The Stacks project

Proof. Immediate from Proposition 45.12.4 applied to the category of smooth projective schemes over $k$, the functor $A : X \mapsto \bigoplus _{i \geq 0} H^{2i}(X)(i)$, and the map $c_1^ H$. $\square$

Proof. Recall that we have an isomorphism

see Chow Homology, Lemma 42.58.1. It is an isomorphism of rings by Chow Homology, Remark 42.56.5. We define $\gamma $ by the formula $\gamma (\alpha ) = ch^ H(\alpha ')$ where $ch^ H$ is as in Lemma 45.14.1 and $\alpha ' \in K_0(\textit{Vect}(X))$ is such that $ch(\alpha ') \cap [X] = \alpha $ in $\mathop{\mathrm{CH}}\nolimits ^*(X) \otimes \mathbf{Q}$.

The construction $\alpha \mapsto \gamma (\alpha )$ is compatible with pullbacks because both $ch^ H$ and taking Chern classes is compatible with pullbacks, see Lemma 45.14.1 and Chow Homology, Remark 42.59.9.

We still have to see that $\gamma $ is graded. Let $\psi ^2 : K_0(\textit{Vect}(X)) \to K_0(\textit{Vect}(X))$ be the second Adams operator, see Chow Homology, Lemma 42.56.1. If $\alpha \in \mathop{\mathrm{CH}}\nolimits ^ i(X)$ and $\alpha ' \in K_0(\textit{Vect}(X)) \otimes \mathbf{Q}$ is the unique element with $ch(\alpha ') \cap [X] = \alpha $, then we have seen in Chow Homology, Section 42.58 that $\psi ^2(\alpha ') = 2^ i \alpha '$. Hence we conclude that $ch^ H(\alpha ') \in H^{2i}(X)(i)$ by Lemma 45.12.5 as desired. $\square$

Proof. The scheme $X$ is smooth and projective over $k$ and hence we have $K_0(X) = K_0(\textit{Vect}(X))$. See Derived Categories of Schemes, Lemmas 36.36.2 and 36.38.5. Let $\alpha \in K_0(\text{Vect}(X))$ be an element whose restriction to $Z$ is $\mathcal{C}_{Z/X}$. By Chow Homology, Lemma 42.56.3 there exists an element $\alpha ^\vee $ which restricts to $\mathcal{C}_{Z/X}^\vee $. By the blow up formula (Chow Homology, Lemma 42.59.11) we have

where $\mathcal{F}$ is the kernel of the surjection $\pi ^*\mathcal{C}_{Z/X} \to \mathcal{C}_{E/X'}$. Observe that $b^*\alpha ^\vee - [\mathcal{O}_{X'}(E)]$ is an element of $K_0(\text{Vect}(X'))$ which restricts to $[\pi ^*\mathcal{C}_{Z/X}^\vee ] - [\mathcal{C}_{E/X'}^\vee ] = [\mathcal{F}^\vee ]$ on $E$. Since capping with Chern classes commutes with $j_*$ we conclude that the above is equal to

in the chow group of $X'$. Hence we see that setting

we get the first relation $\theta \cdot [E] = b^![Z]$ for example by Chow Homology, Lemma 42.62.2. For the second relation observe that

in the chow groups of $E$. To prove that $\pi _*$ of this is equal to $[Z]$ it suffices to prove that the degree of the codimension $r - 1$ cycle $(-1)^{r - 1}c_{r - 1}(\mathcal{F}) \cap [E]$ on the fibres of $\pi $ is $1$. This is a computation we omit. $\square$

Proof. Let $b : X' \to X$ be the blowing up. By (A7) it suffices to show that

By Lemma 45.14.3 we have

Hence because $b^*a$ restricts to zero on $E$ and since $\gamma ([E]) = c^ H_1(\mathcal{O}_{X'}(E))$ we get what we want from (A4). $\square$

Proof. Let $X$ be a nonempty smooth projective scheme over $k$ which is equidimensional of dimension $d$. We will show that the graded $F$-vector space $H^*(X)(d)[2d]$ is a left dual to $H^*(X)$. This will prove what we want by Homology, Lemma 12.17.5. We are going to use axiom (A5) which in particular says that

