# What can be the dimension of a pointless smooth proper Z-scheme?

What is the smallest dimension $d$ such that there is a smooth proper morphism $X \to \operatorname{Spec} \mathbb Z$ of relative dimension $d$, with $X$ nonempty, without a section?

Of course, there must also be such a morphism in every larger dimension - just take $X \times \mathbb P^n$.

As described in this excellent question, $d\geq2$. As described in the accepted answer, $d\leq 6$. We can improve that to $d \leq 5$ by noting that the E7 lattice also produces a nonsingular hypersurface, because the unique potential singular point over $\mathbb F_2$ fails to lie on the hypersurface.

But that still leaves a lot of uncertainty! Can anyone clarify?

Here is an auxiliary question, which I think might prove easier to answer:

What is the smallest dimension of an $X$ satisfying those conditions that is also the flag variety of a reductive group?

• @Jan Hendrik: No such K3 surface exists (by a theorem of Fontaine).
– naf
Nov 7, 2012 at 11:37
• I have one observation regarding Question 2. There are speculations (perhaps even conjectures) of Colliot-Th&eacute;l&egrave;ne that for a geometrically rational variety, or even geometrically rationally connected variety, over a number field, the only obstruction to the Hasse principle is the Brauer-Manin obstruction. For a smooth, projective, geometrically rational scheme over $\mathbb{Z}$, the Brauer-Manin obstruction vanishes. Also there are local points at all non-Archimedean places. So, according to this speculation, there is a rational point if and only if there is a real point. Nov 7, 2012 at 15:24
• My vague argument for question $2$ goes like this: A flag variety of a reductive group is a twist of a $G/P$, $G$ a split reductive group. The automorphism group of that is $G/Z(G)$, so the twist is classified by $H^1(G/Z(G))$. The twist must be nontrivial, else there is a point. There is an exact sequence $H^1(G) \to H^1(G/Z(G)) \to H^2(Z(G))$. For $SL_n$ and $SP_n$, $Z(G)$ is a group of roots of unity, so its cohomology measures ramification. Thus if the variety is unramified, the image in $H^2(Z(G))$ should be trivial. So $H^1(G)$ must be nontrivial. Nov 8, 2012 at 0:21
• But that's also trivial for $SL_n$ and $SP_n$ by a version of Hilbert 90. So the group must be something else - an exceptional lie group or $Spin(n)$, $n\geq 7$. The smallest dimension of a flag variety of one of these groups is $5$. There's a lot of steps there I do not know how to check but perhaps this is correct. Nov 8, 2012 at 0:22
• I think every semisimple group has an empty flag variety - just take a compact real form and look at the set of all Borel subgroups contained in it. Nov 8, 2012 at 15:38