Does smooth and proper over $\mathbb Z$ imply rational? Does smooth and proper over $\mathbb Z$ imply rational?
I think someone told me that this is a standard conjecture. Is it a widely held? held at all? Did someone in particular make this conjecture? write it down somewhere? give evidence for it? Would you like to provide evidence? Why rational and not, say, unirational?
Fontaine's non-digitized letter to Messing shows that if $X$ is smooth and proper over $\mathbb Z$, then the low degree nondiagonal Dolbeault groups vanish $H^q(X_{\mathbb Q}; \Omega^p)=0$ for $p\ne q$,  $p+q\leqslant 3$. An obvious conjecture is to remove the degree restriction. Is this expected? Does this conjecture imply the rationality one? (This hypothesis on a single space does not force rationality: there are fake projective planes that mimic the cohomology of the real projective plane, but they have finite covers that would again be smooth and proper, but which can't have such cohomology by the Atiyah-Bott fixed point formula. a little more here)

What I'm looking for: A citation of a survey would be ideal. Short of that, I would accept a citation of someone making this conjecture (which may well be the paper I cite above). In any event, I would be happy if people provide evidence for or against the conjecture.

Some definitions: propriety is a relative version of compactness; let's just say that the scheme is cut out of projective space by homogeneous equations with coefficients in $\mathbb Z$. Smoothness means that the fibers, that is, the schemes defined by the same equations considered over finite fields (or their algebraic closures) are smooth. This is stronger than regularity of the total space and more like a submersion, so by analogy with Ehresmann's theorem, all the fibers are supposed to be "the same." For example, $\operatorname{Spec}\mathbb Z[i]$ is regular, but the map to $\operatorname{Spec}\mathbb Z$ is not smooth, but ramifies at $(2)$. (Geometrically) rational means that that the field of rational functions with complex coefficients are the same as on $\mathbb CP^n$, namely $\mathbb C(x_1,\ldots,x_n)$. I think this question shows that strict rationality, a fraction field of $\mathbb Q(x_1,\ldots,x_n)$, is too much to ask for.
To complete the list of relevant MO questions, would (geometric) rationality imply a Hasse principle, answering this question?
 A: (This answer has been edited -- it used to say that a finite cover of $\overline M_{g,n}$ gives a counterexample, which no longer seems obvious.)
If you had written "Deligne-Mumford stack" instead of "scheme", then a counterexample would be given by the spaces $\overline M_{g,n}$, which are certainly smooth and proper but far from rational in the large $g$ limit (or, for $g > 0$, in the large $n$ limit). The original references here are, I guess, Deligne (for $\overline M_{1,11}$) and Harris-Mumford (for $\overline M_{25}$). 
Kevin Buzzard's hint with the Ramanujan $\Delta$ function is relevant here; indeed, $H^{11,0}(\overline M_{1,11})$ is nonzero, and the $\ell$-adic Galois representation corresponding to $H^{11,0}\oplus H^{0,11}$ is the representation attached to $\Delta$. 
My answer to the question Is the moduli space of curves defined over the field with one element? contains some more detailed information about these things.
The natural way to find a scheme instead of a stack would then be to consider finite or generically finite covers of $\overline M_{g,n}$ by smooth proper schemes. There are several closely related constructions of such covers in the literature by Looijenga, Boggi-Pikaart, Pikaart-de Jong, Abramovich-Corti-Vistoli, all using some kind of non-abelian level structure on curves, but as far as I can tell none of them work over the integers. 
