Proposition 64.34.1. 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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