**Proof.**
We go by the axioms one by one.

Axiom (A1). We have to show $H^*(\emptyset ) = 0$ and that $(i^*, j^*) : H^*(X \amalg Y) \to H^*(X) \times H^*(Y)$ is an isomorphism where $i$ and $j$ are the coprojections. By the functorial nature of the isomorphisms $H^*(X) \otimes _ F F' \to (H')^*(X_{k'})$ this follows from linear algebra: if $\varphi : V \to W$ is an $F$-linear map of $F$-vector spaces, then $\varphi $ is an isomorphism if and only if $\varphi _{F'} : V \otimes _ F F' \to W \otimes _ F F'$ is an isomorphism.

Axiom (A2). This means that given a morphism $f : X \to Y$ of smooth projective schemes over $k$ and an invertible $\mathcal{O}_ Y$-module $\mathcal{N}$ we have $f^*c_1^ H(\mathcal{L}) = c_1^ H(f^*\mathcal{L})$. This is immediately clear from the corresponding property for $c_1^{H'}$, the commutative diagrams in the lemma, and the fact that the canonical map $V \to V \otimes _ F F'$ is injective for any $F$-vector space $V$.

Axiom (A3). This follows from the principle stated in the proof of axiom (A1) and compatibility of $c_1^ H$ and $c_1^{H'}$.

Axiom (A4). Let $i : Y \to X$ be the inclusion of an effective Cartier divisor over $k$ with both $X$ and $Y$ smooth and projective over $k$. For $a \in H^*(X)$ with $i^*a = 0$ we have to show $a \cup c_1^ H(\mathcal{O}_ X(Y)) = 0$. Denote $a' \in (H')^*(X_{k'})$ the image of $a$. The assumption implies that $(i')^*a' = 0$ where $i' : Y_{k'} \to X_{k'}$ is the base change of $i$. Hence we get $a' \cup c_1^{H'}(\mathcal{O}_{X_{k'}}(Y_{k'})) = 0$ by the axiom for $(H')^*$. Since $a' \cup c_1^{H'}(\mathcal{O}_{X_{k'}}(Y_{k'}))$ is the image of $a \cup c_1^ H(\mathcal{O}_ X(Y))$ we conclude by the princple stated in the proof of axiom (A2).

Axiom (A5). This means that $H^*(\mathop{\mathrm{Spec}}(k)) = F$ and that for $X$ and $Y$ smooth projective over $k$ the map $H^*(X) \otimes _ F H^*(Y) \to H^*(X \times Y)$, $a \otimes b \mapsto p^*(a) \cup q^*(b)$ is an isomorphism where $p$ and $q$ are the projections. This follows from the principle stated in the proof of axiom (A1).

We interrupt the flow of the arguments to show that for every smooth projective scheme $X$ over $k$ the diagram

\[ \xymatrix{ \mathop{\mathrm{CH}}\nolimits ^*(X) \ar[r]_-\gamma \ar[d]_{g^*} & \bigoplus H^{2i}(X)(i) \ar[d] \\ \mathop{\mathrm{CH}}\nolimits ^*(X_{k'}) \ar[r]^-{\gamma '} & \bigoplus (H')^{2i}(X_{k'})(i) } \]

commutes. Observe that we have $\gamma $ as we know axioms (A1) – (A4) already; see Lemma 45.14.2. Also, the left vertical arrow is the one discussed in Chow Homology, Section 42.67 for the morphism of base schemes $g : \mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$. More precisely, it is the map given in Chow Homology, Lemma 42.67.4. Pick $\alpha \in \mathop{\mathrm{CH}}\nolimits ^*(X)$. Write $\alpha = ch(\beta ) \cap [X]$ in $\mathop{\mathrm{CH}}\nolimits ^*(X) \otimes \mathbf{Q}$ for some $\beta \in K_0(\textit{Vect}(X)) \otimes \mathbf{Q}$ so that $\gamma (\alpha ) = ch^{H}(\beta )$; this is our construction of $\gamma $. Since the map of Chow Homology, Lemma 42.67.4 is compatible with capping with Chern classes by Chow Homology, Lemma 42.67.8 we see that $g^*\alpha = ch((X_{k'} \to X)^*\beta ) \cap [X_{k'}]$. Hence $\gamma '(g^*\alpha ) = ch^{H'}((X_{k'} \to X)^*\beta )$. Thus commutativity of the diagram will hold if for any locally free $\mathcal{O}_ X$-module $\mathcal{E}$ of rank $r$ and $0 \leq i \leq r$ the element $c_ i^ H(\mathcal{E})$ of $H^{2i}(X)(i)$ maps to the element $c_ i^{H'}(\mathcal{E}_{k'})$ in $(H')^{2i}(X_{k'})(i)$. Because we have the projective space bundle formula for both $X$ and $X'$ we may replace $X$ by a projective space bundle over $X$ finitely many times to show this. Thus we may assume $\mathcal{E}$ has a filtration whose graded pieces are invertible $\mathcal{O}_ X$-modules $\mathcal{L}_1, \ldots , \mathcal{L}_ r$. See Chow Homology, Lemma 42.43.1 and Remark 42.43.2. Then $c^ H_ i(\mathcal{E}$ is the $i$th elementary symmetric polynomial in $c^ H_1(\mathcal{L}_1), \ldots , c^ H_1(\mathcal{L}_ r)$ and we conclude by our assumption that we have agreement for first Chern classes.

