hcju/stacksqa huggingface dataset:
The lemma is immediate if $k$ has characteristic $0$ or is perfect. Thus we may assume $k$ is an infinite field of characteristic $p > 0$. \medskip\noindent We may assume $X$ is irreducible of dimension $d$. Then $k' = H^0(X, \mathcal{O}_X)$ is a finite separable field extension of $k$ and that $X$ is geometrically integral over $k'$. See Varieties, Lemmas \ref{varieties-lemma-smooth-geometrically-normal}, \ref{varieties-lemma-proper-geometrically-reduced-global-sections}, and \ref{varieties-lemma-baby-stein}. We may and do replace $k$ by $k'$ and assume that $X$ is geometrically integral. \medskip\noindent Let $x \in X$ be a closed point. To prove the lemma we are going to show that $[x] \in \CH_0(X)$ is rationally equivalent to an integer linear combination of classes of closed points whose residue fields are separable over $k$. Choose an ample invertible $\mathcal{O}_X$-module $\mathcal{L}$. Set $$ V = \{s \in H^0(X, \mathcal{L}) \mid s(x) = 0 \} $$ After replacing $\mathcal{L}$ by a power we may assume (a) $\mathcal{L}$ is very ample, (b) $V$ generates $\mathcal{L}$ over $X \setminus x$, (c) the morphism $X \setminus x \to \mathbf{P}(V)$ is an immersion, (d) the map $V \to \mathfrak m_x\mathcal{L}_x/\mathfrak m_x^2\mathcal{L}_x$ is surjective, see Morphisms, Lemma \ref{morphisms-lemma-finite-type-ample-very-ample}, Varieties, Lemma \ref{varieties-lemma-generate-over-complement}, and Properties, Proposition \ref{properties-proposition-characterize-ample}. Consider the set $$ V^d \supset U = \{ (s_1, \ldots, s_d) \in V^d \mid s_1, \ldots, s_d \text{ generate } \mathfrak m_x\mathcal{L}_x/\mathfrak m_x^2\mathcal{L}_x \text{ over }\kappa(x) \} $$ Since $\mathcal{O}_{X, x}$ is a regular local ring of dimension $d$ we have $\dim_{\kappa(x)}(\mathfrak m_x/\mathfrak m_x^2) = d$ and hence we see that $U$ is a nonempty (Zariski) open of $V^d$. For $(s_1, \ldots, s_d) \in U$ set $H_i = Z(s_i)$. Since $s_1, \ldots, s_d$ generate $\mathfrak m_x\mathcal{L}_x$ we see that $$ H_1 \cap \ldots \cap H_d = x \amalg Z $$ scheme theoretically for some closed subscheme $Z \subset X$. By Bertini (in the form of Varieties, Lemma \ref{varieties-lemma-bertini}) for a general element $s_1 \in V$ the scheme $H_1 \cap (X \setminus x)$ is smooth over $k$ of dimension $d - 1$. Having chosen $s_1$, for a general element $s_2 \in V$ the scheme $H_1 \cap H_2 \cap (X \setminus x)$ is smooth over $k$ of dimension $d - 2$. And so on. We conclude that for sufficiently general $(s_1, \ldots, s_d) \in U$ the scheme $Z$ is \'etale over $\Spec(k)$. In particular $H_1 \cap \ldots \cap H_d$ has dimension $0$ and hence $$ [H_1] \cdot \ldots \cdot [H_d] = [x] + [Z] $$ in $\CH_0(X)$ by repeated application of Chow Homology, Lemma \ref{chow-lemma-intersect-properly} (details omitted). This finishes the proof as it shows that $[x] \sim_{rat} - [Z] + [Z']$ where $Z' = H'_1 \cap \ldots \cap H'_d$ is a general complete intersection of vanishing loci of sufficiently general sections of $\mathcal{L}$ which will be \'etale over $k$ by the same argument as before.
