Let F : (Sch)^{opp} —> (Sets) be a functor. Assume that

- F is an fppf sheaf,
- F is relatively representable, i.e., F —> F times F is representable,
- F is limit preserving (i.e., locally of finite presentation),
- F has effective versal deformations, and
- F satisfies openness of versality.

Then, if S is an excellent base scheme, the functor F is an algebraic space. Originally Artin proved this over an excellent Dedekind domain. The excellent base scheme case is discussed for example in Approximation of versal deformations authored by Brian Conrad and myself.

What I want to know is this: Is it really necessary to assume that S is excellent? Can you start with a non-excellent Noetherian scheme and make a counter example? I now think sometimes you can.

Here is a related question. Suppose that X is a scheme and U —> X is a surjective morphism of schemes. Set R = U \times_X U so that we get a groupoid scheme (U, R, s, t, c). Let U/R denote the fppf quotient sheaf. Is it true that the canonical map U/R —> X is an isomorphism? The answer to this is: No! If you let U = Spec(Z[t]) be the normalization of X = Spec(Z[t^2, t^3]), then U/R does not have a good deformation theory. Namely, Spec(A)-valued points of U/R are given by equivalence classes of pairs (A —> B, b) where A —> B is faithfully flat of finite presentation and b in B is an element such that b^2 and b^3 are elements of A. The pairs (A —> B, b) and (A —> B’, b’) are equivalent if there exists a third pair (A —> B”, b”) and A-algebra maps B —> B”, B’ —> B” mapping b^2, b^3 to (b”)^2, (b”)^3 and (b’)^2, (b’)^3 to (b”)^2, (b”)^2. The transformation U/R —> X maps the pair (A —> B, b) to the ring map Z[t^2, t^3] —> A which maps t^2 to b^2 and t^3 to b^3. Let k be a field and let O = (k —> k, 0). We claim the tangent space of this point is zero. Namely, a first order deformation is given by a pair (k[e]/(e^2) —> B, b) where b is in eB. Hence b^2 = b^3 = 0 and so this pair is equivalent to the pair (k[e]/(e^2) —> B, 0) and so also (k[e]/(e^2) —> k[e]/(e^2), 0).

But what if we assume the following: (*) X is locally Noetherian and for every closed point x of X there exists a point u in U mapping to x such that the map on complete local rings O^_{X, x} —> O^_{U, u} has a section? I think in this case F = U/R has a good deformation theory at all closed points, which recovers the complete local ring O^_{X, x}. Moreover, by construction the quotient U/R is an fppf sheaf, and is limit preserving. It is also relatively representable (this is a general fact). I think openness of versality should be OK too (did not check this). So if Artin’s criterion applies then U/R is an algebraic space and the map U/R —> X is a morphism of algebraic spaces of finite type over the base which is bijective on field valued points and induces isomorphisms on complete local rings, which would force it to be an isomorphism.

If so, then this implies in particular that there exists a faithfully flat morphism of finite type X’ —> X which factors through the morphism U —> X!

There exists a Noetherian 1-dimensional local domain A whose residue field has characteristic 0 and whose completion is not reduced. There is a paper by Gabber where he proves that the completion A^ cannot be written as a directed limit of flat A-algebras of finite type. Let S = X = Spec(A). Let U = Spec(C), where C ⊂ A^ is a finite type A-sub algebra such that the map C —> A^ does not factor through any flat finite type A-algebra. Let R = U \times_X U as above. If the discussion above is correct, then the functor F = U/R satisfies all of Artin’s axioms but isn’t an algebraic space. [Edit July 3, 2011: This isn’t quite right, see this post!]

On the other hand, I have a feeling that Artin’s criterion may hold over Noetherian base schemes S such that for any local ring A which is essentially of finite type over S the completion A^ is a directed limit of flat finite type A-algebras. Is there an example to show that this is not the same as asking S to be excellent?

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