Define a map

by multiplying by $\gamma ([\Delta ]) \in H^{2d}(X \times X)(d)$. On the other hand, define a map

by first using pullback $\Delta ^*$ by the diagonal morphism $\Delta : X \to X \times X$ and then using the $F$-linear map $\lambda : H^{2d}(X)(d) \to F$ of axiom (A6) precomposed by the projection $H^*(X)(d) \to H^{2d}(X)(d)$. In order to show that $H^*(X)(d)$ is a left dual to $H^*(X)$ we have to show that the composition of the maps

and

is the identity. If $a \in H^*(X)$ then we see that the composition maps $a$ to

where $q_ i : X \times X \times X \to X$ and $q_{ij} : X \times X \times X \to X \times X$ are the projections, $\Delta _{23} : X \times X \to X \times X \times X$ is the diagonal, and $p_ i : X \times X \to X$ are the projections. The equality holds because $\Delta _{23}^*(q_{12}^*\gamma ([\Delta ]) = \Delta _{23}^*\gamma ([\Delta \times X]) = \gamma ([\Delta ])$ and because $\Delta _{23}^* q_3^*a = p_2^*a$. Since $\gamma ([\Delta ]) \cup p_1^*a = \gamma ([\Delta ]) \cup p_2^*a$ (see below) the above simplifies to

by our choice of $\lambda $ as desired. The second condition $(\epsilon \otimes 1) \circ (1 \otimes \eta ) = \text{id}$ of Categories, Definition 4.43.5 is proved in exactly the same manner.

Note that $p_1^*a$ and $\text{pr}_2^*a$ restrict to the same cohomology class on $\Delta \subset X \times X$. Moreover we have $\mathcal{C}_{\Delta /X \times X} = \Omega ^1_\Delta $ which is the restriction of $p_1^*\Omega ^1_ X$. Hence Lemma 45.14.4 implies $\gamma ([\Delta ]) \cup p_1^*a = \gamma ([\Delta ]) \cup p_2^*a$ and the proof is complete. $\square$

Proof. Axiom (B)(a) is immediate from axiom (A5). Let $X$ and $Y$ be nonempty smooth projective schemes over $k$ equidimensional of dimensions $d$ and $e$. To see that axiom (B)(b) holds, observe that the diagonal $\Delta _{X \times Y}$ of $X \times Y$ is the intersection product of the pullbacks of the diagonals $\Delta _ X$ of $X$ and $\Delta _ Y$ of $Y$ by the projections $p : X \times Y \times X \times Y \to X \times X$ and $q : X \times Y \times X \times Y \to Y \times Y$. Compatibility of $\gamma $ with intersection products then gives that

is the cup product of the pullbacks of $\gamma ([\Delta _ X])$ and $\gamma ([\Delta _ Y])$ by $p$ and $q$. Write

and similarly $\gamma ([\Delta _ X]) = \sum \eta _{X, i}$ and $\gamma ([\Delta _ Y]) = \sum \eta _{Y, i}$. The observation above implies we have

(If our cohomology theory vanishes in negative degrees, which will be true in almost all cases, then only the term for $i = 0$ contributes and $\eta _{X \times Y, 0}$ lies in $H^0(X) \otimes H^0(Y) \otimes H^{2d}(X)(d) \otimes H^{2e}(Y)(e)$ as expected, but we don't need this.) Since $\lambda _ X : H^{2d}(X)(d) \to F$ and $\lambda _ Y : H^{2e}(Y)(e) \to F$ send $\eta _{X, 0}$ and $\eta _{Y, 0}$ to $1$ in $H^*(X)$ and $H^*(Y)$, we see that $\lambda _ X \otimes \lambda _ Y$ sends $\eta _{X \times Y, 0}$ to $1$ in $H^*(X) \otimes H^*(Y) = H^*(X \times Y)$ and the proof is complete. $\square$

Proof. We have $\gamma ([\mathop{\mathrm{Spec}}(k)]) = 1 \in H^*(\mathop{\mathrm{Spec}}(k))$ by construction. Since

the map $\int _{\mathop{\mathrm{Spec}}(k)} = \lambda $ of axiom (A6) must send $1$ to $1$ because we have seen that $\int _{\mathop{\mathrm{Spec}}(k) \times \mathop{\mathrm{Spec}}(k)} = \int _{\mathop{\mathrm{Spec}}(k)} \int _{\mathop{\mathrm{Spec}}(k)}$ in Lemma 45.14.7. $\square$

