Lemma 68.11.1. 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 (68.11.0.1) we have

## 68.11 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 (68.3.0.1) this reduces to the case of an affine morphism of schemes which is treated in Cohomology of Schemes, Lemma 30.5.1.
$\square$

Lemma 68.11.2 (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

\[ g^*R^ if_*\mathcal{F} \longrightarrow R^ if'_*\mathcal{F}', \]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

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 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

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 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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