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\begin{document}
\normalsize
\section{Lecture 10: Affinoid adic spaces I}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Introduction}
The aim of this lecture is to define the topological space underlying affinoid adic spaces, and discuss some essential topological properties as well as a fundamental correspondence in the spirit of the Nullstellensatz. We review the main result from Lecture 9, rephrasing it in a handy way, and define the adic spectrum of a Huber ring $A$ with respect to some additional data attached to $A$. We shall deduce spectrality of the adic spectrum at once from Corollary 9.3.2, and hence, in the last instance, from spectrality of the continuous valuation spectrum $\Cont(A)$.

In the last section we formulate five topics concerning affinoid adic spectra which we shall address in the next lecture.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Pro-constructible loci in $\Cont(A)$}
Let $A$ be an arbitrary commutative ring (no topology). Recall from $\S 9.1$ that for any ideal $J\subset A$ whose radical is the radical of a finitely generated ideal (ie. $\Spec(A)-V(J)$ is qc) we studied the subspace $\Spv(A,J)$ of the valuation spectrum $\Spv(A)$; this consists of the points $v \in\Spv(A)$ admitting no {\em proper} horizontal specialization supported over $\Spec(A) - V(J)$
(a vacuous constraint when $v$ itself lies over $V(J)$, so mainly of interest when $v$ does not lie over
$V(J)$; e.g., $\Spv(A, A)$ is the set of $v$ with no proper horizontal specialization whatsoever,
or equivalently $c\Gamma_v = \Gamma_v$).   In terms of notation introduced in \S9.2, in Definition 9.2.3 we defined
$$\Spv(A,J):=\{v\in\Spv(A)\mid c\Gamma_v(J)=\Gamma_v\}$$
where $c\Gamma_v(J)$ is a convex subgroup of $\Gamma_v$ containing $c\Gamma_v$ whose properties were
summarized in Proposition 9.2.2. Geometrically, the points of $\Spv(A,J)$ not over $V(J)$ are those ``at the interior edge'' 
(relative to horizontal specialization) of
the set of points of $\Spv(A)$ not over $V(J)$.
The assignment $$r:v\mapsto v_{\vert_{c\Gamma_v(J)}}$$ gave a continuous spectral retraction of $\Spv(A)$ onto $\Spv(A,J)$, 
which was used in our deduction of spectrality of $\Spv(A,J)$ from that of $\Spv(A)$.

We now fix a {\em Huber ring} $A$, and a couple of definition $(A_0, I)$, so $A_0\subset A$ is an open bounded subring and $I$ a finitely generated ideal of $A_0$ such that the topology on $A_0$ is the $I$-adic topology. 

Recall the fundamental Theorem 9.3.1:
\begin{thm} \label{continuity} We have:
$$\Cont(A)=\{v\in\Spv(A,A^{00}\cdot A)\mid v(a)<1\ \textup{ for all }\ a\in A^{00}\}.$$
\end{thm}
\begin{rem}
Recall that if $A$ is Tate then $A^{00}$ contains a unit of $A$, and therefore $A^{00}\cdot A=A$. Hence, if $A$ is Tate then 
$$\Cont(A)=\{v\in\Spv(A,A)\mid v(a)<1\ \textup{ for all }\ a\in A^{00}\}.$$
\end{rem}

For general Huber ring $A$ we have 
$$\Spv(A,A^{00}\cdot A)=\Spv(A,I\cdot A)$$
where the ideal $I$ of $A_0$ is finitely generated as such. We fix generators
$\{a_1, \dots, a_n\}$ of $I$ as an ideal of $A_0$, so these are generators
of $I \cdot A$ that are {\em topologically nilpotent} in $A$ (whereas if $A$ is Tate then $I \cdot A$ can also be
generated by the element 1 that is only topologically nilpotent when $\{0\}$ is dense in $A$). 
By Corollary 9.3.3, we have the following concrete description of the continuous valuation spectrum:
$$\Cont(A)=\{v\in\Spv(A)\mid \gamma_i := v(a_i) \in\Gamma_v\ \textup{are all cofinal}, \,\, 
v(a) < 1/\max_i \gamma_i \textup{ for all } a \in A_0\}$$
with the understanding that the condition ``$v(a) < 1/\max_i \gamma_i$'' is vacuously satisfied when $\gamma_i = 0$ for all $i$.

