Lemma 37.64.3. Let $(R, I)$ be a pair consisting of a ring and an ideal $I$ contained in the Jacobson radical. Set $S = \mathop{\mathrm{Spec}}(R)$ and $S_0 = \mathop{\mathrm{Spec}}(R/I)$. Let $f : X \to S$ be proper, flat, and of finite presentation. Denote $X_0 = S_0 \times _ S X$. Let $E \in D(\mathcal{O}_ X)$ be pseudo-coherent. If the derived restriction $E_0$ of $E$ to $X_0$ is $S_0$-perfect, then $E$ is $S$-perfect.

**Proof.**
Choose a finite affine open covering $X = U_1 \cup \ldots \cup U_ n$. For each $i$ we can choose a closed immersion $U_ i \to \mathbf{A}^{d_ i}_ S$. Set $U_{i, 0} = S_0 \times _ S U_ i$. For each $i$ the complex $E_0|_{U_{i, 0}}$ has tor amplitude in $[a_ i, b_ i]$ for some $a_ i, b_ i \in \mathbf{Z}$. Let $x \in X$ be a point. We will show that the tor amplitude of $E_ x$ over $R$ is in $[a_ i - d_ i, b_ i]$ for some $i$. This will finish the proof as the tor amplitude can be read off from the stalks by Cohomology, Lemma 20.45.5.

Since $f$ is proper $f(\overline{\{ x\} })$ is a closed subset of $S$. Since $I$ is contained in the Jacobson radical, we see that $f(\overline{\{ x\} })$ meeting the closed subset $S_0 \subset S$. Hence there is a specialization $x \leadsto x_0$ with $x_0 \in X_0$. Pick an $i$ with $x_0 \in U_ i$, so $x_0 \in U_{i, 0}$. We will fix $i$ for the rest of the proof. Write $U_ i = \mathop{\mathrm{Spec}}(A)$. Then $A$ is a flat, finitely presented $R$-algebra which is a quotient of a polynomial $R$-algebra in $d_ i$-variables. The restriction $E|_{U_ i}$ corresponds (by Derived Categories of Schemes, Lemma 36.3.5 and 36.10.2) to a pseudo-coherent object $K$ of $D(A)$. Observe that $E_0$ corresponds to $K \otimes _ A^\mathbf {L} A/IA$. Let $\mathfrak q \subset \mathfrak q_0 \subset A$ be the prime ideals corresponding to $x \leadsto x_0$. Then $E_ x = K_{\mathfrak q}$ and $K_{\mathfrak q}$ is a localization of $K_{\mathfrak q_0}$. Hence it suffices to show that $K_{\mathfrak q_0}$ has tor amplitude in $[a_ i - d_ i, b_ i]$ as a complex of $R$-modules. Let $I \subset \mathfrak p_0 \subset R$ be the prime ideal corresponding to $f(x_0)$. Then we have

the second equality because $R \to A$ is flat. By our choice of $a_ i, b_ i$ this complex has cohomology only in degrees in the interval $[a_ i, b_ i]$. Thus we may finally apply More on Algebra, Lemma 15.82.9 to $R \to A$, $\mathfrak q_0$, $\mathfrak p_0$ and $K$ to conclude. $\square$

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