Does a scheme have a "separification"? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T17:53:46Z http://mathoverflow.net/feeds/question/404 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/404/does-a-scheme-have-a-separification Does a scheme have a "separification"? Anton Geraschenko 2009-10-13T04:16:20Z 2011-02-16T13:41:47Z <p><strong>Background:</strong> </p> <p>(1) If <strong>C</strong> and <strong>D</strong> are categories and there is a forgetful functor U:<strong>C</strong>&rarr;<strong>D</strong>, then a <em><b>C</b>-ification</em> functor F:<strong>D</strong>&rarr;<strong>C</strong> is an adjoint to U. For example, the (left) adjoint to the forgetful functor from groups to monoids is "groupification" of a monoid, given by formally adjoining inverses. The (left) adjoint to the forgetful functor from presheaves to sheaves is the usual "sheafification" functor.</p> <p>Note that whenever you have a (left adjoint) <strong>C</strong>-ification functor F (whenever you have an adjunction, for that matter), you get a universal property. For any object X&isin;<strong>D</strong>, there is a canonical morphism (called the <em>unit of adjunction</em>) &epsilon;<sub>X</sub>:X&rarr;U(F(X)) with the property that any morphism f:X&rarr;U(Y) factors as f=U(g)\circ &epsilon;<sub>X</sub> for a unique morphism g:F(X)&rarr;Y in <strong>C</strong>.</p> <p>(2) A scheme X is <em>separated</em> if the diagonal morphism X&rarr;XxX is a closed immersion. It is enough to check that the image of the diagonal is closed. Being separated is the algebro-geometric analogue of being hausdorff, which nothing in algebraic geometry ever is. </p> <p><hr /></p> <p>My question is whether there exists a "separification" functor adjoint to the forgetful functor U from the category of separated schemes to the category of schemes. Note that the forgetful functor U does not respect colimits (you can glue together separated schemes to get a non-separated scheme), so it has no hope of having a right adjoint. But U <em>does</em> respect limits (it's enough to show that an arbitrary product of separated schemes is separated and that fiber products of separated schemes are separated), so it might have a left adjoint.</p> <p>To put it another way, given a scheme X, is there a canonically defined separated scheme X<sup>s</sup> and a morphism X&rarr;X<sup>s</sup> so that any morphism from X to a separated scheme factors uniquely through X&rarr;X<sup>s</sup>?</p> <p><hr /></p> <p>Related questions I'd like to know the answer to:</p> <ul> <li>Is there a "relative separification" functor. That is, does an arbitrary morphism of schemes f:X&rarr;Y admit a canonical factorization through a separated morphism f<sup>s</sup>:X'&rarr;Y. This would be analogous to Stein factorization, which I regard as "relative affinification". An arbitrary (quasi-compact and quasi-separated) morphism f:X&rarr;Y canonically factors through the affine morphism Spec<sub>Y</sub>(f<sub>*</sub>O<sub>X</sub>)&rarr;Y</li> <li>Is there a separification functor for algebraic spaces? Is it possible that the separification of a scheme is naturally an algebraic space?</li> <li>Is there a separification functor for algebraic stacks? (An algebraic stack is separated if the diagonal is proper.)</li> </ul> http://mathoverflow.net/questions/404/does-a-scheme-have-a-separification/974#974 Answer by Thanos D. Papaïoannou for Does a scheme have a "separification"? Thanos D. Papaïoannou 2009-10-18T01:42:35Z 2011-02-16T13:41:47Z <p>Carnahan's suggestion is the natural thing to do in the category of topological spaces, but it's unclear if we may execute it in the category of algebraic spaces, since it's unclear if the projections from the closure of the diagonal down to $X$ are always etale. Note that even for topological spaces, quotienting $X\times X$ out by the closure of the diagonal—i.e. quotienting out $X$ by the relation "$x\sim y$ if there is no pair of open neighbourhoods U of $x$ and $V$ of $y$ such that $U$ and $V$ are disjoint"—doesn't necessarily give a Hausdorff topological space since it's not an equivalence relation: it's not transitive!</p> <p>But this is a technical problem: the real reason why there shouldn't exist a separification is that separatedness is a global geometric property and it's difficult to replace a scheme with another scheme for which a global geometric property holds, e.g. it's difficult to construct compactifications of schemes, even though that's relatively simple in the category of topological spaces.</p> <p>At any rate, one way to produce a separification of a scheme is to produce a compactification, e.g. via <a href="http://math.stanford.edu/~conrad/papers/nagataalgspace.pdf" rel="nofollow">Nagata's theorem</a>.</p> http://mathoverflow.net/questions/404/does-a-scheme-have-a-separification/3497#3497 Answer by David Rydh for Does a scheme have a "separification"? David Rydh 2009-10-31T00:39:00Z 2009-10-31T00:39:00Z <p>I think that it is highly unlikely that there exists a separification functor. What does exist is the following:</p> <p><strong>Theorem</strong> (Raynaud-Gruson): Let <em>S</em> be a base scheme and work relative to <em>S</em>. Given a non-separated scheme <em>X</em> of finite type, there is a blow-up (a proper birational morphism) <em>X'</em>-><em>X</em> such that <em>X'</em> admits an étale morphism to a projective scheme <em>Z</em> (in particular a separated scheme).</p> <p>Note that there are non-separated schemes which does not even admit a quasi-finite morphism onto a separated scheme (e.g. take <strong>A</strong>^2 with a double origin and blow-up one of the origins).</p> <p>The Theorem is false as stated for non-locally separated algebraic spaces. There are 3 different solutions to this:</p> <p>A) Take an alteration instead of a modification.</p> <p>B) Replace étale with quasi-finite flat.</p> <p>C) Allow <em>Z</em> to be a proper stack with finite diagonal.</p>