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Let $S$ be a finite type regular integral affine scheme of finite type over $\text{Spec}(\mathbf{Z})$, and $\mathcal{X}\to S$ a smooth projective morphism. Let $\eta$ be the generic point of $S$, $s\in S$ a closed point.

One can construct a specialization map

$$sp: \text{CH}^n_{\rm alg}(\mathcal{X}_{\overline{\eta}})\to \text{CH}^n_{\rm alg}(X_{\overline{s}})$$

(see eg. Fulton's book, $\S$20.3).

It is known that $sp$ is very often not injective.

For any such $\mathcal{X}\to S$, does there exist $s\in S$ closed such that it is surjective?

Is there a class of $\mathcal{X}\to S$ for which this happens?

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    $\begingroup$ If $n = 1$, using the Lefschetz $(1,1)$ theorem one can prove $sp$ has torsion cokernel. Probably surjectivity is too much to hope for. $\endgroup$
    – user87684
    Commented Feb 21, 2018 at 16:13
  • $\begingroup$ What's an example where it is not injective for any $s$? $\endgroup$
    – byu
    Commented Feb 21, 2018 at 16:23
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    $\begingroup$ @Merlin: I don't think the cokernel is always torsion. You can have a family of K3 surfaces, say, with $\operatorname{Pic}(\mathcal{X}_{\bar{\eta }})=\mathbb{Z} $ but $\operatorname{rk} \operatorname{Pic}(\mathcal{X}_{\bar{s }})>1$ for some $s$ (though I believe that the rank is 1 for $s$ general enough). $\endgroup$
    – abx
    Commented Feb 21, 2018 at 16:39
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    $\begingroup$ I believe that can happen where $S$ is the complement of the discriminant hypersurface in the projective space over $\mathbb{Z}/p\mathbb{Z}$ with homogeneous coordinates $\{y_{b,c,d,e} : (b,c,d,e)\in \mathbb{Z}_{\geq 0}^4 , b+c+d+e=4\}$, and where $\mathcal{X}$ is the universal quartic surface in $\mathbb{P}^3_{\mathbb{Z}/p\mathbb{Z}}$. Every smooth quartic over a finite field (i.e., residue field of $S$) has Picard rank at least $2$. I believe that the geometric generic fiber has Picard rank $1$. $\endgroup$
    – Jason Starr
    Commented Feb 21, 2018 at 16:52
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    $\begingroup$ I agree. The induced map between Néron-Severi gps, however, does have torsion cokernel for some $s$ (at least when $k$ has characteristic zero. In positive char one can deduce it from the Tate conjecture for divisors). $\endgroup$
    – user87684
    Commented Feb 21, 2018 at 17:01

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