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Following up on this question which received a negative answer, I wonder if something weaker is true.

We work in the same set-up as the previous question. Let $B$ be a smooth quasi-projective variety over a field of characteristic zero (you may assume these are just the complex numbers). Let $\pi\colon \mathcal{X} \rightarrow B$ be a smooth and projective morphism with geometrically integral fibres. Let $Z$ be a cycle on $\mathcal{X}$, i.e. a $\mathbb{Z}$-linear combination of closed integral subvarieties of $\mathcal{X}$. Suppose in addition that $Z$ is flat over $B$, i.e. each component of the support of $Z$ is flat over $B$.

Question: is the locus $B_{\text{trivial}} = \{ b\in B \mid \text{ the cycle }Z_b \text{ of }\mathcal{X}_b \text{ is rationally trivial}\}$ a countable union of closed subvarieties of $B$?

What if we ask the same question with rationally trivial replaced by algebraically trivial?

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  • $\begingroup$ This is true rationally, i.e., with $\mathbb{Q}$ coefficients, but not true integrally, at least if $b$ is allowed to be any point (not necessarily closed). The problem is that increasing the base field of a variety does not induce an injection on Chow groups (but does rationally). To get an explicit example, consider the same cycle as in Jason Starr's answer to your earlier question but with $X$ there any Enriques surface over $\mathbb{C}$. Then the cycle is trivial over each closed point but not over the generic point (since there is torsion in the Neron-Severi group). $\endgroup$
    – naf
    Commented Jan 31 at 7:34
  • $\begingroup$ Ah that's a great example. Do you have a reference for the statement with $\mathbb{Q}$-coefficients? $\endgroup$
    – Jef
    Commented Jan 31 at 10:32
  • $\begingroup$ The statement with $\mathbb{Q}$ coefficients follows from the fact that all the data can be defined over a countable field, injectivity of the base change map with $\mathbb{Q}$ coefficients, and the fact (which you had already observed in your previous question) that the locus is closed under specialisation (which also holds with $\mathbb{Q}$ coefficients). $\endgroup$
    – naf
    Commented Feb 1 at 2:07

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