Proposition 50.22.1. Let $k$ be a field of characteristic zero. The functor that sends a smooth projective scheme $X$ over $k$ to $H_{dR}^*(X/k)$ is a Weil cohomology theory in the sense of Weil Cohomology Theories, Definition 45.11.4.

## 50.22 A Weil cohomology theory

Let $k$ be a field of characteristic $0$. In this section we prove that the functor

defines a Weil cohomology theory over $k$ with coefficients in $k$ as defined in Weil Cohomology Theories, Definition 45.11.4. We will proceed by checking the constructions earlier in this chapter provide us with data (D0), (D1), and (D2') satisfying axioms (A1) – (A9) of Weil Cohomology Theories, Section 45.14.

Throughout the rest of this section we fix the field $k$ of characteristic $0$ and we set $F = k$. Next, we take the following data

For our $1$-dimensional $F$ vector space $F(1)$ we take $F(1) = F = k$.

For our functor $H^*$ we take the functor sending a smooth projective scheme $X$ over $k$ to $H^*_{dR}(X/k)$. Functoriality is discussed in Section 50.3 and the cup product in Section 50.4. We obtain graded commutative $F$-algebras by Lemma 50.4.1.

For the maps $c_1^ H : \mathop{\mathrm{Pic}}\nolimits (X) \to H^2(X)(1)$ we use the de Rham first Chern class introduced in Section 50.9.

We are going to show axioms (A1) – (A9) hold.

In this paragraph, we are going to reduce the checking of the axioms to the case where $k$ is algebraically closed by using Weil Cohomology Theories, Lemma 45.14.18. Denote $k'$ the algebraic closure of $k$. Set $F' = k'$. We obtain data (D0), (D1), (D2') over $k'$ with coefficient field $F'$ in exactly the same way as above. By Lemma 50.3.5 there are functorial isomorphisms

for $X$ smooth and projective over $k$. Moreover, the diagrams

commute by Lemma 50.9.1. This finishes the proof of the reduction.

Assume $k$ is algebraically closed field of characteristic zero. We will show axioms (A1) – (A9) for the data (D0), (D1), and (D2') given above.

Axiom (A1). Here we have to check that $H^*_{dR}(X \coprod Y/k) = H^*_{dR}(X/k) \times H^*_{dR}(Y/k)$. This is a consequence of the fact that de Rham cohomology is constructed by taking the cohomology of a sheaf of differential graded algebras (in the Zariski topology).

Axiom (A2). This is just the statement that taking first Chern classes of invertible modules is compatible with pullbacks. This follows from the more general Lemma 50.9.1.

Axiom (A3). This follows from the more general Proposition 50.14.1.

Axiom (A4). This follows from the more general Lemma 50.15.6.

Already at this point, using Weil Cohomology Theories, Lemmas 45.14.1 and 45.14.2, we obtain a Chern character and cycle class maps

for $X$ smooth projective over $k$ which are graded ring homomorphisms compatible with pullbacks between morphisms $f : X \to Y$ of smooth projective schemes over $k$.

Axiom (A5). We have $H_{dR}^*(\mathop{\mathrm{Spec}}(k)/k) = k = F$ in degree $0$. We have the Künneth formula for the product of two smooth projective $k$-schemes by Lemma 50.8.2 (observe that the derived tensor products in the statement are harmless as we are tensoring over the field $k$).

Axiom (A7). This follows from Proposition 50.17.4.

Axiom (A8). Let $X$ be a smooth projective scheme over $k$. By the explanatory text to this axiom in Weil Cohomology Theories, Section 45.14 we see that $k' = H^0(X, \mathcal{O}_ X)$ is a finite separable $k$-algebra. It follows that $H_{dR}^*(\mathop{\mathrm{Spec}}(k')/k) = k'$ sitting in degree $0$ because $\Omega _{k'/k} = 0$. By Lemma 50.20.2 we also have $H_{dR}^0(X, \mathcal{O}_ X) = k'$ and we get the axiom.

