This section is the dual of the section on projective dimension.
Definition 15.69.1. Let $R$ be a ring. Let $K$ be an object of $D(R)$. We say $K$ has finite injective dimension if $K$ can be represented by a finite complex of injective $R$-modules. We say $K$ has injective-amplitude in $[a, b]$ if $K$ is isomorphic to a complex with $I^ i$ an injective $R$-module for all $i \in \mathbf{Z}$.
\[ \ldots \to 0 \to I^ a \to I^{a + 1} \to \ldots \to I^{b - 1} \to I^ b \to 0 \to \ldots \]
Clearly, $K$ has bounded injective dimension if and only if $K$ has injective-amplitude in $[a, b]$ for some $a, b \in \mathbf{Z}$. Furthermore, if $K$ has bounded injective dimension, then $K$ is bounded. Here is the obligatory lemma.
Lemma 15.69.2. Let $R$ be a ring. Let $K$ be an object of $D(R)$. Let $a, b \in \mathbf{Z}$. The following are equivalent
$K$ has injective-amplitude in $[a, b]$,
$\mathop{\mathrm{Ext}}\nolimits ^ i_ R(N, K) = 0$ for all $R$-modules $N$ and all $i \not\in [a, b]$,
$\mathop{\mathrm{Ext}}\nolimits ^ i(R/I, K) = 0$ for all ideals $I \subset R$ and all $i \not\in [a, b]$.
Proof. Assume (1). We may assume $K$ is the complex
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
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
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
Denote $K' = (I^ a \to \ldots \to I^{b - 1})$ the corresponding object of $D(R)$. We obtain a distinguished triangle
in $D(R)$. Thus for every ideal $I \subset R$ an exact sequence
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. $\square$
Example 15.69.3. Let $R$ be a Dedekind domain. Then every nonzero ideal $I$ is a finite projective module, see Lemma 15.22.11. Thus $R/I$ has projective dimension $1$. Hence every $R$-module $M$ has injective dimension $\leq 1$ by Lemma 15.69.2. Thus $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) = 0$ for $i \geq 2$ and any pair of $R$-modules $M, N$. It follows that any object $K$ in $D^ b(R)$ is isomorphic to the direct sum of its cohomologies: $K \cong \bigoplus H^ i(K)[-i]$, see Derived Categories, Lemma 13.27.10.
Example 15.69.4. Let $k$ be a field and let $R$ be the ring of dual numbers over $k$, i.e., $R = k[x]/(x^2)$. Denote $\epsilon \in R$ the class of $x$. Let $M = R/(\epsilon )$. Then $M$ is quasi-isomorphic to the complex and $R$ is an injective $R$-module. However one usually does not consider $M$ to have finite injective dimension in this situation. This explains why we consider bounded (in both directions) complexes of injective modules in our definition of bounded injective dimension of objects of $D(R)$.
\[ \ldots \to R \xrightarrow {\epsilon } R \xrightarrow {\epsilon } R \]
Lemma 15.69.5. Let $R$ be a ring. Let $K \in D(R)$.
If $K$ is in $D^ b(R)$ and $H^ i(K)$ has finite injective dimension for all $i$, then $K$ has finite injective dimension.
If $K^\bullet $ represents $K$, is a bounded complex of $R$-modules, and $K^ i$ has finite injective dimension for all $i$, then $K$ has finite injective dimension.
Proof. Omitted. Hint: Apply the spectral sequences of Derived Categories, Lemma 13.21.3 to the functor $F = \mathop{\mathrm{Hom}}\nolimits _ R(N, -)$ to get a computation of $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(N, K)$ and use the criterion of Lemma 15.69.2. $\square$
Lemma 15.69.6. Let $R$ be a Noetherian ring. Let $I \subset R$ be an ideal contained in the Jacobson radical of $R$. Let $K \in D^+(R)$ have finite cohomology modules. Then the following are equivalent
$K$ has finite injective dimension, and
there exists a $b$ such that $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(R/J, K) = 0$ for $i > b$ and any ideal $J \supset I$.
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
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. $\square$
Lemma 15.69.7. Let $(R, \mathfrak m, \kappa )$ be a local Noetherian ring. Let $K \in D^+(R)$ have finite cohomology modules. Then the following are equivalent
$K$ has finite injective dimension, and
$\mathop{\mathrm{Ext}}\nolimits ^ i_ R(\kappa , K) = 0$ for $i \gg 0$.
Proof. This is a special case of Lemma 15.69.6. $\square$
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