\begin{eg}\label{Tate cofinal} The following example clarifies the importance of cofinality over
the condition ``$v(a) < 1$'' when $v$ has higher rank (i.e., not trivial and not of rank 1). 
Assume $A$ is Tate with ring of definition $A_0$, and let $\varpi\in A^{\times}$ be a 
pseudo-uniformizer (i.e., a topologically nilpotent unit). By replacing $\varpi$ with $\varpi^m$ for large enough $m$ we may
and do assume $\varpi \in A_0$, so $I := \varpi A_0$ is an ideal of definition of $A_0$.  Since $v(\varpi) \ne 0$
for all valuation $v$ on $A$ (as $\varpi \in A^{\times}$), we have:
$$\Cont(A)=\{v\in\Spv(A)\mid v(\varpi)\ \textup{is cofinal}, \,\,v(a) < 1/v(\varpi) \textup{ for all } a \in A_0\}.$$
Note that $v$ may well be of rank $>1$, and the difference between having $v(\varpi)<1$ and 
having $v(\varpi)$ be cofinal for $\Gamma_v$ is substantial.  For instance, consider the field 
$$A:=k(u,x) = k(x)(u) = k(u)(x)$$
where we endow $A$ with the ``$x$-adic topology'' to make it a Huber ring (i.e., as the fraction field of the discrete valuation
ring $k(u)[x]_{(x)}$). Consider the valuation ring $$R_v:=k[x]_{(x)}+uk(x)[u]_{(u)}\subset A$$
with associated valuation on $A$ denoted as $v$.  Clearly
$v(x)<1$ since $x \in R_v$, but $v(x)$ is not cofinal for $\Gamma_v=a^{\bZ}\times b^{\bZ}$, with $v(x) = (a,1)$ and 
$v(u)=(1,b)$ for any choice of $0< a,b<1$.  Concretely, $x$ is not topologically nilpotent for $v$.  Thus,
$v$ is {\em not} continuous on $A$ even though $v(x) < 1$ with $x$ a pseudo-uniformizer of the Tate ring $A$. 
This also illustrates the fact that the subring of power-bounded elements in a higher-rank valuation ring
need {\em not} coincide with the valuation ring (since $1/x$ is power-bounded because $1/x^n \in u k(x)[u]_{(u)}$ for all
$n \ge 1$ yet $1/x \not\in R_v$).

One can also make a mixed-characteristic 
analogue of the preceding example by equipping the field $A' := \mathbf{Q}(u)$ with the $p$-adic
topology and working with the valuation $v'$ corresponding
to the rank-2 valuation ring $\mathbf{Z}_{(p)} + u \mathbf{Q}[u]_{(u)}$.
In both cases the two valuations we considered are not continuous, since their values respectively on $x$ and $p$ are not cofinal.
\end{eg}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\paragraph{Loci cut out by inequalities}\ \\
Choose $a\in A^0$ and $v\in\Cont(A)$.  Experience with rank-1 valuations might lead one to
guess that necessarily $v(a)\le 1$ in $\Gamma_v$, but this is false in higher rank:

\begin{eg} \label{rank one} Let $(K,v)$ be a valued field with valuation ring $R$, where $v$ is nontrivial and of rank $>1$. 
Assume that $v$ is microbial; i.e., there exists a topologically nilpotent element of $K^{\times}$, or
equivalently, $K$ is Tate. Note that $v$ is continuous on $K$ by design.

By Proposition 9.1.3 (and its proof), 
there exists a height-1 prime ideal $\fp\subset R$ such that $R_{\fp}$ induces the same topology on $K$, and 
$R_{\mathfrak{p}}$ is a rank-1 valuation ring.   Thus, $K^{0}=R_{\fp}$.  Since $R$ is higher-rank, we have 
$$R\subsetneq R_{\fp}.$$
Any $a \in R_{\fp}-R\subset K-R$ is a power-bounded element of $K$ which does not satisfy $v(a)\le 1$ in $\Gamma_v$.
\end{eg}

