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\begin{document}
\normalsize
\section{Lecture 6: More constructions with Huber rings}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
\subsection{Introduction}
Recall from Definition 5.2.4 that a {\em Huber ring} is a commutative topological ring $A$ equipped with an open subring $A_0$, such that the subspace topology on $A_0$ is $I$-adic for some {\em finitely generated} ideal $I$ of $A_0$. This means that $\{I^n\}_{n\geq 0}$ is a base of open neighborhoods of $0\in A$, and implies that $A$ is a non-archimedean ring (see Definition 5.2.1). We stress that the topology on a Huber ring $A$ need not be separated (i.e., Hausdorff) nor complete. 

Recall from Lecture 5 that, given a Huber ring $A$ with a subring of definition $A_0$ which is $I$-adic with respect to a finitely generated ideal $I$ of $A_0$, its completion was defined to be 
$$A^{\wedge}:=\varprojlim A/I^n$$ where the limit is in the sense of (additive)
topological groups and is seen to carry a natural structure of topological ring whose topology is in fact separated and complete. A subring of definition of $A^{\wedge}$ is
 given by $A_0^{\wedge}$, the usual $I$-adic completion of $A_0$ in the sense
 of commutative algebra, and we saw (by a non-trivial theorem when $A_0$ is not noetherian) that
the inverse limit (same as subspace!)
topology on $A_0^{\wedge}$ is its $I (A_0)^{\wedge}$-adic topology
that is moreover separated and complete. 

This lecture is devoted to introduce a special class of Huber rings which will play the role of Tate algebras (with ``radii'') in rigid geometry. We will make use of the constructions discussed here 
to define a good analogue of ``rational domain'' for adic spaces in the next lecture.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
\subsection{Topologically nilpotent units}
We start off with an instructive class of Huber rings that play an essential technical role in later developments with adic spaces.
Let $K$ be a nontrivially valued field, which is to say a field equipped with a nontrivial 
valuation $v:K\to\Gamma\cup\{0\}$ (perhaps with rank $> 1$; see Remark 2.2.4). 
Let $R\subset K$ be its valuation ring. 

We view $K$ as a topological ring $A$ with base of open neighbourhoods 
of $0$ given by
$$A_{\gamma}:=\{x\in K\mid v(x)<\gamma\},\ \ \gamma\in\Gamma,$$
so $K$ is thereby a non-archimedean ring. Note that $K$ is even a topological field, which is
to say that $K^{\times} = K - \{0\}$ with its subspace topology is a topological group.  Indeed,
since multiplication is continuous (as $K$ is a topological ring) 
we just have to check that inversion is continuous near 1, or in other words
that for any $\gamma \in \Gamma$ there exists $\delta \in \Gamma$ such that if
$v(x - 1) < \delta$ then $v(x^{-1} - 1) < \gamma$.
Without loss of generality $\gamma < 1$, so $v(x) = 1$ and hence
$v(x^{-1} - 1) = v(1 - x)/v(x) = v(x - 1)$.  Thus, we may use $\delta = \gamma$.

The question we wish to address is:
when is $K$ a Huber ring? By definition, we seek an open subring $A_0\subset K$ such that the topology on $A_0$ is $I$-adic,for some finitely generated ideal $I\subset A_0$. Such $A_0$ may well not be the valuation ring $R$, and therefore $I$ may well not be principal! The easy case is when $v$ is a rank-1 valuation, and hence any pseudo-uniformizer $\varpi$ (that is, $\varpi\in R$ with $0<v(\varpi)<1$) will generate an ideal $I=\varpi R$ making $R$ into an open subring of $K$ with the 
$I$-adic topology.  

Suppose we are given $A_0\subset K=:A$ and a nonzero finitely generated ideal $I\subset A_0$ with the desired properties. Since we are assuming $K$ is non-trivially valued, $I$ must be nonzero
since $\{0\}$ is not an open subset of $A$. Therefore, we may
 pick a nonzero element $\varpi\in I$. But $\{I^n\}_{n\ge 0}$ is a base of open neighbourhoods of $0$ in $A$, and they are open ideals of $A_0$, so $\varpi$ is topologically nilpotent in $A$.
 Moreover, $A=K$ is a field and $\varpi$ is nonzero, so $\varpi$ is a topologically nilpotent {\em unit}.

