Lemma 15.91.23 (0924)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 15: More on Algebra
Section 15.91: Derived Completion
Lemma 15.91.23 (cite)

Lemma 15.91.23. Let $A \to B$ be a ring map. Let $I \subset A$ be an ideal. The inverse image of $D_{comp}(A, I)$ under the restriction functor $D(B) \to D(A)$ is $D_{comp}(B, IB)$.

Proof. Using Lemma 15.91.2 we see that $L \in D(B)$ is in $D_{comp}(B, IB)$ if and only if $T(L, f)$ is zero for every local section $f \in I$. Observe that the cohomology of $T(L, f)$ is computed in the category of abelian groups, so it doesn't matter whether we think of $f$ as an element of $A$ or take the image of $f$ in $B$. The lemma follows immediately from this and the definition of derived complete objects. $\square$

Comments (4)

Comment #6421 by Peng DU on July 19, 2021 at 12:52

Need add "in" after "is" in the statement.

Comment #6426 by Johan on July 19, 2021 at 14:42

No, it actually is equal.

Comment #7929 by Peng Du on December 17, 2022 at 11:18

I'm wondering if being derived complete satisfies faithful flat descent: if $A\to B$ is faithful flat, $I$ a finitely generated ideal in $A$ , does the left adjoint to the restriction functor have similar property? That is, for $K\in D(A)$ , we have $K\in D_{comp}(A, I)$ iff $K\otimes_A^{\bf L}B\in D_{comp}(B, IB)$ ?

I'm mainly interested in the case when $B=\prod_{i=1}^{n}A_{f_i}$ , where the $f_i$ 's generates the unit ideal in $A$ .

(BTW, is there a way for me to be noticed if a comment is responded?)

Comment #8175 by Aise Johan de Jong on March 01, 2023 at 11:54

No (and no).

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