Axiom (A6). Suppose given $F$-vector spaces $V$, $W$, an element $v \in V$, and a tensor $\xi \in V \otimes _ F W$. Denote $V' = V \otimes _ F F'$, $W' = W \otimes _ F F'$ and $v'$, $\xi '$ the images of $v$, $\xi $ in $V'$, $V' \otimes _{F'} W'$. The linear algebra principle we will use in the proof of axiom (A6) is the following: there exists an $F$-linear map $\lambda : W \to F$ such that $(1 \otimes \lambda )\xi = v$ if and only if there exists an $F'$-linear map $\lambda ' : W \otimes _ F F' \to F'$ such that $(1 \otimes \lambda ')\xi ' = v'$.

Let $X$ be a nonempty equidimensional smooth projective scheme over $k$ of dimension $d$. Denote $\gamma = \gamma ([\Delta ])$ in $H^{2d}(X \times X)(d)$ (unadorned fibre products will be over $k$). Observe/recall that this makes sense as we know axioms (A1) – (A4) already; see Lemma 45.14.2. We may decompose

\[ \gamma = \sum \gamma _ i, \quad \gamma _ i \in H^ i(X) \otimes _ F H^{2d - i}(X)(d) \]

in the Künneth decomposition. Similarly, denote $\gamma ' = \gamma ([\Delta ']) = \sum \gamma '_ i$ in $(H')^{2d}(X_{k'} \times _{k'} X_{k'})(d)$. By the linear algebra princple mentioned above, it suffices to show that $\gamma _0$ maps to $\gamma '_0$ in $(H')^0(X) \otimes _{F'} (H')^{2d}(X')(d)$. By the compatibility of Künneth decompositions we see that it suffice to show that $\gamma $ maps to $\gamma '$ in

\[ (H')^{2d}(X_{k'} \times _{k'} X_{k'})(d) = (H')^{2d}((X \times X)_{k'})(d) \]

Since $\Delta _{k'} = \Delta '$ this follows from the discussion above.

Axiom (A7). This follows from the linear algebra fact: a linear map $V \to W$ of $F$-vector spaces is injective if and only if $V \otimes _ F F' \to W \otimes _ F F'$ is injective.

Axiom (A8). Follows from the linear algebra fact used in the proof of axiom (A1).

Axiom (A9). Let $X$ be a nonempty smooth projective scheme over $k$ equidimensional of dimension $d$. Let $i : Y \to X$ be a nonempty effective Cartier divisor smooth over $k$. Let $\lambda _ Y$ and $\lambda _ X$ be as in axiom (A6) for $X$ and $Y$. We have to show: for $a \in H^{2d - 2}(X)(d - 1)$ we have $\lambda _ Y(i^*(a)) = \lambda _ X(a \cup c_1^ H(\mathcal{O}_ X(Y))$. By Remark 45.14.6 we know that $\lambda _ X : H^{2d}(X)(d) \to F$ and $\lambda _ Y : H^{2d - 2}(Y)(d - 1)$ are uniquely determined by the requirement in axiom (A6). Having said this, it follows from our proof of axiom (A6) for $H^*$ above that $\lambda _ X \otimes \text{id}_{F'}$ corresponds to $\lambda _{X_{k'}}$ via the given identification $H^{2d}(X)(d) \otimes _ F F' = H^{2d}(X_{k'})(d)$. Thus the fact that we know axiom (A9) for $F'(1), (H')^*, c_1^{H'}$ implies the axiom for $F(1), H^*, c_1^ H$ by a diagram chase. This completes the proof of the lemma.
$\square$

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