How do we recognize topologically nilpotent units in $A=K$?  As a first attempt we might guess that such elements are those nonzero elements $\varpi\in A$ with $v(\varpi)<1$. This is the case when the valuation $v$ is rank-1.  Indeed, the ``archimedean property'' of rank-1 value groups gives that
 if $0<v(\varpi)<1$ then (via an inclusion $\Gamma \hookrightarrow \bR^{\times}_{>0}$) we have $v(\varpi^n)=v(\varpi)^n\to 0$ as $n\to\infty$, and the principal ideal $I=\varpi R$ is moreover open
 since multiplication by $\varpi$ is an automorphism of the topological space $K$
 (as multiplication by $1/\varpi$ is a continuous inverse).  
 
 But if $v$ is a higher-rank
 valuation then the property of $\varpi$ of being a pseudo-uniformizer generally
 does {\em not} imply that 
 $\varpi$ is topologically nilpotent. This is illustrated
 by the rank-2 valuation ring considered in Example 2.3.4 and Example 4.3.3:


\begin{eg} \label{eg: ourfavouritexample} Let $k$ be a field, and $K=k(\!(u)\!)(\!(t)\!)$. Let $w$ be the rank two valuation produced endowing the residue field $k(\!(u)\!)$ of $k(\!(u)\!)[\![t]\!]$ with the $u$-adic valuation, as in Example 2.3.4. Its valuation ring is
$$R:=k[\![u]\!]+t k(\!(u)\!)[\![t]\!],$$
with value group $\Gamma:= a^{\bZ}\times b^{\bZ}$ for real numbers $0<a,b<1$ 
with $a := w(t)$ and $b := w(u)$. The total ordering on $\Gamma$ is lexicographical, where the ``$t$-factor'' comes first. Such ordering arises naturally from the construction explained in $\S 2.3$ and detailed in Example 4.3.3.

Briefly, recall that $\Gamma$ is a split extension of the value group of the $t$-adic valuation on $K$, which is $a^{\bZ}$, and of the $u$-adic valuation on $k(\!(u)\!)$, which is $b^{\bZ}$. Geometrically,
the $t$-adic valuation on $K$ (with valuation ring $k(\!(u)\!)[\![t]\!]$) seen as a point in $\Spv(K)$ is a vertical generization of $w$; i.e., it is obtained as $w_{/_H}$ for the convex subgroup $H:=1\times b^{\bZ}$ of $\Gamma$, as explained in Example 4.3.3. Let us define $v:=w_{/_H}$, and recall that the picture is as follows, with $\eta$ the trivial valuation on $K$:
\begin{displaymath}
\xymatrix{\Spv(K)\ar[dddd] & \Spv(K) \\
			&  \eta \ar@{~>}[d]\\
		 &	v\ar@{~>}[d]\\
			& w \\
		\Spec(K)		& (0) }
\end{displaymath}

That $w$ is a specialization of $v$ means in particular that every open subset of $\Spv(K)$ containing $w$ already contains $v$. We now interpret this in terms of the valuation topologies on $K$.

We claim that the valuation topologies on $K$ arising from $v$ and $w$ {\em coincide}.  
Observe that the unit $t \in K^{\times}$ is topologically nilpotent for both topologies.
Indeed, for $v$ this is clear from the definition, and for $w$ we just have to check that for
any $\gamma = (a^i, b^j)$ there exists a large $n$ so that $(a,1)^n \le \gamma$.
That is, we want $(a^{n-i}, 1/b^j) \le (1,1)$.  Taking $n > i$ ensures $a^{n-i} < 1$, so the lexicographical
ordering implies $(a^{n-i}, 1/b^j) < (1,1)$ regardless of the value of $j$.
Hence, for the valuation rings $R$ and $R_v = k(\!(u)\!)[\![t]\![$ of $w$ and $v$ respectively, the collections
$\{t^n R\}$ and $\{t^n R_v\}$ for $n \ge 0$ are respective bases of opens around 0 for the two topologies.
But $t^{n+1}R_v \subset t^n R \subset t^n R_v$, so the topologies coincide.

Informally, the $w$-adic topology on $K$ coincides with the ``$t$-adic topology''
even though $w$ is a higher-rank valuation.
In particular, although $w(u) < (1,1) = 1_{\Gamma_w}$, 
$u$ is {\em not} topologically nilpotent (a purely topological property) since we can see this in terms of the $v$-adic valuation.
More concretely, $t/u^n \in R$ for all $n \ge 0$, so $w(t) \le w(u)^n$ for all $n \ge 0$.
Similarly we see that $1/u$ is power-bounded in the $w$-adic topology of $K$
(a purely topological notion) yet this lies {\em outside} the $w$-adic valuation ring $R$.
\end{eg}

Example \ref{eg: ourfavouritexample} shows that given a nontrivially-valued field 
$K$ with valuation $v:K\to\Gamma\cup\{0\}$, it is not true that $\varpi\in K^{\times}$ satisfying
$0<v(\varpi)<1$ is topologically nilpotent. Nevertheless, for our favourite choice $K=k(\!(u)\!)(\!(t)\!)$ we can find a topologically nilpotent element $t$ in $K$ and this establishes the Huber condition for $K$.

