We defined the projective dimension of a module in Algebra, Definition 10.109.2.
Definition 15.68.1. Let $R$ be a ring. Let $K$ be an object of $D(R)$. We say $K$ has finite projective dimension if $K$ can be represented by a bounded complex of projective modules. We say $K$ has projective-amplitude in $[a, b]$ if $K$ is quasi-isomorphic to a complex where $P^ i$ is a projective $R$-module for all $i \in \mathbf{Z}$.
\[ \ldots \to 0 \to P^ a \to P^{a + 1} \to \ldots \to P^{b - 1} \to P^ b \to 0 \to \ldots \]
Clearly, $K$ has finite projective dimension if and only if $K$ has projective-amplitude in $[a, b]$ for some $a, b \in \mathbf{Z}$. Furthermore, if $K$ has finite projective dimension, then $K$ is bounded. Here is a lemma to detect such objects of $D(R)$.
Lemma 15.68.2. 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 projective-amplitude in $[a, b]$,
$\mathop{\mathrm{Ext}}\nolimits ^ i_ R(K, N) = 0$ for all $R$-modules $N$ and all $i \not\in [-b, -a]$,
$H^ n(K) = 0$ for $n > b$ and $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(K, N) = 0$ for all $R$-modules $N$ and all $i > -a$, and
$H^ n(K) = 0$ for $n \not\in [a - 1, b]$ and $\mathop{\mathrm{Ext}}\nolimits ^{-a + 1}_ R(K, N) = 0$ for all $R$-modules $N$.
Proof. Assume (1). We may assume $K$ is the complex
where $P^ i$ is a projective $R$-module for all $i \in \mathbf{Z}$. In this case we can compute the ext groups by the complex
and we obtain (2).
Assume (2) holds. Choose an injection $H^ n(K) \to I$ where $I$ is an injective $R$-module. Since $\mathop{\mathrm{Hom}}\nolimits _ R(-, I)$ is an exact functor, we see that $\mathop{\mathrm{Ext}}\nolimits ^{-n}(K, I) = \mathop{\mathrm{Hom}}\nolimits _ R(H^ n(K), I)$. We conclude in particular that $H^ n(K)$ is zero for $n > b$. Thus (2) implies (3).
By the same argument as in (2) implies (3) gives that (3) implies (4).
Assume (4). The same argument as in (2) implies (3) shows that $H^{a - 1}(K) = 0$, i.e., we have $H^ i(K) = 0$ unless $i \in [a, b]$. In particular, $K$ is bounded above and we can choose a a complex $P^\bullet $ representing $K$ with $P^ i$ projective (for example free) for all $i \in \mathbf{Z}$ and $P^ i = 0$ for $i > b$. See Derived Categories, Lemma 13.15.4. Let $Q = \mathop{\mathrm{Coker}}(P^{a - 1} \to P^ a)$. Then $K$ is quasi-isomorphic to the complex
as $H^ i(K) = 0$ for $i < a$. Denote $K' = (P^{a + 1} \to \ldots \to P^ b)$ the corresponding object of $D(R)$. We obtain a distinguished triangle
in $D(R)$. Thus for every $R$-module $N$ an exact sequence
By assumption the term on the right vanishes. By the implication (1) $\Rightarrow $ (2) the term on the left vanishes. Thus $Q$ is a projective $R$-module by Algebra, Lemma 10.77.2. Hence (1) holds and the proof is complete. $\square$
Example 15.68.3. 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 but $M$ does not have finite projective dimension as defined in Algebra, Definition 10.109.2. This explains why we consider bounded (in both directions) complexes of projective modules in our definition of finite projective dimension of objects of $D(R)$.
\[ R \xrightarrow {\epsilon } R \xrightarrow {\epsilon } R \to \ldots \]
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #7927 by Felipe Zaldivar on
Comment #8174 by Aise Johan de Jong on