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I want to prove that $\mathbb P_k^2$ is 'etale simply connected or that every finite morphism $X \to \mathbb P_k^2$ is ramified. Firstly I assume $X$ is regular. So if $X \to \mathbb P_k^2$ is unramified then $\chi(\mathcal O_X)=n \chi(\mathcal O_\mathbb {P^2})=n$. But $\chi(\mathcal O_X)=h^0(\mathcal O_X)-h^1(\mathcal O_X)+h^2(\mathcal O_X)=h^0(\mathcal O_X)-h^1(\mathcal O_X)+h^0(K_X)$ by Serre duality, then I want to say that $h^0(K_X)=0$, but I don't know what theorem should I use, intuitively $h^0(K_{\mathbb P^2})=0$ and $\Omega^1_{X/\mathbb P^2}=0$ should imply $h^0(K_X)=0$. If it's true then $n=\chi(\mathcal O_X)=h^0(\mathcal O_X)-h^1(\mathcal O_X)=1-h^1(\mathcal O_X) \le 1,$ then $n=1$ and since $X$ is regular $X=\mathbb P^2.$ So, my questions:

1) How can it be proven that $h^0(K_X)=0$?

2) What should I do if $X$ is not regular?

UPD. $k$ is algebraically closed then perfect, then $X/k$ is regular iff $X$ is smooth, $\mathbb P^2$ is regular <=> is smooth<=> $\mathbb P^2 \to \operatorname{Spec} k$ is smooth. Every 'etale morphism is smooth, composition of smooth morphisms is smooth, so $X \to \operatorname{Spec} k$ is smooth <=> $X$ is regular. Also pullback of ample sheaf over finite morphism is ample. Finite morphism is always separated, so $X$ is separated, so there is a closed immersion $X \to \mathbb P^n$, so $X$ is regular projective variety.

UPD. $h^0(K_X)=0$ because $f^*(-K_{\mathbb P^2})=-K_X$ since $f$ is 'etale. But $-K_{\mathbb P^2}$ has at least $2$ linearly independent sections, so $f^*(-K_{\mathbb P^2})=-K_X$ has at least $2$ linearly independent sections $s_1,s_2$, too. Then if $h^0(K_X)\ne 0$ it has a section $t\ne 0$, then $ts_1,ts_2$ are two linearly independent sections of $Г(\mathcal O_X)=k$, contradiction.

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  • $\begingroup$ What is $X$? It sometimes happen that $h^0(K_X) \neq 0$. $\endgroup$ – Fredrik Meyer Nov 4 '14 at 20:17
  • $\begingroup$ @FredrikMeyer I'd like to prove at least for X -- smooth projective variety. $\endgroup$ – qwerzzx Nov 4 '14 at 20:29
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This is proved in the fundamental text on the fundamental group by Grothendieck and his collaborators : SGA 1, Exposé XI, Proposition 1.1, page 285.

That $\mathbb P^r_k$ is simply connected (for an algebraically closed field $k$) is proved by induction on $r$:

r=1
This is a consequence of Hurwitz's formula for an étale covering $f:X\to \mathbb P^1_k$ of degree $d$.
Indeed $$2g_X-2=d(2.g_{\mathbb P^1_k}-2)+\text {degree of ramification}=d.(-2)+0=-2d$$ is only possible for $d=1$.
r arbitrary
They use the multiplicativity of $\pi_1$ to show that $\mathbb P^1_k\times...\times \mathbb P^1_k$ is simply connected and conclude that $\mathbb P^r_k$ is simply connected because they have already shown that $\pi_1$ is a birational invariant.

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  • $\begingroup$ Thanks a lot. I found this exercise in lecture notes on the course on Weil's conjectures and 'etale cohomology. That was an exercise right after the definition of $\pi_1$, before basic properties of it. I think that there should be some proof that doesn't use the theory of 'etale fundamental group. $\endgroup$ – qwerzzx Nov 4 '14 at 20:45
  • $\begingroup$ "..in the fundamental text on the fundamental group by Grothendieck..." This guy is so fundamental. So sad he passed away today. $\endgroup$ – Patrick Da Silva Nov 14 '14 at 18:35

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