Proof. It suffices to prove this on generators for $\mathop{\mathrm{CH}}\nolimits _*(P)$. Thus it suffices to prove this for a cycle class of the form $\xi ^ i \cdot p^*\alpha $ where $0 \leq i \leq r - 1$ and $\alpha \in \mathop{\mathrm{CH}}\nolimits _*(X)$. Note that $p_*(\xi ^ i \cdot p^*\alpha ) = 0$ if $i < r - 1$ and $p_*(\xi ^{r - 1} \cdot p^*\alpha ) = \alpha $. On the other hand, we have $\gamma (\xi ^ i \cdot p^*\alpha ) = c^ i \cup p^*\gamma (\alpha )$ and by the projection formula (Lemma 45.9.1) we have

Thus it suffices to show that $p_*c^ i = 0$ for $i < r - 1$ and $p_*c^{r - 1} = 1$. Equivalently, it suffices to prove that $\lambda _ P : H^{2d + 2r - 2}(P)(d + r - 1) \to F$ defined by the rules

satisfies the condition of axiom (A5). This follows from the computation of the class of the diagonal of $P$ in Lemma 45.6.2. $\square$

Proof. We observe that

as cycles on $\mathop{\mathrm{Spec}}(k') \times \mathop{\mathrm{Spec}}(k')$. Our construction of $\gamma $ always sends $[X]$ to $1$ in $H^0(X)$. Thus $1 \otimes 1 = 1 = \sum (\mathop{\mathrm{Spec}}(\sigma ) \times \text{id})^*\gamma ([\Delta ])$. Denote $\lambda : H^0(\mathop{\mathrm{Spec}}(k')) \to F$ the map from axiom (A6), in other words $(\text{id} \otimes \lambda )(\gamma (\Delta )) = 1$ in $H^0(\mathop{\mathrm{Spec}}(k'))$. We obtain

Since $\lambda $ is another name for $\int _{\mathop{\mathrm{Spec}}(k')}$ (Remark 45.14.6) the proof is complete. $\square$

Proof. We will use without further mention that we've constructed our cycle class map $\gamma $ in Lemma 45.14.2 compatible with intersection products and pullbacks and that we've already shown axioms (A), (B), (C)(a), (C)(c), and (C)(d) of Section 45.9, see Lemma 45.14.5, Remark 45.14.6, and Lemmas 45.14.7 and 45.14.8. In particular, we may use (for example) Lemma 45.9.1 to see that pushforward on $H^*$ is compatible with composition and satisfies the projection formula.

Let $f : X \to Y$ be a morphism of nonempty equidimensional smooth projective schemes over $k$. We are trying to show $f_*\gamma (\alpha ) = \gamma (f_*\alpha )$ for any cycle class $\alpha $ on $X$. We can write $\alpha $ as a $\mathbf{Q}$-linear combination of products of Chern classes of locally free $\mathcal{O}_ X$-modules (Chow Homology, Lemma 42.58.1). Thus we may assume $\alpha $ is a product of Chern classes of finite locally free $\mathcal{O}_ X$-modules $\mathcal{E}_1, \ldots , \mathcal{E}_ r$. Pick $p : P \to X$ as in the splitting principle (Chow Homology, Lemma 42.43.1). By Chow Homology, Remark 42.43.2 we see that $p$ is a composition of projective space bundles and that $\alpha = p_*(\xi _1 \cap \ldots \cap \xi _ d \cap \cdot p^*\alpha )$ where $\xi _ i$ are first Chern classes of invertible modules. By Lemma 45.14.9 we know that $p_*$ commutes with cycle classes. Thus it suffices to prove the property for the composition $f \circ p$. Since $p^*\mathcal{E}_1, \ldots , p^*\mathcal{E}_ r$ have filtrations whose successive quotients are invertible modules, this reduces us to the case where $\alpha $ is of the form $\xi _1 \cap \ldots \cap \xi _ t \cap [X]$ for some first Chern classes $\xi _ i$ of invertible modules $\mathcal{L}_ i$.

