I think I have an argument that might work. The goal is to prove that this is impossible. There are some gaps in it.
Let $X$ be a connected smooth proper scheme over $mathbb Z$. Clearly $Gamma(X,mathcal O_X)=mathbb Z$. (If the ring had zero-divsors, it would indicate $X$ reducible, impossible, or $X$ non-reduced, thus ramified, impossible. If it were a ring of integers of a number field it would give ramification at some prime.) Since $H^1(X,mathcal O_X)$ is the tangent space of the Picard scheme, and the Picard scheme is trivial, $H^1(X,mathcal O_X)$ is trivial. (This probably requires smoothness of the Picard scheme. I'm not sure if that holds.) I need to assume that $H^2(X,mathcal O_X)$ is torsion-free. (I would think that smoothness over a scheme should imply locally free higher pushforwards, which over an affine scheme implies torsion-free cohomology, but I don't know. This is true in characteristic 0 by Deligne, but we are obviously not in characteristic 0 here.)
We have the exact sequence $0to mathcal O_X to mathcal O_X to mathcal O_X/pto 0$, with the first map multiplication by $p$. Taking cohomology and filling in what we know, we get
$ 0 to mathbb Z to mathbb Z to H^0(X, mathcal O_X/p) to 0 to 0 to H^1(X,mathcal O_X/p) to H^2(X,mathcal O_X)to H^2(X,mathcal O_X) $
which since those are also the cohomology groups of $X_P$, gives $Gamma(X_p,mathcal O_{X_p})=mathbb F_p$, $H^1(X_p,mathcal O_{X_p})=0$.
Now let $Yto X$ be a cyclic etale cover of degree $p$. Artin-Schreier on $X$ gives $H^1_{et}(X_P,mathbb Z/p)=mathbb Z/p$. Thus there is a unique connected etale degree-$p$ cover of $X_p$, so it's the one you get by tensoring over $mathbb F_p$ with $mathbb F_{p^p}$. Since $Gamma(Y_p,mathcal O_{Y_p})=mathbb F_p$, it is connected, and is not the result of tensoring anything with $mathbb F_{p^p}$. This is a contradiction.
No cyclic etale covers of degree $p$ $implies$ no cyclic etale covers $implies$ no etale covers. (since ever group has a cyclic subgroup.)
No comments:
Post a Comment