$\DeclareMathOperator\Sp{Sp}$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 $\geq 1$.
Noetherianess – Noetherianess of Subalgebras of an Affinoid Algebra
ac.commutative-algebraag.algebraic-geometryarithmetic-geometryp-adic-analysisrigid-analytic-geometry
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Best Answer
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:
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.