The Stacks project

This is a variant of [Lemma 3.5.4, BS]

Lemma 24.35.3. Let $R$ be a ring. Let $f_1, \ldots , f_ r \in R$. Let $K_ n$ be the Koszul complex on $f_1^ n, \ldots , f_ r^ n$ viewed as a differential graded $R$-algebra. Let $(M_ n)$ be an object of $D(\mathbf{N}, (K_ n))$. Then for any $t \geq 1$ we have

\[ R\mathop{\mathrm{lim}}\nolimits (M_ n \otimes _ R^\mathbf {L} K_ t) = R\mathop{\mathrm{lim}}\nolimits (M_ n \otimes _{K_ n}^\mathbf {L} K_ t) \]

in $D(R)$.

Proof. We fix $t \geq 1$. For $n \geq t$ let us denote ${}_ nK_ t$ the differential graded $R$-algebra $K_ t$ viewed as a left differential graded $K_ n$-module. Observe that

\[ M_ n \otimes _ R^\mathbf {L} K_ t = M_ n \otimes _{K_ n}^\mathbf {L} (K_ n \otimes _ R^\mathbf {L} K_ t) = M_ n \otimes _{K_ n}^\mathbf {L} (K_ n \otimes _ R K_ t) \]

Hence by Lemma 24.35.2 it suffices to show that $({}_ nK_ t)$ and $(K_ n \otimes _ R K_ t)$ are pro-isomorphic in the derived category. The multiplication maps

\[ K_ n \otimes _ R K_ t \longrightarrow {}_ nK_ t \]

are maps of left differential graded $K_ n$-modules. Thus to finish the proof it suffices to show that for all $n \geq 1$ there exists an $N > n$ and a map

\[ {}_ NK_ t \longrightarrow {}_ NK_ n \otimes _ R K_ t \]

in $D(K_ N^{opp}, \text{d})$ whose composition with the multiplication map is the transition map (in either direction). This is done in Divided Power Algebra, Lemma 23.12.4 by an explicit construction. $\square$

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Your email address will not be published.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 0GZJ. Beware of the difference between the letter 'O' and the digit '0'.