The Riemann-Hurwitz formula implies that the projective line $\mathbb{P}^1_K$ over any algebraically closed field $K$ is simply connected (i.e., $\pi_1^{et}(\mathbb{P}^1_K) = 1$; equivalently, if $\phi\colon C\to \mathbb{P}^1_{K}$ is finite etale, then $\deg\phi=1$).
For $K=\mathbb{C}$, this follows from the connection with topology and from the fact that the complex plane $\mathbb{A}^1_{\mathbb{C}}$ is contractable.
In positive characteristic the affine line is not simply connected due to Artin-Schreier covers.
My question is whether there is a short proof for this fact in positive characteristic?
Best Answer
You can deduce this from the classification of vector bundles on $\mathbf{P}^1$. Say $f:C \to \mathbf{P}^1$ is a connected finite etale Galois cover of degree $n$. We must show $n=1$.
The sheaf $E := f_* \mathcal{O}_C$ is a rank $n$ vector bundle on $\mathbf{P}^1$, so we can write it as $E \simeq \oplus_{i=1}^n \mathcal{O}(a_i)$ for some integers $a_i$. As $C$ is connected, we have $h^0(\mathbf{P}^1,E) = 1$, so we must have $a_1 = 0$ and $a_i < 0$ for $i > 1$ (after rearrangement). It then follows that after pullback along any finite cover $g:D \to \mathbf{P}^1$ of smooth connected curves, we still have $h^0(D, g^* E) = 1$ as negative line bundles remain negative after pullback along finite maps, and thus cannot acquire sections.
But $f$ was a finite etale cover, so there is some $g:D \to \mathbf{P}^1$ as above with $g^* C \to D$ is just a disjoint union of $n$ copies of $D$ mapping down (in fact, one may take $g = f$ as $f$ was Galois). But then $h^0(D, g^*E) = h^0(D, \oplus_{i=1}^n \mathcal{O}_D) = n$. The only way this can happen is if $n=1$.