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. $\square$

Comment #6421 by Peng DU on

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

Comment #7929 by Peng Du on

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?)

There are also:

• 14 comment(s) on Section 15.91: Derived Completion

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).