Lemma 45.8.3. 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. Let $n$ be an integer invertible in $k$. Then there exists a finite separable extension $k'/k$ such that the pullback of $[x] - [x']$ to $X_{k'}$ is divisible by $n$ in $\mathop{\mathrm{CH}}\nolimits _0(X_{k'})$.

**Proof.**
Let $k'$ be a separable algebraic closure of $k$. Suppose that we can show the the pullback of $[x] - [x']$ to $X_{k'}$ is divisible by $n$ in $\mathop{\mathrm{CH}}\nolimits _0(X_{k'})$. Then we conclude by Lemma 45.8.2. Thus we may and do assume $k$ is separably algebraically closed.

Suppose $\dim (X) > 1$. Let $\mathcal{L}$ be an ample invertible sheaf on $X$. Set

After replacing $\mathcal{L}$ by a power we see that for a general $v \in V$ the corresponding divisor $H_ v \subset X$ is smooth away from $x$ and $x'$, see Varieties, Lemmas 33.47.1 and 33.47.3. To find $v$ we use that $k$ is infinite (being separably algebraically closed). If we choose $s$ general, then the image of $s$ in $\mathfrak m_ x\mathcal{L}_ x/\mathfrak m_ x^2\mathcal{L}_ x$ will be nonzero, which implies that $H_ v$ is smooth at $x$ (details omitted). Similarly for $x'$. Thus $H_ v$ is smooth. By Varieties, Lemma 33.48.3 (applied to the base change of everything to the algebraic closure of $k$) we see that $H_ v$ is geometrically connected. It suffices to prove the result for $[x] - [x']$ seen as an element of $\mathop{\mathrm{CH}}\nolimits _0(H_ v)$. In this way we reduce to the case of a curve.

Assume $X$ is a curve. Then we see that $\mathcal{O}_ X(x - x')$ defines a $k$-rational point $g$ of $J = \underline{\mathop{\mathrm{Pic}}\nolimits }^0_{X/k}$, see Picard Schemes of Curves, Lemma 44.6.7. Recall that $J$ is a proper smooth variety over $k$ which is also a group scheme over $k$ (same reference). Hence $J$ is geometrically integral (see Varieties, Lemma 33.7.13 and 33.25.4). In other words, $J$ is an abelian variety, see Groupoids, Definition 39.9.1. Thus $[n] : J \to J$ is finite étale by Groupoids, Proposition 39.9.11 (this is where we use $n$ is invertible in $k$). Since $k$ is separably closed we conclude that $g = [n](g')$ for some $g' \in J(k)$. If $\mathcal{L}$ is the degree $0$ invertible module on $X$ corresponding to $g'$, then we conclude that $\mathcal{O}_ X(x - x') \cong \mathcal{L}^{\otimes n}$ as desired. $\square$

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