Lemma 30.7.4. Consider a cartesian diagram of schemes

\[ \xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X \ar[d]^ f \\ S' \ar[r]^ g & S } \]

Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $\mathcal{G}$ be a quasi-coherent $\mathcal{O}_{S'}$-module flat over $S$. Assume $f$ is quasi-compact and quasi-separated. For any $i \geq 0$ there is an identification

\[ \mathcal{G} \otimes _{\mathcal{O}_{S'}} g^*R^ if_*\mathcal{F} = R^ if'_*\left((f')^*\mathcal{G} \otimes _{\mathcal{O}_{X'}} (g')^*\mathcal{F}\right) \]

**Proof.**
Let us construct a map from left to right. First, we have the base change map $Lg^*Rf_*\mathcal{F} \to Rf'_*L(g')^*\mathcal{F}$. There is also the adjunction map $\mathcal{G} \to Rf'_*L(f')^*\mathcal{G}$. Using the relative cup product We obtain

\begin{align*} \mathcal{G} \otimes _{\mathcal{O}_{S'}}^\mathbf {L} Lg^*Rf_*\mathcal{F} & \to Rf'_*L(f')^*\mathcal{G} \otimes _{\mathcal{O}_{S'}}^\mathbf {L} Rf'_*L(g')^*\mathcal{F} \\ & \to Rf'_*\left(L(f')^*\mathcal{G} \otimes _{\mathcal{O}_{X'}}^\mathbf {L} L(g')^*\mathcal{F}\right) \\ & \to Rf'_*\left((f')^*\mathcal{G} \otimes _{\mathcal{O}_{X'}} (g')^*\mathcal{F}\right) \end{align*}

where for the middle arrow we used the relative cup product, see Cohomology, Remark 20.28.7. The source of the composition is

\[ \mathcal{G} \otimes _{\mathcal{O}_{S'}}^\mathbf {L} Lg^*Rf_*\mathcal{F} = \mathcal{G} \otimes _{g^{-1}\mathcal{O}_ S}^\mathbf {L} g^{-1}Rf_*\mathcal{F} \]

by Cohomology, Lemma 20.27.4. Since $\mathcal{G}$ is flat as a sheaf of $g^{-1}\mathcal{O}_ S$-modules and since $g^{-1}$ is an exact functor, this is a complex whose $i$th cohomology sheaf is $\mathcal{G} \otimes _{g^{-1}\mathcal{O}_ S} g^{-1}R^ if_*\mathcal{F} = \mathcal{G} \otimes _{\mathcal{O}_{S'}} g^*R^ if_*\mathcal{F}$. In this way we obtain global maps from left to right in the equality of the lemma. To show this map is an isomorphism we may work locally on $S'$. Thus we may and do assume that $S$ and $S'$ are affine schemes.

Proof in case $S$ and $S'$ are affine. Say $S = \mathop{\mathrm{Spec}}(A)$ and $S' = \mathop{\mathrm{Spec}}(B)$ and say $\mathcal{G}$ corresponds to the $B$-module $N$ which is assumed to be $A$-flat. Since $S$ is affine, $X$ is quasi-compact and quasi-separated. We will use a hypercovering argument to finish the proof; if $X$ is separated or has affine diagonal, then you can use a Čech covering. Let $\mathcal{B}$ be the set of affine opens of $X$. By Hypercoverings, Lemma 25.11.4 we can find a hypercovering $K = (I, \{ U_ i\} )$ of $X$ such that each $I_ n$ is finite and each $U_ i$ is an affine open of $X$. By Hypercoverings, Lemma 25.5.3 there is a spectral sequence with $E_2$-page

\[ E_2^{p, q} = \check{H}^ p(K, \underline{H}^ q(\mathcal{F})) \]

converging to $H^{p + q}(X, \mathcal{F})$. Since each $U_ i$ is affine and $\mathcal{F}$ is quasi-coherent the value of $\underline{H}^ q(\mathcal{F})$ is zero on $U_ i$ for $q > 0$. Thus the spectral sequence degenerates and we conclude that the cohomology modules $H^ q(X, \mathcal{F})$ are computed by

\[ \prod \nolimits _{i \in I_0} \mathcal{F}(U_ i) \to \prod \nolimits _{i \in I_1} \mathcal{F}(U_ i) \to \prod \nolimits _{i \in I_2} \mathcal{F}(U_ i) \to \ldots \]

Next, note that the base change of our hypercovering to $S'$ is a hypercovering of $X' = S' \times _ S X$. The schemes $S' \times _ S U_ i$ are affine too and we have

\[ \left((f')^*\mathcal{G} \otimes _{\mathcal{O}_{S'}} (g')^*\mathcal{F}\right) (S' \times _ S U_ i) = N \otimes _ A \mathcal{F}(U_ i) \]

In this way we conclude that the cohomology modules $H^ q(X', (f')^*\mathcal{G} \otimes _{\mathcal{O}_{S'}} (g')^*\mathcal{F})$ are computed by

\[ N \otimes _ A \left( \prod \nolimits _{i \in I_0} \mathcal{F}(U_ i) \to \prod \nolimits _{i \in I_1} \mathcal{F}(U_ i) \to \prod \nolimits _{i \in I_2} \mathcal{F}(U_ i) \to \ldots \right) \]

Since $N$ is flat over $A$, we conclude that

\[ H^ q(X', (f')^*\mathcal{G} \otimes _{\mathcal{O}_{S'}} (g')^*\mathcal{F}) = N \otimes _ A H^ q(X, \mathcal{F}) \]

Since this is the translation into algebra of the statement we had to show the proof is complete.
$\square$

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