# Verdier localisation

$$\newcommand{\Perf}{\operatorname{Perf}}$$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 $$n.$$

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.