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

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

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. $\square$

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