**Proof.**
The implication (1) $\Rightarrow $ (2) is immediate. Assume (2). Say $H^ i(K) = 0$ for $i < a$. Then $\mathop{\mathrm{Ext}}\nolimits ^ i(M, K) = 0$ for $i < a$ and all $R$-modules $M$. Thus it suffices to show that $\mathop{\mathrm{Ext}}\nolimits ^ i(M, K) = 0$ for $i > b$ any finite $R$-module $M$, see Lemma 15.69.2. By Algebra, Lemma 10.62.1 the module $M$ has a finite filtration whose successive quotients are of the form $R/\mathfrak p$ where $\mathfrak p$ is a prime ideal. If $0 \to M_1 \to M \to M_2 \to 0$ is a short exact sequence and $\mathop{\mathrm{Ext}}\nolimits ^ i(M_ j, K) = 0$ for $i > b$ and $j = 1, 2$, then $\mathop{\mathrm{Ext}}\nolimits ^ i(M, K) = 0$ for $i > b$. Thus we may assume $M = R/\mathfrak p$. If $I \subset \mathfrak p$, then the vanishing follows from the assumption. If not, then choose $f \in I$, $f \not\in \mathfrak p$. Consider the short exact sequence

\[ 0 \to R/\mathfrak p \xrightarrow {f} R/\mathfrak p \to R/(\mathfrak p, f) \to 0 \]

The $R$-module $R/(\mathfrak p, f)$ has a filtration whose successive quotients are $R/\mathfrak q$ with $(\mathfrak p, f) \subset \mathfrak q$. Thus by Noetherian induction and the argument above we may assume the vanishing holds for $R/(\mathfrak p, f)$. On the other hand, the modules $E^ i = \mathop{\mathrm{Ext}}\nolimits ^ i(R/\mathfrak p, K)$ are finite by our assumption on $K$ (bounded below with finite cohomology modules), the spectral sequence (15.67.0.1), and Algebra, Lemma 10.71.9. Thus $E^ i$ for $i > b$ is a finite $R$-module such that $E^ i/fE^ i = 0$. We conclude by Nakayama's lemma (Algebra, Lemma 10.20.1) that $E^ i$ is zero.
$\square$

## Comments (2)

Comment #4278 by Bogdan on

Comment #4443 by Johan on