Axiom (A6). Let $X$ be a nonempty smooth projective scheme over $k$ which is equidimensional of dimension $d$. Denote $\Delta : X \to X \times _{\mathop{\mathrm{Spec}}(k)} X$ the diagonal morphism of $X$ over $k$. We have to show that there exists a $k$-linear map

such that $(1 \otimes \lambda )\gamma ([\Delta ]) = 1$ in $H^0_{dR}(X/k)$. Let us write

with $\gamma _ i \in H_{dR}^ i(X/k) \otimes _ k H_{dR}^{2d - i}(X/k)$ the Künneth components. Our problem is to show that there is a linear map $\lambda : H_{dR}^{2d}(X/k) \to k$ such that $(1 \otimes \lambda )\gamma _0 = 1$ in $H^0_{dR}(X/k)$.

Let $X = \coprod X_ i$ be the decomposition of $X$ into connected and hence irreducible components. Then we have correspondingly $\Delta = \coprod \Delta _ i$ with $\Delta _ i \subset X_ i \times X_ i$. It follows that

and moreover $\gamma ([\Delta _ i])$ corresponds to the class of $\Delta _ i \subset X_ i \times X_ i$ via the decomposition

We omit the details; one way to show this is to use that in $\mathop{\mathrm{CH}}\nolimits ^0(X \times X)$ we have idempotents $e_{i, j}$ corresponding to the open and closed subschemes $X_ i \times X_ j$ and to use that $\gamma $ is a ring map which sends $e_{i, j}$ to the corresponding idempotent in the displayed product decomposition of cohomology. If we can find $\lambda _ i : H_{dR}^{2d}(X_ i/k) \to k$ with $(1 \otimes \lambda _ i)\gamma ([\Delta _ i]) = 1$ in $H^0_{dR}(X_ i/k)$ then taking $\lambda = \sum \lambda _ i$ will solve the problem for $X$. Thus we may and do assume $X$ is irreducible.

Proof of Axiom (A6) for $X$ irreducible. Since $k$ is algebraically closed we have $H^0_{dR}(X/k) = k$ because $H^0(X, \mathcal{O}_ X) = k$ as $X$ is a projective variety over an algebraically closed field (see Varieties, Lemma 33.9.3 for example). Let $x \in X$ be any closed point. Consider the cartesian diagram

Compatibility of $\gamma $ with pullbacks implies that $\gamma ([\Delta ])$ maps to $\gamma ([x])$ in $H_{dR}^{2d}(X/k)$, in other words, we have $\gamma _0 = 1 \otimes \gamma ([x])$. We conclude two things from this: (a) the class $\gamma ([x])$ is independent of $x$, (b) it suffices to show the class $\gamma ([x])$ is nonzero, and hence (c) it suffices to find any zero cycle $\alpha $ on $X$ such that $\gamma (\alpha ) \not= 0$. To do this we choose a finite morphism

To see such a morphism exist, see Intersection Theory, Section 43.23 and in particular Lemma 43.23.1. Observe that $f$ is finite syntomic (local complete intersection morphism by More on Morphisms, Lemma 37.59.10 and flat by Algebra, Lemma 10.128.1). By Proposition 50.19.3 we have a trace map

whose composition with the canonical map

is multiplication by the degree of $f$. Hence we see that we get a map

such that $\Theta \circ f^*$ is multiplication by a positive integer. Hence if we can find a zero cycle on $\mathbf{P}^ d_ k$ whose class is nonzero, then we conclude by the compatibility of $\gamma $ with pullbacks. This is true by Lemma 50.11.4 and this finishes the proof of axiom (A6).

Below we will use the following without further mention. First, by Weil Cohomology Theories, Remark 45.14.6 the map $\lambda _ X : H^{2d}_{dR}(X/k) \to k$ is unique. Second, in the proof of axiom (A6) we have seen that $\lambda _ X(\gamma ([x])) = 1$ when $X$ is irreducible, i.e., the composition of the cycle class map $\gamma : \mathop{\mathrm{CH}}\nolimits ^ d(X) \to H_{dR}^{2d}(X/k)$ with $\lambda _ X$ is the degree map.

