The Stacks project

Proof. If $x$ and $x'$ lie on distinct irreducible components of $X$, then the result is obvious. Thus we may $X$ irreducible of dimension $d$. Say $[x] - [x']$ is divisible by $n > 1$ in $\mathop{\mathrm{CH}}\nolimits _0(X)$. We may write $[x] - [x'] = n(\sum m_ i [x_ i])$ in $\mathop{\mathrm{CH}}\nolimits _0(X)$ for some $x_ i \in X$ closed points whose residue fields are separable over $k$ by Lemma 45.8.1. Then

in $H^{2d}(X)(d)$. Denote $i^*, (i')^*, i_ i^*$ the pullback maps $H^0(X) \to H^0(x)$, $H^0(X) \to H^0(x')$, $H^0(X) \to H^0(x_ i)$. Recall that $H^0(x)$ is a finite separable $F$-algebra and that $\int _ x : H^0(x) \to F$ is the trace map (Lemma 45.10.2) which we will denote $\text{Tr}_ x$. Similarly for $x'$ and $x_ i$. Then by Poincaré duality in the form of axiom (A)(b) the equation above is dual to

which takes place in $\mathop{\mathrm{Hom}}\nolimits _ F(H^0(X), F)$. Finally, observe that $i^*$ and $(i')^*$ are surjective as $x$ and $x'$ are $k'$-rational points and hence the compositions $H^0(\mathop{\mathrm{Spec}}(k')) \to H^0(X) \to H^0(x)$ and $H^0(\mathop{\mathrm{Spec}}(k')) \to H^0(X) \to H^0(x')$ are isomorphisms. By Lemma 45.10.5 we conclude that $i^* = (i')^*$ which contradicts the assumption that $\gamma ([x]) \not= \gamma ([x'])$. $\square$