# Clarification

Let me just clarify what I was trying to say in the previous post.

Setup. Let A be a Noetherian local ring. Set S = Spec(A). Denote U the punctured spectrum of A. Let A ⊂ C ⊂ A^* be a finite type A-algebra contained in the completion of A. Set V = Spec(C) ∐ U. Consider the functor F : (Sch/S)^{opp} —> Sets which to a scheme T/S assigns

1. F(T) = {*} = a singleton if there exists an fppf covering {T_i —> T} such that each T_i —> S factors through V.
2. F(T) = ∅ else.

I claim that all of Artin’s criteria are satisfied for this fppf sheaf (details omitted). Moreover, note that both F(Spec(A^*)) = {*} and F(U) = {*} are nonempty.

If Artin’s criteria imply that F is an algebraic space, then, choosing a surjective etale morphism X —> F where X is an affine scheme, we conclude that X is surjective and etale over A (this takes a little argument). Using the definition of F we find a faithfully flat, finite type A-algebra B and an A-algebra map C —> B.

Conversely, if there exists an A-algebra map C —> B with B a faithfully flat, finite type A-algebra, then Spec(B) —> F is a flat, surjective, finitely presented morphism and F is an algebraic space (by a result of Artin we blogged about recently).

This analysis singles out the following condition on a Noetherian local ring A: Every finite type A-algebra C contained in the completion of A should have an A-algebra map to a faithfully flat, finite type A-algebra B. But it isn’t necessary for B to also map into A^*! I missed this earlier when I was thinking about this issue. For example any dvr has this property (but there exist non-excellent dvrs).

Finally, if Artin’s criteria characterize algebraic spaces over Spec(R) for some Noetherian ring R then this property holds for any local ring of any finite type R-algebra. Likely this isn’t a sufficient condition.

## 3 thoughts on “Clarification”

1. Dear Johan,

Just so I understand, this condition is a weaker condition than saying that the completion of $A$ is a (co)limit of finitely generated, flat $A$-algebras; the $A$-algebra map $C\rightarrow B$ does not need to factor the $A$-algebra map $C\rightarrow A^*$.

Best regards,

Jason

• Johan on said:

Yes, exactly.