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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