$\DeclareMathOperator\Sp{Sp}$Let $X=\Sp(A)$ be a connected smooth affinoid rigid space over a discretely valued non-archimedean field $K$. Let $\mathcal{R}$ be a valuation ring of $K$, and fix a uniformizer $\pi$. If $Y=\Sp(B)\subset X$ is a connected affinoid subdomain, we have a morphism of affinoid $K$-algebras $\varphi : A\rightarrow B$. We define $B^{\circ}_{A}$ as the algebra $\varphi^{-1}( B^{\circ})\subset A$. That is, $B^{\circ}_{A}$ is the algebra of rigid functions on $X$ which are power-bounded when restricted to $Y$. This algebra is both open, closed, and integrally closed in $A$. Furthermore, as $X$ is smooth, it contains $A^{\circ}$. I would like to know whether this algebra or its image under $\varphi$ are noetherian. In particular, I am interested in the case where $Y$ is a Laurent subdomain given by the locus on which a rigid function $f\in A$ takes values with valuation $\geq 1$.


1 Answer 1


I think the answer is yes. Let $\pi$ be a uniformizer of $K$. Your ring $R:=B^\circ_A$ satisfies $R[1/\varpi]=A$ and $\widehat{R}$ ($\pi$-adic completion) is isomorphic to $B^\circ$. Therefore the result will follow from the lemma:

Lemma. Let $R$ be a ring, $\pi$ an element of $R$ such that the rings $R[1/\pi]$ and $\widehat{R}$ ($\pi$-adic completion) are noetherian. Then, $R$ is noetherian as well.

Proof. Let $I\subseteq R$ be an ideal. Then $I\cdot R[1/\pi]$ is finitely generated, and we can pick a system of generators $(f_1, \ldots, f_r)$ of $I\cdot R[1/\pi]$ lying in (the image of) $I$. In other words, $I$ contains a finitely generated ideal $J=(f_1, \ldots, f_r)\subseteq I$ such that $J\cdot R[1/\pi] = I\cdot R[1/\pi]$. It suffices now to show that $I':=I/J$ is finitely generated as an ideal in $R':=R/I'$.

Now the ring $R'=R/I'$ satisfies the same conditions as $R$ does, as $R'[1/\pi] = R[1/\pi]/J\cdot R[1/\pi]$ and $\widehat{R}{}' = \widehat{R}/J$. Moreover, the ideal $I'\subseteq R'$ satisfies $I'\cdot R'[1/\pi]=0$, and hence it contains $\pi^n$ for some $n$. It is therefore the preimage in $R'$ of an ideal in $R'/\pi^n = (R/\pi^n)/J$. But $R/\pi^n = \widehat{R}/\pi^n$ is noetherian, so the ideal in $R'/\pi^n$ is finitely generated.

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    $\begingroup$ That works. Fun fact that should be more widely known: Instead of asking $\hat{R}$ to be noetherian, it's enough to ask $R/\pi$ to be noetherian (the two are equivalent, see stacks.math.columbia.edu/tag/05GH). $\endgroup$ Oct 12, 2022 at 9:27
  • $\begingroup$ Dear Piotir Achinger, thank you for the argument. Could you clarify why $\widehat{R}$ is isomorphic to $B^{\circ}$? $\endgroup$ Oct 12, 2022 at 15:45
  • $\begingroup$ I can see that this is the case if, for example, $Y$ is given by the locus of points in which a rigid function $f\in A$ takes values $\leq 1$. However, in the case I am mostly interested in, the image of $A$ is not dense in $B$. Thus, its intersection with $B^{\circ}$ will not be dense in $B^{\circ}$. This intersection is the image of $B^{\circ}_{A}$ under $\varphi$. So, if it is not dense, its $\pi$-adic completion cannot be the whole $B^{\circ}$. Again, thank you for your answer, and please feel free to point out any mistakes in my argument. $\endgroup$ Oct 12, 2022 at 16:20
  • $\begingroup$ If I am not mistaken, this also follows from the following category-theoretic fact: let $\mathcal A$ be an abelian category and $\mathcal A_0\subset\mathcal A$ a localizing subcategory. Then $\mathcal A$ is locally Noetherian if and only if both $\mathcal A_0$ and $\mathcal A/\mathcal A_0$ are locally Noetherian. Here, we take $\mathcal A$ to be the category of $R$-modules, and $\mathcal A_0$ to be the category of $R[\pi^{-1}]$-modules, then $\mathcal A/\mathcal A_0$ is the category of derived $\pi$-complete modules. $\endgroup$
    – Z. M
    Oct 12, 2022 at 17:47
  • $\begingroup$ @FernandoPeñaVázquez You are right, my mistake. Indeed $R/\pi$ may be smaller than $B^\circ/\pi$, but bigger than the image of $A^\circ/\pi$, and it is unclear why it should be finitely generated over $\mathcal{O}_K/\pi$ or noetherian. I will think about this and edit the answer later. $\endgroup$ Oct 13, 2022 at 7:37

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