Lemma 37.68.1. Let $i : X \to X'$ be a finite order thickening of schemes. Let $K' \in D(\mathcal{O}_{X'})$ be an object such that $K = Li^*K'$ is pseudo-coherent. Then $K'$ is pseudo-coherent.

Proof. We first prove $K'$ has quasi-coherent cohomology sheaves. To do this, we may reduce to the case of a first order thickening, see Section 37.2. Let $\mathcal{I} \subset \mathcal{O}_{X'}$ be the quasi-coherent sheaf of ideals cutting out $X$. Tensoring the short exact sequence

$0 \to \mathcal{I} \to \mathcal{O}_{X'} \to i_*\mathcal{O}_ X \to 0$

with $K'$ we obtain a distinguished triangle

$K' \otimes _{\mathcal{O}_{X'}}^\mathbf {L} \mathcal{I} \to K' \to K' \otimes _{\mathcal{O}_{X'}}^\mathbf {L} i_*\mathcal{O}_ X \to (K' \otimes _{\mathcal{O}_{X'}}^\mathbf {L} \mathcal{I})[1]$

Since $i_* = Ri_*$ and since we may view $\mathcal{I}$ as a quasi-coherent $\mathcal{O}_ X$-module (as we have a first order thickening) we may rewrite this as

$i_*(K \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{I}) \to K' \to i_*K \to i_*(K \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{I})[1]$

Please use Cohomology, Lemma 20.51.4 to identify the terms. Since $K$ is in $D_\mathit{QCoh}(\mathcal{O}_ X)$ we conclude that $K'$ is in $D_\mathit{QCoh}(\mathcal{O}_{X'})$; this uses Derived Categories of Schemes, Lemmas 36.10.1, 36.3.9, and 36.4.1.

Assume $K'$ is in $D_\mathit{QCoh}(\mathcal{O}_{X'})$. The question is local on $X'$ hence we may assume $X'$ is affine. Say $X' = \mathop{\mathrm{Spec}}(A')$ and $X = \mathop{\mathrm{Spec}}(A)$ with $A = A'/I$ and $I$ nilpotent. Then $K'$ comes from an object $M' \in D(A')$, see Derived Categories of Schemes, Lemma 36.3.5. Thus $M = M' \otimes _{A'}^\mathbf {L} A$ is a pseudo-coherent object of $D(A)$ by Derived Categories of Schemes, Lemma 36.10.2 and our assumption on $K$. Hence we can represent $M$ by a bounded above complex of finite free $A$-modules $E^\bullet$, see More on Algebra, Lemma 15.64.5. By More on Algebra, Lemma 15.75.3 we conclude that $M'$ is pseudo-coherent as desired. $\square$

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