The Stacks project

Proof. The first equality holds because

The second equality holds because

This ends the proof. $\square$

Proof. This follows from axioms (B)(a) and (B)(b). $\square$

Proof. By axiom (C)(d) we see that $H^0(\mathop{\mathrm{Spec}}(k))$ is nonzero and even $\gamma ([\mathop{\mathrm{Spec}}(k)])$ is nonzero. Since $\mathop{\mathrm{Spec}}(k) \times \mathop{\mathrm{Spec}}(k) = \mathop{\mathrm{Spec}}(k)$ we get

by axiom (B)(a) which implies (look at dimensions) that only $H^0$ is nonzero and moreover has dimension $1$. Thus $F = H^0(\mathop{\mathrm{Spec}}(k))$ via the unique $F$-algebra isomorphism given by mapping $1 \in F$ to $1 \in H^0(\mathop{\mathrm{Spec}}(k))$. Since $[\mathop{\mathrm{Spec}}(k)] \cdot [\mathop{\mathrm{Spec}}(k)] = [\mathop{\mathrm{Spec}}(k)]$ in the Chow ring of $\mathop{\mathrm{Spec}}(k)$ we conclude that $\gamma ([\mathop{\mathrm{Spec}}(k)) \cup \gamma ([\mathop{\mathrm{Spec}}(k)]) = \gamma ([\mathop{\mathrm{Spec}}(k)])$ by axiom (C)(c). Since we already know that $\gamma ([\mathop{\mathrm{Spec}}(k)])$ is nonzero we conclude that it has to be equal to $1$. Finally, we have $\int _{\mathop{\mathrm{Spec}}(k)} 1 = 1$ since $\int _{\mathop{\mathrm{Spec}}(k)} \gamma ([\mathop{\mathrm{Spec}}(k)]) = 1$ by axiom (C)(d). $\square$

Proof. First assume $X$ is nonempty. Observe that $[X]$ is the pullback of $[\mathop{\mathrm{Spec}}(k)]$ by the structure morphism $p : X \to \mathop{\mathrm{Spec}}(k)$. Hence we get $\gamma ([X]) = 1$ by axiom (C)(a) and Lemma 45.9.4. 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.

Let $x \in X$ be a closed point whose residue field $k'$ is separable over $k$, see Varieties, Lemma 33.25.6. Let $i : \mathop{\mathrm{Spec}}(k') \to X$ be the inclusion morphism. Denote $p : X \to \mathop{\mathrm{Spec}}(k)$ is the structure morphism. Observe that $p_*i_*[\mathop{\mathrm{Spec}}(k')] = [k' : k][\mathop{\mathrm{Spec}}(k)]$ in $\mathop{\mathrm{CH}}\nolimits _0(\mathop{\mathrm{Spec}}(k))$. Using axiom (C)(b) twice and Lemma 45.9.4 we conclude that

is nonzero. Thus $i_*\gamma ([\mathop{\mathrm{Spec}}(k)]) \in H^{2d}(X)(d)$ is nonzero (because it maps to something nonzero via $p_*$). This concludes the proof in case $X$ is nonempty.

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. $\square$

Proof. This is true because $1 = \gamma ([X])$ in $H^0(X)$ by Lemma 45.9.5 and then we can apply axiom (C)(b). $\square$

Proof. 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 45.9.3. On the other hand, recall that $\gamma ([\Delta ]) = \Delta _*1$ (Lemma 45.9.6) and hence

by Lemma 45.9.1. On the other hand, we have

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. $\square$

Proof. 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

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 45.9.7 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 45.9.5 we have $H^0(\mathbf{P}^1_ k) \not= 0$. By Poincaré duality, more precisely axiom (A)(b), we have $\beta _0 = \beta _2$. Therefore the lemma holds. $\square$

Proof. If $X$ or $Y$ is empty, then this is true because $H^*(\emptyset ) = 0$ by Lemma 45.9.5. Thus we may assume both $X$ and $Y$ are nonempty.

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.

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 $\mathop{\mathrm{CH}}\nolimits ^0(X \amalg Y)$ and that $[X]$ and $[Y]$ are orthogonal idempotents in $\mathop{\mathrm{CH}}\nolimits ^0(X \amalg Y)$. Thus

is a decomposition into orthogonal idempotents. Here we have used Lemmas 45.9.5 and 45.9.6 and axiom (C)(c). Then we see that

by the projection formula (Lemma 45.9.1) and hence the map is injective.

