The Stacks project

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

\begin{align*} K \otimes _ R^\mathbf {L} \kappa (\mathfrak p_0) & = (K \otimes _ R^\mathbf {L} R/I) \otimes _{R/I}^\mathbf {L} \kappa (\mathfrak p_0) \\ & = (K \otimes _ A^\mathbf {L} A/IA) \otimes _{R/I}^\mathbf {L} \kappa (\mathfrak p_0) \end{align*}

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DJZ. Beware of the difference between the letter 'O' and the digit '0'.