Presenting $\mathbb{Q}[[t]]$ as an explicit colimit of smooth $\mathbb{Q}$-algebras: an explicit example for the Popescu's theorem

By the seminal Popescu's theorem, $R=\mathbb{Q}[[t]]$ is a filtered colimit of smooth $\mathbb{Q}$-algebras. Could you give me a hint: which $\mathbb{Q}$-algebras can yield such a colimit? My problem is that subalgebras of $R$ seem to be 'usually' singular.

Upd. 1. I realized that $R$ actually contains plenty of smooth $\mathbb{Q}$-subalgebras; yet I am not sure that I can prove that $R$ can be presented as their filtered colimit. Is this true; are there any nice references on these matters?

Upd. 2. Possibly, it suffices to consider henselizations of freely generated $\mathbb{Q}$-subalgebras of $R$ here. In any case, I would like to understand presentations of regular complete local rings as 'smooth colimits' better.

• The proof is totally non-constructive (and doesn't even make a directed system on the first pass, let alone go way beyond subalgebras). As a warm-up, one could consider the far more elementary theorem of Lazard: every flat module is a direct limit of finite free modules (generally not submodules, of course). Can we make it explicit for $\mathbf{Z}[[x]]$ as a $\mathbf{Z}[x]$-module? – user76758 Jan 21 '14 at 8:33
• Well, I don't know much about these things; that's why I ask. In any case, it seems quite interesting to know whether subalgebras/submodules are sufficient for these purposes. – Mikhail Bondarko Jan 21 '14 at 8:48
• As far as I know there are only two proofs, the original one by Popescu (and its simplified versions by several people) and the one by Spivakovsky. None of them can be made with subalgebras (I don't even think you can bound the transcendence degree of the algebras in the directed system when starting with an algebra of finite transcendence degree). But ... – Vinteuil Jan 21 '14 at 9:40
• @Vinteuil. According to Görtz-Wedhorn, 'Algebraic Geometry', p. 268, Th. 10.76, Popescu's theorem holds with subalgebras. Is this wrong ? – Damian Rössler Jan 21 '14 at 12:40
• @Rössler. Yes, it is wrong in that generality. However, as I said above, in some special cases that may be sufficient to Bondarko's interests, taking subalgebras suffice. In the paper by Spivakovsky (in JAMS) there is also a section treating the case of subalgebras. – Vinteuil Jan 21 '14 at 13:24

One can prove Popescu's theorem directly in this special case (due to the two strong assumptions -- characteristic $0$ and one-dimensionality -- present here). The basic point is that any 'singular' subalgebra $A \subset R$ may be resolved by a proper birational map $X \to \mathrm{Spec}(A)$ by Hironaka, and that the map $\mathrm{Spec}(R) \to \mathrm{Spec}(A)$ lifts to $X$ by the valuative criterion.
Write $R$ as a filtered colimit of finitely generated $\mathbb{Q}$-subalgebras $A_i \subset R$, so $\mathrm{Spec}(R) = \lim \mathrm{Spec}(A_i)$. Each $A_i$ is a domain, and the map $\mathrm{Spec}(R) \to \mathrm{Spec}(A_i)$ maps the generic point to the generic point. For each $i$, consider the cofiltered projective system $C_i := \{X_{i,j} \to \mathrm{Spec}(A_i)\}$ of all proper maps which are isomorphisms over the generic point of the target. Pullback along $\mathrm{Spec}(A_{i'}) \to \mathrm{Spec}(A_i)$ gives a functor $C_i \to C_{i'}$, so we get a two-variable cofiltered projective system $\{X_{i,j}\}$ of schemes. Also, the map $\mathrm{Spec}(R) \to \mathrm{Spec}(A_i)$ admits a unique lift to $\mathrm{Spec}(R) \to X_{i,j}$ for each $j$ by the valuative criterion for properness. Let $B_{i,j}$ be the local ring of $X_{i,j}$ at the image of the closed point of $\mathrm{Spec}(R)$, so there is a natural map $B_{i,j} \to R$ extending the given map $A_i \to R$. Varying over the projective system, we get a filtered system $\{B_{i,j}\}$ of rings with a map to $R$. One checks that $\mathrm{colim} B_{i,j} \simeq R$; the surjectivity is completely trivial, while injectivity comes from the fact that $B_{i,j}$'s are domains birational to $A_i$ for a cofinal set of $j$'s. Moreover, by resolutions, for a fixed $i$, the $B_{i,j}$'s are smooth over $\mathbb{Q}$ for a cofinal set of $j$'s. Thus, we have expressed $R$ as a filtered colimit of smooth $\mathbb{Q}$-algebras.