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

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 45.9.5) it suffices to prove $i^*$ is surjective after replacing $X$ and $Y$ by $X \times \mathop{\mathrm{Spec}}(k')$ and $Y \times \mathop{\mathrm{Spec}}(k')$ 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 33.25.6) then there is a morphism $Y \times \mathop{\mathrm{Spec}}(k') \to X \times \mathop{\mathrm{Spec}}(k')$ and we find that the proof is complete. $\square$

Proof. This proof is the same as the proof of Lemma 45.7.9; we urge the reader to read the proof of that lemma instead.

We obtain a contravariant functor from the category of smooth projective schemes over $k$ to the category of graded $F$-vector spaces by setting $H^*(X) = G(h(X))$. By assumption we have a canonical isomorphism

compatible with pullbacks. Using pullback along the diagonal $\Delta : X \to X \times X$ we obtain a canonical map

of graded vector spaces compatible with pullbacks. This defines a functorial graded $F$-algebra structure on $H^*(X)$. Since $\Delta $ commutes with the commutativity constraint $h(X) \otimes h(X) \to h(X) \otimes h(X)$ (switching the factors) and since $G$ is a functor of symmetric monoidal categories (so compatible with commutativity constraints), and by our convention in Homology, Example 12.17.4 we conclude that $H^*(X)$ is a graded commutative algebra! Hence we get our datum (D1).

Since $\mathbf{1}(1)$ is invertible in the category of motives we see that $G(\mathbf{1}(1))$ is invertible in the category of graded $F$-vector spaces. Thus $\sum _ i \dim _ F G^ i(\mathbf{1}(1)) = 1$. By assumption we only get something nonzero in degree $-2$. Our datum (D0) is the vector space $F(1) = G^{-2}(\mathbf{1}(1))$. Since $G$ is a symmetric monoidal functor we see that $F(n) = G^{-2n}(\mathbf{1}(n))$ for all $n \in \mathbf{Z}$. It follows that

a formula we will frequently use below.

Let $X$ be a smooth projective scheme over $k$. By Lemma 45.3.1 we have

Applying the functor $G$ this maps into $\mathop{\mathrm{Hom}}\nolimits (G(\mathbf{1}), G(h(X)(r)))$. By taking the image of $1$ in $G^0(\mathbf{1}) = F$ into $G^0(h(X)(r)) = H^{2r}(X)(r)$ we obtain

This is the datum (D2).

Let $X$ be a nonempty smooth projective scheme over $k$ which is equidimensional of dimension $d$. By Lemma 45.3.1 we have

Thus the class of the cycle $[X]$ in $\mathop{\mathrm{CH}}\nolimits _ d(X)$ defines a morphism $h(X)(d) \to \mathbf{1}$. Applying $G$ and taking degree $0$ parts we obtain

This map $\int _ X : H^{2d}(X)(d) \to F$ is the datum (D3).

Let $X$ be a smooth projective scheme over $k$ which is nonempty and equidimensional of dimension $d$. By Lemma 45.4.9 we know that $h(X)(d)$ is a left dual to $h(X)$. Hence $G(h(X)(d)) = H^*(X) \otimes _ F F(d)[2d]$ is a left dual to $H^*(X)$ in the category of graded $F$-vector spaces. Here $[n]$ is the shift functor on graded vector spaces. By Homology, Lemma 12.17.5 we find that $\sum _ i \dim _ F H^ i(X) < \infty $ and that $\epsilon : h(X)(d) \otimes h(X) \to \mathbf{1}$ produces nondegenerate pairings $H^{2d - i}(X)(d) \otimes _ F H^ i(X) \to F$. In the proof of Lemma 45.4.9 we have seen that $\epsilon $ is given by $[\Delta ]$ via the identifications

Thus $\epsilon $ is the composition of $[X] : h(X)(d) \to \mathbf{1}$ and $h(\Delta )(d) : h(X)(d) \otimes h(X) \to h(X)(d)$. It follows that the pairings above are given by cup product followed by $\int _ X$. This proves axiom (A).

