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

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

## 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Your email address will not be published.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.

## Comments (0)