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Do there exist 3 absolutely irreducible homogeneous polynomials in $\mathbb{Z}[a, b, c, d, e, f]$ such that

  1. each one defines a hypersurface in $\mathbb{P}^5_{\mathbb{Z}}$ smooth over $Spec(\mathbb{Z})$
  2. the scheme-theoretic intersection of the hypersurfaces is an integral scheme $S$ whose function field is not isomorphic to $\mathbb{Q}(X, Y)$ and whose morphism $S\rightarrow Spec(\mathbb{Z})$ is smooth projective of relative dimension 2 with a geometrically connected generic fiber?

What is a specific example if they do exist?

From Fontaine's https://www.math.u-psud.fr/~fontaine/Zschemas.pdf it follows that the off-diagonal Hodge numbers of the generic fiber vanish in total degree≤3 (hypersurfaces in $\mathbb{P}^5_{\mathbb{Q}}$ satisfy this condition) so $S_{\mathbb{Q}}$ is not e.g. a K3 surface.

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  • $\begingroup$ Finding interesting varieties that are smooth over $\operatorname{Spec} \mathbf Z$ is very hard; see e.g. this question. For hypersurfaces of fixed degree and dimension, there is a chance that an elementary argument with discriminants can give a negative answer, but this can only give lower bounds on the degree. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Sep 16, 2019 at 20:20
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    $\begingroup$ Hmmm... doesn't Fontaine's theorem also imply that the canonical bundle complete intersection would be negative? Then, the degrees of the three hypersurfaces would have to have sum which is less than 6. So one of them is linear, and then you can reduce to hypersurfaces in P4. $\endgroup$
    – Yosemite Stan
    Commented Sep 16, 2019 at 20:29
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    $\begingroup$ So I think you just need to show that quadrics cannot have everywhere good reduction, and cubic surfaces cannot have everywhere good reduction. Right? Surely this is not so hard. $\endgroup$
    – Yosemite Stan
    Commented Sep 16, 2019 at 20:31

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