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\begin{document}
\normalsize
\section{Lecture 5: Huber rings and continuous valuations}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
\subsection{Introduction}
%Before moving on, we briefly recall the setting we discussed last time. We had a commutative ring $A$, and a valuation
%$$v:A\to\Gamma\cup\{0\}$$
%and we learnt how to produce specializations of $v$ as a point in $\Spv(A)$ by means of two exhaustive procedures, which were ``commutative'' in a suitable sense (see Theorems 4.3.12, 4.4.1 and 4.4.2 of the Lecture 4 notes).
%
%We remark that, provided we are {\em not} in the situation $$v(A)\subseteq\Gamma_{\le 1}$$
%under the correspondence provided by Theorem between $v$-convex primes of $\Spec(A)$ and horizontal specializations of $v\in\Spv(A)$, such specializations are exactly the unique generic points in the fibers of $\Spv(A)\to\Spec(A)$ over the corresponding $v$-convex primes. The reader may refer to Examples 4.3.2 and 4.3.3 in Lecture 4.
%
%As an immediate consequence of the above remark, if $v(A)\nsubseteq\Gamma_{\le 1}$ then the image of $\overline{\{v\}}$ in $\Spec(A)$ is exactly the set of $v$-convex primes in $\Spec(A)$.

The aim of this lecture is to introduce a special class of topological commutative rings, which we shall call Huber rings.
These have associated spaces of valuations satisfying suitable continuity assumptions that will lead us to adic spaces later.
We start by discussing some basic facts about Huber rings.

{\bf Notation}:  If $S, S'$ are subsets of ring, we write $S \cdot S'$ to denote the set of
finite sums of products $ss'$ for $s \in S$ and $s' \in S'$.  We define
$S_1 \cdots S_n$ similarly for subsets $S_1, \dots, S_n$, and likewise
for $S^n$ (with $S_i = S$ for $1 \le i \le n$).  For ideals in a ring this agrees
with the usual notion of product among finitely many such.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Huber rings: preliminaries}
We begin with a definition:
\begin{defin} \label{def: nonarch}
Let $A$ be a topological ring. We say $A$ is {\em non-archimedean} if it admits a base of neighbourhoods of $0$ consisting of subgroups of the additive group $(A,+)$ underlying $A$.
\end{defin}
Typical examples of non-archimedean rings are non-archimedean fields $k$
and more generally $k$-affinoid algebras for such $k$.

\begin{eg}\label{eg: witt}
As another example, let $A:=W(\ccO_F)$, where $\ccO_F$ is the valuation ring of a non-archimedean field of characteristic $p>0$. A base of open neighbourhoods of $0$ required in Definition \ref{def: nonarch} is given by
$$p^nA=\{(0,\cdots, 0,\ccO_F,\ccO_F,\cdots ),\ \textup{zeros in the first}\ n\ \textup{slots}\};$$
the $p$-adic topology on $A$ is given by equipping
 $A = \prod_{n \ge 0} \ccO_F$ with the product topology in which each factor is discrete.

However, $\ccO_F$ naturally comes with its {\em valuation} topology, and using this for
the product topology yields a topology on $A$ that can be described
as follows. Let $\varpi\in\ccO_F$ be a {\em pseudo-uniformizer}; i.e., an element $\varpi \in F$ such that $$0<|\varpi|<1,$$
(so $\ccO_F$ has the $\varpi$-adic topology).  Such an element exists because the valuation on $F$
is non-trivial.  For the Teichm\"uller lift
$$[\varpi]=(\varpi, 0,\cdots , 0)\in A=\prod_{n=1}^{\infty}\ccO_F$$
it is an exercise to check that the $(p,[\varpi])$-adic topology on $A$ is the product topology for the valuation topology on $\ccO_F$.

For work with adic and perfectoid spaces we will need to topologize several algebraic constructions one can perform on $A$, 
such as localizations like $A[1/p]=W(\ccO_F)[1/p]$, or $A[1/[\varpi]]$. We will come to localizations next time.
\end{eg}

The next examples give us motivation to define Huber rings.

\begin{eg} \label{eg: topology on k affinoid}
Let $k$ be a non-archimedean field, and 
$$A:=k\langle X_1,\cdots, X_n\rangle/I$$
be an affinoid Tate $k$-algebra. This has a unique $k$-Banach topology
(defined via the residue norm via any such presentation, keeping in mind
that all ideals in a Tate algebra are closed), though a $k$-Banach algebra norm is not canonical.

