Fix a prime number $p$. Let $A = \overline{\mathbf{Z}_p}$ be the integral closure of the $p$adic integers $\mathbf{Z}_p$ in some fixed algebraic closure of its fraction field, and let $B$ be the $p$adic completion of $A$. Is the map $A \to B$ an inductive limit of smooth morphisms? If $A$ was excellent, this would follow from Artin/Popescu approximation theorems, but $A$ is not even noetherian. Of course, one can ask similar questions much more generally, but this is the case I am interested in.

$\begingroup$ It may be helpful if you indicate why you pose the question (e.g., idle curiosity or something more substantial). For example, some nonarchimedean geometry (most naturally, Berkovich spaces) ensures that for a finite system of polynomial equations over $A$ (or even something more general), any solution in $B$ can be approximated arbitrarily well by a solution in $A$. So if that is your goal then the ArtinPopescu (and "smoothening") viewpoint in such generality is unnecessary. $\endgroup$ – grp Sep 12 '12 at 4:25

1$\begingroup$ I would like to know what the cotangent complex of $B$ relative to $\mathbf{Z}_p$ is. If the question I ask has an affirmative answer, then the relative cotangent complex of $B$ over $A$ is concentrated in degree $0$ (and is moreover a $\mathbf{Q}_p$module). $\endgroup$ – anon Sep 13 '12 at 12:42
This is not an answer to the question in the main body, but the one raised by the OP in the comments.
The goal is to show that $L_{B/A}$ is concentrated in degree $0$ and $\mathbf{Q}_p$module for $A = \overline{\mathbf{Z}_p}$ and $B$ the $p$adic completion of $A$. The map $A \to B$ is flat and an isomorphism after reduction modulo $p$. The flat base change formula for the cotangent complex shows
$$L_{B/A} \otimes_B^L B/p \simeq 0,$$
i.e., that $p$ acts invertibly on $L_{B/A}$. By flat base change again, we get
$$L_{B/A} = L_{B/A}[p^{1}] \simeq L_{B[p^1]/A[p^{1}]}.$$
As $A[p^{1}] \to B[p^{1}]$ is an extension of fields of characteristic $0$, it is enough to show:
Claim: If $E/K$ is an extension of fields of characteristic $0$, then $L_{E/K}$ is an $E$vector space placed in degree $0$.
Proof: By expressing $E$ as a filtered colimit of finitely generated field extensions, as the cotangent complex commtues with such colimits, we may assume $E/K$ is finitely generated, i.e., there exists a finitely generated $K$algebra $A$ which is a domain such that $E$ is the fraction field of $A$. By generic smoothness (as we are in characteristic $0$), we may assume $A$ is smooth over $K$. Then $L_{A/K} = \Omega^1_{A/K}$ is a finite locally free $A$module in degree $0$. Localising at the generic point and using $L_{A/K} \otimes_A E \simeq L_{E/K}$ then proves the claim.
Let me answer (?) your question with a question. What is this question really asking? It seems that it asks something like this. Let's call an Avariety a finitely presented, irreducible, scheme X over A. Now, suppose that we have a Bpoint x ∈ X(B). Assume that the image of x is dense in X. Question: Can we find a morphism Y > X of Avarieties and a Bpoint y of Y which maps to X.
OK, and we can certainly find an alteration Y > X such that Y is the base change of a strictly semistable scheme over a suitable dvr, say R, contained in A. Since Y > X is an alteration and since B is algebraically closed (right?) we can lift x to a y in Y(B). Right?
This already proves something about your cotangent complex because Y is lci over A. Right?

1$\begingroup$ answer_bot, although you are new here, by the looks of it you have a lot of valuable knowledge to offer. The general norm we try to adhere to at MO is to give answers that are straight to the point and carry some semblance of being authoritative, and not get drawn into protracted discussions (the SE model is illsuited to them). So while you seem like you might be fun to talk to in person, it would be better to cut down a tad on the chat and the answering in riddles for now. $\endgroup$ – Todd Trimble♦ Dec 22 '13 at 2:33

$\begingroup$ @ToddTrimble: Do you understand what answer_bot is saying? Speaking as somebody who has many in person conversations with severely confused young mathematicians, often the only correct answer to a confused question is an enlightening question. $\endgroup$ – Jason Starr Mar 23 '14 at 0:56

$\begingroup$ @JasonStarr In a word, no. My message might have been prompted however by the post as it was before I did some editing, where answer_bot was joking around in a way that I utterly failed to understand. If you think that answer_bot has answered enlighteningly with a question, then by all means please vote it up. $\endgroup$ – Todd Trimble♦ Mar 23 '14 at 1:57