\medskip

\begin{rem} \label{rem: height one} We remark that if $K$ is a nontrivially-valued field then the trivial valuation is {\em not} continuous in the sense of Definition 5.3.1 since $\{0\}$ is not open. We also note that in Example \ref{eg: ourfavouritexample} {\em both} $w$ and its vertical generization $v$ are continuous, and the only possible further vertical generization of $v$ is the trivial valuation $\eta$ on 
$K$, which was not continuous. 

In a later lecture we shall discuss the notion of {\em analytic points} in the ``continuous'' valuation spectrum of a Huber ring $A$: these are the $v\in\Cont(A)$ having support $\fp_v\in\Spec(A)$ {\em not} open (so $w$ and $v$ in Example \ref{eg: ourfavouritexample} are both analytic, while $\eta$ is not, since it is not even in $\Cont(A)$). We will see that within the subspace of analytic points of $\Cont(A)$ there are no horizontal specializations, and every analytic point is of rank $\geq 1$, admitting a unique rank-1 vertical generization.  For this reason, rank-1 points will wind up playing an important role in the general theory of adic spaces, despite the abundance of higher-rank points. 
\end{rem}

Here is a converse to the preceding considerations.

\begin{prop} \label{prop: field case} Let $K$ be a nontrivially-valued field with valuation ring $R$, and suppose there exists a topologically nilpotent unit $\varpi\in K^{\times}$. Then $K$ is Huber, with ring of definition $A_0:=R$ and ideal of definition 
$I=\varpi^e A_0$ for $e$ large enough so that $\varpi^e$ lies in the open subring $R$.
\end{prop}

\begin{proof} Replace $\varpi$ by $\varpi^e$ for large enough $e$ such that $\varpi \in R$.
Since $\varpi$ is a unit in $K$, multiplication by $\varpi$ is invertible and continuous, and hence $I^n=\varpi^n R$ is an open ideal of $R$ for all nonnegative integers $n$. We claim $\{\varpi^n R\}_{n\ge 0}$ is a base of neighbourhoods of $0$ in $K$. 

Choose $\gamma$ in the nontrivial value group $\Gamma$ of the valuation on $K$ associated to $R$, so we want to show that for large enough $n\ge 0$, we have
$$v(\varpi^n x)<\gamma$$
for all $x\in R$. We have $v(x)\le 1$, so $v(\varpi^n x) \le v(\varpi^n)$,
and since $\varpi$ is topologically nilpotent we have that $v(\varpi^n)<\gamma$ for sufficiently large $n\ge 0$.  This proves the claim.
\end{proof}

\begin{rem} \label{rem: field case} In the situation of Proposition \ref{prop: field case}, 
by inverting $\varpi$ we recover $K$: 
$$R[1/\varpi]=K.$$
Indeed, choose $x\in K$ so we seek some $n\ge 0$ such that $\varpi^nx\in R$. We may
assume $x\neq 0$. Then we are reduced to showing that $$v(\varpi)^n\leq 1/v(x) = v(1/x)$$
for some large $n$, 
where the right side is an element of $\Gamma$. But this is ensured by the fact that $\varpi$ is topologically nilpotent.
\end{rem}

We have seen that a nontrivially-valued field with higher-rank value group may have its
valuation topology defined by powers of a single element. However, generally that topology is 
{\em not} the adic one defined by powers of the (nonzero) maximal ideal $\fm_R$. Here are two examples (one of which is our favourite usual one):

\begin{eg} \label{eg: max adic gone}

\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(\arabic{enumi})}}
\item Let $K$ be a non-archimedean field (i.e., complete for a nontrivial non-archimedean absolute
value) with valuation ring $R$ such that $\fm_R = \fm_R^e$ for some $e > 1$,
as happens for $\bC_p$ or (the completion of) $\bQ_p^{\rm cyc}$. 
The $\fm_R$-adic topology on $R$ is not even Hausdorff.

\item  Consider $R = k[\![u]\!] + tk(\!(u)\!)[\![t]\!]$ as in Example \ref{eg: ourfavouritexample}, with lexicographically ordered value group $a^{\bZ} \times b^{\bZ}$ for $a,b$ real numbers $0<a,b<1$.
Note that in contrast with the previous example, this one does not have any infinitely
divisible elements in the value group.
We have seen that $u$ lies in the maximal ideal $\fm_R$ of $R$ and is {\em not} topologically nilpotent (and for the same reason $1/u$ is power-bounded in the topology of the valued field 
$\Frac(R) = R[1/t]$), so the topology cannot be $\fm_R$-adic. 

