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Given $i: Z \subset X$, a closed immersion of smooth schemes over some field $k$. Is there an open subscheme $U$ of $X$ such that $Z \cap U$ is non-empty and such that the Gysin map of the Chow groups (CH is the total Chow ring here)

$$i^!: CH(U) \rightarrow CH(Z \cap U)$$

is surjective?

(Variants I would also be interested in, if these are helpful simplifications: $k$ algebraically closed or both groups tensored with $\mathbf Q$, $Z$ a divisor in $X$)

Thank you,

Jakob Scholbach

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1 Answer 1

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I don't think that this is true in general. Take $X=\mathbb{P}^2$, then $CH(X)$ is finitely generated. Moreover, finite generation holds for any open $U\subset X$ by the exact sequence $$CH(X-U)\to CH(X)\to CH(U)\to 0$$ [Fulton, Intersection theory I, 1.8.]

On the other when $Z$ is curve of degree $3$ or more, $CH(Z\cap U)$ is not finitely generated when $k$ is algebraically closed. To see this, observe that $CH(Z)$ contains the rational points of the Jacobian $J(Z)$, which is not finitely generated (it is uncountable if $k=\mathbb{C}$). Now use the above sequence, to conclude that $CH(Z\cap U)$ is also non finitely generated.

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  • $\begingroup$ Never mind about finite generation. As you say, the restriction map on Chow groups from X to U is surjective, and so the same would have be true of the restriction from X to Z. So for any projective variety one would have to have all the Chow groups cyclic, which fails even for many elliptic curves over finite fields, and indeed just about any randomly chosen projective variety. Perhaps the intuition driving this question is that for smooth manifolds, there is a tubular nbd of Z which is diffeomorphic to the normal bundle, which however is false in the algebraic category. $\endgroup$
    – user10849
    Commented Jan 24, 2011 at 18:53

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