The Stacks project

Lemma 15.102.4. Let $R$ be a Noetherian local ring. Let $I \subset R$ be an ideal and let $E$ be a nonzero module over $R/I$. If $R/I$ has finite projective dimension and $E$ has finite projective dimension over $R/I$, then $E$ has finite projective dimension over $R$ and

\[ \text{pd}_ R(E) = \text{pd}_ R(R/I) + \text{pd}_{R/I}(E) \]

Proof. We will use that, for a finite module, having finite projective dimension over $R$, resp. $R/I$ is the same as being a perfect module, see discussion following Definition 15.74.1. We see that $E$ has finite projective dimension over $R$ by Lemma 15.74.7. Thus we can apply Auslander-Buchsbaum (Algebra, Proposition 10.111.1) to see that

\[ \text{pd}_ R(E) + \text{depth}(E) = \text{depth}(R),\quad \text{pd}_{R/I}(E) + \text{depth}(E) = \text{depth}(R/I), \]

and

\[ \text{pd}_ R(R/I) + \text{depth}(R/I) = \text{depth}(R) \]

Note that in the first equation we take the depth of $E$ as an $R$-module and in the second as an $R/I$-module. However these depths are the same (this is trivial but also follows from Algebra, Lemma 10.72.11). This concludes the proof. $\square$

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Your email address will not be published.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 0929. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 15.102.4 as pdf