By the same reasoning,
the same thing happens for {\em any} valuation ring whose maximal ideal contains {\em some} element that is not topologically nilpotent.  Likewise, in all such cases there are power-bounded elements outside the valuation ring and inside the valued fraction field, namely the reciprocal of any element of $\fm_R$ that is not topologically nilpotent.
\end{enumerate}
\end{eg}

In the preceding considerations, we studied a nontrivially-valued field $K$ with valuation ring $R$ and found a topologically nilpotent unit $\varpi\in K^{\times}$ which is sufficient to cover the topology
on $K = R[1/\varpi]$.  This fits into a more general situation related to topologizing 
certain localizations of Huber rings. To explain this, we first prove the following general fact:

\begin{prop} \label{prop: topologizing gen fiber} Let $R$ be a ring, and $\varpi \in R$
an element that is not a zero-divisor. We endow $R$ with the $\varpi$-adic topology. 
Then $R[1/\varpi]$ has a unique structure of topological ring making 
the topological ring $R$ an {\em open} subring.
\end{prop}

\begin{proof}
Since $\varpi$ is not a zero-divisor, multiplication by $\varpi^n$ is injective on $R$.
If there is such a topological ring structure then a neighborhood basis of $0$ is given
by $\varpi^n R$ for all $n \ge 0$, so likewise for all $x \in R[1/\varpi]$
a neighborhood basis is given by $x + \varpi^n R$ for $n \ge 0$.

Conversely, let's define a subset $U \subset R[1/\varpi]$ to be {\em open}
if for every $u \in U$ we have $u + \varpi^n R \subset U$ for sufficiently large $n$
(depending on $u$).  This is clearly a topology on $R[1/\varpi]$
makes $R$ an open subspace with its $\varpi$-adic topology
and that makes $R[1/\varpi]$ a topological group under addition, and the only issue
is to check that multiplication is continuous.  For $x, y \in R[1/\varpi]$
we can pick $m \ge 0$ such that $x, y \in \varpi^{-m}R$, so for $n \ge m+N$ with $N > 0$ we have
$$(x + \varpi^n r)(y + \varpi^n r') \in xy + \varpi^N R,$$
giving the desired continuity.
\end{proof}

We give a non-example which we have already encountered.

\begin{eg} \label{eg: unit disc gen fiber} Let $R:=\bZ_p[\![x]\!]$, with the $(p,x)$-adic topology. We claim there does not exist any extension of the $(p,x)$-adic topology on $R$ to a topological ring structure on $R[1/p]$ making $R$ into an open subring with the given topology, or even to a topological group
structure making $R$ an open subgroup.  Indeed, if there were then the openness of
the inclusion of $R$ into $(1/p)R$ would imply that $pR$ is open in $R$ for the $(p,x)$-adic topology, which is not true.  (The elements $x^n$ converge to 0 for the $(p,x)$-adic topology but
$x^n \not\in pR$ for all $n\ge 0$.)
%
%Looking back to Example \ref{eg: ourfavouritexample}, the picture is, analogously, as follows. Let us call $\eta$ and $\omega$ the trivial valuations on $\bF_p(\!(x)\!)$ and $\bQ_p$ respectively, and $v_x$ and $v_p$ the $x$-adic valuation on $\bF_p(\!(x)\!)$ and the $p$-adic valuation on $\bQ_p$ respectively. Recall that $\Spv(\bF_p)$ consists in just the trivial valuation, and looking at Example 4.3.2, the picture is the following:
%\begin{displaymath}
%\xymatrix{\Spv(\bZ_p[\![x]\!])\ar[ddd] & \Spv(\bF_p(\!(x)\!)) & \Spv(\bF_p) & \Spv(\bQ_p)\\
%			&  \eta \ar@{~>}[d]\ar@{~>}[r] & \overline{\eta}=\overline{\omega}\ar@{=}[d] & \omega\ar@{~>}[l]\ar@{~>}[d]\\
%		 &	v_x \ar@{~>}[r]  & 			\overline{v_x}=\overline{v_p} & v_p\ar@{~>}[l]\\
%		\Spec(\bZ_p[\![x]\!])	& (p)	& (p,x) & (x) }
%\end{displaymath}
%Notice we do not discuss (now) the ``generic fiber'' of $\Spv(\bZ_p[\![x]\!])$, as its
%analogue in the setting of adic spaces
% is {\em not} going to be affinoid! Let us call $\alpha$ the unique point in $\Spv(\bF_p)$, 
% to which all the other points in the picture specialize. The prime ideal $(p,x)$ in 
% $\bZ_p[\![x]\!]$ is the unique open prime ideal, and $(p)$ and $(x)$ are both not open. Due to Remark \ref{rem: height one} we expect to have unique rank-1 points corresponding to continuous 
% valuations on $\bF_p(\!(x)\!)$ and $\bQ_p$ respectively, and indeed we (trivially) do. 
\end{eg}