Axiom (A9). Let $Y \subset X$ be a nonempty smooth divisor on a nonempty smooth equidimensional projective scheme $X$ over $k$ of dimension $d$. We have to show that the diagram

commutes where $\lambda _ X$ and $\lambda _ Y$ are as in axiom (A6). Above we have seen that if we decompose $X = \coprod X_ i$ into connected (equivalently irreducible) components, then we have correspondingly $\lambda _ X = \sum \lambda _{X_ i}$. Similarly, if we decompoese $Y = \coprod Y_ j$ into connected (equivalently irreducible) components, then we have $\lambda _ Y = \sum \lambda _{Y_ j}$. Moreover, in this case we have $\mathcal{O}_ X(Y) = \otimes _ j \mathcal{O}_ X(Y_ j)$ and hence

in $H_{dR}^2(X/k)$. A straightforward diagram chase shows that it suffices to prove the commutativity of the diagram in case $X$ and $Y$ are both irreducible. Then $H_{dR}^{2d - 2}(Y/k)$ is $1$-dimensional as we have Poincar'e duality for $Y$ by Weil Cohomology Theories, Lemma 45.14.5. By axiom (A4) the kernel of restriction (left vertical arrow) is contained in the kernel of cupping with $c^{dR}_1(\mathcal{O}_ X(Y))$. This means it suffices to find one cohomology class $a \in H_{dR}^{2d - 2}(X)$ whose restriction to $Y$ is nonzero such that we have commutativity in the diagram for $a$. Take any ample invertible module $\mathcal{L}$ and set

Then we know that $a|_ Y = c^{dR}_1(\mathcal{L}|_ Y)^{d - 1}$ and hence

by our description of $\lambda _ Y$ above. This is a positive integer by Chow Homology, Lemma 42.41.4 combined with Varieties, Lemma 33.45.9. Similarly, we find

Since we know that $c_1(\mathcal{O}_ X(Y)) \cap [X] = [Y]$ more or less by definition we have an equality of zero cycles

on $X$. Thus these cycles have the same degree and the proof is complete.

**Proof.**
In the discussion above we showed that our data (D0), (D1), (D2') satisfies axioms (A1) – (A9) of Weil Cohomology Theories, Section 45.14. Hence we conclude by Weil Cohomology Theories, Proposition 45.14.17.

Please don't read what follows. In the proof of the assertions we also used Lemmas 50.3.5, 50.9.1, 50.15.6, 50.8.2, 50.20.2, and 50.11.4, Propositions 50.14.1, 50.17.4, and 50.19.3, Weil Cohomology Theories, Lemmas 45.14.18, 45.14.1, 45.14.2, and 45.14.5, Weil Cohomology Theories, Remark 45.14.6, Varieties, Lemmas 33.9.3 and 33.45.9, Intersection Theory, Section 43.23 and Lemma 43.23.1, More on Morphisms, Lemma 37.59.10, Algebra, Lemma 10.128.1, and Chow Homology, Lemma 42.41.4. $\square$

Remark 50.22.2. In exactly the same manner as above one can show that Hodge cohomology $X \mapsto H_{Hodge}^*(X/k)$ equipped with $c_1^{Hodge}$ determines a Weil cohomology theory. If we ever need this, we will precisely formulate and prove this here. This leads to the following amusing consequence: If the betti numbers of a Weil cohomology theory are independent of the chosen Weil cohomology theory (over our field $k$ of characteristic $0$), then the Hodge-to-de Rham spectral sequence degenerates at $E_1$! Of course, the degeneration of the Hodge-to-de Rham spectral sequence is known (see for example [Deligne-Illusie] for a marvelous algebraic proof), but it is by no means an easy result! This suggests that proving the independence of betti numbers is a hard problem as well and as far as we know is still an open problem. See Weil Cohomology Theories, Remark 45.11.5 for a related question.

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