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Recall that a quartic surface in $\mathbb{P}^3_\mathbb{C}$ has $N = 35$ coefficients. Let $U$ be the open subset of $\mathbb{P}^{N-1}$ parametrising smooth quartic surfaces and let $Q \to U$ be the corresponding family of all smooth quartic surfaces. Let $Pic_{Q/U}$ denote the Picard scheme of this family.

What is $Pic_{Q/U}$?

Picard groups of quartic surfaces can behave quite erratically, but my guess is that $Pic_{Q/U} \cong \mathbb{Z}$ is a constant group scheme, generated by $\mathcal{O}(1)$. It would be nice to have confirmation of this, ideally with a proof.

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    $\begingroup$ Formation of the Picard scheme commutes with base change, so the fibers of $Pic_{Q/U} \to U$ are complicated. $\endgroup$
    – Angelo
    Commented Jul 14, 2020 at 19:56
  • $\begingroup$ Thanks this makes sense. Still, I'm struggling to come up with a flat map $V \to U$ with $V$ connected such that $Pic_{Q/U}(V) \neq \mathbb{Z}$, since the genetic quartic surface has Picard group $\mathbb{Z}$. $\endgroup$
    – Daniel Loughran
    Commented Jul 14, 2020 at 20:32
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    $\begingroup$ There is no such flat map; the special fibers of $Pic_{Q/U} \to U$ are concentrated on proper subvarieties of $U$. $\endgroup$
    – Angelo
    Commented Jul 15, 2020 at 10:20
  • $\begingroup$ Yes I finally get it now. I think I was getting confused between the small and big fppf site; now it all makes sense, thanks! $\endgroup$
    – Daniel Loughran
    Commented Jul 16, 2020 at 9:30

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