Suppose $f:X \to Y$ is a finite morphism with $X$ and $Y$ being affine varieties, such that $X$ is unirational. In fact $X$ is more than unirational, it is the image of a morphism from a zariski open subset of $\mathbb P^n$. Are there any results that allows one to conclude the finite generation of the Chow ring $A^{*}(X) \otimes \mathbb Q$, if it is known that the chow ring $A^{*}(Y) \otimes \mathbb Q$ is finitely generated, are there any counter examples to this assertion?

I am assuming that the varieties are defined over $\mathbb C$.


1 Answer 1


This is not true. Take $\bar{X}=$ a smooth cubic threefold in $\Bbb{P}^4$, $\bar{Y}=\Bbb{P}^3$, $f:\bar{X}\rightarrow \bar{Y}$ the projection from a general point of $\Bbb{P}^4$. Now let $H$ be a general hyperplane in $\Bbb{P}^3$, and $S:=f^{-1}(H)$; this is a smooth hyperplane section of $\bar{X}$. Put $Y:=\bar{Y}\smallsetminus H$ and $X:=\bar{X}\smallsetminus S$. Then all your hypotheses hold; consider the localization exact sequence $$A^*(S)\rightarrow A^*(\bar{X})\rightarrow A^*(X)\rightarrow 0\ .$$ $A^*(\bar{X})$ is huge (it contains the intermediate Jacobian of $\bar{X}$, isomorphic to $(\Bbb{R}/\Bbb{Z})^{10}$ as a group), and quotienting by the small group $A^*(S)$ (isomorphic to $\Bbb{Z}^{9}$) does not improve the situation.

  • $\begingroup$ In your example, what is the surjective morphism from a Zariski open subset of $\mathbb{P}^3$ to $X$? Of course you can find a generically 2-to-1 morphism from a rational variety $P$ to $X$. However, I worry that if you excise a subvariety of $P$ to make it isomorphic to an open subset of $\mathbb{P}^3$, then you will also have to excise a subvariety of $X$ that will kill most of its Chow groups. $\endgroup$ Nov 4, 2013 at 12:24
  • $\begingroup$ Actually I didn't address that part of the question. You are probably right, having such a surjective morphism is a very strong assumption. $\endgroup$
    – abx
    Nov 4, 2013 at 19:25

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