Lemma 15.77.1. Let $R$ be a ring and let $\mathfrak p \subset R$ be a prime. Let $K$ be pseudo-coherent and bounded below. Set $d_ i = \dim _{\kappa (\mathfrak p)} H^ i(K \otimes _ R^\mathbf {L} \kappa (\mathfrak p))$. If there exists an $a \in \mathbf{Z}$ such that $d_ i = 0$ for $i < a$, 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 $D(R_ f)$. In particular $K \otimes _ R^\mathbf {L} R_ f$ is perfect.

Proof. After decreasing $a$ we may assume that also $H^ i(K^\bullet ) = 0$ for $i < a$. By Lemma 15.76.4 after replacing $R$ by $R_ f$ for some $f \in R$, $f \not\in \mathfrak p$ we can write $K^\bullet = \tau _{\leq a - 1}K^\bullet \oplus \tau _{\geq a}K^\bullet$ in $D(R)$ with $\tau _{\geq a}K^\bullet$ perfect. Since $H^ i(K^\bullet ) = 0$ for $i < a$ we see that $\tau _{\leq a - 1}K^\bullet = 0$ in $D(R)$. Hence $K^\bullet$ is perfect. Then we can conclude using Lemma 15.75.6. $\square$

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).