The Stacks project

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 [Drinfeld-number] 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

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


\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 $

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


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

Comments (3)

Comment #15 by Emmanuel Kowalski on

A few typos: (1) I don't understand what " as behave " means (just after the definition of ); (2) On the following line, is written instead of ; (3) On the item "DV implies", "of" is written instead of "if"; (4) (Not really a typo...) In the proof, is used and then replaced by .

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 and 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...)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03W6. Beware of the difference between the letter 'O' and the digit '0'.