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15.68 Projective dimension

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

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

where $P^ i$ is a projective $R$-module for all $i \in \mathbf{Z}$.

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

  1. $K$ has projective-amplitude in $[a, b]$,

  2. $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(K, N) = 0$ for all $R$-modules $N$ and all $i \not\in [-b, -a]$,

  3. $H^ n(K) = 0$ for $n > b$ and $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(K, N) = 0$ for all $R$-modules $N$ and all $i > -a$, and

  4. $H^ n(K) = 0$ 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

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

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

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

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

\[ \ldots \to 0 \to Q \to P^{a + 1} \to \ldots \to P^ b \to 0 \to \ldots \]

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

\[ K' \to K \to Q[-a] \to K'[1] \]

in $D(R)$. Thus for every $R$-module $N$ an exact sequence

\[ \mathop{\mathrm{Ext}}\nolimits ^{-a}(K', N) \to \mathop{\mathrm{Ext}}\nolimits ^1(Q, N) \to \mathop{\mathrm{Ext}}\nolimits ^{1 - a}(K, N) \]

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

\[ R \xrightarrow {\epsilon } R \xrightarrow {\epsilon } R \to \ldots \]

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 $D(R)$.

Comments (2)

Comment #7927 by Felipe Zaldivar on

On Definition 15.68.1, it says "We say K as projective-amplitude" , it must be: " We say K has projective-amplitude" (change "as" to "has")

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