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 $\CH_0(X_{k'})$. Then we conclude by Lemma \ref{lemma-chow-limit}. Thus we may and do assume $k$ is separably algebraically closed. \medskip\noindent Suppose $\dim(X) > 1$. Let $\mathcal{L}$ be an ample invertible sheaf on $X$. Set $$ V = \{s \in H^0(X, \mathcal{L}) \mid s(x) = 0\text{ and }s(x') = 0 \} $$ 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 \ref{varieties-lemma-generate-over-complement} and \ref{varieties-lemma-bertini}. 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 \ref{varieties-lemma-connectedness-ample-divisor} (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 $\CH_0(H_v)$. In this way we reduce to the case of a curve. \medskip\noindent Assume $X$ is a curve. Then we see that $\mathcal{O}_X(x - x')$ defines a $k$-rational point $g$ of $J = \underline{\Pic}^0_{X/k}$, see Picard Schemes of Curves, Lemma \ref{pic-lemma-picard-pieces}. 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 \ref{varieties-lemma-geometrically-connected-criterion} and \ref{varieties-lemma-smooth-geometrically-normal}). In other words, $J$ is an abelian variety, see Groupoids, Definition \ref{groupoids-definition-abelian-variety}. Thus $[n] : J \to J$ is finite \'etale by Groupoids, Proposition \ref{groupoids-proposition-review-abelian-varieties} (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.
It clearly suffices to show that the kernel of flat pullback $\CH_i(X) \to \CH_i(X_{k'})$ by $\pi : X_{k'} \to X$ is torsion for any finite extension $k'/k$. This is clear because $\pi_* \pi^* \alpha = [k' : k] \alpha$ by Chow Homology, Lemma \ref{chow-lemma-finite-flat}.
First assume $X$ is nonempty. Observe that $[X]$ is the pullback of $[\Spec(k)]$ by the structure morphism $p : X \to \Spec(k)$. Hence we get $\gamma([X]) = 1$ by axiom (C)(a) and Lemma \ref{lemma-base}. Let $X' \subset X$ be an irreducible component. By functoriality it suffices to show $1 \not = 0$ in $H^0(X')$. Thus we may and do assume $X$ is irreducible, and in particular nonempty and equidimensional, say of dimension $d$. To see that $1 \not = 0$ it suffices to show that $H^*(X)$ is nonzero. \medskip\noindent Let $x \in X$ be a closed point whose residue field $k'$ is separable over $k$, see Varieties, Lemma \ref{varieties-lemma-smooth-separable-closed-points-dense}. Let $i : \Spec(k') \to X$ be the inclusion morphism. Denote  $p : X \to \Spec(k)$ is the structure morphism. Observe that $p_*i_*[\Spec(k')] = [k' : k][\Spec(k)]$ in $\CH_0(\Spec(k))$. Using axiom (C)(b) twice and Lemma \ref{lemma-base} we conclude that $$ p_*i_*\gamma([\Spec(k')]) = \gamma([k' : k][\Spec(k)]) = [k' : k] \in F = H^0(\Spec(k)) $$ is nonzero. Thus $i_*\gamma([\Spec(k)]) \in H^{2d}(X)(d)$ is nonzero (because it maps to something nonzero via $p_*$). This concludes the proof in case $X$ is nonempty. \medskip\noindent Finally, we consider the case of the empty scheme. Axiom (B)(a) gives $H^*(\emptyset) \otimes H^*(\emptyset) = H^*(\emptyset)$ and we get that $H^*(\emptyset)$ is either zero or $1$-dimensional in degree $0$. Then axiom (B)(a) again shows that $H^*(\emptyset) \otimes H^*(X) = H^*(\emptyset)$ for all smooth projective schemes $X$ over $k$. Using axiom (A)(b) and the nonvanishing of $H^0(X)$ we've seen above we find that $H^*(X)$ is nonzero in at least two degrees if $\dim(X) > 0$. This then forces $H^*(\emptyset)$ to be zero.
