1
$\begingroup$

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!

$\endgroup$
9
  • $\begingroup$ 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
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ @LSpice Thanks for the pointers! $\endgroup$ Jul 16, 2021 at 12:28

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks a lot for putting it together so succinctly. Much appreciated! $\endgroup$ Jul 16, 2021 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.