\begin{rem} \label{rem: powerbounded} 
We recall Proposition 5.4.13, saying that given a Huber ring $A$, the subring of power-bounded elements $A^0$ is the filtered direct limit of the bounded subrings of $A$ (that is, the subrings of definition of $A$, by Proposition 5.4.8). Example \ref{eg: ourfavouritexample} gives an example of Huber ring $A$ (the field $K$) such that our ring of definition $A_0$ (that is, $R$ in the example) is {\em strictly} contained in $A^0$. In fact, since $w(u^n)>w(t)$, $1/u$ is power-bounded, but it is not 
contained in $R$.
\end{rem}





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
\subsection{Topologizing localizations}

In rigid-analytic geometry over a non-archimedean field $k$, for a $k$-affinoid algebra $A$ we
define the relative Tate algebra
$$A\langle X_1,\cdots,X_n\rangle=\{\sum_I a_I X^I\mid a_I\to 0\ \textup{as}\ |I|\to\infty\}$$
that is equal to $A\fotimes_k k\langle X_1\cdots, X_n\rangle$ (where by $X$ we mean $(X_1,\cdots, X_n)$). Note that $A\langle X_1,\cdots,X_n\rangle$ is the universal example of a ``complete'' non-archimedean topological $A$-algebra $B$ equipped with an ordered $n$-tuple of power-bounded elements $b_1,\cdots, b_n$. By ``complete'' we mean that all Cauchy nets converge, and we implicitly include the separatedness assumption.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
\paragraph{Rigid-analytic rational subdomains}\ \\

Letting again $A$ be a $k$-affinoid algebra, let $t_1,\cdots,t_n$ be units in $A$. 
Recall that $A$ has an intrinsic Banach topology, though the ``sup-norm'' is only such a norm
when $A$ is reduced (and in the non-reduced case it is not a norm at all, since it vanishes
on nonzero nilpotent elements).  The Banach $A$-algebra
$$(*)\ \ \ A\left\langle\frac{X_1}{t_1},\cdots,\frac{X_n}{t_n}\right\rangle$$
corresponds to a rational subdomain of $\Sp\ A\langle X_1,\cdots, X_n\rangle$, namely 
$$\{(s, x)\in\Sp(A)\times\bA^{n,{\rm{an}}}_k\mid |x_i|\leq|t_i(s)|\ \textup{for all}\ i\}$$
where by $x$ we mean $(x_1,\cdots, x_n)$, and by $\bA^{n,{\rm{an}}}_k$ we mean the {\em analytic} affine $n$-space over $k$.
Informally, this is a family of closed polyndiscs parameterized by ${\rm{Sp}}(A)$,
with fiber over $s \in {\rm{Sp}}(A)$ having polyradii $\{|t_i(s)|\}$.

Explicitly, $(*)$ is given by
$$\left\{\sum_Ia_IX^I\mid a_It^I\to 0\ \textup{as}\ |I|\to\infty\right\}.$$
This can be characterized in purely topological terms: it is the complete non-archimedean $A$-algebra $f:A\to B$ equipped with an ordered $n$-tuple of elements 
$b_1,\cdots, b_n\in B$ such that $b_i/f(t_i)$ is power-bounded for all $i$.






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
\paragraph{A slightly different variant}\ \\

Let now $f_1,\cdots, f_n$ and $g$ be elements of $A$ with {\em no common zero}. Then we let
$$A\left\langle\frac{f_1}{g},\cdots,\frac{f_n}{g}\right\rangle:=\frac{A\langle X_1,\cdots, X_n\rangle}{\langle gX_i-f_i\rangle}.$$
Notice that since the $f_i$'s and $g$ have no common zero, $g$ is a unit in $A\left\langle\frac{f_1}{g},\cdots,\frac{f_n}{g}\right\rangle$ (as a zero of $g$ in the MaxSpec of this $k$-affinoid algebra
would project to a point in ${\rm{Sp}}(A)$ that is a common zero of the $f_i$'s, a contradiction). 
We can interpret $A\left\langle\frac{f_1}{g},\cdots,\frac{f_n}{g}\right\rangle$ as the universal 
complete non-archimedean $A$-algebra in which $g$ is a unit and $f_i/g$ is power-bounded for
all $i$.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
\paragraph{Finally, back to Huber rings}\ \\
Let $A$ be a non-archimedean ring with {\em no} Hausdorff or completeness assumptions. We now address the problem of finding suitable universal mapping properties for analogous constructions as for the ones recalled above, but for ``Tate algebras'' over Huber rings.  (The case
of relative rational domains will be addressed next time.)  For technical
reasons, it really simplifies matters to first carry out constructions with universal topologies
in the absence of completions, and then to bring in completions only afterwards.

