Proposition 15.94.2. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. The functor which sends $K \in D(A)$ to the derived limit $K' = R\mathop{\mathrm{lim}}\nolimits ( K \otimes _ A^\mathbf {L} A/I^ n )$ is the left adjoint to the inclusion functor $D_{comp}(A) \to D(A)$ constructed in Lemma 15.91.10.
Proof. Say $(f_1, \ldots , f_ r) = I$ and let $K_ n^\bullet $ be the Koszul complex with respect to $f_1^ n, \ldots , f_ r^ n$. By Lemma 15.91.18 it suffices to prove that
\[ R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} K_ n^\bullet ) = R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} A/(f_1^ n, \ldots , f_ r^ n) ) = R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} A/I^ n ). \]
By Lemma 15.94.1 the pro-objects $\{ K_ n^\bullet \} $ and $\{ A/(f_1^ n, \ldots , f_ r^ n)\} $ of $D(A)$ are isomorphic. It is clear that the pro-objects $\{ A/(f_1^ n, \ldots , f_ r^ n)\} $ and $\{ A/I^ n\} $ are isomorphic. Thus the map from left to right is an isomorphism by Lemma 15.87.12. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)