\begin{rem} \label{non existence height 1} It is not true that height-1 prime ideals always exist in a 
general valuation ring $R$ that is not a field! For example, let us consider the abelian group
$$\Gamma:=  (a^{\bZ})^{\oplus I},\ \ 0<a<1$$
where $I$ is $\bZ$ itself, and $\Gamma$ is correspondingly ordered lexicographically. 
Concretely, an element $\gamma\in\Gamma$ is a doubly-infinite sequence of powers 
$$\gamma=(a^{c_n})_{n\in\bZ}$$
where $c_n=0$ for all but finitely many $n\in\bZ$. For any $1\neq\gamma\in\Gamma$, let $n_0$ be the minimal integer such that $c_{n_0}\neq 0$. If a {\em convex} subgroup of $\Gamma$ contains $\gamma$ then it must contain the entire $n_0$-component $a^{\bZ}$
and so contains the group of elements $(a^{d_n})$ where $d_n=0$ for $n<n_0$. This latter subgroup is manifestly convex, and it is
not hard to deduce from this that these are exactly the {\em nontrivial proper} convex subgroups of $\Gamma$.  (Consider
the quotient of $\Gamma$ modulo the direct sum of the factors for $n > n_0$.)

Since we can move ``arbitrarily far to the left'', among {\em proper} convex subgroups of $\Gamma$ there are no maximal elements with respect to inclusion. Every totally ordered abelian group $G$ is the value group of a 
suitable valuation $v$ on the fraction field of the group algebra $k[G]$ (see \cite[Exer.\,33, Chapter 5]{AM}, which
is reminiscent of our study of $K=k(\!(u)\!)(\!(x)\!)$ as a rank-2 valued field). Thus,
the valuation ring $R_v$  has no height-1 prime ideal due to the inclusion-reversing 
bijective correspondence between the set of nonzero 
prime ideals of $R$ and the set of proper convex subgroups of $\Gamma$ \cite[Prop.\,2.14]{Wed}.
\end{rem}

Let us now prove the following:
\begin{lemma} \label{leq constructible} For any element $a\in A$, the locus
$$\{v\in\Cont(A)\mid v(a)\le 1\ \textup{in}\ \Gamma_v\}\subseteq \Cont(A)$$
is constructible.
\end{lemma}

The preceding discussion shows that if $a \in A^0$ then this constructible subset can be a proper subset
due to the behavior at higher-rank valuations!

\begin{proof} Let $J\subset A$ be any ideal whose radical is the radical of a finitely generated ideal of $A$, and call $X:=\Spv(A,J)$. 
(The case of interest to us is $J = A^{00} \cdot A$.)
By Proposition 9.2.5(2), for every $s\in A$ and nonempty finite subset $T\subset A$ such that 
$$J\subset\rad(T\cdot A),$$
the subset
$$X(T/s)=\{v \in X\,|\,v(t)\le v(s)\neq 0, t\in T\}$$
is a qc open of $X$.  

Now set $J = A^{00} \cdot A$. We know that the subset $\Cont(A)\subset X$ 
is proconstructible, and the above gives that $X(\{a,1\}/1)$ is a qc open subset of $X$. The following diagram of sets is cartesian:
\begin{displaymath}
\xymatrix{\Cont(A)\ \ar@{^(->}[r] & \Spv(A,I)\\
		\{v\in\Cont(A)\mid v(a)\le 1\}\ \ar@{^(->}[u]\ar@{^(->}[r] & X(\{a,1\}/1)\ar@{^(->}[u]}
\end{displaymath}
Hence, we are reduced to proving that for a spectral space $Y$, qc open subset $U \subset Y$, 
and proconstructible subset $Z$ $($so $Z$ is spectral$)$, the open subset $U \cap Z \subset Z$ is
qc and hence constructible in $Z$.   

Endowing $Y$ and $Z$ with the constructible topology,
$$Z_{\rm cons}\hookrightarrow Y_{\rm cons}$$
is a continuous injection between compact Hausdorff spaces, and hence is a closed map. In particular $Z$ is closed in $Y$ in the constructible topology, and being $U$ qc, $Z\cap U$ is closed in $Z$ in the constructible topology. Since $Z_{\rm cons}$ is compact Hausdorff, $Z\cap U$ is compact for the constructible topology. It follows $Z\cap U$ is qc for the initial topology on $Z$
(which has far fewer open sets than for the constructible topology on $Z$ in general). 
\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Huber pairs and their adic spectrum}