Consider the subring 
$$\ccO_k\{X_1,\cdots,X_n\}$$
of $k\langle X_1\cdots X_n\rangle$ consisting of power series $\sum a_I X^I$
with coefficients in the valuation ring $\ccO_k$ of $k$ such that $a_I \rightarrow 0$ as $|\!|I|\!| \rightarrow \infty$.
Note that for a pseudo-uniformizer $\varpi$, this subring is
the $\varpi$-adic completion of $\ccO_k[X_1,\cdots, X_n]$ (even when $k$ is not discretely-valued, so $\ccO_k$ is not noetherian). 

Let $A_0$ be the image in $A$ of $\ccO_k\{X_1,\cdots, X_n\}$;
this is the quotient of $\ccO_k\{X_1,\cdots,X_n\}$ by $I\cap\ccO_k\{X_1,\cdots,X_n\}$. Note that 
 $A_0$ is an open subring of $A$ with respect to the subspace topology (quotient maps of Banach spaces are open).

The topology on $A_0$ is the $\varpi$-adic topology, and the topology on $A$ is ``controlled'' by $A_0$, in the sense that a basis of neighbourhoods of $0$ in $A$ is given by $\varpi^n A_0$, for $n\geq 1$.
\end{eg}

We are now ready to give the following:
\begin{defin}\label{def: Huber rings} A {\em Huber ring} $A$ is a topological ring admitting an open subring $A_0\subset A$ on which the subspace topology is the $I$-adic topology, for some finitely generated ideal $I$ of $A_0$. We call any such $A_0$ a {\em ring of definition} for the topology on $A$, or more briefly a {\em ring of definition}, and a couple $(A_0, I)$ will be called a {\em couple of definition}.
\end{defin}


Although Definition \ref{def: Huber rings} may make it seem that the topology on such rings $A$ is tightly linked to the assignment of a specific ring of definition $A_0$, and of a finitely generated ideal in this latter, we will see that this is not really a problem.
Let us first give a number of preliminary remarks:

\begin{rem} \label{rem: bunch of remarks}
\ 
\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(\arabic{enumi})}}
\item Let $a\in A$. A base of open neighbourhoods of $\{a\}$ in $A$ is given by
$$\{a+I^n\}_{n\ge 0}$$
where $I^n$ is viewed as an ideal of $A_0$, for each $n\ge 0$.
\item We fix a couple $(A_0, I)$ as prescribed in Definition \ref{def: Huber rings}. Assigning separatedness and completeness assumption on $A$ amounts to assigning them to the $I$-adic topology on $A_0$. We {\em do not} assume the topology on $A$ is separated neither complete! It is useful to avoid this assumption in order to handle algebraic constructions on $A$.
\item In \cite{H1}, Huber gives the following definition: a ring $A$ is a Huber ring ($f$-adic ring, in the terminology adopted in \cite{H1}) if there exists a subset $U$ of $A$ and a finite subset $T\subseteq U$ such that $\{U^n \mid n\ge 1\}$ is a fundamental system of neighborhoods of $0$ in $A$ and such that $T\cdot U = U^2\subseteq U$.\\
The advantage of such definition is that it does not mention a couple $(A_0,I)$ as done in Definition \ref{def: Huber rings}, which gives us initial motivation to believe that the assignment of a {\em specific} ring of definition $A_0\subset A$ and the $I$-adic topology on $A_0$ for some {\em specific} finitely generated ideal $I$ of $A_0$, is not so tightly related to the assignment of a Huber ring. 
See \cite[Prop. 6.1]{Wed} for a proof of the equivalence of the above definition with Definition \ref{def: Huber rings}.
\end{enumerate}
\end{rem}






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Digression: continuity of valuations}
\begin{defin} \label{def: continuous valuations} Consider a valuation 
$$v:A\to\Gamma\cup\{0\}$$ 
on a non-archimedean topological ring, and assume $\Gamma$ is generated by $v(A - {\rm{supp}}(v))$ (i.e.,
$\Gamma$ is the ``minimal'' value group for $v$, so $\Gamma$ is intrinsic to $v$ as a point of
$\Spv(A)$).  We say $v$ is {\em continuous} if for all elements $a\in A$ and 
$\gamma\in\Gamma$ the subsets of $A$
$$\{x\in A\mid v(x-a)<\gamma\}$$
are open.  (It is equivalent to consider such conditions with just $a=0$.)
\end{defin}

Note that a trivial valuation is continuous if and only if its support is an open prime ideal, since $\{v < 1\}$ is the support
of $v$ when $\Gamma_v = 1$.

\begin{rem} \label{rem: curiosity} Provided that $\Gamma \ne 1$, it is equivalent to require that the subsets
$$A_{\le \gamma} := \{x\in A\mid v(x)\le \gamma\}$$
are open for all $\gamma\in\Gamma$.
Indeed, defining $A_{<\gamma}:$ analogously, 
if $A_{<\gamma}$ is open then the additive subgroup $A_{\le \gamma}$ is open, and conversely if $\Gamma$ contains
elements $\delta < \gamma$ then 
$$A_{<\gamma}=\bigcup_{\delta<\gamma}A_{\le\delta}.$$
Such $\delta$ exist precisely because we assumed $\Gamma \ne 1$.

