Lemma 67.11.2 (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 65.30.4. Note that flatness of $g$ and $g'$ is equivalent to flatness of the morphisms of small étale ringed sites, see Morphisms of Spaces, Lemma 65.30.9. 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 67.8.2. 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 67.5.1). 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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 073K. Beware of the difference between the letter 'O' and the digit '0'.