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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does a (connected?) projective (or just complete?) variety over a finite field have cardinality congruent to 1 mod the size of the field?

Do (connected?) affine varieties have cardinality a power of the size of the field?

I think the answers to these questions are supposed to be yes, but I suspect some sort of better formulation of the question (missing hypotheses perhaps) is needed. Maybe there needs to be some mention of rational points over an algebraic closure. Feel free to fill such things in.

It would be nice if there was any sort of converse, like "if a variety is connected and has cardinality 1 mod |K|, then it is complete", but I'm less sure something like this exists.

share|cite|improve this question
I admit that I haven't read this too carefully, but you know about Chevalley-Warning, right? This sounds a lot like what you want. – Dylan Moreland Jun 22 '12 at 0:03
@DylanMoreland, thanks that sounds like the right direction, especially for the affine case. – Jack Schmidt Jun 22 '12 at 0:09
Looks like the affine statement is pretty wrong, and the projective one needs a load of hypotheses. Connected and has an algebraic cell decomposition looks good for now. – Jack Schmidt Jun 22 '12 at 0:42
up vote 2 down vote accepted

Nope. The general case is very far from the case of projective and affine space. Already this is false for elliptic curves; in general we only have the Hasse bound $$q + 1 - 2 \sqrt{q} \le |E(\mathbb{F}_q)| \le q + 1 + 2 \sqrt{q}$$

and neither lower nor upper bound can be improved conditional on the Sato-Tate conjecture. See also the Weil conjectures.

In the affine case a simple counterexample is the variety $xy = 1$, which has cardinality $q - 1$ over $\mathbb{F}_q$.

In the projective case, any curve of genus $0$ with at least one point over $\mathbb{F}_q$ is isomorphic to $\mathbb{P}^1$ and hence has $q + 1$ points over $\mathbb{F}_q$, but it may happen that it has no such points, e.g. take the projective closure of $x^4 + y^4 = -1$ over $\mathbb{F}_5$.

share|cite|improve this answer
What genus does y=1/x have? I am considering that "genus 0" is one of the missing hypotheses. – Jack Schmidt Jun 22 '12 at 0:14
@Jack: it has genus $0$, but it is missing more than one point at infinity. (One of them corresponds to the horizontal asymptote and one of them corresponds to the vertical asymptote.) – Qiaochu Yuan Jun 22 '12 at 0:15
Hrm. Maybe the affine case just isn't true. I was worried about disjoint unions of points earlier (so added connected), but this allows us to remove points one by one. In the projective case, does genus 0 help? Counting zeros is saner projectively right? Bezout's theorem or something. – Jack Schmidt Jun 22 '12 at 0:17
@Jack: I added an edit addressing this. Are you only interested in curves? – Qiaochu Yuan Jun 22 '12 at 0:20
Usually yes, but this time I'm actually mostly concerned with large dimensions, and am only looking for an affirmative result (so trying to find sufficient hypotheses). They should include flag manifolds, and some of their other-group variations. – Jack Schmidt Jun 22 '12 at 0:21

Your Answer


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.