We go by the axioms one by one. \medskip\noindent 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. \medskip\noindent 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 $V$. \medskip\noindent Axiom (A3). This follows from the principle stated in the proof of axiom (A1) and compatibility of $c_1^H$ and $c_1^{H'}$. \medskip\noindent 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). \medskip\noindent Axiom (A5). This means that $H^*(\Spec(k)) = F$ 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). \medskip\noindent We interrupt the flow of the arguments to show that for every smooth projective scheme $X$ over $k$ the diagram $$ \xymatrix{ \CH^*(X) \ar[r]_-\gamma \ar[d]_{g^*} & \bigoplus H^{2i}(X)(i) \ar[d] \\ \CH^*(X_{k'}) \ar[r]^-{\gamma'} & \bigoplus (H')^{2i}(X_{k'})(i) } $$ commutes. Observe that we have $\gamma$ as we know axioms (A1) -- (A4) already; see Lemma \ref{lemma-cycle-classes}. Also, the left vertical arrow is the one discussed in Chow Homology, Section \ref{chow-section-change-base} for the morphism of base schemes $g : \Spec(k') \to \Spec(k)$. More precisely, it is the map given in Chow Homology, Lemma \ref{chow-lemma-pullback-base-change}. Pick $\alpha \in \CH^*(X)$. Write $\alpha = ch(\beta) \cap [X]$ in $\CH^*(X) \otimes \mathbf{Q}$ 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 \ref{chow-lemma-pullback-base-change} is compatible with capping with Chern classes by Chow Homology, Lemma \ref{chow-lemma-pullback-base-change-chern-classes} 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 \ref{chow-lemma-splitting-principle} and Remark \ref{chow-remark-the-proof-shows-more}. 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. \medskip\noindent 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 $(1 \otimes \lambda')\xi' = v'$. \medskip\noindent 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 \ref{lemma-cycle-classes}. We may decompose $$ \gamma = \sum \gamma_i, \quad \gamma_i \in H^i(X) \otimes_F H^{2d - i}(X)(d) $$ in the K\"unneth 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 K\"unneth 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. \medskip\noindent 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. \medskip\noindent Axiom (A8). Follows from the linear algebra fact used in the proof of axiom (A1). \medskip\noindent 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 \ref{remark-trace} 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.