# Bound for number of points on surface over $\mathbb{F}_p$

I know of the bound for the number of points on an elliptic curve over a finite field: $$|\# E(\mathbb{F}_q) - q - 1| < 2\sqrt{q}$$ where this includes the point at infinity. I have been told that there are higher analogues of this formula which involve $q^{(k-1)/2}$ (instead of $q^{1/2}$ here), but I know very little about the theory so I'm not exactly sure what the $k$ is and all that. Can any one help me out with the formula for this? Thanks!

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The general form that this bound takes is known under the term Weil conjectures, which is a theorem of Deligne. Here is how it goes:

if $X$ is non-singular n-dimensional projective variety over $\mathbb{F}_q$ of dimension $n$, (an elliptic curve is the special case $n=1$), then you collect information about the number of points of $X$ over $\mathbb{F}_{q^m}$ for all $m$ in the generating function $$\zeta(X, s) = \exp\left(\sum_{m = 1}^\infty \frac{N_m}{m} (q^{-s})^m\right).$$ This is known as the zeta function of $X$.

The Weil conjectures then say that $\zeta(X, s)$ is a rational function of $T = q^{−s}$ and can be written as $$\prod_{i=0}^{2n} P_i(q^{-s})^{(-1)^{i+1}} = \frac{P_1(T)\dotsb P_{2n-1}(T)}{P_0(T)\dotsb P_{2n}(T)},$$ where each $P_i(T)$ is an integral polynomial that factors over $\mathbb{C}$ as $\prod_j (1 - \alpha_{i,j}T)$. If $X$ came from a projective variety defined over a number field with good reduction at $p=\text{char } \mathbb{F}_q$, then the degree of the $i$-th polynomial is the $i$-th Betty number of $X$. Moreover, $P_0(T) = 1 − T$ and $P_{2n}(T) = 1 − q^nT$. The Riemann hypothesis for these varieties, which is part of Weil's conjectures, says that $$|\alpha_{i,j}| = q^{i/2},$$ and this is the source of the Hasse bound you just cited. Here is what happens in your special case:

Let's assume for the moment that $X$ is a curve, so $n=1$. The degree of $P_1$ is twice the genus $g$ of the curve. If you take the logarithmic derivative of the zeta function and do some rearranging, you will find that $$N_m = 1 + q^m - (\alpha_{1,1}^m+\cdots+\alpha_{1,2g}^m),$$ which gives you the bound $|N_m - q^m - 1|\leq 2g\sqrt{q}^m$ that you quoted in the special case $g=1$.

If $X$ is a higher dimensional variety, then you can try to do the same manipulations, but I believe that the expression for $N_m$ will be less clean. Nevertheless, you should get something to the effect that $N_m = q^{nm} + O(q^{(2n-1)m/2})$. So your $k$ is my $2n$.

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Dear Alex, I think that the bound in the last line actually goes back to Lang--Weil (and in particular, predates Deligne). (You can get it by fibreing a higher dimensional variety by curves, or something similar, and using the Hasse--Weil bound for curves.) Best wishes, –  Matt E Apr 21 '11 at 5:47
@Matt Dear Matthew, thank you, that's very interesting! If you have time to write your comment up as an answer at some point or to provide a reference, I would be very interested. –  Alex B. Apr 21 '11 at 6:43
Dear Alex, See e.g. Theorem 2.1 of these notes by Mustata (where you can also find a reference to the original paper). Best wishes, –  Matt E Apr 21 '11 at 7:22
@Matt Dear Matt, wonderful, thank you! Best wishes, –  Alex B. Apr 21 '11 at 8:35

I think what's going on is if you have an equation in $k$ variables over a field of $q$ elements then you expect it to have $q^{k-1}$ solutions and the error term, under suitable hypotheses, is some constant multiple of $q^{(k-1)/2}$.

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