Assume $\alpha = c_1(\mathcal{L}_1) \cap \ldots \cap c_1(\mathcal{L}_ t) \cap [X]$ for some invertible modules $\mathcal{L}_ i$ on $X$. Let $\mathcal{L}$ be an ample invertible $\mathcal{O}_ X$-module. For $n \gg 0$ the invertible $\mathcal{O}_ X$-modules $\mathcal{L}^{\otimes n}$ and $\mathcal{L}_1 \otimes \mathcal{L}^{\otimes n}$ are both very ample on $X$ over $k$, see Morphisms, Lemma 29.39.8. Since $c_1(\mathcal{L}_1) = c_1(\mathcal{L}_1 \otimes \mathcal{L}^{\otimes n}) - c_1(\mathcal{L}^{\otimes n})$ this reduces us to the case where $\mathcal{L}_1$ is very ample. Repeating this with $\mathcal{L}_ i$ for $i = 2, \ldots , t$ we reduce to the case where $\mathcal{L}_ i$ is very ample on $X$ over $k$ for all $i = 1, \ldots , t$.

Assume $k$ is infinite and $\alpha = c_1(\mathcal{L}_1) \cap \ldots \cap c_1(\mathcal{L}_ t) \cap [X]$ for some very ample invertible modules $\mathcal{L}_ i$ on $X$ over $k$. By Bertini in the form of Varieties, Lemma 33.47.3 we can successively find regular sections $s_ i$ of $\mathcal{L}_ i$ such that the schemes $Z(s_1) \cap \ldots \cap Z(s_ i)$ are smooth over $k$ and of codimension $i$ in $X$. By the construction of capping with the first class of an invertible module (going back to Chow Homology, Definition 42.24.1), this reduces us to the case where $\alpha = [Z]$ for some nonempty smooth closed subscheme $Z \subset X$ which is equidimensional.

Assume $\alpha = [Z]$ where $Z \subset X$ is a smooth closed subscheme. Choose a closed embedding $X \to \mathbf{P}^ n$. We can factor $f$ as

Since we know the result for the second morphism by Lemma 45.14.9 it suffices to prove the result when $\alpha = [Z]$ where $i : Z \to X$ is a closed immersion and $f$ is a closed immersion. Then $j = f \circ i$ is a closed embedding too. Using the hypothesis for $i$ and $j$ we win.

We still have to prove the lemma in case $k$ is finite. We urge the reader to skip the rest of the proof. Everything we said above continues to work, except that we do not know we can choose the sections $s_ i$ cutting out our $Z$ over $k$ as $k$ is finite. However, we do know that we can find $s_ i$ over the algebraic closure $\overline{k}$ of $k$ (by the same lemma). This means that we can find a finite extension $k'/k$ such that our sections $s_ i$ are defined over $k'$. Denote $\pi : X_{k'} \to X$ the projection. The arguments above shows that we get the desired conclusion (from the assumption in the lemma) for the cycle $\pi ^*\alpha $ and the morphism $f \circ \pi : X_{k'} \to Y$. We have $\pi _*\pi ^*\alpha = [k' : k] \alpha $, see Chow Homology, Lemma 42.15.2. On the other hand, we have

by the projection formula for our cohomology theory. Observe that $\pi $ is a projection (!) and hence we have $\pi _*(1) = \int _{\mathop{\mathrm{Spec}}(k')}(1) 1$ by Lemma 45.9.2. Thus to finish the proof in the finite field case, it suffices to prove that $\int _{\mathop{\mathrm{Spec}}(k')}(1) = [k' : k]$ which we do in Lemma 45.14.10. $\square$

Proof. If $l = 1$ or $l = n - 1$ then $p$ is a projective bundle and the result follows from Lemma 45.14.9. In general there exists a morphism

such that both $h$ and $p \circ h$ are compositions of projective space bundles. Namely, denote $\mathbf{G}(1, 2, \ldots , l; n)$ the partial flag variety. Then the morphism

is a compostion of projective space bundles and similarly the structure morphism $\mathbf{G}(1, 2, \ldots , l; n) \to \mathop{\mathrm{Spec}}(k)$ is of this form. Thus we may set $Y = X \times \mathbf{G}(1, 2, \ldots , l; n)$. Since every cycle on $X \times \mathbf{G}(l, n)$ is the pushforward of a cycle on $Y$, the result for $Y \to X$ and the result for $Y \to X \times \mathbf{G}(l, n)$ imply the result for $p$. $\square$

Proof. By Lemma 45.14.11 it suffices to show that $i_*(1) = \gamma ([Z])$ if $i : Z \to X$ is a closed immersion of nonempty smooth projective equidimensional schemes over $k$. Say $Z$ has codimension $r$ in $X$. Let $\mathcal{L}$ be a sufficiently ample invertible module on $X$. Choose $n > 0$ and a surjection

