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Let $X$ be a smooth projective variety over $\mathbf{Q}$, and $\mathcal{X}$ a smooth projective model over $\mathbf{Z}[1/N]$ for $N$ large enough.

Call $\eta$ the generic point $\text{Spec}(\mathbf{Q}) \to\text{Spec}(\mathbf{Z}[1/N]) =: S$, and $s\in S$ any closed point.

There is a specialization map

$$\text{NS}(X_{\overline{\eta}})\to \text{NS}(X_{\overline{s}})$$

Does there exist $s$ such that it is an isomorphism?

By the proper and smooth base change theorem in $\ell$-adic cohomology, it is injective, so the issue is surjectivity.

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    $\begingroup$ In this paper: projecteuclid.org/euclid.ant/1513730132, Charles shows that if $X$ is a K3 surface over a number field, then the Picard rank of any reduction is bounded below by the Picard rank of $X$ + another quantity. This quantity is defined in terms of the transcendental lattice of $X$ and can be non-zero. If it is, then there can be no $s$ such that the specialization map is an isomorphism. $\endgroup$ – Keerthi Madapusi Pera Mar 17 '18 at 1:40
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    $\begingroup$ I think you already get examples for every projective, smooth morphism of relative dimension $1$, $f:\mathcal{Y}\to \text{Spec}\ \mathbb{Z}[1/N]$, whose geometric fibers are connected curves of genus $g>0$. Define $\mathcal{X}$ to be the fiber product $\mathcal{Y}\times \mathcal{Y}$. The graph of the Frobenius morphism of $\mathcal{Y}_s$ appear to give an extra divisor class on $\mathcal{X}_s$ for every closed point $s\in \text{Spec}\ \mathbb{Z}[1/N]$. $\endgroup$ – Jason Starr Mar 17 '18 at 12:03
  • $\begingroup$ You are right, Jason, already in genus 1! If the generic fiber of f is an elliptic curve without complex multiplication, then the Picard rank of any (good) reduction will be larger because elliptic curves over finite fields always have complex multiplication. See Milldenhall's paper in DMJ 67 (1992). $\endgroup$ – Bruno Kahn Sep 19 '18 at 16:29

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