Lemma 69.11.1. Let $S$ be a scheme. Let $f : X \to Y$ be an affine morphism of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. In this case $f_*\mathcal{F} \cong Rf_*\mathcal{F}$ is a quasi-coherent sheaf, and for every diagram (69.11.0.1) we have
69.11 Cohomology and base change, I
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. Suppose further that $g : Y' \to Y$ is a morphism of algebraic spaces over $S$. Denote $X' = X_{Y'} = Y' \times _ Y X$ the base change of $X$ and denote $f' : X' \to Y'$ the base change of $f$. Also write $g' : X' \to X$ the projection, and set $\mathcal{F}' = (g')^*\mathcal{F}$. Here is a diagram representing the situation:
Here is the simplest case of the base change property we have in mind.
Proof. By the discussion surrounding (69.3.0.1) this reduces to the case of an affine morphism of schemes which is treated in Cohomology of Schemes, Lemma 30.5.1. $\square$
Lemma 69.11.2 (Flat base change). Let $S$ be a scheme. Consider a cartesian diagram of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module with pullback $\mathcal{F}' = (g')^*\mathcal{F}$. Assume that $g$ is flat and that $f$ is quasi-compact and quasi-separated. For any $i \geq 0$
the base change map of Cohomology on Sites, Lemma 21.15.1 is an isomorphism
if $Y = \mathop{\mathrm{Spec}}(A)$ and $Y' = \mathop{\mathrm{Spec}}(B)$, then $H^ i(X, \mathcal{F}) \otimes _ A B = H^ i(X', \mathcal{F}')$.
Proof. The morphism $g'$ is flat by Morphisms of Spaces, Lemma 67.30.4. Note that flatness of $g$ and $g'$ is equivalent to flatness of the morphisms of small étale ringed sites, see Morphisms of Spaces, Lemma 67.30.9. Hence we can apply Cohomology on Sites, Lemma 21.15.1 to obtain a base change map
To prove this map is an isomorphism we can work locally in the étale topology on $Y'$. Thus we may assume that $Y$ and $Y'$ are affine schemes. Say $Y = \mathop{\mathrm{Spec}}(A)$ and $Y' = \mathop{\mathrm{Spec}}(B)$. In this case we are really trying to show that the map
is an isomorphism where $X_ B = \mathop{\mathrm{Spec}}(B) \times _{\mathop{\mathrm{Spec}}(A)} X$ and $\mathcal{F}_ B$ is the pullback of $\mathcal{F}$ to $X_ B$. In other words, it suffices to prove (2).
Fix $A \to B$ a flat ring map and let $X$ be a quasi-compact and quasi-separated algebraic space over $A$. Note that $g' : X_ B \to X$ is affine as a base change of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$. Hence the higher direct images $R^ i(g')_*\mathcal{F}_ B$ are zero by Lemma 69.8.2. Thus $H^ p(X_ B, \mathcal{F}_ B) = H^ p(X, g'_*\mathcal{F}_ B)$, see Cohomology on Sites, Lemma 21.14.6. Moreover, we have
where $\underline{A}$, $\underline{B}$ denotes the constant sheaf of rings with value $A$, $B$. Namely, it is clear that there is a map from right to left. For any affine scheme $U$ étale over $X$ we have
hence the map is an isomorphism. Write $B = \mathop{\mathrm{colim}}\nolimits M_ i$ as a filtered colimit of finite free $A$-modules $M_ i$ using Lazard's theorem, see Algebra, Theorem 10.81.4. We deduce that
The first equality because $g'_*\mathcal{F}_ B = \mathcal{F} \otimes _{\underline{A}} \underline{B}$ as seen above. The second because $\otimes $ commutes with colimits. The third equality because cohomology on $X$ commutes with colimits (see Lemma 69.5.1). The fourth equality because $M_ i$ is finite free (i.e., because cohomology commutes with finite direct sums). The fifth because $\otimes $ commutes with colimits. The sixth by choice of our system. $\square$
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