0
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ 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 $L/K$. $\endgroup$ Commented May 2 at 3:36

1 Answer 1

0
$\begingroup$

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 $|X|_{\sup}=p^{-1/2} \notin |K|$.

$\endgroup$

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.