On the other hand, when $\Gamma = 1$ then $A_{\le 1} = A$ and $A_{< 1} = {\rm{supp}}(v)$, and the former
is always open whereas the latter may not be so.
\end{rem}

In Lecture 7 we shall study the ``continuous valuation spectrum'' of a commutative non-archimedean topological ring $A$: 
$$\Cont(A):=\{v\in\Spv(A)\mid v\ \textup{continuous}\}\subset\Spv(A).$$
This is functorial in $A$ in the sense that for a continuous map $f:A' \rightarrow A$ between non-archimedean
rings and $v \in \Cont(A)$ the valuation $v' := v \circ f \in \Spv(A')$ is continuous.  Indeed,
the value group $\Gamma'$ of $v'$ is a subgroup of the value group $\Gamma$ of $v$
and for any $\gamma' \in \Gamma'$ clearly $A'_{< \gamma'} = f^{-1}(A_{< \gamma'})$ is open.
Huber rings are non-archimedean rings, and for such rings $A$
we will see later that $\Cont(A)$ endowed with the subspace topology is a spectral space (hence qcqs).
The quasi-compactness properties of the ``adic spectum'' ${\rm{Spa}}(A)$ 
of Huber rings $A$ (to be defined later, using some additional structure on $A$
that will have to be incorporated into the notation) will rest on this fact.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Boundedness}
We recall that given a non-archimedean field $k$, and a $k$-affinoid algebra $A$, the set
$$A^0:=\{a\in A\mid\|a\|_{\rm sup}\le 1\}\subset A$$
consists of precisely the power-bounded elements of $A$ (that is, those elements the set of whose powers is bounded in $A$
 in terms of the Banach topology on $A$).
 Provided that $A$ is {\em reduced}, the sup-norm is a $k$-Banach algebra norm and thus $A^0$ is its unit ball.
 (For reduced $A$,
 whether or not this norm arises as the residue norm for a presentation of $A$ as a quotient of a Tate algebra is a subtle question;
 it amounts to asking if $A^0$ is topologically of finite type over $\ccO_k$.   One obvious obstruction is the possibility
 that there may exist $a \in A$ such that $|\!|a|\!|_{\rm{sup}} \not\in |k|$.  This is the only obstruction
 when $k$ is ``stable'' \cite[6.4.3/1]{bgr}.  See \cite[3.6.1]{bgr} for examples of non-stable $k$ and sufficient criteria
 for stability; stability holds in the discretely-valued case.)
 
Let $A_0\subset A$ be a ring of definition. Then from a quotient presentation of $A$ as in Example \ref{eg: topology on k affinoid}:
$$A\simeq k\langle T\rangle /I;$$
here, $T$  is a finite set of indeterminates.
Note that boundedness for a subset of $A$ 
is equivalent to being contained in some $\varpi^{-n}A_0$ for some sufficiently large positive integer $n$.
\begin{rem} Any ring of definition $A_0$ for a $k$-affinoid algebra $A$ is contained in $A^0$. We 
will give an analogue of this for Huber rings, but first we have to define boundedness without the crutch
of a Banach space structure (since it is {\em important} to consider Huber rings for which $I$ might not
be able to be taken as principal; e.g., $\mathbf{Z}_p[\![x]\!]$ with the $(p, x)$-adic topology).
\end{rem}

\begin{defin}\label{def: bounded} Let $A$ be a Huber ring. A subset $\Sigma\subset A$ is {\em bounded} if for every open neighbourhood 
$U$ of $0$ in $A$ there exists an open neighbourhood $V$ of $0$ in $A$ such that 
$$V\cdot\Sigma\subset U.$$
$($Here $V \cdot \Sigma$ denotes the additive span of the set of products $vs$ for $v \in V$ and $s \in \Sigma$,
and since we can shrink $U$ to be an additive subgroup of $A$ it is equivalent to demand that for every $U$
there is some $V$ such that $vs \in U$ for all $v \in V$ and $s \in \Sigma$.$)$

We say that $\Sigma$ is {\em power-bounded} if the set of products
$$\{s_1\cdot \cdots \cdot s_n\mid s_i\in\Sigma, i=1,\cdots n, n\ge 1\}$$
is bounded. For example, $\{a\}$ is power-bounded with $a \in A$ precisely when the set
$\{a, a^2, a^3, \dots\}$ is bounded. 
\end{defin}

