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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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  • $\begingroup$ Could you provide us with a bit of motivation, background, etc. to this question? $\endgroup$
    – Kevin H. Lin
    Commented Dec 23, 2009 at 1:43
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    $\begingroup$ 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. $\endgroup$
    – Bjorn Poonen
    Commented Dec 23, 2009 at 2:27
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    $\begingroup$ 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 ;-) $\endgroup$
    – Kevin Buzzard
    Commented Dec 23, 2009 at 10:22
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    $\begingroup$ 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? $\endgroup$
    – Greg Kuperberg
    Commented Dec 23, 2009 at 17:23
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    $\begingroup$ @James Borger: I disagree. If Q is the split quadric in P^7, then H^6(Q) is two dimensional, and only one dimension is generated by restriction from P^7. The same should be true for the E_8 quadric, right? $\endgroup$
    – David E Speyer
    Commented Dec 25, 2009 at 14:59

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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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    $\begingroup$ 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. $\endgroup$
    – moonface
    Commented Dec 23, 2009 at 15:53
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    $\begingroup$ 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. $\endgroup$
    – Bjorn Poonen
    Commented Dec 24, 2009 at 2:57
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    $\begingroup$ Oh, I see what Bjorn is saying. The point is that if all the partial derivates vanish you can't deduce that f vanishes, because the usual trick doen't work for a degree 2 equation in characteristic 2. So my "singular point" might not lie on the variety! $\endgroup$
    – Kevin Buzzard
    Commented Dec 24, 2009 at 15:01
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    $\begingroup$ Very nice. It would be even nicer if someone can come up with an example having non-trivial fundamental group. $\endgroup$
    – Minhyong Kim
    Commented Dec 25, 2009 at 13:30
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    $\begingroup$ As Minhyong pointed out to me via email, this has nothing to do with sections: this is just another question! Is there a smooth proper variety over Z with non-trivial fund gp? Take the product of this with a smooth proper variety with no sections if you want one with no sections. The problem is that no-one will see this question ;-) Minhyong---you should ask another question! Maybe flag varieties do the job again. $\endgroup$
    – Kevin Buzzard
    Commented Dec 26, 2009 at 9:47

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