This is essentially a game of appropriately 
topologizing the polynomial ring $A[X_1,\cdots, X_n]$, which we shall 
denote as $A[X]$ for brevity. In the sequel 
we shall indeed always write $X$ for $(X_1,\cdots, X_n)$ when this will not cause confusion. 

The idea is to describe the topology on $A[X]$
in terms of a base of neighbourhoods around $0$ in $A$. We define:
$$U[X]:=\left\{\sum_{\rm finite} u_I X^I\mid u_I\in U\right\}\subset A[X]$$
as {\em additive subgroups} of $A[X]$, where $U$ runs through elements of a base of open neighbourhoods of $0$ in $A$. We therefore wonder if we can make $U[X]$ into a neighbourhood base around $0\in A[X]$ in order to generate a topological ring structure on $A[X]$.

We first observe that $\{f+ U[X]\}_U$ is indeed a neighbourhood base around $\{f\}$ in $A[X]$, for all $f\in A[X]$, $U$ running through a neighbourhood base of 
additive open subgroups of $A$. The only issue 
is to ensure continuity of the multiplicative structure on $A[X]$.  More precisely:

\begin{lemma} \label{lemma: multiplicationok} Let $R$ be a commutative ring, and $\Sigma:=\{H_{\alpha}\}$ be a collection of additive subgroups of $R$. These are a base of neighbourhoods for a $($necessarily unique$)$ non-archimedean ring structure on $R$ if and only if:
\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(\arabic{enumi})}}
\item for all $H, H'\in\Sigma$, $H\cap H'$ contains some $H''$ in $\Sigma$, 
\item for all $H\in\Sigma$, there exists $H'\in\Sigma$ such that $H\cdot H'\subseteq H$, 
\item for all $r\in R$ and all $H\in\Sigma$, there exists $H'\in\Sigma$ such that $rH'\subset H$.
\end{enumerate}
\end{lemma}

We remark that point $(1)$ in the Lemma just encodes the fact that the $H_{\alpha}$'s are a base of neighbourhoods of $0$ for an additive topological group structure on $R$. Point $(2)$ just ensures that the multiplication of $R$ is continuous at $0$, and point $(3)$ handles products of translates of the form
$$(r+x)\cdot(r'+x')=rr'+rx'+r'x+xx',$$
where $rr'$ is already ``small'' by point $(2)$ but $rx'+r'x+xx'$ must be handled.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
\paragraph{Application: (uncompleted!) relative Tate algebras}\ \\

Let $A$ be a non-archimedean topological ring, and let $T_1,\cdots, T_n\subset A$ be finitely many finite nonempty subsets. For example $T_i$ may be a singleton $\{t_i\}$ for all $i$.

We define, for every multi-index $I\in\bN^n$:
$$T^I:=T_1^{i_1}\cdot \cdots \cdot T_n^{i_n}:=\{\xi_1\cdot\cdots\cdot \xi_n\mid \xi_j\in T_j^{i_j},\ j=1,\cdots, n\}$$
where $T_j^{i_j}$ is the set of all $i_j$-fold products (allowing repetitions) among elements in $T_j$.

\begin{defin} \label{defin: notation} We set
$$A[X]_T:=A[X_1,\cdots, X_n]$$
as an $A$-algebra, equipped with a neighbourhood base of $0$ given by
$$U[TX]:=\left\{\sum_{\rm finite} a_I X^I\mid a_I\in T^I\cdot U\right\}$$
for open additive subgroups $U\subset A$, where by $T^I\cdot U$ we mean the set of all finite sums of monomials $\xi\cdot u$, for $\xi\in T^I$ and $u\in U$.
\end{defin}

\begin{rem}
In Definition \ref{defin: notation}, one should think of the classical rigid setting, where (under
mild hypotheses) the {\em completion}
$A[X]_T$ is
 going to correspond to the adic space of points $v$
 with the property that $v(x_it_{ij})\leq 1$ for all $t_{ij}\in T_i$ and all $i$.
 \end{rem}
 
 \begin{rem} \label{rem: horrendousnotation}
We warn the reader that the notation $U[TX]$ used to denote
$$\left\{\sum_{\rm finite} a_I X^I\mid a_I\in T^IU \mbox{ for all } I\right\}$$
is different from that in \cite{Wed}, where the same set is denoted by $U_{[X]}$
(suppressing the mention of $T$). 
Thus, when referring to \cite{Wed} the reader should
 first read \cite[Rem. 5.4.7]{Wed} and keep in mind this Remark.
\end{rem}


