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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.

flag	asked Feb 13 2012 at 20:13 David Steinberg 844●1●5●16

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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

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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

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The term you actually want is "essentially small" (equivalent to a small category). – Qiaochu Yuan Feb 14 2012 at 2:52

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out of curiosity: how does the Freyd-Mitchell embedding make life much easier? – Yosemite Sam Feb 14 2012 at 11:59

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S. Carnahan · Accepted Answer · 2012-02-14 02:09:09Z UTC

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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.

answered Feb 14 2012 at 2:09

S. Carnahan♦
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Thank you. Do you know of a reference for the proof? – David Steinberg Feb 14 2012 at 3:22

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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

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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

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

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