Proposition 63.34.1. There exists a finite set $x_1, \ldots , x_ n$ of closed points of $X$ such that set of all frobenius elements corresponding to these points topologically generate $\pi _1(X)$.

Proof. Pick $N\gg 0$ and let

$\{ x_1, \ldots , x_ n\} = {\text{ set of all closed points of} \atop X \text{ of degree} \leq N\text{ over } k}$

Let $\Gamma \subset \pi _1(X)$ be as in the variant statement for these points. Assume $\Gamma \neq \pi _1(X)$. Then we can pick a normal open subgroup $U$ of $\pi _1(X)$ containing $\Gamma$ with $U \neq \pi _1(X)$. By R.H. for $X$ our set of points will have some $x_{i_1}$ of degree $N$, some $x_{i_2}$ of degree $N - 1$. This shows $\deg : \Gamma \to \widehat{\mathbf{Z}}$ is surjective and so the same holds for $U$. This exactly means if $Y \to X$ is the finite étale Galois covering corresponding to $U$, then $Y_{\overline{k}}$ irreducible. Set $G = \text{Aut}(Y/X)$. Picture

$Y \to ^ G X,\quad G = \pi _1(X)/U$

By construction all points of $X$ of degree $\leq N$, split completely in $Y$. So, in particular

$\# Y(k_ N)\geq (\# G)\# X(k_ N)$

Use R.H. on both sides. So you get

$q^ N+1+2g_ Yq^{N/2}\geq \# G\# X(k_ N)\geq \# G(q^ N+1-2g_ Xq^{N/2})$

Since $2g_ Y-2 = (\# G)(2g_ X-2)$, this means

$q^ N + 1 + (\# G)(2g_ X - 1) + 1)q^{N/2}\geq \# G (q^ N + 1 - 2g_ Xq^{N/2})$

Thus we see that $G$ has to be the trivial group if $N$ is large enough. $\square$

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

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

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