# On the noetherianess of some subalgebras of an affinoid algebra

$$\DeclareMathOperator\Sp{Sp}$$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 $$\geq 1$$.

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.

• 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). Oct 12, 2022 at 9:27
• Dear Piotir Achinger, thank you for the argument. Could you clarify why $\widehat{R}$ is isomorphic to $B^{\circ}$? Oct 12, 2022 at 15:45
• 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. Oct 12, 2022 at 16:20
• 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.
– Z. M
Oct 12, 2022 at 17:47
• @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. Oct 13, 2022 at 7:37