Lemma 15.79.2. Let $R$ be a Noetherian ring. If $R$ is a strong generator for $D_{perf}(R)$, then $R$ is regular of finite dimension.

**Proof.**
Assume $D_{perf}(R) = \langle R \rangle _ n$ for some $n \geq 1$. For any finite $R$-module $M$ we can choose a complex

of finite free $R$-modules with $H^ i(P) = 0$ for $i = -n, \ldots , - 1$ and $M \cong \mathop{\mathrm{Coker}}(d^{-1})$. Note that $P$ is in $D_{perf}(R)$. For any $R$-module $N$ we can compute $\mathop{\mathrm{Ext}}\nolimits ^ n_ R(M, N)$ the finite free resolution $P$ of $M$, see Algebra, Section 10.71 and compare with Derived Categories, Section 13.27. In particular, the sequence above defines an element

and for any element $\overline{\xi }$ in $\mathop{\mathrm{Ext}}\nolimits ^ n_ R(M, N)$ there is a $R$-module map $\varphi : \mathop{\mathrm{Coker}}(d^{-n - 1}) \to N$ such that $\varphi $ maps $\xi $ to $\overline{\xi }$. For $j = 1, \ldots , n - 1$ consider the complexes

with $\mathop{\mathrm{Coker}}(d^{-n - 1})$ in degree $-n$ and $P^ t$ in degree $t$. We also set $K_ n = \mathop{\mathrm{Coker}}(d^{-n - 1})[n]$. Then we have maps

which induce vanishing maps on cohomology. By Lemma 15.79.1 since $P \in D_{perf}(R) = \langle R \rangle _ n$ we find that the composition of this maps is zero in $D(R)$. Since $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(P, K_ n) = \mathop{\mathrm{Hom}}\nolimits _{K(R)}(P, K_ n)$ by Derived Categories, Lemma 13.19.8 we conclude $\xi = 0$. Hence $\mathop{\mathrm{Ext}}\nolimits ^ n_ R(M, N) = 0$ for all $R$-modules $N$, see discussion above. It follows that $M$ has projective dimension $\leq n - 1$ by Algebra, Lemma 10.109.8. Since this holds for all finite $R$-modules $M$ we conclude that $R$ has finite global dimension, see Algebra, Lemma 10.109.12. We finally conclude by Algebra, Lemma 10.110.8. $\square$

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