# Simple example of a perfect complex not isomorphic to a strictly perfect complex?

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").
• If I recall correctly you need a scheme without an ample family of line bundles, and this rules out all quasi-projective examples. Jan 7 '17 at 18:25
• 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. Jan 7 '17 at 18:26
• oh shoot. Then let me revise. Jan 7 '17 at 18:44
• 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. Jan 7 '17 at 21:11