Lemma 30.22.1. Let $A$ be a Noetherian ring and set $S = \mathop{\mathrm{Spec}}(A)$. Let $f : X \to S$ be a proper morphism of schemes. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module flat over $S$. Then

1. $R\Gamma (X, \mathcal{F})$ is a perfect object of $D(A)$, and

2. for any ring map $A \to A'$ the base change map

$R\Gamma (X, \mathcal{F}) \otimes _ A^{\mathbf{L}} A' \longrightarrow R\Gamma (X_{A'}, \mathcal{F}_{A'})$

is an isomorphism.

Proof. Choose a finite affine open covering $X = \bigcup _{i = 1, \ldots , n} U_ i$. By Lemmas 30.7.1 and 30.7.2 the Čech complex $K^\bullet = {\check C}^\bullet (\mathcal{U}, \mathcal{F})$ satisfies

$K^\bullet \otimes _ A A' = R\Gamma (X_{A'}, \mathcal{F}_{A'})$

for all ring maps $A \to A'$. Let $K_{alt}^\bullet = {\check C}_{alt}^\bullet (\mathcal{U}, \mathcal{F})$ be the alternating Čech complex. By Cohomology, Lemma 20.23.6 there is a homotopy equivalence $K_{alt}^\bullet \to K^\bullet$ of $A$-modules. In particular, we have

$K_{alt}^\bullet \otimes _ A A' = R\Gamma (X_{A'}, \mathcal{F}_{A'})$

as well. Since $\mathcal{F}$ is flat over $A$ we see that each $K_{alt}^ n$ is flat over $A$ (see Morphisms, Lemma 29.25.2). Since moreover $K_{alt}^\bullet$ is bounded above (this is why we switched to the alternating Čech complex) $K_{alt}^\bullet \otimes _ A A' = K_{alt}^\bullet \otimes _ A^{\mathbf{L}} A'$ by the definition of derived tensor products (see More on Algebra, Section 15.59). By Lemma 30.19.2 the cohomology groups $H^ i(K_{alt}^\bullet )$ are finite $A$-modules. As $K_{alt}^\bullet$ is bounded, we conclude that $K_{alt}^\bullet$ is pseudo-coherent, see More on Algebra, Lemma 15.64.17. Given any $A$-module $M$ set $A' = A \oplus M$ where $M$ is a square zero ideal, i.e., $(a, m) \cdot (a', m') = (aa', am' + a'm)$. By the above we see that $K_{alt}^\bullet \otimes _ A^\mathbf {L} A'$ has cohomology in degrees $0, \ldots , n$. Hence $K_{alt}^\bullet \otimes _ A^\mathbf {L} M$ has cohomology in degrees $0, \ldots , n$. Hence $K_{alt}^\bullet$ has finite Tor dimension, see More on Algebra, Definition 15.66.1. We win by More on Algebra, Lemma 15.74.2. $\square$

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