
## 15.65 Projective dimension

We defined the projective dimension of a module in Algebra, Definition 10.108.2.

Definition 15.65.1. 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 finite complex of projective modules. We say $K$ as 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 bounded projective dimension if and only if $K$ has projective-amplitude in $[a, b]$ for some $a, b \in \mathbf{Z}$. Furthermore, if $K$ has bounded projective dimension, then $K$ is bounded. Here is the obligatory lemma.

Lemma 15.65.2. 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 $i \not\in [-b, -a]$.

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 that $H^ n(K)$ is zero for $n \not\in [a, b]$. In particular, $K$ is bounded above and we can choose a quasi-isomorphism

$P^\bullet \to K$

with $P^ i$ projective (for example free) for all $i \in \mathbf{Z}$ and $P^ i = 0$ for $i > b$. See Derived Categories, Lemma 13.16.5. Let $Q = \mathop{\mathrm{Coker}}(P^{a - 1} \to P^ a)$. Then $K$ is quasi-isomorphic to the complex

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

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

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

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 10.76.2. $\square$

Example 15.65.3. 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 10.108.2. This explains why we consider bounded (in both directions) complexes of projective modules in our definition of bounded projective dimension of objects of $D(R)$.

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