If $\Sigma, \Sigma' \subset A$ are bounded subsets then so is the additive span $\Sigma \cdot \Sigma'$ of the set of products $ss'$
for $s \in \Sigma$ and $s' \in \Sigma'$.  Indeed, pick an open $U \subset A$, so there exists an open $V \subset A$ around 0 
such that $V \cdot \Sigma \subset U$.
There is also an open $V' \subset A$ around 0 such that $V' \cdot \Sigma' \subset V$, so $V' \cdot (\Sigma \cdot \Sigma) \subset
(V' \cdot \Sigma') \cdot \Sigma \subset V \cdot \Sigma \subset U$.  We will often use this fact without comment.
By the same reasoning, if $\Sigma_1, \dots, \Sigma_n$ is a finite collection of bounded
subsets of $A$ then the additive span $\Sigma_1 \cdots \Sigma_n$ of the set of products $\prod s_i$ for $s_i \in \Sigma_i$
is also bounded.  In particular, $\Sigma^n$ is bounded for each $n \ge 1$ when $\Sigma$ is bounded;
power-boundedness is of course a much stronger condition on $\Sigma$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\paragraph{Boundedness and continuous maps of Huber rings}\ \\
In contrast with the case of morphisms of $k$-affinoid algebras in classical rigid geometry (where all maps
are bounded between Banach spaces),
not all continuous homomorphisms between Huber rings respect topological properties such as boundedness! 

The issue is that if $f:A\to B$ is a continuous map between Huber rings, it is not clear how to control open neighbourhoods of $0$ in $B$ using just $f(A)$. 
This makes it hard to control boundedness of $f(\Sigma)$ in $B$ for a bounded subset $\Sigma$ in $A$.
(One case where all works well is when $f$ is surjective and $B$ has the quotient topology from $A$.)
Here is an example:

\begin{eg}
Let $A=\bQ_p$ with the $(0)$-adic topology, and let $B=A$ with the $p$-adic topology. So $A$ has the discrete topology, and it is Huber because it is trivially adic, with ring of definition $A_0=A$. Any ring homomorphism $A\to B$ is therefore continuous, but it cannot be adic. Call $f$ the set-theoretic identity map $A\to B$. Then
$f(A)=B$, which is not bounded in $B$ while $A$ is bounded in $A$.
\end{eg}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\paragraph{Adic homomorphisms of Huber rings}\ \\

Here is a particularly nice class of continous homomorphisms between Huber rings, called {\em adic},  which avoids the problem
in the preceding example (and is naturally motivated by experience with formal schemes as well).

\begin{defin} \label{def: adic maps}
Let $A$ and $B$ be Huber rings. A ring homomorphism $f:A\to B$ is called {\em adic} if 
there exist rings of definitions $A_0$ of $A$ and $B_0$ of $B$
such that $f(A_0)\subset B_0$ and for some ideal of definition $I\subset A_0$ the ideal $f(I)B_0$ is an ideal of definition in $B_0$.
\end{defin}

Any adic map of Huber rings is trivially continuous.  It is also easy to check (exercise!) that
if $f:A \rightarrow B$ is adic and $A_0$ and $B_0$ are rings of definition such that
$f(A_0) \subset B_0$ then $f(I)B_0$ is an ideal of definition of $B_0$ for {\em any} ideal of definition $I$ of $A_0$;
this is very important in practice.

For any continuous map of Huber rings it is immediate from the definitions 
that there exist rings of definition $A_0\subset A$ and $B_0\subset B$, and ideals of definition $I\subset A_0$ and $J\subset B_0$, such that $f(A_0)\subset B_0$ and $f(I)\subset J$. However, it is not true in general that $f(I)B_0$ is still an ideal of definition. For a homomorphism between two Huber rings, being adic is a strictly stronger condition that just being continuous.  By design, an adic map between Huber
rings carries bounded sets to bounded sets.

