The Stacks project

Proof. One implication was shown in Lemma 15.65.1. Assume for any filtered colimit $M = \mathop{\mathrm{colim}}\nolimits M_ i$ of $R$-modules we have $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^ n_ R(K, M_ i) = \mathop{\mathrm{Ext}}\nolimits ^ n_ R(K, M)$ for $n < -m$ and $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^{-m}_ R(K, M_ i) \to \mathop{\mathrm{Ext}}\nolimits ^{-m}_ R(K, M)$ is injective. We will show $K$ is $m$-pseudo-coherent.

Let $t$ be the maximal integer such that $H^ t(K)$ is nonzero. We will use induction on $t$. If $t < m$, then $K$ is $m$-pseudo-coherent by Lemma 15.64.7. If $t \geq m$, then since $\mathop{\mathrm{Hom}}\nolimits _ R(H^ t(K), M) = \mathop{\mathrm{Ext}}\nolimits ^{-t}_ R(K, M)$ we conclude that $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ R(H^ t(K), M_ i) \to \mathop{\mathrm{Hom}}\nolimits _ R(H^ t(K), M)$ is injective for any filtered colimit $M = \mathop{\mathrm{colim}}\nolimits M_ i$. This implies that $H^ t(K)$ is a finite $R$-module by Algebra, Lemma 10.11.1. Choose a finite free $R$-module $F$ and a surjection $F \to H^ t(K)$. We can lift this to a morphism $F[-t] \to K$ in $D(R)$ and choose a distinguished triangle

in $D(R)$. Then $H^ i(L) = 0$ for $i \geq t$. Moreover, the long exact sequence of $\mathop{\mathrm{Ext}}\nolimits $ associated to this distinguished triangle shows that $L$ inherts the assumption we made on $K$ by a small argument we omit. By induction on $t$ we conclude that $L$ is $m$-pseudo-coherent. Hence $K$ is $m$-pseudo-coherent by Lemma 15.64.2. $\square$