**Proof.**
Assume (1). We may assume $K$ is the complex

\[ \ldots \to 0 \to I^ a \to I^{a + 1} \to \ldots \to I^{b - 1} \to I^ b \to 0 \to \ldots \]

where $I^ i$ is a injective $R$-module for all $i \in \mathbf{Z}$. In this case we can compute the ext groups by the complex

\[ \ldots \to 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(N, I^ a) \to \ldots \to \mathop{\mathrm{Hom}}\nolimits _ R(N, I^ b) \to 0 \to \ldots \]

and we obtain (2). It is clear that (2) implies (3).

Assume (3) holds. Choose a nonzero map $R \to H^ n(K)$. Since $\mathop{\mathrm{Hom}}\nolimits _ R(R, -)$ is an exact functor, we see that $\mathop{\mathrm{Ext}}\nolimits ^ n_ R(R, K) = \mathop{\mathrm{Hom}}\nolimits _ R(R, H^ n(K)) = H^ n(K)$. We conclude that $H^ n(K)$ is zero for $n \not\in [a, b]$. In particular, $K$ is bounded below and we can choose a quasi-isomorphism

\[ K \to I^\bullet \]

with $I^ i$ injective for all $i \in \mathbf{Z}$ and $I^ i = 0$ for $i < a$. See Derived Categories, Lemma 13.15.5. Let $J = \mathop{\mathrm{Ker}}(I^ b \to I^{b + 1})$. Then $K$ is quasi-isomorphic to the complex

\[ \ldots \to 0 \to I^ a \to \ldots \to I^{b - 1} \to J \to 0 \to \ldots \]

Denote $K' = (I^ a \to \ldots \to I^{b - 1})$ the corresponding object of $D(R)$. We obtain a distinguished triangle

\[ J[-b] \to K \to K' \to J[1 - b] \]

in $D(R)$. Thus for every ideal $I \subset R$ an exact sequence

\[ \mathop{\mathrm{Ext}}\nolimits ^ b(R/I, K') \to \mathop{\mathrm{Ext}}\nolimits ^1(R/I, J) \to \mathop{\mathrm{Ext}}\nolimits ^{1 + b}(R/I, K) \]

By assumption the term on the right vanishes. By the implication (1) $\Rightarrow $ (2) the term on the left vanishes. Thus $J$ is a injective $R$-module by Lemma 15.55.4.
$\square$

## Comments (2)

Comment #4276 by Bogdan on

Comment #4441 by Johan on