Recall that $\Delta^* : H^*(X \times X) \to H^*(X)$ is equal to the cup product map $H^*(X) \otimes_F H^*(X) \to H^*(X)$, see Remark \ref{remark-replace-cup-product}. On the other hand, recall that $\gamma([\Delta]) = \Delta_*1$ (Lemma \ref{lemma-push-unit}) and hence $$ \int_{X \times X} \gamma([\Delta]) \cup a \otimes b = \int_{X \times X} \Delta_*1 \cup a \otimes b = \int_X a \cup b $$ by Lemma \ref{lemma-pushforward}. On the other hand, we have $$ \int_{X \times X} (\sum e_{i, j} \otimes e'_{2d -i , j}) \cup a \otimes b = \sum (\int_X a \cup e_{i, j})(\int_X e'_{2d - i, j} \cup b) $$ by axiom (B)(b); note that we made two switches of order so that the sign for each term is $1$. Thus if we choose $a$ such that $\int_X a \cup e_{i, j} = 1$ and all other pairings equal to zero, then we conclude that $\int_X e'_{2d - i, j} \cup b = \int_X a \cup b$ for all $b$, i.e., $e'_{2d - i, j} = a$. This proves the lemma.
Let $x \in \mathbf{P}^1_k$ be a $k$-rational point. Observe that $\Delta = \text{pr}_1^*x + \text{pr}_2^*x$ as divisors on $\mathbf{P}^1_k \times \mathbf{P}^1_k$. Using axiom (C)(a) and additivity of $\gamma$ we see that $$ \gamma([\Delta]) = \text{pr}_1^*\gamma([x]) + \text{pr}_2^*\gamma([x]) = \gamma([x]) \otimes 1 + 1 \otimes \gamma([x]) $$ in $H^*(\mathbf{P}^1_k \times \mathbf{P}^1_k) = H^*(\mathbf{P}^1_k) \otimes_F H^*(\mathbf{P}^1_k)$. However, by Lemma \ref{lemma-class-diagonal} we know that $\gamma([\Delta])$ cannot be written as a sum of fewer than $\sum \beta_i$ pure tensors where $\beta_i = \dim_F H^i(\mathbf{P}^1_k)$. Thus we see that $\sum \beta_i \leq 2$. By Lemma \ref{lemma-unit} we have $H^0(\mathbf{P}^1_k) \not = 0$. By Poincar\'e duality, more precisely axiom (A)(b), we have $\beta_0 = \beta_2$. Therefore the lemma holds.
If $X$ or $Y$ is empty, then this is true because $H^*(\emptyset) = 0$ by Lemma \ref{lemma-unit}. Thus we may assume both $X$ and $Y$ are nonempty. \medskip\noindent We first show that the map is injective. First, observe that we can find morphisms $X' \to X$ and $Y' \to Y$ of smooth projective schemes so that $X'$ and $Y'$ are equidimensional of the same dimension and such that $X' \to X$ and $Y' \to Y$ each have a section. Namely, decompose $X = \coprod X_d$ and $Y = \coprod Y_e$ into open and closed subschemes equidimensional of dimension $d$ and $e$. Then take $X' = \coprod X_d \times \mathbf{P}^{n - d}$ and $Y' = \coprod Y_e \times \mathbf{P}^{n - e}$ for some $n$ sufficiently large. Thus pullback by $X' \amalg Y' \to X \amalg Y$ is injective (because there is a section) and it suffices to show the injectivity for $X', Y'$ as we do in the next