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Let $R$ be (assumed to be commutative, Noetherian) a regular local ring. Let $A$ be a direct limit of $R$-smooth algebras, such that the transition maps are $R$-étale.

Let $U= Spec(B)$ be an affine open subscheme of $Spec(A)$.

Further, assume that A and B are Noetherian (since it might happen that A is not necessarily Noetherian as noted at Are essentially smooth schemes noetherian?).

Is it true that $B$ can be written as a direct limit of $R$-smooth algebras with transition maps $R$-étale.

By Popescu's desingularization theorem, it follows that $B$ is a direct limit of $R$-smooth algebras. But I suppose $R$-étale transition maps may not be guaranteed.

Also, can we put further restrictions on the base ring $R$, so that such a statement as above would be true?

Any comments are much appreciated!

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  • $\begingroup$ What does R-etale mean? $\endgroup$ Jul 15, 2021 at 8:35
  • $\begingroup$ @PiotrAchinger I mean by that is that if $C_i$ and $C_j$ are $R$-smooth rings that are terms in the direct limit, the transition map $\phi_{ij}: C_i \to C_j$ is etale morphism of rings over $R$. $\endgroup$ Jul 15, 2021 at 8:43
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    $\begingroup$ If $B=A_f$ and $A=\varinjlim A_i$, then wlog $f\in A_0$, and then we have $B=\varinjlim (A_i)_f$. In general, ${\rm Spec}(B)$ is the union of finitely many ${\rm Spec}(A_f)$’s, and you should be able to conclude. $\endgroup$ Jul 15, 2021 at 8:44
  • $\begingroup$ I thought a bit and didn't manage to complete the argument. In general, if $U$ is the union of $\operatorname{Spec}(A[f_j^{-1}])$ for $j=1, \ldots, n$, then again assuming $f_j\in A_0$ for all $j$, we can take $U_i\subseteq \operatorname{Spec}(A_i)$ to be the union of $\operatorname{Spec}(A_i[f_j^{-1}])$. Then it is easy to check that $B=\varinjlim B_i$ where $B_i = \mathcal{O}(U_i)$. So we are done once we know that the $U_i$ are affine for $i\gg 0$, but I was unable to prove this. $\endgroup$ Jul 16, 2021 at 7:35
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    $\begingroup$ @LSpice Thanks for the pointers! $\endgroup$ Jul 16, 2021 at 12:28

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Turning the comments into an answer (CW). Write $A=\varinjlim_{i\in I} A_i$ and let $X=\operatorname{Spec}(A)$, $X_i=\operatorname{Spec}(A_i)$ and $U=\operatorname{Spec}(B)\subseteq X$. Every point $x\in U$ has an open neighborhood of the form $\operatorname{Spec}(A[f^{-1}])\subseteq U$ for some $f\in A$. Since $U$ is quasi-compact, we have $U=\bigcup_{j=1}^n \operatorname{Spec}(A[f^{-1}_j])$ for some $f_1, \ldots, f_n\in A$. In particular, we have an exact sequence $$ 0\to B \to \prod_j A[f_j^{-1}] \to \prod_{j,k} A[(f_j f_{k})^{-1}]. $$ Changing the index set $I$, we may assume that it has a smallest element $0$ and that $f_1, \ldots, f_n\in A_0$. Let $U_i\subseteq X_i$ denote the union of the opens $\operatorname{Spec}(A_i[f_j^{-1}])$ for $j=1, \ldots, n$. Writing $B_i=\mathcal{O}(U_i)$, we then have short exact sequences
$$ 0\to B_i \to \prod_j A_i[f_j^{-1}] \to \prod_{j,k} A_i[(f_j f_{k})^{-1}]. $$ Since for $f\in A_0$, we have $A[f^{-1}]= \varinjlim_i A_i[f^{-1}]$, and because colimit is exact, taking the colimit of the above exact sequences and comparing with the previous one we obtain $$ B \simeq \varinjlim B_i. $$ Now each $U_i$ is smooth and the maps $U_i\to U_{i'}$ are etale for $i\geq i'$. So we can conclude if we show that the $U_i$ are affine for $i\gg 0$. But this follows from SP Tag 01Z6.

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  • $\begingroup$ Thanks a lot for putting it together so succinctly. Much appreciated! $\endgroup$ Jul 16, 2021 at 12:45

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