Lemma 15.96.4. Let $A$ be a ring, let $\mathfrak p \subset A$ be a prime ideal, and let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ be a bounded complex of finite free $A$-modules. If $H^ i(M^\bullet )_\mathfrak p$ is free for all $i$, then $I_ i(M^\bullet , f)_\mathfrak p$ is a principal ideal and in fact generated by a power of $f$ for all $i$.

Proof. We may assume $A$ is local with maximal ideal $\mathfrak p$ by Lemma 15.96.3. We may also replace $M^\bullet$ with a quasi-isomorphic complex by Lemma 15.96.1. By our assumption on the freeness of cohomology modules we see that $M^\bullet$ is quasi-isomorphic to the complex whose term in degree $i$ is $H^ i(M^\bullet )$ with vanishing differentials, see for example Derived Categories, Lemma 13.27.9. In other words, we may assume the differentials in the complex $M^\bullet$ are all zero. In this case it is clear that $I_ i(M^\bullet , f) = (f^{r_ i})$ is principal. $\square$

## 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Your email address will not be published.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).

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 0GSV. Beware of the difference between the letter 'O' and the digit '0'.