parapgrah. \medskip\noindent Let us show the map is injective when $X$ and $Y$ are equidimensional of the same dimension $d$. Observe that $[X \amalg Y] = [X] + [Y]$ in $\CH^0(X \amalg Y)$ and that $[X]$ and $[Y]$ are orthogonal idempotents in $\CH^0(X \amalg Y)$. Thus $$ 1 = \gamma([X \amalg Y] = \gamma([X]) + \gamma([Y]) = i_*1 + j_*1 $$ is a decomposition into orthogonal idempotents. Here we have used Lemmas \ref{lemma-unit} and \ref{lemma-push-unit} and axiom (C)(c). Then we see that $$ a = a \cup 1 = a \cup i_*1 + a \cup j_*1 = i_*(i^*a) + j_*(j^*a) $$ by the projection formula (Lemma \ref{lemma-pushforward}) and hence the map is injective. \medskip\noindent We show the map is surjective. Write $e = \gamma([X])$ and $f = \gamma([Y])$ viewed as elements in $H^0(X \amalg Y)$. We have $i^*e = 1$, $i^*f = 0$, $j^*e = 0$, and $j^*f = 1$ by axiom (C)(a). Hence if $i^* : H^*(X \amalg Y) \to  H^*(X)$ and $j^* : H^*(X \amalg Y) \to H^*(Y)$ are surjective, then so is $(i^*, j^*)$. Namely, for $a, a' \in H^*(X \amalg Y)$ we have $$ (i^*a, j^*a') = (i^*(a \cup e + a' \cup f), j^*(a \cup e + a' \cup f)) $$ By symmetry it suffices to show $i^* : H^*(X \amalg Y) \to  H^*(X)$ is surjective. If there is a morphism $Y \to X$, then there is a morphism $g : X \amalg Y \to X$ with $g \circ i = \text{id}_X$ and we conclude. To finish the proof, observe that in order to prove $i^*$ is surjective, it suffices to do so after tensoring by a nonzero graded $F$-vector space. Hence by axiom (B)(b) and nonvanishing of cohomology (Lemma \ref{lemma-unit}) it suffices to prove $i^*$ is surjective after replacing $X$ and $Y$ by $X \times \Spec(k')$ and $Y \times \Spec(k')$ for some finite separable extension $k'/k$. If we choose $k'$ such that there exists a closed point $x \in X$ with $\kappa(x) = k'$ (and this is possible by Varieties, Lemma \ref{varieties-lemma-smooth-separable-closed-points-dense}) then there is a morphism $Y \times \Spec(k') \to X \times \Spec(k')$ and we find that the proof is complete.
Denote $i : X \to X \amalg Y$ and $j : Y \to X \amalg Y$ be the coprojections. By Lemma \ref{lemma-weil-additive} the map $(i^*, j^*) : H^*(X \amalg Y) \to H^*(X) \times H^*(Y)$ is an isomorphism. The statement of the lemma means that under the isomorphism $(i^*, j^*) : H^{2d}(X \amalg Y)(d) \to H^{2d}(X)(d) \oplus H^{2d}(Y)(d)$ the map $\int_X + \int_Y$ is tranformed into $\int_{X \amalg Y}$. This is true because $$ \int_{X \amalg Y} a = \int_{X \amalg Y} i_*(i^*a) + j_*(j^*a) = \int_X i^*a + \int_Y j^*a $$ where the equality $a = i_*(i^*a) + j_*(j^*a)$ was shown in the proof of Lemma \ref{lemma-weil-additive}.
