# Comparing the definitions of perfect complexes on algebraic stacks and schemes

Let $X$ be an algebraic stack. In Perfect complexes on algebraic stacks (4.1) a perfect complex $P \in \mathsf{D}_{\mathrm qc}(X)$ is defined to be a complex such that, for any smooth morphism $\operatorname{Spec}(A) \to X$ (where $A$ is a commutative ring), the complex $\mathbf{R}\Gamma(X,P_{|\operatorname{Spec}(A)})$ is a perfect complex of $A$-modules.

On the other hand, if $X$ is just a scheme, then a complex $P \in \mathsf{D}_{\mathrm qc}(X)$ is called perfect if it is quasi-isomorphic to a bounded complex of locally free sheaves of finite rank.

Question: if $X$ is a scheme, how can I prove that the two definitions agree? Is there a reference with a proof of this fact?

• Thomason-Trobaugh? – Jason Starr Oct 10 '16 at 17:21
• Perhaps I have misunderstood, but you seem to be asking why the property of being a perfect complex is preserved by smooth pullbacks. Given a bounded complex of vector bundles on a scheme $X$, their pullback under any smooth affine map $Spec(A) \to X$ is a bounded complex of locally free $A$-modules. On the other hand, the property of being a bounded complex of vector bundles can be checked locally (even in the Zariski topology). Thus the two definitions agree. – Sam Gunningham Oct 10 '16 at 19:14
• I think at least part of the OP's question comes from the fact that the OP has the wrong definition of perfect complex in general. In general (as Leo Alonso says), a perfect complex is only \emph{locally} (in say the Zariski topology) quasi-iso. to a bounded complex of vector bundles. For "nice" schemes, it's a theorem (probably in Thomason-Trobaugh), that one gets a global quasi-iso. I learned about it from remark 1.7 of arxiv.org/pdf/math/0302304v2.pdf, which should give suitable references. – Daniel Pomerleano Oct 11 '16 at 9:52