Some examples and non-examples of adic maps:
\begin{eg} \label{eg: boundedness} 
\ 
\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(\arabic{enumi})}}
\item Consider the inclusion $f:\bZ_p\to\bZ_p[\![x]\!]$, with the $p$-adic and $(p,x)$-adic topologies respectively. 
The map $f$ is manifestly continuous, but not adic.
\item On the other hand, if we consider, analogously, $f$ to be the inclusion $\bZ_p\to\bZ_p\{x\}$, then $f$ {\em is} adic, as the topology on $\bZ_p\{x\}$ is the $p$-adic topology.
\end{enumerate}
\end{eg}


\begin{rem} \label{rem: anticipation} Next time we shall study the following
special class of Huber rings:  a Huber ring $A$ is {\em Tate} if there exists a topologically nilpotent unit $u\in A$
(beware that if $u$ lies in a ring of definition $A_0$ then $u$ cannot be a unit
of $A_0$ when $A \ne 0$, as there are no topologically nilpotent units in $A_0$ when $A_0 \ne 0$; why not?). For example, if $A$ is an affinoid algebra over some non-archimedean field $k$, then any pseudo-uniformizer $\varpi$ is a topologically nilpotent unit. 
Tate rings are the class of Huber rings which exhibit behavior akin to Banach algebras.
\end{rem}
If $f:A\to B$ is a continuous homomorphism of {\em Tate} rings then it is in fact adic! We prove a little more.
\begin{prop} Let $f:A\to B$ be a continuous homomorphism of Huber rings. If $A$ is Tate then so is $B$, and $f$ is adic.
\end{prop}
\begin{proof} The proof is immediate from the definitions! If we let $A_0$ and $B_0$ be rings of definitions such that $f(A_0)\subset B_0$ (given by the continuity assumption on $f$) and if $u\in A$ is a topologically nilpotent unit, then so is $f(u)$, thus proving that $B$ is Tate. 

Since $A_0$ is open, we may replace $u$ by some power and assume $u\in A_0$, so that $f(u)\in B_0$. 
Multiplication by the {\em unit} $u$ on $A$ is a homeomorphism, so 
the ideal $uA_0$ in $A_0$ is {\em open} yet its powers
tend to 0 (since $u$ is topologically nilpotent and $A_0$ has its topology defined by powers
of an ideal).   Thus, $uA_0$ is an ideal of definition of $A_0$. By the same reasoning, $f(u)B_0$ is an ideal of definition of $B_0$.
But this latter ideal is $f(uA_0)B_0$, so $f$ is indeed adic.
\end{proof}


\begin{prop} \label{prop: bounded and def} Let $A$ be a Huber ring. An open subring $A_0\subset A$ is a ring of definition if and only if $A_0\subset A$ is bounded.
\end{prop}
\begin{proof} See \cite[Prop.\,1]{H1} or \cite[Lemma 6.2]{Wed} for proofs. The proof makes use of the equivalent definition of Huber ring introduced in Remark \ref{rem: bunch of remarks}.
\end{proof}

\begin{rem} Let $A$ be a Huber ring, and $(A_0,I)$ be given as usual. Then for any {\em open} subring $B\subset A$
we have $B\supset I^n$ for sufficiently large positive integers $n$. Such containement does {\em not} necessarily mean that $I^n$ is an ideal of $B$: it will certainly be an ideal of $A_0$, and the containement is purely set-theoretic.
\end{rem}

As a corollary of Proposition \ref{prop: bounded and def}, if $A_0$ and $A_1$ are two distinct rings of definition for the Huber ring $A$
then we can produce more rings of definition from them. 
More precisely, we have the following Corollary whose proof is left as an exercise to the reader:
\begin{cor} If $A_0$ and $A_1\subset A$ are rings of definition for $A$, then so are $A_0\cap A_1$ and $A_0\cdot A_1$, where this latter ``product ring'' is the subring of $A$ generated by the finite sums of products of elements of $A_0$ and $A_1$ respectively.
\end{cor}
(If $I_j \subset A_j$ is an ideal of definition then it isn't clear if $I_0 \cap I_1$ is finitely generated,
so we don't know if it is an ideal of definition of $A_0 \cap A_1$; however, this open intersection is obviously bounded,
so Proposition \ref{prop: bounded and def} applies.  For the open 
subring $A_0 \cdot A_1$ that is visibly bounded, the finitely generated ideals $I_0 \cdot A_1$ and $A_0 \cdot I_1$
are easily seen to be open and in fact ideals of definition.)

\begin{rem} \label{rem: Z plus I trick} Let $A$ be a commutative ring endowed with the $I$-adic topology with respect to some nonzero proper finitely generated ideal $I$ (so $A$ is Huber). Then a somewhat bizarre ring of definition for $A$ is $\bZ+I$. Indeed,
say $A$ is a Huber ring, with ring of definition $A_0$ and ideal of definition $I\subset A_0$. Then $\bZ+I\subset A_0$  is an open subring of $A$, and it is trivially bounded, so by Proposition \ref{prop: bounded and def} it is a ring of definition for $A$.
\end{rem}

