The Stacks project

Email from Kovacs of 23/02/2018.

Lemma 15.97.6 (Kollár-Kovács). Let $I$ be an ideal of a Noetherian ring $A$. Let $K \in D(A)$. Set $K_ n = K \otimes _ A^\mathbf {L} A/I^ n$. Assume for all $i \in \mathbf{Z}$ we have

  1. $H^ i(K)$ is a finite $A$-module, and

  2. the system $H^ i(K_ n)$ satisfies Mittag-Leffler.

Then $\mathop{\mathrm{lim}}\nolimits H^ i(K)/I^ nH^ i(K)$ is equal to $\mathop{\mathrm{lim}}\nolimits H^ i(K_ n)$ for all $i \in \mathbf{Z}$.

Proof. Recall that $K^\wedge = R\mathop{\mathrm{lim}}\nolimits K_ n$ is the derived completion of $K$, see Proposition 15.94.2. By Lemma 15.94.4 we have $H^ i(K^\wedge ) = \mathop{\mathrm{lim}}\nolimits H^ i(K)/I^ nH^ i(K)$. By Lemma 15.87.4 we get short exact sequences

\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{i - 1}(K_ n) \to H^ i(K^\wedge ) \to \mathop{\mathrm{lim}}\nolimits H^ i(K_ n) \to 0 \]

The Mittag-Leffler condition guarantees that the left terms are zero (Lemma 15.87.1) and we conclude the lemma is true. $\square$

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Your email address will not be published.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



View Lemma 15.97.6 as pdf