Now we apply the construction in Lemma \ref{leq constructible} many times:

\begin{defin} \label{Spa} For a subset $\Sigma \subset A$, define 
$$\Spa(A,\Sigma):=\{v\in\Cont(A)\mid v(a)\le 1 \textup{ for all }\ a\in\Sigma\}$$
and call it the {\em adic spectrum} of the pair $(A,\Sigma)$.
\end{defin}

As a consequence of Lemma \ref{leq constructible}, $\Spa(A,\Sigma)$ is proconstructible in $\Cont(A)$ for all subsets $\Sigma\subset A$, and since this latter is spectral, we deduce that $\Spa(A,\Sigma)$ is spectral.


\begin{rem} \label{Spa spectral} By the ultrametric inequality, for every topologically nilpotent element $f\in A^{00}$ we have that $v(a)\le 1$ if and only if $v(a+f)\le 1$. In particular
$\Spa(A, \Sigma) = \Spa(A, \Sigma + A^{00})$ inside $\Cont(A)$.
\end{rem}

The subring $\bZ[\Sigma + A^{00}] \subset A$ generated by any subset $\Sigma\subset A$ together with $A^{00}$
is an open subring of $A$, and due to the ultrametric inequality clearly
$$\Spa(A,\Sigma) = \Spa(A, \bZ[\Sigma + A^{00}]).$$
For the same reason, if $B\subset A$ is a subring and $v:A\to\Gamma_v\cup\{0\}$ is a valuation on $A$ such that $v(b)\le 1$ in 
$\Gamma_v$ for all $b \in B$ then the integral closure $\widetilde{B}$ of $B$ in $A$ (which contains
all nilpotent elements of $A$!) satisfies
$$v(\widetilde{b})\le 1$$
for all $\widetilde{b} \in \widetilde{B}$, as we see
by applying the ultrametric inequality to a {\em monic} polynomial relation for $\overline{b}$ over $B$, 
much as how we show that a valuation ring is integrally closed in its fraction field.  (Note that such
$\widetilde{B}$ is generally {\em not} normal as a ring in the sense of commutative algebra when $A$ is not
a normal ring.)  Thus, 
$\Spa(A, \Sigma) = \Spa(A, A^+)$ where $A^+$ is the open integrally closed subring of $A$
given by the integral closure of $\bZ[\Sigma+A^{00}]$ in $A$.

\begin{defin}\label{Huber pair} A {\em Huber pair} is $(A, A^+)$ where
$A$ is a Huber ring and $A^+\subset A^0$ is an open subring of $A$
that is integrally closed in $A$. The {\em adic spectrum} of a Huber pair $(A,A^+)$ is 
$$\Spa(A,A^+),$$
and is denoted $\Spa(A)$ when $A^+$ is understood from the context.
\end{defin}

\begin{rem}\label{a0sup} In \cite{H1} and \cite[$\S 7.3$]{Wed}, Huber pairs are called ``affinoid rings''.  The requirement
$A^+ \subset A^0$ will play an important technical role beginning in our study of rational domains in \S11.4
and adic spectra of completions in \S11.5 (it is lurking in
the proofs of results we will invoke from \cite[3.9-3.11]{H1} and in the very statement of Proposition 11.6.1).
\end{rem}

Before discussing examples, let us give one more definition and discuss an important correspondence, which the reader may regard as to be in the spirit of an adic version of the algebraic ``Nullstellensatz''.
%Hum.

\begin{defin} \label{Nullstellensatz like} Given a Huber ring $A$, we define
$$\ccG_A:=\{A^+\subset A\ \textup{open integrally closed subring}\}$$
and
$$\ccF_A:=\{\textup{proconstructible sets of}\ \Cont(A)\ \textup{of the form}\ \Spa(A,\Sigma),\ \Sigma\subset A\ \textup{subset}\}.$$
\end{defin}

Note that  we do {\em not} require $A^+ \subset A^0$ (in contrast with the situation in Remark \ref{a0sup}).

\begin{thm}\label{Nullstellensatz} We have a bijective inclusion-reversing correspondence:
$$\ccG_A\xrightarrow{\sigma} \ccF_A$$
sending an open subring of integral elements $A^+\subset A$ to $\Spa(A,A^+)$, with inverse given by the assigning
to any such proconstructible subset $\mathcal{V} \subset \Cont(A)$ the subring
$$\{a\in A\mid v(a)\le 1 \mbox{ for all } v\in \mathcal{V}\}.$$
Moreover, if $A^+\subset A^0$ then $\Spa(A,A^+)$ contains all rank-$1$ points of $\Cont(A)_{\rm{an}}$ 
and all trivial valuations of $\Cont(A)$.
\end{thm}

