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I'm looking for the simplest possible example (one that's easy to remember) for the situation described in the title. More precisely I'm looking for the following example:

A (probably has to be singular) algebraic surface $X$ (2-dimensional, reduced, integral, finite type over $k$ algebraically closed) and a complex of coherent sheaves $\mathcal{F}^\bullet$ on $X$ whose image in $D(X)$ is perfect but not quasi-isomorphic to a strictly perfect complex. (Bonus points for a simple toric surface example).


Definitions:

  1. A strictly perfect complex is an object in the derived category $D(X)$ isomorphic to a bounded complex of vector bundles
  2. A perfect complex is an object in the derived category $M \in D(X)$ satisfying that for every point $x \in X$ there's a open zariski neighborhood $x \in U$ over which $M_U \in D(U)$ is perfect ("locally strictly perfect").
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    $\begingroup$ If I recall correctly you need a scheme without an ample family of line bundles, and this rules out all quasi-projective examples. $\endgroup$ – Denis Nardin Jan 7 '17 at 18:25
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    $\begingroup$ I am pretty sure that on every quasi-projective scheme over a field, e.g., your curve above, every perfect object is strictly perfect. This should be in SGA 6, but you might first look in Thomason-Trobaugh. $\endgroup$ – Jason Starr Jan 7 '17 at 18:26
  • $\begingroup$ oh shoot. Then let me revise. $\endgroup$ – Saal Hardali Jan 7 '17 at 18:44
  • $\begingroup$ Any such example will fail to have the "resolution property". If memory serves, there is an example of a (non-separated) scheme failing the resolution property in the article of Edidin, Hassett, Kresch, and Vistoli. $\endgroup$ – Jason Starr Jan 7 '17 at 21:11
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Perhaps it is silly to post this as an answer. My answer to your other question also answers this question.
Can the homological dimension of a coherent sheaf explode along a formal deformation? (is the resolution property hereditary for formal deformations?)
That example is smooth, integral, and 2-dimensional over a field. However, it is not separated.

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  • $\begingroup$ Thanks for reminding me about this! I was very worried about this issue once. Now that I know that there's a dichotomy (ala the other question) I'm much more relaxed. $\endgroup$ – Saal Hardali May 17 '17 at 13:09

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