This gives a morphism $g : Z \to \mathbf{G}(n - r, n)$ to the Grassmannian over $k$, see Constructions, Section 27.22. Consider the composition

Pushforward along the second morphism is compatible with classes of cycles by Lemma 45.14.12. The conormal sheaf $\mathcal{C}$ of the closed immersion $Z \to X \times \mathbf{G}(n - r, n)$ sits in a short exact sequence

See More on Morphisms, Lemma 37.11.13. Since $\mathcal{C}_{Z/X} \otimes \mathcal{L}|_ Z$ is the pull back of a finite locally free sheaf on $\mathbf{G}(n - r, n)$ we conclude that the class of $\mathcal{C}$ in $K_0(Z)$ is the pullback of a class in $K_0(X \times \mathbf{G}(n - r, n))$. Hence we have the property for $Z \to X \times \mathbf{G}(n - r, n)$ and we conclude. $\square$

Proof. We may replace $k''$ by its normal closure over $k$ which is Galois over $k$, see Fields, Lemma 9.21.5. Then $k''$ is Galois over $k'$ as well, see Fields, Lemma 9.21.4. We deduce we have an isomorphism

This produces an isomorphism $\coprod _\sigma \mathop{\mathrm{Spec}}(k'') \to \mathop{\mathrm{Spec}}(k') \times \mathop{\mathrm{Spec}}(k'')$ which on cohomology produces the isomorphism

where $\pi : \mathop{\mathrm{Spec}}(k'') \to \mathop{\mathrm{Spec}}(k')$ is the morphism corresponding to the inclusion of $k'$ in $k''$. We conclude the lemma is true by taking $a'' = 1$. $\square$

Proof. We may replace $X$ by a connected component of $X$ (some details omitted). Thus we may assume $X$ is connected and hence irreducible. Set $k' = \Gamma (X, \mathcal{O}_ X) = \Gamma (X', \mathcal{O}_{X'})$; we omit the proof of the equality. Choose a closed point $x' \in X'$ which isn't contained in the exceptional divisor and whose residue field $k''$ is separable over $k$; this is possible by Varieties, Lemma 33.25.6. Denote $x \in X$ the image (whose residue field is equal to $k''$ as well of course). Consider the diagram

The class of the diagonal $\Delta = \Delta _ X$ pulls back to the class of the “diagonal point” $\delta _ x : x \to x \times X$ and similarly for the class of the diagonal $\Delta '$. On the other hand, the diagonal point $\delta _ x$ pulls back to the diagonal point $\delta _{x'}$ by the left vertical arrow. Write $\gamma ([\Delta ]) = \sum \eta _ i$ with $\eta _ i \in H^ i(X) \otimes H^{2d - i}(X)(d)$ and $\gamma ([\Delta ']) = \sum \eta '_ i$ with $\eta '_ i \in H^ i(X') \otimes H^{2d - i}(X')(d)$. The arguments above show that $\eta _0$ and $\eta '_0$ map to the same class in

We have $H^0(\mathop{\mathrm{Spec}}(k')) = H^0(X) = H^0(X')$ by axiom (A8). By Lemma 45.14.14 this common value maps injectively into $H^0(x')$. We conclude that $\eta _0$ maps to $\eta '_0$ by the map

This means that $\int _ X$ is equal to $\int _{X'}$ composed with the pullback map. This proves the lemma. $\square$

Proof. Let $i : Z \to X$ be as in Lemma 45.14.13. Consider the diagram

Let $\theta \in \mathop{\mathrm{CH}}\nolimits ^{r - 1}(X')$ be as in Lemma 45.14.3. Then $\pi _*j^!\theta = [Z]$ in $\mathop{\mathrm{CH}}\nolimits _*(Z)$ implies that $\pi _*\gamma (j^!\theta ) = 1$ by Lemma 45.14.9 because $\pi $ is a projective space bundle. Hence we see that

We have $j_*(1) = \gamma ([E])$ by (A9). Thus this is equal to

Since $b_*(1) = 1$ by Lemma 45.14.15 the proof is complete. $\square$

Proof. We have axioms (A), (B) and (C)(a), (C)(c), and (C)(d) of Section 45.9 by Lemmas 45.14.5, 45.14.7, and 45.14.8. We have axiom (C)(b) by Lemma 45.14.16. Finally, the additional condition of Definition 45.11.4 holds because it is the same as our axiom (A8). $\square$

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

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

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

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. $\square$