The Stacks project

[nilpotence]

Lemma 45.8.5 (Voevodsky). 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. For $n$ large enough the class of the zero cycle

\[ ([x] - [x']) \times \ldots \times ([x] - [x']) \in \mathop{\mathrm{CH}}\nolimits _0(X^ n) \]

is torsion.

Proof. If we can show this after base change to the algebraic closure of $k$, then the result follows over $k$ because the kernel of pullback is torsion by Lemma 45.8.4. 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 $x'$. See proof of Lemma 45.8.3. Hence we may assume $X$ is a curve.

Assume $X$ is a curve and $k$ is algebraically closed. Write $S^ n(X) = \underline{\mathrm{Hilb}}^ n_{X/k}$ with notation as in Picard Schemes of Curves, Sections 44.2 and 44.3. 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 44.3.4.

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 31.17.6 or Divisors, Lemma 31.17.7 there is a norm for $\pi $. Set $\mathcal{N} = \text{Norm}_\pi (\mathcal{L}_1)$, see Divisors, Lemma 31.17.2. We have

\[ \pi ^*\mathcal{N} = (\mathcal{L}_1 \otimes \ldots \otimes \mathcal{L}_ n)^{\otimes (n - 1)!} \]

in $\mathop{\mathrm{Pic}}\nolimits (X^ n)$ 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 $\mathop{\mathrm{CH}}\nolimits _0(S^ n(X))$. Observe that $c_1(\mathcal{L}_ i) \cap c_1(\mathcal{L}_ i) = 0$ because $\mathcal{L}_ i$ is pulled back from a curve, see Chow Homology, Lemma 42.33.6. 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.

There is a canonical morphism

\[ f : S^ n(X) \longrightarrow \underline{\mathrm{Pic}}^ n_{X/k} \]

See Picard Schemes of Curves, Lemma 44.6.7. For $n \geq 2g - 1$ this morphism is a projective space bundle (details omitted; compare with the proof of Picard Schemes of Curves, Lemma 44.6.7). The invertible sheaf $\mathcal{N}$ is trivial on the fibres of $f$, see below. Thus by the projective space bundle formula (Chow Homology, Lemma 42.35.2) we see that $\mathcal{N} = f^*\mathcal{M}$ for some invertible module $\mathcal{M}$ on $\underline{\mathrm{Pic}}^ n_{X/k}$. Of course, then we see that

\[ c_1(\mathcal{N})^ n = f^*(c_1(\mathcal{M})^ n) \]

is zero because $n > g = \dim (\underline{\mathrm{Pic}}^ n_{X/k})$ and we can use Chow Homology, Lemma 42.33.6 as before.

We still have to show that $\mathcal{N}$ is trivial on a fibre $F$ of $f$. Since the fibres of $f$ are projective spaces and since $\mathop{\mathrm{Pic}}\nolimits (\mathbf{P}^ m_ k) = \mathbf{Z}$ (Divisors, Lemma 31.28.5), this can be shown by computing the degree of $\mathcal{N}$ on a line contained in the fibre. Instead we will prove it by proving that $\mathcal{N}$ is algebraically equivalent to zero. First we claim there is a connected finite type scheme $T$ over $k$, an invertible module $\mathcal{L}'$ on $T \times X$ and $k$-rational points $p, q \in T$ such that $\mathcal{M}_ p \cong \mathcal{O}_ X$ and $\mathcal{M}_ q = \mathcal{L}$. Namely, since $\mathcal{L} = \mathcal{O}_ X(x - x')$ we can take $T = X$, $p = x'$, $q = x$, and $\mathcal{L}' = \mathcal{O}_{X \times X}(\Delta ) \otimes \text{pr}_2^*\mathcal{O}_ X(-x')$. Then we let $\mathcal{L}'_ i$ on $T \times X^ n$ for $i = 1, \ldots , n$ be the pullback of $\mathcal{L}'$ by $\text{id}_ T \times \text{pr}_ i : T \times X^ n \to T \times X$. Finally, we let $\mathcal{N}' = \text{Norm}_{\text{id}_ T \times \pi }(\mathcal{L}'_1)$ on $T \times S^ n(X)$. By construction we have $\mathcal{N}'_ p = \mathcal{O}_{S^ n(X)}$ and $\mathcal{N}'_ q = \mathcal{N}$. We conclude that

\[ \mathcal{N}'|_{T \times F} \]

is an invertible module on $T \times F \cong T \times \mathbf{P}^ m_ k$ whose fibre over $p$ is the trivial invertible module and whose fibre over $q$ is $\mathcal{N}|_ F$. Since the euler characteristic of the trivial bundle is $1$ and since this euler characteristic is locally constant in families (Derived Categories of Schemes, Lemma 36.32.2) we conclude $\chi (F, \mathcal{N}^{\otimes s}|_ F) = 1$ for all $s \in \mathbf{Z}$. This can happen only if $\mathcal{N}|_ F \cong \mathcal{O}_ F$ (see Cohomology of Schemes, Lemma 30.8.1) and the proof is complete. Some details omitted. $\square$


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