# Open affine subscheme of a direct limit of smooth algebras

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!

• What does R-etale mean? Jul 15, 2021 at 8:35
• @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 $R$. Jul 15, 2021 at 8:43
• 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. Jul 15, 2021 at 8:44
• 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. Jul 16, 2021 at 7:35
• @LSpice Thanks for the pointers! Jul 16, 2021 at 12:28

## 1 Answer

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.

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