Let $\mathcal{X}$ be an algebraic space. Can it happen that there does not exist a map $\mathcal{X} \to X$ with $X$ a scheme that is initial for maps from $\mathcal{X}$ to schemes? Are there reasonable conditions (e.g. finite type) that we can put on $\mathcal{X}$ so that there does exist such a universal map to a scheme?
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asked Nov 8, 2009 at 3:58
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$\begingroup$ Isn;t there always a map to $\mathrm{Spec}(\Gamma(X,\mathcal{O}_X))$? This would be an initial object for morphisms to affine schemes. $\endgroup$– KapilCommented Apr 17, 2022 at 4:08
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$\begingroup$ One may perhaps formulate a modification (pun intended!) to the question. Given a reduced irreducible algebraic space $\mathcal{X}$ is there a morphism to a scheme $Y$ such that the function field of $Y$ is the same as the function field of $\mathcal{X}$. Is there a theorem of Artin on this question? $\endgroup$– KapilCommented Apr 17, 2022 at 4:18
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I think the answer is almost certainly yes, an algebraic space can fail to have a universal map to a scheme (a "schemification"). I don't have a proof, but I think I know the right place to look for one (besides David Rydh's immediate surroundings).
If we can find two maps of schemes which do not have a coequalizer in the category of schemes, but do have a coequalizer in the category of algebraic spaces, then the coequalizer algebraic space will not have a schemification.
Consider Hironaka's example of a non-projective proper variety (see page 15 of Knutson's Algebraic Spaces). It has an action of ℤ/2 for which there is an algebraic space quotient. But in the category of schemes, there is no geometric quotient for this action. The question is whether there is a categorical quotient in this case.
answered Nov 9, 2009 at 21:31
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$\begingroup$ I wondered about that example when reading about algebraic spaces. Does anyone know the answer to this question? $\endgroup$– mdelandCommented Nov 11, 2009 at 14:47