# Smooth proper scheme over Z

Does every smooth proper morphism $X \to \operatorname{Spec} \mathbf{Z}$ with $X$ nonempty have a section?

EDIT [Bjorn gave additional information in a comment below, which I am recopying here. -- Pete L. Clark]

Here are some special cases, according to the relative dimension $d$. If $d=0$, a positive answer follows from Minkowski's theorem that every nontrivial finite extension of $\mathbf{Q}$ ramifies at at least one prime. If $d=1$, it is a consequence (via taking the Jacobian) of the theorem of Abrashkin and Fontaine that there is no nonzero abelian scheme over $\mathbf{Z}$, together with (for the genus $0$ case) the fact that a quaternion algebra over $\mathbf{Q}$ split at every finite place is trivial.

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Could you provide us with a bit of motivation, background, etc. to this question? –  Kevin H. Lin Dec 23 '09 at 1:43
I don't know of any examples of smooth proper varieties over $\mathbf{Z}$, except those constructed in some simple way from the flag varieties of Chevalley groups. Does anyone know other interesting examples? Also, the answer to your question is certainly NO if one replaces the integers by the ring of integers in a suitable number field. Do you know some variant of the question that might remain true? –  moonface Dec 23 '09 at 1:54
Here are some special cases, according to the relative dimension $d$. If $d=0$, a positive answer follows from Minkowski's theorem that every nontrivial finite extension of $\mathbf{Q}$ ramifies at at least one prime. If $d=1$, it is a consequence (via taking the Jacobian) of the theorem of Abrashkin and Fontaine that there is no nonzero abelian scheme over $\mathbf{Z}$, together with (for the genus $0$ case) the fact that a quaternion algebra over $\mathbf{Q}$ split at every finite place is trivial. –  Bjorn Poonen Dec 23 '09 at 2:27
I think your deleting your answer represents a net loss to the community! What do you think the etale cohomology of the "E_8 hypersurface" looks like? Note that I get around Hasse Principle issues by letting the variety have no real points ;-) –  Kevin Buzzard Dec 23 '09 at 10:22
It looks like a great question with a great answer. Can someone help me out by explaining what it means for a scheme over $\mathbb{Z}$ to be smooth, and to have a section? –  Greg Kuperberg Dec 23 '09 at 17:23

Hey Bjorn. Let me try for a counterexample. Consider a hypersurface in projective $N$-space, defined by one degree 2 equation with integral coefficients. When is such a gadget smooth? Well the partial derivatives are all linear and we have $N+1$ of them, so we want some $(N+1)$ times $(N+1)$ matrix to have non-zero determinant mod $p$ for all $p$, so we want the determinant to be +-1. The determinant we're taking is that of a symmetric matrix with even entries down the diagonal (because the derivative of $X^2$ is $2X$) and conversely every symmetric integer matrix with even entries down the diagonal comes from a projective quadric hypersurface. So aren't we now looking for a positive-definite (to stop there being any Q-points or R-points) even unimodular lattice?
So in conclusion I think that the hypersurface cut out by the quadratic form associated in this way to e.g. the $E_8$ lattice or the Leech lattice gives a counterexample!
Nice example! Stupid generalization: this is a flag variety of a reductive group over $\mathbf{Z}$; one gets a similar example from any such that is compact over $\mathbf{R}$, see Gross' paper for a list. –  moonface Dec 23 '09 at 15:53
That's beautiful, Kevin! One comment: If the determinant is $\pm 1$, then the projective quadric hypersurface is smooth over $\mathbf{Z}$, but not conversely. For example, if $f(x_1,\ldots,x_8)$ is the $E_8$ quadratic form, then $f(x_1,\ldots,x_8)+x_9^2=0$ in $\mathbf{P}^8$ works too. –  Bjorn Poonen Dec 24 '09 at 2:57