Does a scheme have a “separification”?

Background:

(1) If C and D are categories and there is a forgetful functor U:CD, then a C-ification functor F:DC 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.

Note that whenever you have a (left adjoint) C-ification functor F (whenever you have an adjunction, for that matter), you get a universal property. For any object X∈D, there is a canonical morphism (called the unit of adjunction) εX:X→U(F(X)) with the property that any morphism f:X→U(Y) factors as f=U(g)\circ εX for a unique morphism g:F(X)→Y in C.

(2) A scheme X is separated if the diagonal morphism X→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.

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 does 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.

To put it another way, given a scheme X, is there a canonically defined separated scheme Xs and a morphism X→Xs so that any morphism from X to a separated scheme factors uniquely through X→Xs?

Related questions I'd like to know the answer to:

• Is there a "relative separification" functor. That is, does an arbitrary morphism of schemes f:X→Y admit a canonical factorization through a separated morphism fs:X'→Y. This would be analogous to Stein factorization, which I regard as "relative affinification". An arbitrary (quasi-compact and quasi-separated) morphism f:X→Y canonically factors through the affine morphism SpecY(f*OX)→Y
• Is there a separification functor for algebraic spaces? Is it possible that the separification of a scheme is naturally an algebraic space?
• Is there a separification functor for algebraic stacks? (An algebraic stack is separated if the diagonal is proper.)
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 Minor corrections: X needs to be in D. Then the unit sends X to U(F(X)), and the morphism g is in C. – S. Carnahan♦ Oct 13 2009 at 5:04 @Scott: corrected. Thanks. – Anton Geraschenko♦ Oct 13 2009 at 5:28 Minor note: maybe it should be "separation" instead of "separification" – S. Carnahan♦ Oct 13 2009 at 19:08 I like "separification". I've also heard people propose "sheafication" instead of "sheafification". – Anton Geraschenko♦ Oct 18 2009 at 2:20 Your proof that $U$ respects limits does not work. Here is a different approach: Since affine schemes are separated, it is enough to prove the following If $\{X \to X_i\}$ is a cone of schemes such that for all affine schemes $T$ the induced map $Hom(T,X) \to lim_i Hom(T,X_i)$ is bijective, then this is already true for all schemes $T$. But this is easy because both sides are sheaves in $T$ with respect to the Zariski Topology. – Martin Brandenburg Jun 14 2011 at 15:29

I think that it is highly unlikely that there exists a separification functor. What does exist is the following:

Theorem (Raynaud-Gruson): Let S be a base scheme and work relative to S. Given a non-separated scheme X of finite type, there is a blow-up (a proper birational morphism) X'->X such that X' admits an étale morphism to a projective scheme Z (in particular a separated scheme).

Note that there are non-separated schemes which does not even admit a quasi-finite morphism onto a separated scheme (e.g. take A^2 with a double origin and blow-up one of the origins).

The Theorem is false as stated for non-locally separated algebraic spaces. There are 3 different solutions to this:

A) Take an alteration instead of a modification.

B) Replace étale with quasi-finite flat.

C) Allow Z to be a proper stack with finite diagonal.

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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!

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.

At any rate, one way to produce a separification of a scheme is to produce a compactification, e.g. via Nagata's theorem.

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Actually, even in the category of topological spaces, if you quotient by the natural relation (two points are equivalent if they aren't "hausdorff-like distinugishable"), you don't always get a hausdorff space. You may have to repeat the procedure many times before you get the hausdorffification of the space. – Anton Geraschenko Oct 18 2009 at 2:12
Being affine is also a global property (every scheme is locally affine), but there does exist an affinification functor. – Anton Geraschenko Oct 18 2009 at 2:16
Dear Anton, I edited my answer so that it holds under your very correct observations. – Thanos D. Papaïoannou Oct 18 2009 at 21:23