When is the power-bounded subring top. of finite type?

Very naive question here. Let $$K$$ be a complete nonarchimedean field, $$A$$ a reduced affinoid $$K$$-algebra. When is the power-bounded subring $$A^\circ$$ topologically of finite type, in the sense that we can find a surjection $$\mathcal{O}_K \left\langle X_1,\dots,X_n \right\rangle \twoheadrightarrow A^\circ$$? The answer is "always" if $$K$$ is discretely valued, or $$K$$ is stable and the value group $$|K^\ast|$$ is divisible (in particular if $$K$$ is algebraically closed). This follows from some stuff in sections 6.3-6.4 of Bosch-Guntzer-Remmert, which in fact proves much more.

Surely the answer is not "always", but I'm having an annoyingly difficult time constructing a counterexample. I suspect that if $$p>2$$ and $$K=\widehat{\mathbf{Q}_p(p^{1/p^\infty})}$$, the power-bounded subring of $$A= K \left\langle X,\frac{X^2}{p} \right\rangle$$ is not topologically of finite type, but I wasn't quite able to prove it. Any help would be appreciated!

• You perhaps already know this, but one result of Grauert--Remmert in the build-up to the reduced fiber theorem is that if $A$ is a geometrically reduced affinoid $K$-algebra then $A_L^\circ$ is always topologically of finite type over $\mathcal{O}_L$ for some finite extension $L/K$. Commented May 2 at 3:36

It turns out I was being silly. Corollary 6.4.3/6 in BGR implies that for $$K$$ NOT discretely valued, $$A^\circ$$ is top. finite type iff there exists a distinguished surjection $$\alpha: K \left\langle X_1,\dots,X_n \right\rangle \twoheadrightarrow A$$, i.e. a surjection such that the sup norm $$|\cdot|_{sup}$$ on $$A$$ coincides with the residue norm $$|\cdot|_\alpha$$. But the existence of a distinguished surjection implies that $$|A|_{sup} = |K|$$, so any $$A$$ with $$|A|_{sup} \supsetneq |K|$$ gives a counterexample. In particular, the ring I wrote down is a counterexample indeed, because $$|X|_{\sup}=p^{-1/2} \notin |K|$$.