The Stacks project

Proof. In case $X$ is separated, choose an affine open covering $\mathcal{U} : X = \bigcup _{i \in I} U_ i$ and recall that

see Lemma 30.2.6. Let $\mathcal{V} : Y = \bigcup _{i \in I} g^{-1}(U_ i)$ be the corresponding affine open covering of $Y$. The opens $V_ i = g^{-1}(U_ i) = U_ i \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(B)$ are affine and $Y$ is separated. Thus we obtain

We claim the map of Čech complexes

is an isomorphism. Namely, as $B$ is finitely presented as an $A$-module we see that tensoring with $B$ over $A$ commutes with products, see Algebra, Proposition 10.89.3. Thus it suffices to show that the maps $\Gamma (U_{i_0 \ldots i_ p}, \mathcal{F}) \otimes _ A B \to \Gamma (V_{i_0 \ldots i_ p}, \mathcal{G})$ are isomorphisms which follows from Lemma 30.5.1. Since $A \to B$ is flat, the same thing remains true on taking cohomology.

In the general case we argue in exactly the same way using affine open covering $\mathcal{U} : X = \bigcup _{i \in I} U_ i$ and the corresponding covering $\mathcal{V} : Y = \bigcup _{i \in I} V_ i$ with $V_ i = g^{-1}(U_ i)$ as above. We will use the Čech-to-cohomology spectral sequence Cohomology, Lemma 20.11.5. 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 $E_2^{p, q} = \check{H}^ p(\mathcal{V}, \underline{H}^ q(\mathcal{G}))$ which converges to $H^{p + q}(Y, \mathcal{G})$. Since the intersections $U_{i_0 \ldots i_ p}$ are separated, the result of the previous paragraph gives isomorphisms $\Gamma (U_{i_0 \ldots i_ p}, \underline{H}^ q(\mathcal{F})) \otimes _ A B \to \Gamma (V_{i_0 \ldots i_ p}, \underline{H}^ q(\mathcal{G}))$. Using that $- \otimes _ A B$ commutes with products and is exact, we conclude that $\check{H}^ p(\mathcal{U}, \underline{H}^ q(\mathcal{F})) \otimes _ A B \to \check{H}^ p(\mathcal{V}, \underline{H}^ q(\mathcal{G}))$ is an isomorphism. Using that $A \to B$ is flat we conclude that $H^ i(X, \mathcal{F}) \otimes _ A B \to H^ i(Y, \mathcal{G})$ is an isomorphism for all $i$ and we win. $\square$