Quasi-separatedness for Algebraic Spaces - MathOverflow most recent 30 from http://mathoverflow.net2013-06-20T12:06:29Zhttp://mathoverflow.net/feeds/question/16381http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/16381/quasi-separatedness-for-algebraic-spacesQuasi-separatedness for Algebraic SpacesDaniel Bergh2010-02-25T09:45:18Z2010-03-18T14:12:40Z
<p>I'm reading Knutson's book on algebraic spaces, and I stumbled over the quasi-separatedness axiom in his definition of algebraic spaces (Definition 1.1, Chapter II). He defines an algebraic space A as a sheaf on the site of schemes with the étale topology satisfying:</p>
<p>I) Local representability. There exists a representable étale covering $U \rightarrow A$, $U$ a scheme.<br>
II) Quasi-separatedness. The map $U \times_A U \rightarrow U \times U$ is quasi-compact.</p>
<p>In a technical remark (1.9) at the end of the section, he argues that the quasi-separatedness assumption is needed for the existence of fibre products in the categeory of algebraic spaces. As I see it, this is wrong. For instance, the proof of existence of fibre products for schemes given i Hartshorne carries over to algebraic spaces without problems, even if we just assume local representability.</p>
<p>This makes me wonder, would it be more natural to take only the local representability as requirement for algebraic spaces, or do we run into problems later on?</p>
<p>Sometimes one sees informal definitions of algebraic spaces as the closure of schemes under étale equivalence relations in the category of étale sheaves. In what sence is this true? Here it seems to me that we really do need the quasi-separatedness axiom (or something similar) since we need an étale equivalence relations $R \rightarrow U \times U$ to satisfy effective descent in order to get local representability for its quotient.</p>
http://mathoverflow.net/questions/16381/quasi-separatedness-for-algebraic-spaces/16391#16391Answer by Chris Schommer-Pries for Quasi-separatedness for Algebraic SpacesChris Schommer-Pries2010-02-25T12:48:33Z2010-02-25T12:48:33Z<p>This issue or question came up indirectly in a couple previous posts, which I think you might like to look at. There is indeed a notion of algebraic space which is more general and doesn't require quasi-separatedness (see below). The first such question was Anton's post: <a href="http://mathoverflow.net/questions/8918/is-an-algebraic-space-group-always-a-scheme" rel="nofollow">Is an Algebraic Space Group Always a Scheme?</a>
In that post he asked whether a group object in algebraic spaces is necessarily a scheme. It turns out that the answer depends very heavily on whether the definition of algebraic space requires quasi-sep. or not. If it requires it, then the answer is yes. If not then there are counter examples, which I learned by asking this question <a href="http://mathoverflow.net/questions/9043/why-is-this-not-an-algebraic-space" rel="nofollow">Why is This Not an Algebraic Space?</a> (the object in question is a group object in non-quasi-separated algebraic spaces, which is not a scheme). </p>
<p>When I learned the definition of algebraic space (which was some time ago in Martin Olsson class on Stacks at UC Berkeley) it didn't include Quasi-Sep. Here is the definition we used, which I looked up in <a href="http://math.berkeley.edu/~anton/index.php?m1=writings" rel="nofollow">Anton's wonderful collection of notes</a>:</p>
<p><strong>Definition</strong>: An <em>algebraic space</em> over S is a functor $X : (Sch/S)^{op} \to Set$ such
that</p>
<ol>
<li>X is a sheaf on the big etale topology on S,</li>
<li>$\Delta : X → X \times_S X$ is representable, and </li>
<li>there exists an S-scheme $U \to S$ and a surjective etale morphism $U \to X$ (surjective as a map of sheaves).</li>
</ol>
http://mathoverflow.net/questions/16381/quasi-separatedness-for-algebraic-spaces/16483#16483Answer by BCnrd for Quasi-separatedness for Algebraic SpacesBCnrd2010-02-26T07:08:16Z2010-03-18T14:12:40Z<p>One can certainly make the basic definitions, and the real issue is to show that the definition "works" using any etale map from a scheme. More precisely, the real work is to show that a weaker definition actually gives a good notion: rather than assume representability of the diagonal, it suffices that $R := U \times_X U$ is a scheme for <em>some</em> scheme $U$ equipped with an etale representable map $U \rightarrow X$. Or put in other terms, we have to show that if $U$ is any scheme and $R \subset U \times U$ is an 'etale subsheaf which is an etale equivalence relation then the quotient sheaf $U/R$ for the big etale site actually has diagonal representable in schemes. Indeed, sometimes we want to <em>construct</em> an algebraic space as simply a $U/R$, so we don't want to have to check "by hand" the representability of the diagonal each time. </p>
<p>That being done, then the question is: which objects give rise to a theory with nice theorems? For example, can we always define an associated topological space whose generic points and so forth give good notions of connectedness, open behavior with respect to fppf maps, etc.? (The definition of "point" needs to be modified from what Knutson does, though equivalent to his definition in the q-s case.) The truth is that once the theory is shown to "make sense" without q-s, it turns out that to prove interesting results one has to assume something extra, the simplest version being the following weaker version of q-s: $X$ is "Zariski-locally q-s" in the sense that $X$ is covered by "open subspaces" which are themselves q-s. This is satisfied for all schemes, which is nice. (There are other variants as well.)</p>
<p>In the stacks project of deJong, as well as the appendix to my paper with Lieblich and Olsson on Nagata compactification for algebraic spaces, some of the weird surprises coming out of the general case (allowing objects not Zariski-locally q-s) are presented. (In that appendix we also explain why the weaker definition given above actually implies the stronger definition as in Chris' answer. This was surely known to certain experts for a long time.) </p>