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I am wondering whether, if a triangulated category $\mathcal{D}$ has a full strong exceptional collection (infinite), it is triangle-equivalent to the bounded derived category of finitely generated ...

For a smooth projective variety $X$, Serre duality gives an exact autoequivalence on the derived category : $$ S_X : D^\flat(X) \to D^\flat(X), \hspace{3em} S_X(-) = - \otimes \omega_X[\dim X] $$ ...

[ALEXEY ELAGIN AND VALERY A. LUNTS, p.4.] Recall that triangulated category $\text{Perf}(A)$ is defined as the full thick triangulated subcategory of $D(A)$ generated by the dg $A$-module $A$. [...

Notations and background. Let $R$ be a commutative noetherian local ring and let $D(R)$ denote the derived category of the category of R-modules. A strictly perfect complex on $R$ is a bounded complex ...