## 63.35 How many points are there really?

If the genus of the curve is large relative to $q$, then the main term in the formula $\# X(k) = q - \sum \omega _ i + 1$ is not $q$ but the second term $\sum \omega _ i$ which can (a priori) have size about $2g_ X\sqrt{q}$. In the paper the authors Drinfeld and Vladut show that this maximum is (as predicted by Ihara earlier) actually at most about $g\sqrt{q}$.

Fix $q$ and let $k$ be a field with $k$ elements. Set

$A(q) = \limsup _{g_ X \to \infty } \frac{\# X(k)}{g_ X}$

where $X$ runs over geometrically irreducible smooth projective curves over $k$. With this definition we have the following results:

• RH $\Rightarrow A(q)\leq 2\sqrt{q}$

• Ihara $\Rightarrow A(q)\leq \sqrt{2q}$

• DV $\Rightarrow A(q)\leq \sqrt{q}-1$ (actually this is sharp if $q$ is a square)

Proof. Given $X$ let $w_1, \ldots , w_{2g}$ and $g = g_ X$ be as before. Set $\alpha _ i = \frac{w_ i}{\sqrt{q}}$, so $|\alpha _ i| = 1$. If $\alpha _ i$ occurs then $\overline{\alpha }_ i = \alpha _ i^{-1}$ also occurs. Then

$N = \# X(k) \leq X(k_ r) = q^ r + 1 - (\sum _ i \alpha _ i^ r) q^{r/2}$

Rewriting we see that for every $r \geq 1$

$-\sum _ i \alpha _ i^ r \geq Nq^{-r/2} - q^{r/2} - q^{-r/2}$

Observe that

$0 \leq |\alpha _ i^ n +\alpha _ i^{n-1} +\ldots +\alpha _ i +1|^2 = (n + 1) + \sum _{j = 1}^ n (n + 1 - j) (\alpha _ i^ j + \alpha _ i^{-j})$

So

\begin{align*} 2g(n+1) & \geq - \sum _ i \left(\sum _{j = 1}^ n (n+1-j)(\alpha _ i^ j +\alpha _ i^{-j})\right)\\ & =-\sum _{j = 1}^ n (n+1-j)\left(\sum _ i\alpha _ i^ j +\sum _ i\alpha _ i^{-j}\right) \end{align*}

Take half of this to get

\begin{align*} g(n+1)& \geq - \sum _{j = 1}^ n (n+1-j)(\sum _ i\alpha _ i^ j)\\ & \geq N\sum _{j = 1}^ n (n+1-j)q^{-j/2}-\sum _{j = 1}^ n (n+1-j)(q^{j/2}+q^{-j/2}) \end{align*}

This gives

$\frac{N}{g}\leq \left(\sum _{j = 1}^ n \frac{n+1-j}{n+1}q^{-j/2} \right)^{-1} \cdot \left( 1 + \frac{1}{g} \sum _{j = 1}^ n \frac{n + 1 - j}{n + 1}(q^{j/2} + q^{-j/2}) \right)$

Fix $n$ let $g\to \infty$

$A(q)\leq \left(\sum _{j = 1}^ n \frac{n+1-j}{n+1}q^{-j/2}\right)^{-1}$

So

$A(q)\leq \mathop{\mathrm{lim}}\nolimits _{n\to \infty }(\ldots ) = \left(\sum _{j = 1}^\infty q^{-j/2}\right)^{-1}=\sqrt{q}-1$
$\square$

Comment #15 by Emmanuel Kowalski on

A few typos: (1) I don't understand what "$X$ as behave $k=\mathbf{F}_q$" means (just after the definition of $A(q)$); (2) On the following line, $g_x$ is written instead of $g_X$; (3) On the item "DV implies", "of" is written instead of "if"; (4) (Not really a typo...) In the proof, $r$ is used and then replaced by $n$.

Comment #22 by Johan on

I tried to improve the typesetting and overal exposition a tiny bit. But this is in the part of the chapter on \'etale morphisms that we still have to rewrite completely. I think this was at the end of my lectures on \'etale cohomology where I started going faster and faster and maybe the note takers somehow couldn't keep up because of course my lectures were crystal clear!

Anyhow, are you claiming the thing with the $r$ and $n$ was/is a mathematical error? That would get you a stacks project T-shirt! I didn't completely understand your comment...

Comment #23 by Emmanuel Kowalski on

No, I don't see anything here that counts as a mathematical error, the r/n thing was only about typography and possible confusion... (Though I would of course love to have a Stacks Project t-shirt... But I may be able to find (tiny) mathematical mistakes if I continue reading...)

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