The property for a topological ring of ``being a Huber ring'' is inherited by open subrings, as explained in the following:
\begin{prop} \label{prop: open subring of Huber} It $A$ is a Huber ring, so is any {\em open} subring $B\subset  A$.
\end{prop}
\begin{proof} We pick a couple $(A_0, I)$, with $I$ a finitely generated ideal of definition in $A_0$. Since $B$ is open, there is a sufficiently large integer $n > 0$ such that $I^n\subset B$. Consider the pair $(B\cap A_0, I^n)$.  Since
 $I^n$ is an ideal of $A_0$ contained in $B$, it is certainly an additive subgroup of $B\cap A_0$. Moreover, $BI^n\subseteq B$ and $(B\cap A_0)I^n\subseteq A_0$, so $(B\cap A_0)I^n\subseteq B\cap A_0$. Hence, $I^n$ is an open ideal in $B \cap A_0$, and its powers go
 to 0 (as we can check in $A_0$, so its adic topology is the one on $B \cap A_0$.  But it is finitely generated
 {\em as an ideal of $B \cap A_0$}?  Not so clear!  However, if $f_1, \dots, f_r$ is a finite
 set of generators of $I$ as an ideal of $A_0$ then every element of $I^{2n}$ is a linear combination of
 $n$-fold products among the $f_j$'s with coefficients in $I^n \subset B \cap A_0$, so the finite set of such $n$-fold
 products generates an ideal $J$ of $B \cap A_0$ sandwiched between $I^{2n}$ and $I^n$.  This finitely generated
 ideal does the job.
\end{proof}

We now prove a few important facts concerning Huber rings and subrings of definition.  The second part of the following
result provides an ample supply of rings of definition (in view of their characterization as the open bounded subrings).

\begin{thm} \label{thm: bdedness} Let $A$ be a Huber ring, and let $A^0$ be the subset of power-bounded elements in $A$.
\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(\arabic{enumi})}}
\item $A^0$ is a subring of $A$, and it is the union of all the rings of definition of $A$.
\item Let $B\subset A$ be a bounded subring, and let $B'$ be an open subring of $A$ containing $B$. Then there exists a ring of definition $A_0$ of $A$ which is contained in $B'$ and contains $B$.
\end{enumerate}
\end{thm}
\begin{proof}
By Proposition \ref{prop: bounded and def}, for (1) it suffices to show that $A^0$ is the union of all bounded subrings of $A$. 
Explicitly, we just have to check that if $A_0$ is a ring of definition and $a \in A$ is power-bounded then the open subring $A_0[a]$ 
is bounded (hence also is a ring of definition).  The set of powers of $a$ is bounded, so we just need to check that
if $\Sigma$ and $\Sigma'$ are bounded subsets of $A$ then the set $\Sigma \cdot \Sigma'$ of sums of pairwise
products is bounded.  There is a base of open additive subgroups around 0, so the boundedness of $\Sigma \cdot \Sigma'$ is clear.

The proof of $(2)$ is left to the reader as an exercise!
\end{proof}

A consequence of Theorem \ref{thm: bdedness} is that as soon as $A^0$ is a bounded subring of $A$, then it is a ring of definition
(as it is obviously open). It may well happen, however, that $A^0$ is not bounded. For example, when $A$ is not reduced: let $A=\bQ_p[\varepsilon]$, with $\varepsilon^2=0$. Then 
$$A^0=\bZ_p\oplus \bQ_p\cdot\varepsilon,$$
which is not bounded (this will be the reason why perfectoid algebras over a perfectoid field will be reduced).

The case in which $A^0$ is bounded is quite important (it will indeed include the case of perfectoid algebras) and it deserves a name:
\begin{defin} \label{def: uniformity} A Huber ring $A$ is called {\em uniform} if $A^0$ is bounded in $A$.
\end{defin}
Uniformity turns out to be a quite handy condition. In \cite{BuVe}, K. Buzzard and A. Verberkmoes show that
a ``localizable'' version of uniformity can be used to prove sheaf-like properties of (pre-)adic affinoid spaces over uniform Tate rings without Noetherian conditions! The reader may refer to \cite[Cor. 5 and Thm. 7]{BuVe}; we will address this later.






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Robustness of the Huber condition}
The Huber condition (we shall often name this way the condition on a commutative ring $A$ of ``being Huber'') behaves very well under many operations on topological rings. We shall now discuss completions and tensor products; localizations and ``polynomial rings''
will be treated next time.





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\paragraph{Completions of Huber rings}\ \\
We consider a Huber ring $A$, with a ring of definition $A_0$ and a finitely generated ideal of definition $I\subset A_0$. 
Note that $A_0$ is an open subring of $A$, and $I$ is an ideal of $A_0$ but not necessarily of $A$, 
so usually it does not make sense to take the $I$-completion of the {\em ring} $A$ in the sense of commutative algebra. 

