I have been starting to learn about about chow groups. I don't know much yet, so hopefully the following is trivial: :-)
For a smooth (projective, if you like) variety $X$ over a field $k$ I will write $CH_n(X)$ for the Chow group of $n$-cycles up to rational equivalence. One of the first things one learns is that these groups in general in no sense satisfy the/a Künneth formula. However, I have been wondering if this might be true in dimension zero, or at least what could be said in this case.
Note that the projections $X_1 \times X_2 \to X_i$ are flat (over a field) so we get a natural map $CH_i(X_1) \otimes CH_j(X_2) \to CH_{i+j}(X_1 \times X_2)$ (flat pullback together with intersection product). Künneth formulas compare the left and right hand side of these maps.
To simplify things (and preclude some arithmetic peculiarities I do not want to focus on), assume that $k = \overline{k}$ is algebraically closed. Then $CH_0(X)$ is the quotient of the free abelian group on the rational points of $X$ by rational equivalence. It follows that $CH_0(X_1) \otimes CH_0(X_2) \to CH_0(X_1 \times X_2)$ is surjective.
My question: What can be said about the kernel of this map?
I have a feeling that assuming $X_1, X_2$ proper may help. Indeed the equivalence relation on zero-cycles has to do with closed integral curves $C \subset X_1 \times X_2.$ The projections $p_i: X_1 \times X_2 \to X_i$ are closed and so $p_i(C)$ is either a point or a closed curve. Presumably the new relations come from the case when $p_i(C)$ are both true curves ... not sure how to proceed.