Is the bounded derived category of coherent sheaves of a variety a small category?

The question is in the title.

I am trying to apply the Mitchell (Freyd-Mitchell?) embedding theorem, which states that for every small abelian category $A$, there exists a ring $R$ such that A embeds into the category $R$-mod. The derived category is not abelian, of course, but I have a particular subcategory that is abelian, and life would be easiest if the derived category was smal, so that the subcategory was small and abelian.

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Of course it is, you can easily find a (transfinite) upper bound for the 'number' of objects – Fernando Muro Feb 13 2012 at 20:18
Fernando -- there may be an issue there, similar to the fact that the category of finite-dimensional vector spaces over a field $k$ is not itself small but is equivalent to the small category with $k^n$'s as objects. – algori Feb 13 2012 at 20:33
Presumably small should be understood up to equivalence. I agree with Fernando, although writing down the details would entail some work: Choose a finite affine cover $\{Spec A_i\}$. Then a bounded complex of coherent sheaves is given by a collection of finitely presented $A_i$ modules $M_i^\bullet$ plus patching data. A morphism is ... – Donu Arapura Feb 14 2012 at 0:03
The term you actually want is "essentially small" (equivalent to a small category). – Qiaochu Yuan Feb 14 2012 at 2:52
out of curiosity: how does the Freyd-Mitchell embedding make life much easier? – Yosemite Sam Feb 14 2012 at 11:59

The answer to your question really depends on what you mean by the word "the". An unhelpful answer is that the coherent sheaves over any variety form a proper class (hence "no"). A more useful answer is (as mentioned in the comments) that there exists a small category that is equivalent to any category that can be reasonably called the bounded derived category of coherent sheaves (hence "yes").

Furthermore, the construction of such a category can be accomplished without the use of replacement. In particular, the category lives in the same ZC universe (i.e., $V_\alpha$ for $\alpha$ a not-necessarily-inaccessible limit ordinal greater than $\omega$ - see e.g., Wikipedia or the set theory section of the Stacks project) as the defining field.

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Thank you. Do you know of a reference for the proof? – David Steinberg Feb 14 2012 at 3:22
No, but it is straightforward. If you set $\kappa$ to be the smallest infinite cardinal containing your base field, then any variety is covered by finitely many spectra of finitely generated rings, also of cardinality $\kappa$. Each complex of coherent sheaves is then encoded by a finite collection of finitely generated modules and arrows describing gluing data and differentials. Up to isom. all modules can be realized by subquotients of a fixed infinite direct sum of copies of a ring (size $\kappa$), and the arrows are maps of such sets ($\leq 2^{2^\kappa}$ total choices). Done. – S. Carnahan Feb 14 2012 at 6:56
Oops, not quite done. For any pair of complexes of size bounded by $\kappa$, the set of ordinary homs is bounded by $2^\kappa$. Localization yields equiv. classes of finite zig-zags of maps, so the hom group between two objects in the derived category is also bounded in size by $2^\kappa$. Total maps in category is then bounded by $2^{2^\kappa}$. – S. Carnahan Feb 14 2012 at 8:14