 2 on p824 and the statements: "an abstract model... can be projective or affine ... our corollaries show that as $\nu$ increases indefinitely, the number of simple points on $V$ in $k_{\nu}$ also increases indefinitely".

Improve this answer
$\endgroup$
1
  • $\begingroup$ Hey, you're right, I'd not noticed that statement. How embarrassing. Thank you. $\endgroup$
    – Martin Bright
    Commented Oct 29, 2013 at 14:51
3
$\begingroup$

If you know that your algebraic variety is given (or can easily be given, or some open subset can) by a bounded number of polynomials in a bounded number of variables of bounded degree, and you don't care about actual constancy of the Betti numbers, you can combine easily the trace formula expression of D. Petersen's answer with bounds for Betti numbers as in Katz ("Sums of Betti Numbers in Arbitrary Characteristic", Finite Fields and Their Applications 7, 29-44 (2001)).

A convenient reference for the trace formula would be SGA 4 1/2, for instance.

Improve this answer
$\endgroup$
2
$\begingroup$

This follows from the Weil conjectures. Namely, for an arbitrary algebraic variety $Y$ one has $$ \# Y(\mathbf F_q) =\sum_i (-1)^i \mathrm{tr}\left(\mathrm{Frob}_q \mid H^i_c(\overline Y, \mathbf Q_\ell) \right). $$ If $Y$ is $d$-dimensional and geometrically irreducible, then $H^{2d}_c(\overline Y,\mathbf Q_\ell)$ is $1$-dimensional and of Tate type (it's spanned by the fundamental class) of weight $2d$. Since all $H^i_c$ have weight $\leq i$, this term dominates for $q$ large enough.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thank you. Two questions: what reference contains all of these statements? and what sort of uniformity statement is there, i.e. in what sort of families do these cohomology groups stay constant? Is there some kind of generalisation of smooth base change? $\endgroup$
    – Martin Bright
    Commented Oct 29, 2013 at 13:03
  • $\begingroup$ If you don't want to refer to the original SGA (which I haven't read) you can cite the book of Freitag and Kiehl. If you really want the cohomology groups to stay constant you have to impose pretty restrictive conditions on your families (e.g. smooth and proper). But in order to make this kind of argument work you only need an upper bound on $\dim H^i_c$ in the family, and this you get with almost no hypotheses at all. $\endgroup$
    – Dan Petersen
    Commented Oct 29, 2013 at 14:42
  • $\begingroup$ Namely, I think if $f \colon X \to Y$ is a finite type morphism of noetherian $k$-schemes then $R^i f_! \mathbf Q_\ell$ is constructible by a finiteness theorem from SGA, and then the ranks of the stalks $H^i_c(X_y)$ are bounded over $Y$. $\endgroup$
    – Dan Petersen
    Commented Oct 29, 2013 at 14:42
  • $\begingroup$ Excellent, thanks for the clarification. $\endgroup$
    – Martin Bright
    Commented Oct 29, 2013 at 14:53
  • 1
    $\begingroup$ Shouldn't it be mentioned somewhere that the weight bound on $H^i_c$ is not in any of the references mentioned so far (e.g., in the formulation of the OP, it seems clear that Deligne's Weil I paper is what is being referenced)? So it should be said that Deligne's Weil II paper is what provides the asserted weight bound without smoothness or properness hypotheses (this is not in F&K or any SGA). $\endgroup$
    – Marguax
    Commented Oct 29, 2013 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.