**Proof.**
Using Cohomology, Lemma 20.17.1 in (1) is allowed since $g'$ is flat by Morphisms, Lemma 29.25.8. Having said this, part (1) follows from part (2). Namely, part (1) is local on $S'$ and hence we may assume $S$ and $S'$ are affine. In other words, we have $S = \mathop{\mathrm{Spec}}(A)$ and $S' = \mathop{\mathrm{Spec}}(B)$ as in (2). Then since $R^ if_*\mathcal{F}$ is quasi-coherent (Lemma 30.4.5), it is the quasi-coherent $\mathcal{O}_ S$-module associated to the $A$-module $H^0(S, R^ if_*\mathcal{F}) = H^ i(X, \mathcal{F})$ (equality by Lemma 30.4.6). Similarly, $R^ if'_*\mathcal{F}'$ is the quasi-coherent $\mathcal{O}_{S'}$-module associated to the $B$-module $H^ i(X', \mathcal{F}')$. Since pullback by $g$ corresponds to $- \otimes _ A B$ on modules (Schemes, Lemma 26.7.3) we see that it suffices to prove (2).

Let $A \to B$ be a flat ring homomorphism. Let $X$ be a quasi-compact and quasi-separated scheme over $A$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Set $X_ B = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(B)$ and denote $\mathcal{F}_ B$ the pullback of $\mathcal{F}$. We are trying to show that the map

\[ H^ i(X, \mathcal{F}) \otimes _ A B \longrightarrow H^ i(X_ B, \mathcal{F}_ B) \]

(given by the reference in the statement of the lemma) is an isomorphism.

In case $X$ is separated, choose an affine open covering $\mathcal{U} : X = U_1 \cup \ldots \cup U_ t$ and recall that

\[ \check{H}^ p(\mathcal{U}, \mathcal{F}) = H^ p(X, \mathcal{F}), \]

see Lemma 30.2.6. If $\mathcal{U}_ B : X_ B = (U_1)_ B \cup \ldots \cup (U_ t)_ B$ we obtain by base change, then it is still the case that each $(U_ i)_ B$ is affine and that $X_ B$ is separated. Thus we obtain

\[ \check{H}^ p(\mathcal{U}_ B, \mathcal{F}_ B) = H^ p(X_ B, \mathcal{F}_ B). \]

We have the following relation between the Čech complexes

\[ \check{\mathcal{C}}^\bullet (\mathcal{U}_ B, \mathcal{F}_ B) = \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \otimes _ A B \]

as follows from Lemma 30.5.1. Since $A \to B$ is flat, the same thing remains true on taking cohomology.

In case $X$ is quasi-separated, choose an affine open covering $\mathcal{U} : X = U_1 \cup \ldots \cup U_ t$. We will use the Čech-to-cohomology spectral sequence Cohomology, Lemma 20.11.5. The reader who wishes to avoid this spectral sequence can use Mayer-Vietoris and induction on $t$ as in the proof of Lemma 30.4.5. The spectral sequence has $E_2$-page $E_2^{p, q} = \check{H}^ p(\mathcal{U}, \underline{H}^ q(\mathcal{F}))$ and converges to $H^{p + q}(X, \mathcal{F})$. Similarly, we have a spectral sequence with $E_2$-page $E_2^{p, q} = \check{H}^ p(\mathcal{U}_ B, \underline{H}^ q(\mathcal{F}_ B))$ which converges to $H^{p + q}(X_ B, \mathcal{F}_ B)$. Since the intersections $U_{i_0 \ldots i_ p}$ are quasi-compact and separated, the result of the second paragraph of the proof gives $\check{H}^ p(\mathcal{U}_ B, \underline{H}^ q(\mathcal{F}_ B)) = \check{H}^ p(\mathcal{U}, \underline{H}^ q(\mathcal{F})) \otimes _ A B$. Using that $A \to B$ is flat we conclude that $H^ i(X, \mathcal{F}) \otimes _ A B \to H^ i(X_ B, \mathcal{F}_ B)$ is an isomorphism for all $i$ and we win.
$\square$

## Comments (8)

Comment #936 by correction_bot on

Comment #957 by Johan on

Comment #2343 by Daniel on

Comment #2412 by Johan on

Comment #4682 by Nicolas Müller on

Comment #4807 by Johan on

Comment #5458 by Du on

Comment #5676 by Johan on