We now determine precisely when Definition \ref{defin: notation}
gives a topological ring structure on $A[X]_T$, which is to say 
when $A[X]_T$ satisfies the conditions in Lemma \ref{lemma: multiplicationok} with 
$\Sigma:=\{U[TX]\}_U$ for $U$ running over additive subgroups of $A$ forming a 
neighbourhood base around 0 for the topology on $A$.

Property $(1)$ always holds.  Indeed, for all $U, V$ open additive subgroups of $A$ we have 
$$U[TX]\cap V[TX]\supset (U\cap V)[TX].$$
Likewise, (2) always holds.  This requires showing that if $U$ is an 
open additive subgroup of $A$ then there exists 
an open additive subgroup $V\subset A$ such that $$V[TX]\cdot V[TX]\subset U[TX].$$
Consider $V$ for which $V^2\subset U$.  Since 
$T^I\cdot T^J=T^{I+J}$ (with clear meaning of notation), for such $V$
we just need to check that $vX^{I_0} \cdot V[TX] \subset U[TX]$
for any $I_0$ and $v \in V$,  which is obviously true.

We now check exactly when $(3)$ holds. 
Fix $f\in A[X]$ and $U\subset A$. We seek a $V\subset A$ such that $$fV[TX]\subset U[TX].$$
Additivity is not an issue, so without loss of generality we can assume $f=aX^I$ for some $a\in A$ and some multi-index $I\in\bN^n$. (This is why it is technically convenient that we focus now
on topologizing merely a polynomial ring, postponing consideration of completions
until later.)

Since $A[X]$ is commutative and $A$ is a topological ring, we can actually absorb $a$ in $V$ and we are reduced to the case $f(X)=X^{I_0}$ for some fixed multi-index $I_0\in\bN^n$. By iteration,
we may even assume $f(X)=X_{i_0}$ for some fixed $i_0\in\{1,\cdots, n\}$. 
For every open additive subgroup $U\subset A$ we therefore seek a $V\subset A$ 
with the property that
$$X_{i_0}V[TX]\subset U[TX].$$

The typical element of $V[TX]$ will be, by definition, of the form
$$\sum _I a_I X^I,$$
where the sun runs over finitely many multi-indexes, and $a_I\in T^IV$ for such $I$'s. We therefore seek $V$ such that for all such $a_I$'s we get
$$\sum_I a_I X^{e_{i_0}+I}\in U[TX]$$
where $e_{i_0}$ is the standard vector with coordinates $0$ at every slot but the $i_0$th, which is $1$. 

Since additivity is handled by definition of the $V[TX]$'s, we are reduced to analyzing each
$a_I X^{e_{i_0}+I}$ separately. It is therefore necessary and sufficient
to find $V\subset A$ such that
whenever $a_I\in T^IV$ we have
$$a_IX^{e_{i_0}+I}\in U[TX].$$
This is equivalent to saying that for all $I$
$$T^IV\subset T_{i_0}T^IU,$$
which is equivalent to requiring that for every additive open subgroup
$U\subset A$ there is an additive open subgroup $V\subset A$ such that
$$V\subset T_{i_0}U,$$
which is to say $T_{i_0} U$ is {\em open} for all such $U$ and all $i_0$. 
If we iterate we can translate the condition of $A[X]_T$ being a (non-archimedean) topological ring to 
exactly the openness of $T^I\cdot U$ for all $I$ and all $U$ (or equivalently  the openness
of $T_{i_0} U$ for all $i_0$ and all $U$). 

