Algebraic Geometry – Smooth Proper Scheme Over Z

ag.algebraic-geometrynt.number-theory

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.

Best Answer

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!

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