\medskip

Before we discuss the proof of this theorem, we record some interesting consequences.
Suppose $A^+ \subset A^0$, so $\Spa(A,A^+)$
contains all rank-1 points and all trivial valuations in 
$\Cont(A)$.  By Proposition 9.1.3 and the discussion in \S9.1, each fiber of $\Spa(A, A^+) \rightarrow \Spec(A)$
that is not empty has a unique generic point:  it is a trivial valuation for open primes in $A$ and it
is a rank-1 valuation for non-open primes of $A$.  
(It can happen that some fibers over $\Spec(A)$ are empty yet $A \ne 0$; e.g., 
even the entire space $\Spa(A,A^+)$ can be empty
with $A \ne 0$. We will address this issue next time.)

Since all specialization (equivalently,
generization) relations inside $\Cont(A)_{\rm{an}}$ are vertical (by Proposition 8.3.10), the same holds in the subspace
$\Spa(A, A^+)_{\rm{an}}$.   Thus, 
every {\em analytic} point $v\in\Spa(A,A^+)$ admits a unique ``minimal-rank'' generization $\eta_v$ 
in the space $\Spa(A, A^+)_{\rm{an}}$ of analytic points.  This $\eta_v$
is a rank-1 point with the same support as $v$ and it defines the unique rank-1
valuation on $\kappa(v)$ yielding the {\em same} topology as $v$ (due to how $\eta_v$ is built).
This is especially useful when $A$ is Tate, as all perfectoid algebras will be, as then 
{\em all} points of $\Cont(A)$ are analytic.

On the other hand, if $v$ is a non-analytic
point then $\mathfrak{p}_v$ is open and hence {\em all} valuations on ${\rm{Frac}}(A/\mathfrak{p}_v)$
are continuous on $A$.  The fiber of $\Spa(A,A^+) \rightarrow \Spec(A)$ over an open prime ideal $\fp$
is therefore the Riemann--Zariski space $\Spv(\kappa(\fp), A^+)$ of valuations
on $\kappa(\fp)$ whose valuation ring contains the image of $A^+$.  This 
has a unique generic point, namely the trivial valuation on $\kappa(\fp)$, and it does {\em not}
recover the $v$-topology  on $\kappa(\fp) = \kappa(v)$ unless $v$ is itself a trivial valuation.


\begin{proof}  We first check that all rank-1 analytic points $v$ of $\Cont(A)$ satisy $v(a) \le 1$ for all $a \in A^0$.
(This is false without analyticity: consider a discrete Huber ring $A$ that is the fraction field of a discrete valuation ring,
so $A = A^0$ due to the discreteness of the impose topology.) 
Let $(A_0,I)$ be a pair of definition for $A$ as usual, so by continuity 
$v(t) < 1$ for all $t \in I$. By analyticity, $v(I)$ is nonzero.  Choose $t \in I$
such that $v(t) \ne 0$.  Topological nilpotence of elements of $I$ and continuity of $v$ force $v(s) < 1$ for
all $s \in I$. Power-boundedness of $a$
provides a large integer $m$ so that $a^j I^m \subseteq I$ for all $j \ge 1$.
Hence, $v(a)^j v(t)^m < 1$ for all $j$. Thus, $v(a)^j < 1/v(t)^m$ in the value group of $v$
for all $j > 0$.  Since $v$ is a rank-1 valuation, this forces $v(a) \le 1$, as desired.

The only difficulty for the bijectivity aspect of $\sigma$ is to show that if $a\notin A^+$ then there exists 
$u\in\Spa(A,A^+)$ such that $u(a)>1$.  
Before worrying about continuity of valuations, we try to build {\em some} valuation on $A$ such that 
the valuation ring in the corresponding residue field contains the image of $A^+$ but {\em not} the image of $a$.
This is essentially a problem of building appropriate prime ideals not containing $a$
and then using the fact that local subrings of domains are always dominated by valuation rings.
Horizontal specialization will be used at the end to handle continuity issues.

