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
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
By construction all points of $X$ of degree $\leq N$, split completely in $Y$. So, in particular
Use R.H. on both sides. So you get
Since $2g_ Y-2 = (\# G)(2g_ X-2)$, this means
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
Rewriting we see that for every $r \geq 1$
Observe that
So
Take half of this to get
This gives
Fix $n$ let $g\to \infty $
So
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