The Stacks project

Lemma 15.70.6. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime. Let $K \in D(R)$ be perfect. Set $d_ i = \dim _{\kappa (\mathfrak p)} H^ i(K \otimes _ R^\mathbf {L} \kappa (\mathfrak p))$. Then $d_ i < \infty $ and only a finite number are nonzero. Then there exists an $f \in R$, $f \not\in \mathfrak p$ and a complex

\[ \ldots \to 0 \to R_ f^{\oplus d_ a} \to R_ f^{\oplus d_{a + 1}} \to \ldots \to R_ f^{\oplus d_{b - 1}} \to R_ f^{\oplus d_ b} \to 0 \to \ldots \]

representing $K \otimes _ R^\mathbf {L} R_ f$ in $D(R_ f)$.

Proof. Observe that $K \otimes _ R^\mathbf {L} \kappa (\mathfrak p)$ is perfect as an object of $D(\kappa (\mathfrak p))$, see Lemma 15.69.9. Hence only a finite number of $d_ i$ are nonzero and they are all finite. Applying Lemma 15.70.5 we get a complex representing $K$ having the desired shape over the local ring $R_\mathfrak p$. We have $R_\mathfrak p = \mathop{\mathrm{colim}}\nolimits R_ f$ for $f \in R$, $f \not\in \mathfrak p$ (Algebra, Lemma 10.9.9). We conclude by Lemma 15.69.16. Some details omitted. $\square$


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Your email address will not be published.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