We can write $X = \Spec(k')$ where $k'$ is a finite separable algebra over $k$. Observe that $\deg(X \to \Spec(k)) = [k' : k]$. Choose a finite Galois extension $k''/k$ containing each of the factors of $k'$. (Recall that a finite separable $k$-algebra is a product of finite separable field extension of $k$.) Set $\Sigma = \Hom_k(k', k'')$. Then we get $$ k' \otimes_k k'' = \prod\nolimits_{\sigma \in \Sigma} k'' $$ Setting $Y = \Spec(k'')$ axioms (B)(a) and Lemma \ref{lemma-weil-additive} give $$ H^*(X) \otimes_F H^*(Y) = \prod\nolimits_{\sigma \in \Sigma} H^*(Y) $$ as graded commutative $F$-algebras. By Lemma \ref{lemma-unit} the $F$-algebra $H^*(Y)$ is nonzero. Comparing dimensions on either side of the displayed equation we conclude that $H^*(X)$ sits only in degree $0$ and $\dim_F H^0(X) = [k' : k]$. Applying this to $Y$ we get $H^*(Y) = H^0(Y)$. Since $$ H^0(X) \otimes_F H^0(Y) = H^0(Y) \times \ldots \times H^0(Y) $$ as $F$-algebras, it follows that $H^0(X)$ is a separable $F$-algebra because we may check this after the faithfully flat base change $F \to H^0(Y)$. \medskip\noindent The displayed isomorphism above is given by the map $$ H^0(X) \otimes_F H^0(Y) \longrightarrow \prod\nolimits_{\sigma \in \Sigma} H^0(Y),\quad a \otimes b \longmapsto \prod\nolimits_\sigma \Spec(\sigma)^*a \cup b $$ Via this isomorphism we have $\int_{X \times Y} = \sum_\sigma \int_Y$ by Lemma \ref{lemma-trace-disjoint-union}. Thus $$ \int_X a = \text{pr}_{1, *}(a \otimes 1) = \sum \Spec(\sigma)^*a $$ in $H^0(Y)$; the first equality by Lemma \ref{lemma-pr2star} and the second by the observation we just made. Choose an algebraic closure $\overline{F}$ and a $F$-algebra map $\tau : H^0(Y) \to \overline{F}$. The isomorphism above base changes to the isomorphism $$ H^0(X) \otimes_F \overline{F} \longrightarrow \prod\nolimits_{\sigma \in \Sigma} \overline{F},\quad a \otimes b \longmapsto \prod\nolimits_\sigma \tau(\Spec(\sigma)^*a) b $$ It follows that $a \mapsto \tau(\Spec(\sigma)^*a)$ is a full set of embeddings of $H^0(X)$ into $\overline{F}$. Applying $\tau$ to the formula for $\int_X a$ obtained above we conclude that $\int_X$ is the trace map. By Lemma \ref{lemma-unit} we have $\gamma([X]) = 1$. Finally, we have $\int_X \gamma([X]) = \deg(X \to \Spec(k))$ because $\gamma([X]) = 1$ and the trace of $1$ is equal to $[k' : k]$
Let $x$ be a closed point of $X$ whose residue field is separable over $k$. View $x$ as a scheme and denote $i : x \to X$ the inclusion morphism. To avoid confusion denote $\gamma' : \CH_0(x) \to H^0(x)$ the cycle class map for $x$. Then we have $$ \int_X \gamma([x]) = \int_X \gamma(i_*[x]) = \int_X i_*\gamma'([x]) = \int_x \gamma'([x]) = \deg(x \to \Spec(k)) $$ The second equality is axiom (C)(b) and the third equality is the definition of $i_*$ on cohomology. The final equality is Lemma \ref{lemma-dim-0}. This proves the lemma because $\CH_0(X)$ is generated by the classes of points $x$ as above by Lemma \ref{lemma-generated-by-separable}.
