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I can prove a certain statement for any scheme that can be presented as the limit of an essentially affine (filtering) projective system of smooth varieties over a perfect field such the connecting morphisms are dominant. In this text I only treat schemes that are excellent separated of finite Krull dimension. So, I have the following questions.

  1. Is there a shorter description of schemes that can be presented as limits of this sort (either of all ones, or of excellent separated of finite Krull dimension)?

  2. Is there an interesting subclass in the class of all limit schemes of this sort? I don't want to restrict myself to affine schemes.

  3. If no nice answers to questions 1 and 2 will be given, I will need a term for limits of this sort. Any suggestions?:)

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Here is an answer for the affine case. Assume $f:A\to B$ is a homomorphism of Noetherian rings. Then $B$ is a filtered colimit of smooth (finitely generated) $A$-algebras iff $f$ is regular (flat with geometrically regular fibers). This is due to Popescu and Spivakovsky; see for instance Teissier's Bourbaki talk
http://www.math.jussieu.fr/~teissier/documents/Approx.BBk.pdf
and the references therein (the above result is Thm. 1.1).
If $A=k$ is a field, this says that $B$ is a colimit of smooth $k$-algebras iff it is geometrically regular over $k$. If $k$ is perfect (e.g. the prime field!) you can remove "geometrically". The field case might possibly be simpler than the general case.

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  • $\begingroup$ Thank you! Are you sure that the connecting morphisms are dominant? Also, do you think that anything is known in the non-affine case? $\endgroup$
    – Mikhail Bondarko
    Commented Mar 6, 2013 at 9:09
  • $\begingroup$ I didn't say the morphisms were dominant. That would mean $B$ is the direct limit of its smooth subalgebras, a much stronger condition, it seems. And I know nothing about the non-affine case. $\endgroup$
    – Laurent Moret-Bailly
    Commented Mar 6, 2013 at 9:36
  • $\begingroup$ I'm sorry! I need the connecting morphisms to be dominant, but I forgot to write about this. $\endgroup$
    – Mikhail Bondarko
    Commented Mar 6, 2013 at 10:07

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