Lemma 15.77.1. Let $R$ be a ring. The full subcategory $D_{perf}(R) \subset D(R)$ of perfect objects is the smallest strictly full, saturated, triangulated subcategory containing $R = R[0]$. In other words $D_{perf}(R) = \langle R \rangle$. In particular, $R$ is a classical generator for $D_{perf}(R)$.

Proof. To see what the statement means, please look at Derived Categories, Definitions 13.6.1 and 13.36.3. It was shown in Lemmas 15.73.4 and 15.73.5 that $D_{perf}(R) \subset D(R)$ is a strictly full, saturated, triangulated subcategory of $D(R)$. Of course $R \in D_{perf}(R)$.

Recall that $\langle R \rangle = \bigcup \langle R \rangle _ n$. To finish the proof we will show that if $M \in D_{perf}(R)$ is represented by

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

with $M^ i$ finite projective, then $M \in \langle R \rangle _{b - a + 1}$. The proof is by induction on $b - a$. By definition $\langle R \rangle _1$ contains any finite projective $R$-module placed in any degree; this deals with the base case $b - a = 0$ of the induction. In general, we consider the distinguished triangle

$M_ b[-b] \to M^\bullet \to \sigma _{\leq b - 1}M^\bullet \to M_ b[-b + 1]$

By induction the truncated complex $\sigma _{\leq b - 1}M^\bullet$ is in $\langle R \rangle _{b - a}$ and $M_ b[-b]$ is in $\langle R \rangle _1$. Hence $M^\bullet \in \langle R \rangle _{b - a + 1}$ by definition. $\square$

## Comments (2)

Comment #5075 by eye on

The base case of the induction is $b-a=0$

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