On the other hand, we {\em can} do so for $A_0$, though such completion may be quite wild, due to lack of Noetherianity.
We want to formulate a reasonable sense of $I$-adic completion of $A$, in which $A_0^{\wedge}$ 
should be a natural candidate for a subring of definition.  This is handled by:
\begin{defin} \label{def: completion} Let $A$ be a Huber ring, $A_0$ a subring of definition with the $I$-adic topology, for $I$ a finitely generated ideal of $A_0$. We let
$$A^{\wedge}:=\varprojlim A/I^n$$
be the $I$-adic completion of $A$ in the sense of {\em topological groups}. 
This has a natural structure of topological ring (exercise!), and we define the {\em completion} of the Huber ring $A$ to be $A^{\wedge}$.
\end{defin}
\begin{rem} \label{rem: completion ring definition} It is not clear from Definition \ref{def: completion} what a ring of definition of $A^{\wedge}$ might actually be, and especially what topology this may have. 
Consider $A_0^{\wedge}$, the commutative algebra $I$-adic completion of $A_0$.
This is visibly an open subring of $\widehat{A}$ but  $A_0$ may not be Noetherian, so it is not 
clear if the topology on $A_0^{\wedge}$ is the $I$-adic topology.
\end{rem}

Thanks to the finite generation of $I$, all works out well:

\begin{thm} \label{thm: non Noetherian miracle} Let $A$ be a Huber ring, and $(A_0,I)$ be a couple of definition. Then, with notation as in Definition \ref{def: completion}:
\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(\arabic{enumi})}}
\item The open subring $A_0^{\wedge}\hookrightarrow A^{\wedge}$ has the $I(A_0^{\wedge})$-adic topology and 
for all $n\ge 1$ the natural map
$$A_0/I^n\xrightarrow{\simeq} A_0^{\wedge}/I^n(A_0^{\wedge})$$
is an isomorphism; in particular, $\widehat{A}_0$ is $I$-adically separated and complete.
\item The natural map of rings
$$A\otimes_{A_0}A_0^{\wedge}\rightarrow A^{\wedge}$$
is an isomorphism.
\end{enumerate}
\end{thm}
\begin{proof} The hard part of the proof is point $(1)$, for which we refer the reader to \cite[Lemma 10.91.7]{Stacks}.
(which also treats arbitrary modules over $A$, and uses a method entirely
different from the Artin--Rees technique in commutative algebra). 
The proof of $(2)$ is a little bit tricky, and is explained in \cite[Lemma 1.6]{H1}.
\end{proof}





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\paragraph{Tensor products of Huber rings}\ \\
The setting is now the following: let $A, B$, and $C$ be Huber rings, with {\em adic} morphisms $f:B\to A$ and $g:B\to C$.
We may choose a couple of definition $(B_0, J)$ for $B$ and subrings of definition $A_0\subset A$ and $C_0\subset C$
such that $f(B_0)\subset A_0$ and $f(B_0)\subset C_0$.  Note that $f(J)A_0$ and 
$f(J)C_0$ are automatically ideals of definition in $A$ and $C$ respectively, since $f$ and $g$ are adic. 

We have:
\begin{thm} \label{thm: tensor product} Let $J\subset B_0\subset B$, $A_0\subset A$ and $C_0\subset C$ be as above. Then
equipping the subring $$\textup{image}(A_0\otimes_{B_0}C_0\to A\otimes_BC)$$
of $A\otimes_BC$ with the $J$-adic topology and declaring it to be open makes $A\otimes_BC$ into a Huber ring.
This Hube ring satisfies the expected 
universal property in the category of topological rings with continuous homomorphisms.
\end{thm}

By means of the universal property one sees that the construction of this topology on $A \otimes_B C$ is independent of
the choices of compatible rings of definition. 

The next example shows that assuming in Theorem \ref{thm: tensor product} that $f$ and $g$ are adic maps is necessary
in order that the construction makes sense with appropriate topological properties.

\begin{eg} \label{eg: adic necessary} 
Let $A=\bZ_p[\![x]\!]$ be given the $(p,x)$-adic topology
and define $C=\bQ_p$ and $B=\bZ_p$ with their usual topologies. Set $A_0 = A$, $B_0 = C_0 = \bZ_p$, and $J = (p)$.
Take $f$ and $g$ to be the natural inclusions, so $f:B\to A$ is not adic (as remarked in Example \ref{eg: boundedness}).
Let's make the tensor product $A\otimes_BC = \bZ_p[\![x]\!][1/p]$ into a topological ring in which 
$\bZ_p[\![x]\!]=\textup{image}(A_0\otimes_{B_0}C_0\to A\otimes_BC)$ is given the $J$-adic topology.
Then the natural map $A \rightarrow A \otimes_B C$ isn't even continuous (e.g., $x$ is topologically nilpotent in $A$ but no
power lies in the image of $J(A_0 \otimes_{B_0} C_0)$).

