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An interesting invariant of a rationally chain-connected variety $X/k$ is the exponent of the group, $$ \ker{(\mathrm{CH}_0(X_K) \xrightarrow{\deg} \mathbb{Z})} $$ where $K = k(X)$ is the function field. This is finite and called the torsion of $X$. This is equal to the smallest integer $n$ so that $n [\Delta_X]$ has a decomposition in $\mathrm{CH}_{\dim{X}}(X \times X)$ meaning we can write, $$ n [\Delta_X] = [X] \times [z] + Z $$ where $z$ is a zero cycle and $Z$ is a cycle supported on $D \times X$ where $D \subset X$ is a proper closed subscheme.

Let $X_{N,n,p} \subset \mathbb{P}^N_{\overline{\mathbb{F}}_p}$ be the Fermat hypersurface defined by, $$ x_0^n + \cdots + x_{N+1}^n = 0 $$ over $\overline{\mathbb{F}}_p$. It is know that when $p^\nu \equiv -1 \mod n$ for some $\nu$ and $N \ge 2$ then $X_{N,n,p}$ is unirational.

Therefore, in this case, $X_{N,n,p}$ is a fortiori rationally chain-connected. Is anything known about the torsion of $X_{N,n,p}$?

If there is a rational map $f : \mathbb{P}^n \to X$ then $\mathrm{Tor}(X) \mid \deg{f}$. Since in the case $n = p+1$ there is a degree $p$ map $\mathbb{P}^n \to X_{2,p+1,p}$ and I expect the torsion to be nonzero since the variety is not stably rational I would expect $\mathrm{Tor}(X_{2,p+1,p}) = p$.

What is known in general?

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  • $\begingroup$ Is $K$ the function field of $X$? $\endgroup$
    – naf
    Commented Apr 23, 2023 at 9:44
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    $\begingroup$ You might look at Totaro’s “Hypersurfaces that are not stably rational” Lemma 2.2 $\endgroup$ Commented Apr 23, 2023 at 17:28
  • $\begingroup$ @YosemiteStan thank you for the reference. This indeed confirms that the torsion is divisible by $p$ which I expect from examples. It would be interesting if there is a way to probe the order that $p$ divides it. $\endgroup$
    – Ben C
    Commented Apr 23, 2023 at 17:51

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