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Mar 10, 2015 at 22:04 comment added Fernando Muro I now realize of the absurdity of my second comments and the following one.
Mar 10, 2015 at 16:52 comment added Leo Alonso @DylanWilson post this as an answer. This is exactly what I would say.
Mar 10, 2015 at 16:05 comment added Dylan Wilson (and the reference for my entire comment, almost verbatim, is Thomason's classification of triangulated subcategories paper)
Mar 10, 2015 at 16:01 comment added Dylan Wilson Here's something that is true: On any scheme with an ample family of line bundles (e.g. quasi-projective, or affine, or separated regular Noetherian) every perfect complex is quasi-isomorphic to a strictly perfect complex. I'm not sure if that's what you meant by "is" in your claim, but the reference is SGA6 II.2.2.8 or Thomason-Trobaugh 2.3.1.
Mar 10, 2015 at 16:00 comment added Fernando Muro @ZhaotingWei, I wasn't thinking of gluing, just on the fact that being trivial and being locally free are local properties (the second one for obvious reasons).
Mar 10, 2015 at 14:05 comment added Zhaoting Wei @FernandoMuro Maybe that's not true, at least not so obvious, since for open set $U$, $V$, the complex of free sheaves $\mathcal{E}^{\bullet}_U$ and $\mathcal{E}^{\bullet}_V$ are quite different. Indeed they are quasi-isomorphic on $U\cap V$ but this does not guarantee that they glue together into a complex of locally free sheaves on $U\cup V$.
Mar 10, 2015 at 10:16 comment added Fernando Muro I meant perfect $\Rightarrow$ strictly perfect.
Mar 10, 2015 at 6:48 comment added Fernando Muro It looks like if strictly perfect implies perfect on quasi-compact schemes on the nose, and affine schemes are quasi-compact. Please, correct me itf I'm missing something.
Mar 10, 2015 at 2:43 comment added Zhaoting Wei Thank you for pointing out! That is a typo and I have made correction in the definition of strictly perfect complex.
Mar 10, 2015 at 2:42 history edited Zhaoting Wei CC BY-SA 3.0
In the definition of strictly perfect complex, "free sheaves" has been replaced by "locally free sheaves".
Mar 9, 2015 at 23:26 comment added Steven Landsburg I might be mistaken, but I'd have thought that the usual definition of "strictly perfect" (as found in, say, SGA6) would replace your "complex of finite rank free sheaves" with a complex of finite rank locally free sheaves. This would rule out Fernando's counterexample.
Mar 9, 2015 at 22:40 comment added Fernando Muro Your statement is false in general, a finitely generated non-free projective module concentrated in a fixed degree is a counterexample.
Mar 9, 2015 at 22:12 history asked Zhaoting Wei CC BY-SA 3.0