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Let $X$ be a scheme of finite type over $\mathbb C$. One might be interested in morphisms in the derived category $D(X)$ of coherent sheaves on $X$, that are morphisms $f:E^\bullet\to F^\bullet$ of complexes of vector bundles. However, these are morphisms in $D(X)$, thus not so easy (for me) to handle. I know that when $X$ has enough locally frees, one can choose a representative of $f$ such that $f$ is an actual map of complexes. This was just my motivation for the following question:

When does $X$, i.e. Coh$(X)$, have enough locally frees?

Thanks!

Brenin
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    Check out exercise III.6.8 in Hartshorne which says that if $X$ is noetherian, integral, separated, and locall factorial then it has enough locally frees. Also check out this MO question: http://mathoverflow.net/questions/25122/are-schemes-that-have-enough-locally-frees-necessarily-separated. – Dori Bejleri Oct 27 '14 at 19:52
  • @DoriBejleri: Thanks, that helped! – Brenin Oct 28 '14 at 18:35

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When $X$ has enough locally free sheaves, one says that $X$ has the resolution property. With this keyword you will find lots of literature. It is an open problem if every (semi)separated scheme of finite type over a field has the resolution property; at least no counterexamples are known. A simple example which illustrates the separated assumption is the affine plane with doubled origin, since here every locally free sheaf is free.

Related: math.SE/849958

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Martin Brandenburg
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