Lemma 58.88.4. Let $X$ be a quasi-compact and quasi-separated scheme. Let $E \in D^+(X_{\acute{e}tale})$ and $K \in D^+(\mathbf{Z})$. Then

$R\Gamma (X, E \otimes _\mathbf {Z}^\mathbf {L} \underline{K}) = R\Gamma (X, E) \otimes _\mathbf {Z}^\mathbf {L} K$

Proof. Say $H^ i(E) = 0$ for $i \geq a$ and $H^ j(K) = 0$ for $j \geq b$. We may represent $K$ by a bounded below complex $K^\bullet$ of torsion free $\mathbf{Z}$-modules. (Choose a K-flat complex $L^\bullet$ representing $K$ and then take $K^\bullet = \tau _{\geq b - 1}L^\bullet$. This works because $\mathbf{Z}$ has global dimension $1$. See More on Algebra, Lemma 15.65.2.) We may represent $E$ by a bounded below complex $\mathcal{E}^\bullet$. Then $E \otimes _\mathbf {Z}^\mathbf {L} \underline{K}$ is represented by

$\text{Tot}(\mathcal{E}^\bullet \otimes _\mathbf {Z} \underline{K}^\bullet )$

Using distinguished triangles

$\sigma _{\geq -b + n + 1}K^\bullet \to K^\bullet \to \sigma _{\leq -b + n}K^\bullet$

and the trivial vanishing

$H^ n(X, \text{Tot}(\mathcal{E}^\bullet \otimes _\mathbf {Z} \sigma _{\geq -a + n + 1}\underline{K}^\bullet ) = 0$

and

$H^ n(R\Gamma (X, E) \otimes _\mathbf {Z}^\mathbf {L} \sigma _{\geq -a + n + 1}K^\bullet ) = 0$

we reduce to the case where $K^\bullet$ is a bounded complex of flat $\mathbf{Z}$-modules. Repeating the argument we reduce to the case where $K^\bullet$ is equal to a single flat $\mathbf{Z}$-module sitting in some degree. Next, using the stupid trunctions for $\mathcal{E}^\bullet$ we reduce in exactly the same manner to the case where $\mathcal{E}^\bullet$ is a single abelian sheaf sitting in some degree. Thus it suffices to show that

$H^ n(X, \mathcal{E} \otimes _\mathbf {Z} \underline{M}) = H^ n(X, \mathcal{E}) \otimes _\mathbf {Z} M$

when $M$ is a flat $\mathbf{Z}$-module and $\mathcal{E}$ is an abelian sheaf on $X$. In this case we write $M$ is a filtered colimit of finite free $\mathbf{Z}$-modules (Lazard's theorem, see Algebra, Theorem 10.81.4). By Theorem 58.51.3 this reduces us to the case of finite free $\mathbf{Z}$-module $M$ in which case the result is trivially true. $\square$

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