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Let $X$ be a smooth variety over a field $k$ and I'd like to think about $CH_0(X)$ the 0-Chow group i.e. the group of rational equivalent classes of 0-cycles. I'm wondering if there is any reasonable formulation to make sense of "family"/"moduli" of rational equivalent classes of 0-cycles, forming an fppf sheaf over $\text{Spec} k$? If there is such a sheaf, is there any chance the sheaf is actually representable?

A quick literature search shows some work on certain formulations of "Chow schemes". However, they seem to parametrize the 0-cycles, instead of the rational equivalent classes. But this seems to suggest that now we only need to construct a subsheaf of principal 0-cycles and then we can consider the quotient sheaf.

A known example: if X is in addition one dimensional and proper, we can speak of the Picard scheme which represents the relative Picard functor. I would really like to know what happens when the dimension goes higher.

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When $X$ is a smooth projective variety over an algebraically close field $k$, then $CH_0(X)_{tors} = Alb(X)_{tors}$, a result due to Rojtman in https://www.jstor.org/stable/1971109?seq=1#metadata_info_tab_contents in characteristic zero, and with variants by Bloch (http://www.numdam.org/item/CM_1979__39_1_107_0/) and Milne (http://www.numdam.org/item/CM_1982__47_3_271_0/) for torsion away and at the characteristic respectively.

In general and not restricted to torsion, even for the simplest case of surfaces over $\mathbb{C}$, this is not really possible. Mumford proved that as soon as $p_g > 0$ the albanese map is not, in general, injective https://projecteuclid.org/euclid.kjm/1250523940 but it is expected that the map should be injective when $p_g = 0$ and this is known as Bloch's conjecture which has been verified for many surfaces but it still pen in general.

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  • $\begingroup$ Thanks. I did know of the results you mentioned, but I don't think they do answer my questions? $\endgroup$ – wkf Jan 26 at 20:32
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    $\begingroup$ Well if the albanese is not injective then the motive is not finite dimensional and hence zero cycles are not representable. I forgot the reference but it's in papers of Guletskii, for example. $\endgroup$ – Elden Elmanto Jan 26 at 21:57
  • $\begingroup$ Hi Elden, sorry but I don't think I understand your argument. I tried to look at a few papers of Guletskii but I failed to locate where he mentioned such an argument or the nonrepresentability. Could you please explain a little further or point me to the reference? Many thanks! $\endgroup$ – wkf Jan 27 at 16:32

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