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Let me start with the following Lemma.

$\textbf{Lemma}$ Let $A$ be a Tate ring, and let $f\colon A\to B$ be a finite $A$-algebra. Then there is a unique way to topologise $B$ turning it into a Huber ring, making the structure morphism $A\to B$ continuous and such that there exists rings of definition $A_0$, $B_0$ of $A$ and $B$, respectively, with $f(A_0)\subset B_0$ such that the induced map $A_0\to B_0$ is finite.

$\textbf{Proof.}$ $\textit{Existence}$: Give $B$ the canonical $A$-module topology, i.e. give it the quotient topology for any surjection $A^n\twoheadrightarrow B$ of $A$-modules (this does not depend on the choice). Let $A_0$ be a ring of definition and let $\pi\in A_0\cap A^\times$ be a pseudo-uniformizer. Choose a finite number of generators $\{b_i\}$ of $B$ as an $A$-module s.t. $1$ is one of them, with each $b_i$ being integral over $A_0$. Denote by $M$ the $A_0$-submodule of $B$ generated by $\{b_i\}$. Then the integrality allows to choose $m\gg 0$ such that $B_0:=A_0[\pi^mb_1,\ldots,\pi^m b_n]\subset M$. Since also $\pi^m M\subset B_0$, the subspace topology on $B_0$ is the $\pi$-adic topology. Thus, $B$ is a Huber ring with ring of definition $B_0$. Clearly the structure morphism is continuous and the induced map $A_0\to B_0$ is finite as $B_0$ is integral and of finite type over $A_0$.

$\textit{Uniqueness}$: Since $B_0$ is finite over $A_0$, there exists $m\gg 0$ such that $\pi^m B_0\subset M$. On the other hand, since the structure morphism is continuous, $\pi$ is topologically nilpotent in $B$ and so $\pi^N M\subset B_0$ for $N\gg 0$. This shows that the topology on $B$ is the one uniquely determined by making $M$ with the $\pi$-adic topology an open subgroup, which is precisely the canonical topology.

  1. I am wondering if the condition on the existence of $A_0,B_0$ making $A_0\to B_0$ finite is automatically satisfied knowing that the structure morphism is continuous and finite étale. Namely, in Remark 7.2 of these notes by Matthew Morrow, the uniqueness in case $A\to B$ being finite étale is stated without any finiteness condition on the level of rings of definition.

  2. In Huber's book "Étale Cohomology of Rigid Analytic Varieties and Adic Spaces" it says in $(1.4.2)$ that for a (complete) Huber ring (it does not say Tate) and $A\to B$ a finite algebra, the canonical $A$-module topology on $B$ turns it into a Huber ring. My proof above clearly makes use of the existence of a topologically nilpotent unit, and I would like to know what the general argument looks like. Huber refers to $(3.12.10)$ of his "Bewertungsspektrum und rigide Geometrie", which is unfortunately not available to me.

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  • $\begingroup$ If you look at Scholze's notes on condensed mathematics, he gives a proof of this using solid rings. $\endgroup$ Aug 11 '20 at 5:33
  • $\begingroup$ I don't think this Remark 7.2 is correct as stated: if you start with a non-uniform Tate ring $A$, then we can let $B=A$ as a ring but instead take the topology such that $A^\circ$ is a ring of definition for $B$ (with the same pseudouniformizer). This gives us a different topology on $B$ and the map $A\to B$ is continuous. $\endgroup$ Jan 11 at 8:49
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I think I can prove statement 2. (I too do not have access to this book "Bewertungsspektrum und rigide Geometrie" and was trying to look it up...)

Let $B$ be generated as an $A$-module by $b_1,\dots,b_n$. Write $b_ib_j=\sum a_{ijk}b_k$ for $a_{ijk}\in A$. Pick a small enough ideal of definition $I$ of $A_0$ such that $Ia_{ijk}\subset A_0$ for all $i,j,k$. Then we see that $$ B_0=A_0+Ib_1+\dots+Ib_n $$ is a subring of $B$.

Now note that if $U\subset A^n$ is open additive subgroup containing $0$, then $q^{-1}(q(U))$ is an additive subgroup of $A^n$ containing $U$, so it is open, and thus $q(U)$ is open. We now see that $B_0$ is an additive subgroup of $B$ containing the open additive subgroup $q(I\times\dots \times I)$, so $B_0$ is open.

Now if $V\subset B$ were an open neighborhood of $0$, then $q^{-1}(V)$ contains $I^N\times \dots \times I^N$ for some large $N$, so $U$ contains $q(q^{-1}(V))=I^Nb_1+\dots+I^Nb_n$. Write $1=a_1b_1+\dots+a_nb_n$ for $a_i\in A$. Pick a large enough $M$ such that $M>N-1$ and $I^Ma_i\subset I^N$. Then $$ I^MB_0 = I^MA_0+I^{M+1}b_1+\dots+I^{M+1}b_n \subset I^Nb_1+\dots+I^Nb_n\subset V. $$ We conclude that the ideals $(IB_0)^m$ form an open neighborhood basis of $0$ in $B$. Thus $B$ is a Huber ring with ring of definition $B_0$ and ideal of definition $IB_0$.

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