$\newcommand{\Perf}{\operatorname{Perf}}$This is a toy example that I want to understand, I will be grateful for any help. Given a ring $R$ and $A=\Perf(R)$ the category of perfect complexes over $R$ . Suppose that $m\in\Perf(R)$ and $B$ the smallest thick subcategory generated by $ m $. Let $C$ be the verdier quotient $ C=\Perf(R)/B$.

I want to understand $C$ concretely up to equivalence of triangulated categories.

Is it correct that $C$ Is equivalent to a triangulated subcategory $ D$ of $A$ where $d\in D$ iff the graded abelian groups $$ \operatorname{Hom}_A^n(m,d)=0$$ for any integer $n.$


1 Answer 1


No, not in general. This is true if you pass to "big" categories, so $\mathrm{Mod}(R)$ instead of Perf, and you take the smallest localizing subcategory containing $m$, but in general wrong at the level of small categories.

So what you can do is look at the full subcategory of $\mathrm{Mod}(R)$ of these objects, and take the compact objects therein (which need not be compact in $\mathrm{Mod}(R)$ !).

There are many counterexamples, such as $R= \mathbb Z$ and $m = \mathbb Z/p$. In that case the quotient is equivalent to $\mathrm{Perf}(\mathbb Z[1/p])$ (and the localization functor is base-change), and you can clearly see that no object of $\mathrm{Perf}(\mathbb Z)$ has $\mathbb Z[1/p]$ as its (graded) endomorphism ring.


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