Lemma 15.78.1. 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 15.74.4 and 15.74.5 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

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

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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #5075 by eye on

Comment #5288 by Johan on

There are also: