Does a scheme have a "separification"?

Question

Background:

(1) If C and D are categories and there is a forgetful functor U:C→D, then a C-ification functor F:D→C 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) ε_X:X→U(F(X)) with the property that any morphism f:X→U(Y) factors as f=U(g)\circ ε_X 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.

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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 X^s and a morphism X→X^s so that any morphism from X to a separated scheme factors uniquely through X→X^s?

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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 f^s: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 Spec_Y(f_*O_X)→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.)

Answer 1

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) $X'\to X$ 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 do not even admit a quasi-finite morphism onto a separated scheme (e.g. take $\mathbf A^2$ 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.

Answer 2

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.

Answer 3

Here is an interesting situation to think about, which might produce a counterexample:

Take two copies of $\mathbb{P}^2$ and consider the standard quadratic (Cremona) birational transformation $\tau\colon (x:y:z) \mapsto (u:v:w)=(yz:xz:xy)$ which restricts to an isomorphism between the open sets $\{xyz\neq 0\}$ of one $\mathbb{P}^2$ and $\{uvw\neq 0\}$ of the other; now call $X$ the scheme obtained by gluing the two $\mathbb{P}^2$ together by identifying these two open sets by means of $\tau$. In other words, $X$ is the projective plane with a triangle doubled, but doubled in two different ways.

If we take the closure of the diagonal $\Delta \subseteq X\times X$, it contains every pair where the first part lies on the line $x=0$ (of the first $\mathbb{P}^2$), say, and the second is the point $v=w=0$ (of the second $\mathbb{P}^2$); this closure is not an equivalence relation, but the equivalence relation it generates identifies all the points of the lines $x=0$, $y=0$, $z=0$ and $u=0$, $v=0$ and $w=0$.

So if we try to "separify" $X$ by somehow quotienting by the closure of the diagonal, we are led to contract the three lines $x=0$, $y=0$ and $z=0$ in $\mathbb{P}^2$ to a single point. Now I believe this cannot be done (except by contracting everything to a point). This suggests that there is no universal morphism from $X$ to a separated scheme (or else that this separification should be a point, which is suspicious).

Answer 4

Doesn't a separification exist for trivial reasons?

Given a scheme $X$, consider the category of separated schemes $Y$ together with a morphism $X \to Y$. View this as a diagram, and take the inverse limit. This means we take the product of all schemes $Y$, and set $X^s$ equal to the intersection of the closed graphs of each arrow in this category.

Then every morphism from $X$ to a separated scheme factors through $X^s$. However this is unique because, were there two different morphisms, there equalizer would be a nonempty closed subscheme of $X^s$ which itself is separated and admits a map from $Y$, which is impossible because $X^s$ would have to lie in a map of this graph.

Then we get functoriality the obvious way.

The only issue is set-theoretic. However somehow I think we only need consider maps to separated schemes of bounded cardinalities.

There's likely some obvious flaw in this thinking. If not, I think the examples David and Gro-Tsen have come up with conclusively show that there is no geometrically natural construction of the separification.