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Let $X$ be a smooth projective variety over a field $k$. By (one) definition, the Chow group of zero-cycles $CH_0(X)$ is universally trivial if, for every field extension $k \subset K$, the degree map $CH_0(X_K) \rightarrow \mathbf Z$ is an isomorphism.

This idea has been an important ingredient in some major recent advances in birational geometry. In particular, B. Totaro showed ("Hypersurfaces that are not stably rational", available here) that if $X$ is a very general hypersurface of degree $d$ in $\mathbf P^{n+1}_\mathbf C$ and $d \geq 2 \lceil (n+2)/3 \rceil$, then $CH_0(X)$ is not universally trivial, and hence $X$ is not stably rational.

Of course, the significant cases are when $X$ is Fano, that is, when $d \leq n+1$. Here $X$ is rationally connected, so $CH_0(X) \cong \mathbf Z$; Totaro's theorem implies that after some base extension $\mathbf C \hookrightarrow F$, the Chow group $CH_0(X_F)$ is bigger than $\mathbf Z$.

Question: For $X$ and $d$ as above, what kind of field extension $\mathbf C \hookrightarrow F$ makes $CH_0(X_F)$ bigger than $\mathbf Z$?

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  • $\begingroup$ One choice that works is $F = \mathbb{C}(X)$, the function field of $X$. $\endgroup$ Feb 4, 2016 at 12:00
  • $\begingroup$ Dear @JasonStarr, thanks for your comment. Can you explain why the base-change to that field has bigger Chow group of zero-cycles? And what goes wrong if the degree of $X$ is less than $2n/3$ or so? $\endgroup$
    – Cyclist
    Feb 4, 2016 at 12:06
  • $\begingroup$ The paper by Totaro is beautiful, and the papers by Koll'ar and Voisin that it builds upon are also beautiful. You should just read those papers. For your first question: inside $X\times_{\text{Spec}\ (\mathbb{C})} X$ consider the cycles of the diagonal $\Delta$ and $X\times\{ x_0 \}$ for some $x_0\in X(\mathbb{C})$. Now restrict on the first factor of $X\times_{\text{Spec}\ \mathbb{C}} X$ to $\text{Spec}\ \mathbb{C}(X)$. If the pullback $0$-cycles were rationally equivalent, then that would give an integral decomposition of the diagonal in $X\times_{\text{Spec}\ \mathbb{C}} X$. $\endgroup$ Feb 4, 2016 at 12:13
  • $\begingroup$ OK. I will take your advice and read those papers. Thanks again! $\endgroup$
    – Cyclist
    Feb 4, 2016 at 12:42

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