# Images and completion

Here is a question I have been struggling with for the last couple of weeks.

Question: Let A be a Noetherian henselian local ring. Let A —> B be (a) local ring map of local rings, (b) essentially of finite type, (c) the residue field extension is trivial, and (d) injective. Is the map A^ —> B^ of completions injective?

If the answer is “yes”, then this somehow tells us that (very very roughly) “taking scheme theoretic image commutes with completion”.

The answer is yes if A is in addition excellent. But I would like to know if it is also true in general. It is very possible that there exists a simple counter example, it is also possible that it is true for trivial reasons. The most vexing aspect of this question to me is that I cannot even decide whether it should be true or not. Please leave a comment if you have any references, comments, or suggestions. Thanks!

[Edit on August 22, 2010: I finally figured out that this is wrong. Namely, take A to be the example of Ogoma. It is a normal henselian Noetherian local domain whose completion is k[[x, y, z, w]]/(yz, yw). So the completion is the union of a nonsingular 3 dimensional component and a nonsingular 2 dimensional component. Let C be an affine chart of the blow up of A at its maximal ideal. The special fiber of C has two irreducible components (a plane and a line). Let B be the localization of C at a maximal ideal which is a point on one of them but not the other. Then clearly the completion of B “picks out” one of the irreducible components of the completion of A.]

## 10 thoughts on “Images and completion”

1. I have only a very trivial observation. Since B is essentially of finite type, you can find a sequence of local subrings A = B_0 contained in B_1 contained in … contained in B_n = B such that each B_{i+1} is essentially generated over B_i by a single element. So, using induction on n, it suffices to prove your result in the case that B is essentially generated by a single element, i.e., B = localization at a maximal ideal of A[x]/.

2. Last symbol should have been A[x] modulo finitely generated ideal.

• This doesn’t quite work literally as you say unfortunately (at least I do not see it). What you can reduce to is where B is the quotient of A{x} where A{x} means the henselization of A[x] at the maximal ideal (m_A, x).

3. Also “maximal ideal” should have been “prime ideal”.

4. You’re right. I was thinking of “essentially finite type” for a local homomorphism of local rings, not for a local homomorphism of Henselian local rings.

5. Have you tried this on Ferrand-Raynaud’s examples? They satisfy (a), (c) and (d) and the map between the completions is not injective. They don’t seem likely to satisfy (b) though since B is the normalization of A in those examples.

• I think you mean the paper “Fibres formelles d’un anneau local noeth\’erien”. To see how bad nonexcellent Noetherian local rings can get this paper is an excellent resource. For example they construct a 2 dimensional local domain whose completion has an embedded point of dimension 1. But you are right that these examples do not give a counter example to the question. Namely, in their examples B is the integral closure of A and is local. Hence if B is essentially of finite type over A then it is finite over A. But I can prove that the answer to the question is “yes” if A —> B is finite…

• . . . If A –> B is finite, I believe the results follows from the fact that the completion of B equals the tensor product of B over A with the completion of A. Since the completion of A is flat over A, tensoring with the completion preserves injections. (I realize this argument is obvious.)

• Yeah, that’s right. This is also related to why one needs A to be henselian.

Suppose that A is the local ring of a node P on an irreducible curve C over an algebraically closed field k. Let C’ —> C be the normalization, and let Q be one of the two points over P. Let B be the local ring of C’ at Q. Then A —> B is injective but on completions you get k[[x, y]]/(xy) —> k[[x]] which is not injective. This is why the condition that A be henselian is necessary!

I realize this is obvious to you Jason:)