Equality on the right. We have $[\Delta] \cdot [\Delta] = \Delta_*(\Delta^![\Delta])$ (Chow Homology, Lemma \ref{chow-lemma-intersect-regularly-embedded}). Since $\Delta_*$ preserves degrees of $0$-cycles it suffices to compute the degree of $\Delta^![\Delta]$. The class $\Delta^![\Delta]$ is given by capping $[\Delta]$ with the top Chern class of the normal sheaf of $\Delta \subset X \times X$ (Chow Homology, Lemma \ref{chow-lemma-gysin-fundamental}). Since the conormal sheaf of $\Delta$ is $\Omega_{X/k}$ (Morphisms, Lemma \ref{morphisms-lemma-differentials-diagonal}) we see that the normal sheaf is equal to the tangent sheaf $\mathcal{T}_{X/k} = \SheafHom_{\mathcal{O}_X}(\Omega_{X/k}, \mathcal{O}_X)$ as desired. \medskip\noindent Equality on the left. By Lemma \ref{lemma-degrees-cycles} we have \begin{align*} \deg([\Delta] \cdot [\Delta]) & = \int_{X \times X} \gamma([\Delta]) \cup \gamma([\Delta]) \\ & = \int_{X \times X} \Delta_*1 \cup \gamma([\Delta]) \\ & = \int_{X \times X} \Delta_*(\Delta^*\gamma([\Delta])) \\ & = \int_X \Delta^*\gamma([\Delta]) \end{align*} We have used Lemmas \ref{lemma-push-unit} and \ref{lemma-pushforward}. Write $\gamma([\Delta]) = \sum  e_{i, j} \otimes e'_{2d - i , j}$ as in Lemma \ref{lemma-class-diagonal}. Recalling that $\Delta^*$ is given by cup product (Remark \ref{remark-replace-cup-product}) we obtain $$ \int_X \sum\nolimits_{i, j} e_{i, j} \cup e'_{2d - i, j} = \sum\nolimits_{i, j} \int_X e_{i, j} \cup e'_{2d - i, j} = \sum\nolimits_{i, j} (-1)^i = \sum (-1)^i\beta_i $$ as desired.
We may write $A = \prod A_j$ as a finite product of local Artinian $F$-algebras $(A_j, \mathfrak m_j, \kappa_j)$, see Algebra, Lemma \ref{algebra-lemma-finite-dimensional-algebra} and Proposition \ref{algebra-proposition-dimension-zero-ring}. Denote $A' = \prod \kappa_j$ where the product is over those $j$ such that $\kappa_j/k$ is separable. Then each of the maps $\sigma, \sigma', \sigma_i$ factors over the map $A \to A'$. After replacing $A$ by $A'$ we may assume $A$ is a finite separable $F$-algebra. \medskip\noindent Choose an algebraic closure $\overline{F}$. Set $\overline{A} = A \otimes_F \overline{F}$, $\overline{F}' = F' \otimes_F \overline{F}$, and $\overline{F}_i = F_i \otimes_F \overline{F}$. We can base change $\sigma$, $\sigma'$, $\sigma_i$ to get $\overline{F}$ algebra maps $\overline{A} \to \overline{F}'$ and $\overline{A} \to \overline{F}_i$. Moreover $\text{Tr}_{\overline{F}'/\overline{F}}$ is the base change of $\text{Tr}_{F'/F}$ and similarly for $\text{Tr}_{F_i/F}$. Thus we may replace $F$ by $\overline{F}$ and we reduce to the case discussed in the next paragraph. \medskip\noindent Assume $F$ is algebraically closed and $A$ a finite separable $F$-algebra. Then each of $A$, $F'$, $F_i$ is a product of copies of $F$. Let us say an element $e$ of a product $F \times \ldots \times F$ of copies of $F$ is a minimal idempotent if it generates one of the factors, i.e., if $e = (0, \ldots, 0, 1, 0, \ldots, 0)$. Let $e \in A$ be a minimal idempotent. Since $\sigma$ and $\sigma'$  are surjective, we see that $\sigma(e)$ and $\sigma'(e)$ are minimal idempotents or zero. If $\sigma \not = \sigma'$, then we can choose a minimal idempotent $e \in A$ such that $\sigma(e) = 0$ and $\sigma'(e) \not = 0$ or vice versa. Then $\text{Tr}_{F'/F}(\sigma(e)) = 0$ and $\text{Tr}_{F'/F}(\sigma'(e)) = 1$ or vice versa. On the other hand, $\sigma_i(e)$ is an idempotent and hence $\text{Tr}_{F_i/F}(\sigma_i(e)) = r_i$ is an integer. We conclude that $$ -1 = \sum n m_i r_i = n (\sum m_i r_i) \quad\text{or}\quad 1 = \sum n m_i r_i = n (\sum m_i r_i) $$ which is impossible.
