0
$\begingroup$

Let $X$ be a smooth projective rational variety over a field $k$. Let $CH^i(X)$ denote the Chow group of codimension $i$ algebraic cycles on $X$ modulo rational equivalence. What can one say about Chow groups $X$? Are they torsion-free?

$\endgroup$
7
$\begingroup$

You cannot say much. Suppose $X$ is obtained by blowing up $\mathbb{P}^n$ along a smooth subvariety $V$, say of codimension $c$; then $CH^p(X)=CH^p(\mathbb{P}^n)\oplus CH^{p-1}(V)\oplus\ldots \oplus CH^{p+1-c}(V)$. Thus the Chow groups of $X$ look like those of $V$, which are arbitrary. All you can say for a rational variety $X$ of dimension $n$ is that $CH^1(X)=\mathrm{Pic}(X)$ is a free finitely generated abelian group, and $CH^n(X)=\mathbb{Z}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.