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?


2 Answers 2


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 $n=1$.


You can apply the following statement to $X = \mathbb{P}^1_K$ and $L = O(1)$ when $K$ is a separably closed field.

Let $L$ be a line bundle on a reduced connected scheme $X$ such that $H^{0}(X,\mathcal{O}_X)$ is a separably closed field. Assume that any vector bundle on $X$ is of the form $\oplus_{i=1}^{n} L^{\otimes \lambda_i}$ for some integers $\lambda_1 \geq \dots \geq \lambda_n$, and that $L^{\lambda}$ has no nonzero global section for $\lambda < 0$. Then $X$ is simply connected.

Indeed, let $\mathcal{A}$ be a finite etale $\mathcal{O}_X$-algebra. Then $\mathcal{A} = \oplus_{i=1}^{n} L^{\otimes \lambda_i}$ as an $\mathcal{O}_X$-module for some integers $\lambda_1 \geq \dots \geq \lambda_n$. Because $\mathcal{A}$ has a global section (the unit) one must have $\lambda_1 \geq 0$.

If $\lambda_1 > 0$ then the morphism $L^{2 \lambda_1} \rightarrow \mathcal{A} \otimes_{\mathcal{O}_X} \mathcal{A} \rightarrow \mathcal{A} \rightarrow L^{\lambda_i}$ must be zero for each $i$ since $2 \lambda_1 > \lambda_i$. The the multiplication of $\mathcal{A}$ is $0$ on the $L^{ \lambda_1}$-factor, contradicting the reducedness of $\mathcal{A}$. Thus $\lambda_1 = 0$.

Since $\mathcal{A}$ is a finite etale $\mathcal{O}_X$-algebra the discriminant morphism $$ \mathrm{det}(\mathcal{A})^{\otimes 2} \rightarrow \mathcal{O}_X $$ is an isomorphism, so that $\sum_i \lambda_i =0$. Since $\lambda_{i} \leq 0$ for all $i$ we get that $\lambda_i = 0$ and thus that $\mathcal{A}$ is a trivial vector bundle. In particular the morphism $$ H^0(X,\mathcal{A}) \otimes_{H^0(X,\mathcal{O}_X)} \mathcal{O}_X \rightarrow \mathcal{A} $$ is an isomorphism. In particular $H^0(X,\mathcal{A})$ is a finite etale $H^0(X,\mathcal{O}_X)$-algebra, hence a trivial one, so that $\mathcal{A}$ is a trivial finite etale $\mathcal{O}_X$-algebra.


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