Voevodsky's fun nilpotency thing

my.bib

@@ -6168,6 +6168,17 @@ @INCOLLECTION{Talpo-Vistoli
       YEAR = {2013},
 }
 
+@ARTICLE{nilpotence,
+    AUTHOR = {Voevodsky, V.},
+     TITLE = {A nilpotence theorem for cycles algebraically equivalent to zero},
+   JOURNAL = {Internat. Math. Res. Notices},
+      YEAR = {1995},
+    NUMBER = {4},
+     PAGES = {187--198},
+      ISSN = {1073-7928},
+       URL = {https://doi.org/10.1155/S1073792895000158},
+}
+
 @ARTICLE{Voevodsky,
 	AUTHOR = {Vladimir Voevodsky},
 	TITLE = {Homology of schemes},

weil.tex

@@ -1817,6 +1817,110 @@ \section{Cycles over non-closed fields}
 $\ell$th root in $\Pic(X)$. This is what we had to show.
 \end{proof}
 
+\begin{lemma}
+\label{lemma-smash-nilpotence}
+\begin{reference}
+\cite{nilpotence}
+\end{reference}
+Let $k$ be a field. Let $X$ be a geometrically irreducible
+smooth projective scheme over $k$. Let $x, x' \in X$ be $k$-rational points.
+Then there exists an $n \geq 1$ such that
+$$
+([x] - [x']) \times \ldots \times ([x] - [x']) \in
+A_0(X^n)
+$$
+is torsion.
+\end{lemma}
+
+\begin{proof}
+If we can show this after base change to the algebraic closure of $k$,
+then the result follows over $k$. (The kernel of pullback on zero cycles
+is torsion.) Hence we may and do assume $k$ is algebraically closed.
+Using Bertini we can choose a smooth curve $C \subset X$ passing through
+$x$ and $x'$. Hence we may assume $X$ is a curve.
+
+\medskip\noindent
+Write $S^n(X) = \underline{\Hilbfunctor}^n_{X/k}$ with notation as in
+Picard Schemes of Curves, Sections \ref{pic-section-hilbert-scheme-points}
+and \ref{pic-section-divisors}. There is a canonical morphism
+$$
+\pi : X^n \longrightarrow S^n(X)
+$$
+which sends the $k$-rational point $(x_1, \ldots, x_n)$ to the $k$-rational
+point corresponding to the divisor $[x_1] + \ldots + [x_n]$ on $X$.
+There is a faithful action of the symmetric group $S_n$ on $X^n$.
+The morphism $\pi$ is $S_n$-invariant and the fibres of $\pi$ are
+$S_n$-orbits (set theoretically). Finally, $\pi$ is finite flat of
+degree $n!$, see Picard Schemes of Curves, Lemma
+\ref{pic-lemma-universal-object}.
+
+\medskip\noindent
+Let $\alpha_n$ be the zero cycle on $X^n$ given by the formula in the
+statement of the lemma. Let $\mathcal{L} = \mathcal{O}_X(x - x')$. Then
+$c_1(\mathcal{L}) \cap [X] = [x] - [x']$. Thus
+$$
+\alpha_n = c_1(\mathcal{L}_1) \cap \ldots \cap c_1(\mathcal{L}_n) \cap [X^n]
+$$
+where $\mathcal{L}_i = \text{pr}_i^*\mathcal{L}$ and $\text{pr}_i : X^n \to X$
+is the $i$th projection. By either
+Divisors, Lemma \ref{divisors-lemma-finite-locally-free-has-norm} or
+Divisors, Lemma \ref{divisors-lemma-norm-in-normal-case}
+there is a norm for $\pi$. Set
+$$
+\mathcal{N} = \text{Norm}_\pi(\mathcal{L}_1)
+$$
+See Divisors, Lemma \ref{divisors-lemma-norm-invertible}. We have
+$$
+\pi^*\mathcal{N} =
+(\otimes_{i = 1, \ldots, n} \mathcal{L}_i)^{\otimes (n - 1)!}
+$$
+in $\Pic(X^n)$ by a calculation. Deails omitted; hint: this follows from
+the fact that
+$\text{Norm}_\pi : \pi_*\mathcal{O}_{X^n} \to \mathcal{O}_{S^n(X)}$
+composed with the natural map $\pi_*\mathcal{O}_{S^n(X)} \to \mathcal{O}_{X^n}$
+is equal to the product over all $\sigma \in S_n$ of the action of
+$\sigma$ on $\pi_*\mathcal{O}_{X^n}$. Consider
+$$
+\beta_n = c_1(\mathcal{N})^n \cap [S^n(X)]
+$$
+in $A_0(S^n(X))$. Observe that $c_1(\mathcal{L}_i) \cap c_1(\mathcal{L}_i) = 0$
+because $\mathcal{L}_i$ is pulled back from a curve. Thus we see that
+\begin{align*}
+\pi^*\beta_n
+& =
+((n - 1)!)^n
+(\sum\nolimits_{i = 1, \ldots, n} c_1(\mathcal{L}_i))^n \cap [X^n] \\
+& =
+((n - 1)!)^n n^n 
+c_1(\mathcal{L}_1) \cap \ldots \cap c_1(\mathcal{L}_n) \cap [X^n] \\
+& =
+(n!)^n \alpha_n
+\end{align*}
+Thus it suffices to show that $\beta_n$ is torsion.
+
+\medskip\noindent
+There is a canonical morphism
+$$
+f : S^n(X) \longrightarrow \underline{\Picardfunctor}^n_{X/k}
+$$
+See Picard Groups of Curves, Lemma \ref{pic-lemma-picard-pieces}.
+For $n \geq 2g - 1$ this morphism is a projective space bundle.
+The invertible sheaf $\mathcal{N}$ is trivial on the
+fibres of $f$ (hint: because $\mathcal{L}$ is algebraically
+equivalent to zero, so is $\mathcal{N}$ and the fibres are
+projective spaces). Thus by the projective space bundle formula
+(Chow Homology, Lemma \ref{chow-lemma-chow-ring-projective-bundle})
+we see that $\mathcal{N} = f^*\mathcal{M}$ for some invertible
+module $\mathcal{M}$ on $\underline{\Picardfunctor}^n_{X/k}$.
+Of course, then we see that
+$$
+c_1(\mathcal{N})^n = f^*(c_1(\mathcal{M})^n)
+$$
+is zero because $n > g = \dim(\underline{\Picardfunctor}^n_{X/k})$.
+This finishes the proof.
+\end{proof}
+
+