We now specialize to our case of interest: $A$ a Huber ring. 
In this case we will show that $T \cdot U$ is open 
under a natural hypothesis that is reminscent of the condition ``no common
zeros'' used to define rational domains in the rigid-analytic case (since an open ideal
in an affinoid algebra is necessary the entire ring):

\begin{prop}\label{prop: huberok} Let $A$ be a Huber ring, and $T$ a finite subset of $A$ generating an open ideal.  Then $T \cdot U$ is open in $A$
for all open subgroups $U \subset A$.
\end{prop}

\begin{proof} The ideal generated by $T$ is $T \cdot A$, and this is open by hypothesis.  But if $S$ and $S'$ are open subsets of $A$ around 0 then the
additive subgroup 
$S \cdot S'$ of finite sums of products $ss'$ for $s \in A$ and $s' \in S'$ is open because $A$ is Huber. (Indeed, if $A_0$ is a ring of definition and $I$
is an ideal of definition in $A_0$ then by openness of $S$ and $S'$ there are large $n, n' > 0$ such that $I^n \subset S$ and $I^{n'} \subset S'$.
Hence, the additive subgroup $S \cdot S'$ contains $I^n \cdot I^{n'} = I^{n+n'}$ and therefore is a neighborhood of 0 in $A$, so it is open.)  Thus,
$(T \cdot A)^n = T^n \cdot A$ is open for all $n \ge 1$ (where $T^n$ means the set of $n$-fold products among elements of $T$).

Now fix $n > 0$ and an open subgroup $U$ of $A$, and we aim to show that the additive subgroup $T^n \cdot U$ of $A$ is open, or equivalently
that it contains an open set around 0.   
Choose a ring of definition $A_0$, and choose an ideal of definition $I$ of $A_0$ that is contained in the open subset $T^n \cdot A$ of $A$ around 0. (Namely, pick some ideal of definition of $A_0$ and then replace it with a very large power.)  We seek a large $e$ so that $I^e \subset T^n \cdot A$.
By definition of ``ideal of definition'', $I$ has a {\em finite} set $G$ of generators. Thus, 
$G \subset T^n \cdot A$ and hence we can find a {\em finite} subset $F \subset A$ such that $G \subset T^n \cdot F$ (the set of products of elements of $F$ against $n$-fold products among elements of $T$).  By finiteness of $F$ and openness of $U$ around 0, we can choose $m > 0$ large such that $F \cdot I^m \subset U$.  Thus,
$$T^n \cdot U \supseteq T^n \cdot F \cdot I^m \supseteq G \cdot I^m = I^{m+1}$$
(the final equality by definition of $G$ as a generating set for $I$ as an ideal of the ring $A_0$).
\end{proof}
This concludes the proof that the assignement of $\Sigma:=\{U[TX]\}_U$, for $U$ running over the open additive subgroups of $A$, defines a base of neighbourhoods for a non-archimedean topological ring structure on $A[X]_T$, for $A$ a Huber ring.

We have shown the following:
\begin{prop} \label{prop: umproperty} Let $A$ be a Huber ring, and 
$T = \{T_1, \dots, T_n\}$
a collection of non-empty finite subsets of $A$ such that $T_{i_0} U$ is open
for all open additive subgroups $U \subset A$ and all $i_0$.
Then the non-archimedean $A$-algebra $A[X]_T$ is initial among
non-archimedean $A$-algebras $f:A\to B$
equipped with an ordered $n$-tuple of $b_1,\cdots, b_n\in B$ such that the elements
$$f(t_{ij})b_i\in B$$
are power-bounded for all $i$ and all $t_{ij}\in T_i$.
\end{prop}

\begin{proof}
The essential point is to check that $A[X]_T$ satisfies the asserted power-boundedness
property; the rest is then easy to check (and left to the reader).  That is, 
for all $i$ and all $t_{ij}\in T_i$ we claim that the element 
$$t_{ij}X_i\in A[X]_T$$
is power-bounded. For all $m \ge 1$ we have 
$$(t_{ij}X_i)^m\in T^{me_i}X^{me_i}$$
for any $i$ and any $t_{ij}\in T_i$. On the other hand, 
$$T^{I_0}X^{I_0}U[TX]\subset U[TX]$$
for all $I_0$ and all open additive subgroup
$U\subset A$. We apply this to $I_0=me_i$ for $m \ge 0$ to conclude. 
\end{proof}

A non-example:
\begin{eg} \label{eg: nonex}
Let $A:=\bZ_p[\![x]\!]$, equipped with the $(p,x)$-adic topology, and let $T:=\{p\}$. Then $T$ does not generate an open ideal, and calling $I:=(p,x)$, none of the additive subgroups $pI^nA\subset A$, $n\ge 0$, is $I$-adically open.
\end{eg}

Next time we will pass to completions, record the universal property as such (relative to maps
from $A$ to {\em complete} non-archimedean rings), and then especially carry
out analogous constructions for certain rings of fractions (recovering
a Huber-ring variant of the formation
of rational domains in rigid-analytic geometry, this being
the key reason we consider $T_1, \dots, T_n$ that are finite non-empty sets
and not merely singletons inside $A$). 
%
%\small
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\end{document}