Lemma 68.11.2 (Flat base change). Let $S$ be a scheme. Consider a cartesian diagram of algebraic spaces

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

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 $i \geq 0$

1. the base change map of Cohomology on Sites, Lemma 21.15.1 is an isomorphism

$g^*R^ if_*\mathcal{F} \longrightarrow R^ if'_*\mathcal{F}',$
2. 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 66.30.4. Note that flatness of $g$ and $g'$ is equivalent to flatness of the morphisms of small étale ringed sites, see Morphisms of Spaces, Lemma 66.30.9. Hence we can apply Cohomology on Sites, Lemma 21.15.1 to obtain a base change map

$g^*R^ pf_*\mathcal{F} \longrightarrow R^ pf'_*\mathcal{F}'$

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

$H^ p(X, \mathcal{F}) \otimes _ A B \longrightarrow H^ p(X_ B, \mathcal{F}_ B)$

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 68.8.2. 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

$g'_*\mathcal{F}_ B = \mathcal{F} \otimes _{\underline{A}} \underline{B}$

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

\begin{align*} g'_*\mathcal{F}_ B(U) & = \mathcal{F}_ B(\mathop{\mathrm{Spec}}(B) \times _{\mathop{\mathrm{Spec}}(A)} U) \\ & = \Gamma (\mathop{\mathrm{Spec}}(B) \times _{\mathop{\mathrm{Spec}}(A)} U, (\mathop{\mathrm{Spec}}(B) \times _{\mathop{\mathrm{Spec}}(A)} U \to U)^*\mathcal{F}|_ U) \\ & = B \otimes _ A \mathcal{F}(U) \end{align*}

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

\begin{align*} H^ p(X, g'_*\mathcal{F}_ B) & = H^ p(X, \mathcal{F} \otimes _{\underline{A}} \underline{B}) \\ & = H^ p(X, \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F} \otimes _{\underline{A}} \underline{M_ i}) \\ & = \mathop{\mathrm{colim}}\nolimits _ i H^ p(X, \mathcal{F} \otimes _{\underline{A}} \underline{M_ i}) \\ & = \mathop{\mathrm{colim}}\nolimits _ i H^ p(X, \mathcal{F}) \otimes _ A M_ i \\ & = H^ p(X, \mathcal{F}) \otimes _ A \mathop{\mathrm{colim}}\nolimits _ i M_ i \\ & = H^ p(X, \mathcal{F}) \otimes _ A B \end{align*}

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 68.5.1). 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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