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 $\CH_0(X)$. We may write $[x] - [x'] = n(\sum m_i [x_i])$ in $\CH_0(X)$ for some $x_i \in X$ closed points whose residue fields are separable over $k$ by Lemma \ref{lemma-generated-by-separable}. Then $$ \gamma([x]) - \gamma([x']) = n (\sum m_i \gamma([x_i])) $$ 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 \ref{lemma-dim-0}) which we will denote $\text{Tr}_x$. Similarly for $x'$ and $x_i$. Then by Poincar\'e duality in the form of axiom (A)(b) the equation above is dual to $$ \text{Tr}_x \circ i^* - \text{Tr}_{x'} \circ (i')^* = n(\sum m_i \text{Tr}_{x_i} \circ i_i^*) $$ which takes place in $\Hom_F(H^0(X), F)$. Finally, observe that $i^*$ and $(i')^*$ are surjective as $x$ and $x'$ are $k'$-rational points and hence the compositions $H^0(\Spec(k')) \to H^0(X) \to H^0(x)$ and $H^0(\Spec(k')) \to H^0(X) \to H^0(x')$ are isomorphisms. By Lemma \ref{lemma-algebra-relations} we conclude that $i^* = (i')^*$ which contradicts the assumption that $\gamma([x]) \not = \gamma([x'])$.
By Lemma \ref{lemma-generated-by-separable} it suffices to show: given closed points $x, x' \in X$ whose residue fields are separable over $k$ we have $\deg(x') \gamma([x]) = \deg(x) \gamma([x'])$. \medskip\noindent 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 $Y = X \times_{\Spec(k')} \Spec(k'')$ 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 $\CH_0(X)$. By compatibility with pushforwards given in axiom (C)(b) it suffices to prove $\gamma([y]) = \gamma([y'])$ in $\CH^{2d}(Y)(d)$. This reduces us to the discussion in the next paragraph. \medskip\noindent Assume $x$ and $x'$ are $k'$-rational points. By Lemma \ref{lemma-divide-difference-points} 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 $Y = X \times_{\Spec(k')} \Spec(k'')$. (Note that $y, y' \in Y$ are $k''$-rational points.) By Lemma \ref{lemma-relations-classes-points} 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 $x'$.
There exists an integral closed subscheme $Z \subset X$ of the same dimension as $Y$ mapping onto $Y$. Thus $f_*[Z] = m[Y]$ for some $m > 0$. Then $f_* \gamma([Z]) = m \gamma([Y]) = m$ in $H^*(Y)$ because of Lemma \ref{lemma-unit}. Hence by the projection formula (Lemma \ref{lemma-pushforward}) we have $f_*(f^*a \cup \gamma([Z])) = m a$ and we conclude.
We will use the results of Lemma \ref{lemma-dim-0} without further mention. Write $$ k' \otimes_k k'' = k'' \times l $$ for some finite separable $k'$-algebra $l$. Write $F' = H^0(\Spec(k'))$, $F'' = H^0(\Spec(k''))$, and $G = H^0(\Spec(l))$. Since $\Spec(k') \times \Spec(k'') = \Spec(k'') \amalg \Spec(l)$ we deduce from axiom (B)(a) and Lemma \ref{lemma-weil-additive} that we have $$ F' \otimes_F F'' = F'' \times G $$ The map from left to right identifies $F''$ with $F' \otimes_{F'} F''$. By the same token we have $$ H^*(X) \otimes_F F'' = H^*(X \times_{\Spec(k')} \Spec(k'')) \times H^*(X \times_{\Spec(k')} \Spec(l)) $$ as modules over $F' \otimes_F F'' = F'' \times G$. This proves the lemma.