Lemma 15.79.5. Let $R$ be a Noetherian regular ring of finite dimension. Then $R$ is a strong generator for the full subcategory $D_{perf}(R) \subset D(R)$ of perfect objects.

Proof. We will use that an object $K$ of $D(R)$ is perfect if and only if $K$ is bounded and has finite cohomology modules, see Lemma 15.74.14. Strong generators of triangulated categories are defined in Derived Categories, Definition 13.36.3. Let $d = \dim (R)$.

Let $K \in D_{perf}(R)$. We will show $K \in \langle R \rangle _{d + 1}$. By Algebra, Lemma 10.110.8 every finite $R$-module has projective dimension $\leq d$. We will show by induction on $0 \leq i \leq d$ that if $H^ n(K)$ has projective dimension $\leq i$ for all $n \in \mathbf{Z}$, then $K$ is in $\langle R \rangle _{i + 1}$.

Base case $i = 0$. In this case $H^ n(K)$ is a finite $R$-module of projective dimension $0$. In other words, each cohomology is a projective $R$-module. Thus $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(H^ n(K), H^ m(K)) = 0$ for all $i > 0$ and $m, n \in \mathbf{Z}$. By Derived Categories, Lemma 13.27.9 we find that $K$ is isomorphic to the direct sum of the shifts of its cohomology modules. Since each cohomology module is a finite projective $R$-module, it is a direct summand of a direct sum of copies of $R$. Hence by definition we see that $K$ is contained in $\langle R \rangle _1$.

Induction step. Assume the claim holds for $i < d$ and let $K \in D_{perf}(R)$ have the property that $H^ n(K)$ has projective dimension $\leq i + 1$ for all $n \in \mathbf{Z}$. Choose $a \leq b$ such that $H^ n(K)$ is zero for $n \not\in [a, b]$. For each $n \in [a, b]$ choose a surjection $F^ n \to H^ n(K)$ where $F^ n$ is a finite free $R$-module. Since $F^ n$ is projective, we can lift $F^ n \to H^ n(K)$ to a map $F^ n[-n] \to K$ in $D(R)$ (small detail omitted). Thus we obtain a morphism $\bigoplus _{a \leq n \leq b} F^ n[-n] \to K$ which is surjective on cohomology modules. Choose a distinguished triangle

$K' \to \bigoplus \nolimits _{a \leq n \leq b} F^ n[-n] \to K \to K'$

in $D(R)$. Of course, the object $K'$ is bounded and has finite cohomology modules. The long exact sequence of cohomology breaks into short exact sequences

$0 \to H^ n(K') \to F^ n \to H^ n(K) \to 0$

by the choices we made. By Algebra, Lemma 10.109.9 we see that the projective dimension of $H^ n(K')$ is $\leq \max (0, i)$. Thus $K' \in \langle R \rangle _{i + 1}$. By definition this means that $K$ is in $\langle R \rangle _{i + 1 + 1}$ as desired. $\square$

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).