# Cover of $\Bbb P^1_k$ ($k$ sep. closed) unramified away from $\infty$ and tamely ramified at $\infty$

I'm reading a paper where the authors claim that for a separably closed field $k$ of characteristic $p>0$, there's no cover of $\Bbb P^1_k$ unramified away from $\infty$ and tamely ramified over $\infty$, but I can't see why this happens. Does anyone knows?

I also found another paper with a similar claim but no more details: "[...] But over any algebraically closed field, the projective line has no non-trivial connected covers that are unramified away from infinity and tamely ramified over infinity."

• @Vincius: With covers, do you mean a morphism $f:X \rightarrow \mathbb{P}^1_k$ such that the map f is finite and separable and $X$ is a curve (smooth, geometrically connected)? Or what do you mean exactly by a cover? – user101036 Jan 13 '15 at 12:35
• @user101036, yes. The paper actually talks about a finite separable extension K/k(P^1) with the properties above and concludes K=k(P^1). – Vinicius M. Jan 13 '15 at 12:45
• @Vincius M : I posted an answer which should work if I interpreted your question correctly. – user101036 Jan 13 '15 at 13:03

This follows from Hurwitz theorem. Indeed, let $X$ be a smooth, geometrically connected and projective curve and $f: X \rightarrow \mathbb{P}^1_k$ be a finite and separable morphism and suppose that $f$ is étale outside of $\infty$ and tamely ramified over it. Say that the degree of $f$ is $n$ and the genus of $X$ is g. Then Hurwitz gies us that $$2g-2 = -2n+ \sum_{x \in f^{-1}(\infty)} (e_x -1)[k(x):k].$$ Now, we see that $$\sum_{x \in f^{-1}(\infty)} e_x = n,$$ so we have $$2g-2=-n-\sum_{x \in f^{-1}(\infty)} [k(x):k].$$ However, we know that $g \geq 0$ and since the RHS is always negative, we must have that $g=0$, so $$-2= -n-\sum_{x \in f^{-1}(\infty)}[k(x):k].$$ The only way to make this equation true is if $n=1$ and there is precisely one point in the preimage. So it is of degree 1 and thus an isomorphism.
I used tamely ramified at precisely one place, and this is not true if the cover is not tamely ramified over $\infty$. For a counterexample, you should look at Artin-Schreier coverings.