Sometimes general theory is "good" at showing that a functor is representable by an algebraic spaces (e.g., Hilbert functors, Picard functors, coarse moduli spaces, etc). What sort of general techniques are there to show that an algebraic space is a scheme? Sometimes it's possible to identify your algebraic space with something "else" (e.g. it "comes" from GIT as is the case for the moduli space of curves), but are there other methods?
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3 Answers
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One result along those lines is that that any algebraic space which has a quasi-finite morphism to a scheme is itself a scheme.
More precisely, if $f\colon X\to Y$ is a separated, finite type, quasi-finite morphism of algebraic spaces, then the Stein factorization $X\to \mathrm{Spec}_Y(f_*\mathcal{O}_X)\to Y$ is an open immersion followed by an affine morphism, so $f$ is quasi-affine. In particular, if $Y$ is a scheme, so is $X$.
References
This is Proposition 3.1 of Quot Functors for Deligne-Mumford Stacks or Théorème A.2 of Champs algébriques.
I learned this from Martin Olsson; it's Corollary 17.8 in my notes from his stacks course. It's Theorem 7.2.10 in the book he's since written covering the material from that course. Jason Starr points out that there's a mistake in Lemma 6.2.9 which propagates to this theorem (I haven't checked this), and suggested the two alternative references above.
answered Nov 8, 2009 at 1:23
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$\begingroup$ Cool! Then one might wonder how to produce "enough line bundles" on X to start trying to make these maps to something projective, for example. Or simply how to make them in general. $\endgroup$– mdelandCommented Nov 8, 2009 at 1:49
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$\begingroup$ @Anton: the link to your notes doesn't work. Are these still available somewhere? $\endgroup$ Commented Apr 26, 2021 at 20:20
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2$\begingroup$ @katalaveino Thanks for the ping. I've updated the link. $\endgroup$ Commented Apr 26, 2021 at 22:49
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2$\begingroup$ I am only posting this because I noticed this answer has recently been updated. There is a mistake in Lemma 6.2.9, which I believe propagates to the proof of Stein factorization for algebraic spaces. It would be better to use Proposition 3.1 of my article with Olsson. It would be best to use the correct source: Theoreme A.2, p. 198 of Laumon and Moret-Bailly. $\endgroup$ Commented Apr 27, 2021 at 19:52
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1$\begingroup$ @JasonStarr Thanks! I've just added those references as well. $\endgroup$ Commented Apr 29, 2021 at 1:11
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Another example: the Nakai-Moishezon theorem says that a divisor D on X is ample iff for every curve C on D, $D \cdot C > 0$ and $D^2 > 0$. This holds also for an algebraic space.
As an application, you can show for instance that the coarse space of $\bar{M_g}$, the Deligne-Mumford compactification of the moduli stack of smooth genus g curves, is represented by a projective variety. The point is that Artin's representibility theorem tells you that the coarse space exists as an algebraic space, and you can then use Nakai-Moishezon to show that it has an ample line bundle. This is cool because it avoids GIT.
answered Nov 8, 2009 at 3:51
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$\begingroup$ Ah, cool! Doing a little research I see that Koll\'ar has done exactly this in his paper "Projectivity of Complete Moduli". You let $D$ to be a power of the relative dualizing sheaf. Then for curves $C$ not contained in the boundary it's more or less clear the Nakai-Moishezon condition is satisfied and for curves contained in the boundary you work a little harder but it's still true. He also applies this to compactifications of surfaces of general type. Are there other known applications of this method? $\endgroup$– mdelandCommented Nov 8, 2009 at 15:36
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1$\begingroup$ One is David Smyth's thesis. Its similar though, he gives alternative compactificaitons of $\bar{M_{g,n}}$ and shows that they have projetive coarse moduli spaces (here GIT doesn't work). $\endgroup$ Commented Nov 8, 2009 at 15:59
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One example: an algebraic space $X$ is a scheme iff $X_{\text{red}}$ is a scheme.
One application of this is that a quotient (I may be missing an adjective or two here) by a reductive group over an Artin ring is a scheme. Call the quotient $X$. Then it is easy to prove that $X$ is an algebraic space. On the other hand when your Artin ring is a field the classical theory of reductive groups tells you that the quotient is a scheme, i.e. $X_{\text{red}}$ is a scheme, and you can conclude that $X$ is in fact a scheme.
answered Nov 8, 2009 at 2:26
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$\begingroup$ How do you show that X is a scheme if X_{red} is? I remember we worked this out before, but I don't remember how. I feel like the first thing to do (assuming X is noetherian) is to reduce to showing that a square-zero thickening of a scheme is a scheme. $\endgroup$ Commented Nov 8, 2009 at 3:57
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$\begingroup$ I believe this is in Knutson's book. $\endgroup$ Commented Nov 8, 2009 at 8:20
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$\begingroup$ If $X$ is noetherian and locally separated, this is Corollary III.3.6 of Knutson's book Algebraic Spaces. $\endgroup$ Commented Feb 3, 2015 at 3:43
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$\begingroup$ Apparently the noetherian and locally separated hypotheses are not needed: mathoverflow.net/q/195528. $\endgroup$ Commented Feb 3, 2015 at 5:57