Can an algebraic space fail to have a unviersal map to a scheme? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T14:21:18Z http://mathoverflow.net/feeds/question/4587 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4587/can-an-algebraic-space-fail-to-have-a-unviersal-map-to-a-scheme Can an algebraic space fail to have a unviersal map to a scheme? David Zureick-Brown 2009-11-08T03:58:20Z 2009-11-09T21:31:16Z <p>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?</p> http://mathoverflow.net/questions/4587/can-an-algebraic-space-fail-to-have-a-unviersal-map-to-a-scheme/4776#4776 Answer by Anton Geraschenko for Can an algebraic space fail to have a unviersal map to a scheme? Anton Geraschenko 2009-11-09T21:31:16Z 2009-11-09T21:31:16Z <p>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).</p> <p>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.</p> <p>Consider Hironaka's example of a non-projective proper variety (see page 15 of Knutson's <a href="http://www.springerlink.com/content/l25540tml3284222" rel="nofollow">Algebraic Spaces</a>). It has an action of &#8484;/2 for which there is an algebraic space quotient. But in the category of schemes, there is no <em>geometric quotient</em> for this action. The question is whether there is a <em>categorical quotient</em> in this case.</p>