Lemma 76.13.2. In Situation 76.13.1 consider

$K = R\mathop{\mathrm{lim}}\nolimits _{D_\mathit{QCoh}(\mathcal{O}_ X)}(K_ n) = DQ_ X(R\mathop{\mathrm{lim}}\nolimits _{D(\mathcal{O}_ X)} K_ n)$

Then $K$ is in $D^-_{\mathit{QCoh}}(\mathcal{O}_ X)$.

Proof. The functor $DQ_ X$ exists because $X$ is quasi-compact and quasi-separated, see Derived Categories of Spaces, Lemma 74.19.1. Since $DQ_ X$ is a right adjoint it commutes with products and therefore with derived limits. Hence the equality in the statement of the lemma.

By Derived Categories of Spaces, Lemma 74.19.4 the functor $DQ_ X$ has bounded cohomological dimension. Hence it suffices to show that $R\mathop{\mathrm{lim}}\nolimits K_ n \in D^-(\mathcal{O}_ X)$. To see this, let $U \to X$ be étale with $U$ affine. Then there is a canonical exact sequence

$0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{m - 1}(U, K_ n) \to H^ m(U, R\mathop{\mathrm{lim}}\nolimits K_ n) \to \mathop{\mathrm{lim}}\nolimits H^ m(U, K_ n) \to 0$

by Cohomology on Sites, Lemma 21.23.2. Since $U$ is affine and $K_ n$ is pseudo-coherent (and hence has quasi-coherent cohomology sheaves by Derived Categories of Spaces, Lemma 74.13.6) we see that $H^ m(U, K_ n) = H^ m(K_ n)(U)$ by Derived Categories of Schemes, Lemma 36.3.5. Thus we conclude that it suffices to show that $K_ n$ is bounded above independent of $n$.

Since $K_ n$ is pseudo-coherent we have $K_ n \in D^-(\mathcal{O}_{X_ n})$. Suppose that $a_ n$ is maximal such that $H^{a_ n}(K_ n)$ is nonzero. Of course $a_1 \leq a_2 \leq a_3 \leq \ldots$. Note that $H^{a_ n}(K_ n)$ is an $\mathcal{O}_{X_ n}$-module of finite presentation (Cohomology on Sites, Lemma 21.45.7). We have $H^{a_ n}(K_{n - 1}) = H^{a_ n}(K_ n) \otimes _{\mathcal{O}_{X_ n}} \mathcal{O}_{X_{n - 1}}$. Since $X_{n - 1} \to X_ n$ is a thickening, it follows from Nakayama's lemma (Algebra, Lemma 10.20.1) that if $H^{a_ n}(K_ n) \otimes _{\mathcal{O}_{X_ n}} \mathcal{O}_{X_{n - 1}}$ is zero, then $H^{a_ n}(K_ n)$ is zero too (argue by checking on stalks for example; small detail omitted). Thus $a_{n - 1} = a_ n$ for all $n$ and we conclude. $\square$

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