The Stacks project

Proof. By Lemma 45.8.1 it suffices to show: given closed points $x, x' \in X$ whose residue fields are separable over $k$ we have $\deg (x') \gamma ([x]) = \deg (x) \gamma ([x'])$.

We first reduce to the case of $k'$-rational points. Let $k''/k'$ be a Galois extension such that $\kappa (x)$ and $\kappa (x')$ embed into $k''$ over $k$. Set $Y = X \times _{\mathop{\mathrm{Spec}}(k')} \mathop{\mathrm{Spec}}(k'')$ and denote $p : Y \to X$ the projection. By our choice of $k''/k'$ there exists a $k''$-rational point $y$, resp. $y'$ on $Y$ mapping to $x$, resp. $x'$. Then $p_*[y] = [k'' : \kappa (x)][x]$ and $p_*[y'] = [k'' : \kappa (x')][x']$ in $\mathop{\mathrm{CH}}\nolimits _0(X)$. By compatibility with pushforwards given in axiom (C)(b) it suffices to prove $\gamma ([y]) = \gamma ([y'])$ in $\mathop{\mathrm{CH}}\nolimits ^{2d}(Y)(d)$. This reduces us to the discussion in the next paragraph.

Assume $x$ and $x'$ are $k'$-rational points. By Lemma 45.8.3 there exists a finite separable extension $k''/k'$ of fields such that the pullback $[y] - [y']$ of the difference $[x] - [x']$ becomes divisible by an integer $n > 1$ on $Y = X \times _{\mathop{\mathrm{Spec}}(k')} \mathop{\mathrm{Spec}}(k'')$. (Note that $y, y' \in Y$ are $k''$-rational points.) By Lemma 45.10.6 we have $\gamma ([y]) = \gamma ([y'])$ in $H^{2d}(Y)(d)$. By compatibility with pushforward in axiom (C)(b) we conclude the same for $x$ and $x'$. $\square$