Axiom (B) follows from the assumption that $G$ is compatible with tensor structures and our construction of the cup product above.

Axiom (C). Our construction of $\gamma $ takes a cycle $\alpha $ on $X$, interprets it a correspondence $a$ from $\mathop{\mathrm{Spec}}(k)$ to $X$ of some degree, and then applies $G$. If $f : Y \to X$ is a morphism of nonempty equidimensional smooth projective schemes over $k$, then $f^!\alpha $ is the pushforward (!) of $\alpha $ by the correspondence $[\Gamma _ f]$ from $X$ to $Y$, see Lemma 45.3.6. Hence $f^!\alpha $ viewed as a correspondence from $\mathop{\mathrm{Spec}}(k)$ to $Y$ is equal to $a \circ [\Gamma _ f]$, see Lemma 45.3.1. Since $G$ is a functor, we conclude $\gamma $ is compatible with pullbacks, i.e., axiom (C)(a) holds.

Let $f : Y \to X$ be a morphism of nonempty equidimensional smooth projective schemes over $k$ and let $\beta \in \mathop{\mathrm{CH}}\nolimits ^ r(Y)$ be a cycle on $Y$. We have to show that

for all $c \in H^*(X)$. Let $a, a^ t, \eta _ X, \eta _ Y, [X], [Y]$ be as in Lemma 45.3.9. Let $b$ be $\beta $ viewed as a correspondence from $\mathop{\mathrm{Spec}}(k)$ to $Y$ of degree $r$. Then $f_*\beta $ viewed as a correspondence from $\mathop{\mathrm{Spec}}(k)$ to $X$ is equal to $a^ t \circ b$, see Lemmas 45.3.6 and 45.3.1. The displayed equality above holds if we can show that

is equal to

This follows immediately from Lemma 45.3.9. Thus we have axiom (C)(b).

To prove axiom (C)(c) we use the discussion in Remark 45.7.2. Hence it suffices to prove that $\gamma $ is compatible with exterior products. Let $X$, $Y$ be nonempty smooth projective schemes over $k$ and let $\alpha $, $\beta $ be cycles on them. Denote $a$, $b$ the corresponding correspondences from $\mathop{\mathrm{Spec}}(k)$ to $X$, $Y$. Then $\alpha \times \beta $ corresponds to the correspondence $a \otimes b$ from $\mathop{\mathrm{Spec}}(k)$ to $X \otimes Y = X \times Y$. Hence the requirement follows from the fact that $G$ is compatible with the tensor structures on both sides.

Axiom (C)(d) follows because the cycle $[\mathop{\mathrm{Spec}}(k)]$ corresponds to the identity morphism on $h(\mathop{\mathrm{Spec}}(k))$. This finishes the proof of the lemma. $\square$

Proof. The proof of this lemma is the same as the proof of Lemma 45.7.10; we urge the reader to read the proof of that lemma instead.

By Lemma 45.4.8 it suffices to construct a functor $G$ on the category of smooth projective schemes over $k$ with morphisms given by correspondences of degree $0$ such that the image of $G(c_2)$ on $G(\mathbf{P}^1_ k)$ is an invertible graded $F$-vector space.

Let $X$ be a smooth projective scheme over $k$. There is a canonical decomposition

into open and closed subschemes such that $X_ d$ is equidimensional of dimension $d$. By Lemma 45.9.9 we have correspondingly

If $Y$ is a second smooth projective scheme over $k$ and we similarly decompose $Y = \coprod Y_ e$, then

As well we have $X \otimes Y = \coprod X_ d \otimes Y_ e$ in the category of correspondences. From these observations it follows that it suffices to construct $G$ on the category whose objects are equidimensional smooth projective schemes over $k$ and whose morphisms are correspondences of degree $0$. (Some details omitted.)

Given an equdimensional smooth projective scheme $X$ over $k$ we set $G(X) = H^*(X)$. Observe that $G(X) = 0$ if $X = \emptyset $ (Lemma 45.9.5). Thus maps from and to $G(\emptyset )$ are zero and we may and do assume our schemes are nonempty in the discussions below.

