Lemma 50.8.1. In the situation above there is a canonical isomorphism
of complexes of $f^{-1}\mathcal{O}_ S$-modules.
An important feature of de Rham cohomology is that there is a Künneth formula.
Let $a : X \to S$ and $b : Y \to S$ be morphisms of schemes with the same target. Let $p : X \times _ S Y \to X$ and $q : X \times _ S Y \to Y$ be the projection morphisms and $f = a \circ p = b \circ q$. Here is a picture
In this section, given an $\mathcal{O}_ X$-module $\mathcal{F}$ and an $\mathcal{O}_ Y$-module $\mathcal{G}$ let us set
The bifunctor $(\mathcal{F}, \mathcal{G}) \mapsto \mathcal{F} \boxtimes \mathcal{G}$ on quasi-coherent modules extends to a bifunctor on quasi-coherent modules and differential operators of finite over over $S$, see Morphisms, Remark 29.33.3. The differentials of the de Rham complexes $\Omega ^\bullet _{X/S}$ and $\Omega ^\bullet _{Y/S}$ are differential operators of order $1$ over $S$ by Modules, Lemma 17.30.5. Thus it makes sense to consider the complex
Please see the discussion in Derived Categories of Schemes, Section 36.24.
Lemma 50.8.1. In the situation above there is a canonical isomorphism of complexes of $f^{-1}\mathcal{O}_ S$-modules.
Proof. We know that $ \Omega _{X \times _ S Y/S} = p^*\Omega _{X/S} \oplus q^*\Omega _{Y/S} $ by Morphisms, Lemma 29.32.11. Taking exterior powers we obtain
by elementary properties of exterior powers. These identifications determine isomorphisms between the terms of the complexes on the left and the right of the arrow in the lemma. We omit the verification that these maps are compatible with differentials. $\square$
Set $A = \Gamma (S, \mathcal{O}_ S)$. Combining the result of Lemma 50.8.1 with the map Derived Categories of Schemes, Equation (36.24.0.2) we obtain a cup product
On the level of cohomology, using the discussion in More on Algebra, Section 15.63, we obtain a canonical map
We note that the construction above indeed proceeds by first pulling back and then taking the cup product.
Lemma 50.8.2. Assume $X$ and $Y$ are smooth, quasi-compact, with affine diagonal over $S = \mathop{\mathrm{Spec}}(A)$. Then the map is an isomorphism in $D(A)$.
Proof. By Morphisms, Lemma 29.34.12 the sheaves $\Omega ^ n_{X/S}$ and $\Omega ^ m_{Y/S}$ are finite locally free $\mathcal{O}_ X$ and $\mathcal{O}_ Y$-modules. On the other hand, $X$ and $Y$ are flat over $S$ (Morphisms, Lemma 29.34.9) and hence we find that $\Omega ^ n_{X/S}$ and $\Omega ^ m_{Y/S}$ are flat over $S$. Also, observe that $\Omega ^\bullet _{X/S}$ is a locally bounded. Thus the result by Lemma 50.8.1 and Derived Categories of Schemes, Lemma 36.24.1. $\square$
There is a relative version of the cup product, namely a map
in $D(\mathcal{O}_ S)$. The construction combines Lemma 50.8.1 with the map Derived Categories of Schemes, Equation (36.24.0.1). The construction shows that this map is given by the diagram
Here the first arrow uses the units $\text{id} \to Rp_* p^{-1}$ and $\text{id} \to Rq_* q^{-1}$ of adjunction as well as the identifications $Rf_* p^{-1} = Ra_* Rp_* p^{-1}$ and $Rf_* q^{-1} = Rb_* Rq_* q^{-1}$. The second arrow is the relative cup product of Cohomology, Remark 20.28.7. The third arrow is the map sending a derived tensor product of complexes to the totalization of the tensor product of complexes. The final equality is Lemma 50.8.1. This construction recovers on global section the construction given earlier.
Lemma 50.8.3. Assume $X \to S$ and $Y \to S$ are smooth and quasi-compact and the morphisms $X \to X \times _ S X$ and $Y \to Y \times _ S Y$ are affine. Then the relative cup product is an isomorphism in $D(\mathcal{O}_ S)$.
Proof. Immediate consequence of Lemma 50.8.2. $\square$
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