Suppose instead we try to make $A \otimes_B C$ into a topological ring
so that $A_0 \otimes_{B_0} C_0 = A_0 = \bZ_p[\![x]\!]$ is an open subring in which (inspired by the case of
completed tensor product topologies used to build fiber products of formal schemes) the topology is defined by powers of
$I \otimes_{B_0} C_0 + A_0 \otimes_{B_0} J = I = (p,x)$.  That is, we want
$\bZ_p[\![x]\!][1/p]$ to be a topological ring in which $\bZ_p[\![x]\!]$ with its $(p,x)$-adic topology is an open subring.
There is no such topological ring structure on $A \otimes_B C$!  Indeed, were there such a structure then multiplication by
the unit $p$ would be an automorphism, so $p\bZ_p[\![x]\!]$ would have to be open in $\bZ_[\![x]\!]$
with its $(p,x)$-adic topology.  But that openness is false.  

From a geometric point of view, the problem
is that ${\rm{Spa}}(\bZ_p[\![x]\!]) \times_{{\rm{Spa}}(\bZ_p)} {\rm{Spa}}(\bQ_p)$ 
should correspond to the Berthelot ``generic fiber'' $\{|x| < 1\}$
of the formal unit disc over $\bZ_p$, but this open disc is not affinoid and hence its
structure should not be controlled by a ring.   This is a prelude to the fact that fiber products
among adic spaces do not exist in the same degree of generality that
they do for schemes (or for formal schemes or rigid-analytic spaces), and when they do exist
their construction entails some subtleties.  
\end{eg}






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\paragraph{Localizations of Huber rings}\ \\
We will discuss localizations of Huber rings in detail next time, providing in the process an abundant
supply of new Huber rings from old ones (much like passing to rational domains in rigid-analytic
geometry products many new affinoid algebras from old ones).

Let us end by setting up the context via motivation from the case $A$ is an affinoid algebra over some non-archimedean field $k$. We recall that a rational domain in $X:=\Sp(A)$ is given as the locus $U$ of those points of $X$ over which a given collection of $f_1,\cdots,f_n\in A$ satisfies
$$|f_1|\leq g,\cdots , |f_n|\leq g$$
where $g\in A$, and $f_1,\cdots,f_n,g$ {\em have no common zero}. (Among other things,
the absence of a common zero avoids any ``pinching'' in how this locus
lies in $X$; e.g., it underlies why such a locus is actually an open subset
of $X$ for the naive totally disconnected topology on classical geometric points, and also
underlies why mapping into such loci can be classified by an affinoid algebra,
due to a ``minimum modulus principle'' for a finite collection of analytic functions without a common
zero on an affinoid.)

More precisely, calling
$$(*)\ \ \ A\left\langle\frac{f_i}{g}\right\rangle:=\frac{A\langle X_1,\cdots, X_n\rangle}{\langle g X_i-f_i\rangle}$$
we have
$$U=\Sp\ A\left\langle\frac{f_i}{g}\right\rangle,$$
and the Banach $A$-algebra
$A \langle f_i/g \rangle$ is intrinsically attached to the subset $U \subset X$ in the sense that
for any $k$-affinoid $B$ equipped with a (necessarily continuous)
$k$-algebra homomorphism $f:A \rightarrow B$,
the map ${\rm{Sp}}(f)$ factors through $U \subset X$ set-theoretically if and only if $f$ factors
through $A\langle f_i/g \rangle$ as an $A$-algebra (in which case such a factorization is unique).

The condition on $f_1,\cdots, f_n,g$ not having common zeros being rephrased in the condition:
$$1\in Ag+\sum_{i=1}^n Af_i.$$
Given a Huber ring $A$, and $f_1,\cdots, f_n,g\in A$, the construction of a suitable analogue of $(*)$ will be in the same spirit
but we will impose the condition that the ideal generated by $g$ and the $f_i$'s is merely {\em open}.
(In a $k$-affinoid algebra, the only open ideal is the entire ring; this follows from precisely the ``Tate ring'' property!) The significance of the condition ``generates an open ideal'' will be addressed
next time, when we carry out appropriate definitions for Huber rings and even general
non-archimedean rings without the crutch of Banach topologies (or even a Tate-ring hypothesis).

\small
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\end{document}