Given a correspondence $c \in \text{Corr}^0(X, Y)$ between nonempty equidmensional smooth projective schemes over $k$ we consider the map $G(c) : G(X) = H^*(X) \to G(Y) = H^*(Y)$ given by the rule

It is clear that $G(c)$ is additive in $c$ and hence $\mathbf{Q}$-linear. Compatibility of $\gamma $ with pullbacks, pushforwards, and intersection products given by axioms (C)(a), (C)(b), and (C)(c) shows that we have $G(c' \circ c) = G(c') \circ G(c)$ if $c' \in \text{Corr}^0(Y, Z)$. Namely, for $a \in H^*(X)$ we have

with obvious notation. The first equality follows from the definitions. The second equality holds because $\text{pr}^{23, *}_2 \circ \text{pr}^{12}_{2, *} = \text{pr}^{123}_{23, *} \circ \text{pr}^{123, *}_{12}$ as follows immediately from the description of pushforward along projections given in Lemma 45.9.2. The third equality holds by Lemma 45.9.1 and the fact that $H^*$ is a functor. The fourth equalith holds by axiom (C)(a) and the fact that the gysin map agrees with flat pullback for flat morphisms (Chow Homology, Lemma 42.59.5). The fifth equality uses axiom (C)(c) as well as Lemma 45.9.1 to see that $\text{pr}^{23}_{3, *} \circ \text{pr}^{123}_{23, *} = \text{pr}^{13}_{3, *} \circ \text{pr}^{123}_{13, *}$. The sixth equality uses the projection formula from Lemma 45.9.1 as well as axiom (C)(b) to see that $ \text{pr}^{123}_{13, *} \gamma (\text{pr}^{123, *}_{23}c' \cdot \text{pr}^{123, *}_{12}c) = \gamma (\text{pr}^{123}_{13, *}( \text{pr}^{123, *}_{23}c' \cdot \text{pr}^{123, *}_{12}c))$. Finally, the last equality is the definition.

To finish the proof that $G$ is a functor, we have to show identities are preserved. In other words, if $1 = [\Delta ] \in \text{Corr}^0(X, X)$ is the identity in the category of correspondences (Lemma 45.3.3), then we have to show that $G([\Delta ]) = \text{id}$. This follows from the determination of $\gamma ([\Delta ])$ in Lemma 45.9.7 and Lemma 45.9.2. This finishes the construction of $G$ as a functor on smooth projective schemes over $k$ and correspondences of degree $0$.

By Lemma 45.9.4 we have that $G(\mathop{\mathrm{Spec}}(k)) = H^*(\mathop{\mathrm{Spec}}(k))$ is canonically isomorphic to $F$ as an $F$-algebra. The Künneth axiom (B)(a) shows our functor is compatible with tensor products. Thus our functor is a functor of symmetric monoidal categories.

We still have to check that the image of $G(c_2)$ on $G(\mathbf{P}^1_ k) = H^*(\mathbf{P}^1_ k)$ is an invertible graded $F$-vector space (in particular we don't know yet that $G$ extends to $M_ k$). By Lemma 45.9.8 we only have nonzero cohomology in degrees $0$ and $2$ both of dimension $1$. We have $1 = c_0 + c_2$ is a decomposition of the identity into a sum of orthogonal idempotents in $\text{Corr}^0(\mathbf{P}^1_ k, \mathbf{P}^1_ k)$, see Example 45.3.7. Further we have $c_0 = a \circ b$ where $a \in \text{Corr}^0(\mathop{\mathrm{Spec}}(k), \mathbf{P}^1_ k)$ and $b \in \text{Corr}^0(\mathbf{P}^1_ k, \mathop{\mathrm{Spec}}(k))$ and where $b \circ a = 1$ in $\text{Corr}^0(\mathop{\mathrm{Spec}}(k), \mathop{\mathrm{Spec}}(k))$, see proof of Lemma 45.4.4. Thus $G(c_0)$ is the projector onto the degree $0$ part. It follows that $G(c_2)$ must be the projector onto the degree $2$ part and the proof is complete. $\square$

Proof. Given $G$ as in (2) by setting $H^*(X) = G(h(X))$ we obtain data (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C), see Lemma 45.9.10 and its proof.

Conversely, given data (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C) we get a functor $G$ as in (2) by the construction of the proof of Lemma 45.9.11.

We omit the detailed proof that these constructions are inverse to each other. $\square$