We claim that 
$$1/a\notin A^+[1/a]^{\times}\subset A_a,$$
or in other words $a \not\in A^+[1/a]$ inside $A_a$.  (Keep in mind that we are assuming $a \not\in A^+$.) 
Suppose to the contrary that $a$ lies in this subring of $A_a$, so $a = f/a^n$ inside $A_a$ for some $n \ge 0$ and $f \in A^+$.
Hence, $a^n - f = 0$ in $A_a$, or equivalently $a^m(a^n - f) = 0$ in $A$ for some $m \ge 0$.  
But then $a^{n+m} - f a^{m}  = 0$ in $A$, 
so we have an integral dependence relation for $a$ over $A^+$, forcing $a \in A^+$, contrary to our hypothesis.

 It follows that there exists a prime ideal $\fp$ of $A^+[1/a]$ containing $1/a$. Choose a minimal prime $\fq \subset\fp$ of $A^+[1/a]$, and consider a valuation ring $R\subset\kappa_a(\fp):=\Frac(A^+[1/a]/\fq)$ dominating the local domain $(A^+[1/a]/\fq)_{\fp/\fq}$. This $R$ is attached to a valuation on $\kappa_a(\fp)$ which yields on $A^+[1/a]$ a valuation $v$ whose support is exactly $\fq$, and such that $v(A^+)\le 1$ and $v(\fp)<1$ (the latter due to the local domination). As a consequence, $v(1/a)<1$. 
 
 Localizing the {\em inclusion} $A^+[1/a] \hookrightarrow A_a$ at the minimal prime $\fq$ of $A^+[1/a]$
 yields an injection of rings $A^+[1/a]_{\fq} \hookrightarrow (A_a)_{\fq}$, forcing $(A_a)_{\fq}$ to be nonzero.  
 Thus, there is a prime ideal 
 $\mathfrak{Q}$ of $A_a$ whose contraction to $A^+[1/a]$ lies inside $\fq$ and hence is equal to $\fq$
 due to the minimality of $\fq$.  The resulting injection of domains $A^+[1/a]/\fq \hookrightarrow A_a/\mathfrak{Q}$
 yields a corresponding injection between their fraction fields.
 But valuations on fields always extend to valuations on extension fields (since any local ring
 inside a field is dominated by a maximal one with full fraction field, and such maximal local rings are valuation rings
 by Zorn's Lemma).
 Thus, we can extend $v$ to a valuation $v'$ on $A_a$.
 
 Restriction of $v'$ yields a valuation $w$ on $A$ satisfying $w(a) > 1$ and $w(A^+) \le 1$, but 
 in view of the extremely abstract mechanism by which $w$ was constructed it is probably not continuous!
 To fix this problem, we will work with the horizontal specialization
 $u = w_{\vert_{c\Gamma_w}}$. Note that $u$ has no proper horizontal specialization at all, so it
 certainly lies in $\Spv(A, A^{00} \cdot A)$.  Hence, by our ``algebraic'' characterization of continuity of valuations
 on Huber rings, $u$ is continuous provided that $u(x) < 1$ for all $x \in A^{00}$.
 Since $u(y) \le w(y)$ for all $y \in A$ (in fact, $u(y) = w(y)$ if $w(y) \in c\Gamma_w$, and  $u(y) = 0$ otherwise),
 we have $u(A^+) \le 1$ and $u(a) = w(a) > 1$ (as $c\Gamma_w$ contains $w(A)_{\ge 1}$).
 Hence, it suffices to show $w(x) < 1$ for every $x \in A^{00}$. 
 Topological nilpotence of $x$ in $A$ and openness of $A^+$ in $A$
 gives $x^m a \in A^+$ for large $m$. (Recall that $a \not\in A^+$, so such membership
 in $A^+$ is a nontrivial condition on $x^m$.).  Thus, $w(x^m a) \le 1$, so $w(x)^m \le 1/w(a) < 1$ and hence $w(x) < 1$.
 We conclude that $u \in \Spa(A,A^+)$ and $u(a) > 1$.
 \end{proof}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Some loose ends for affinoid adic spectra}
In the next lecture we will address the following topics that quickly come to mind upon
thinking about $\Spa(A,A^+)$:

\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(\arabic{enumi})}}
\item Give a concrete description of a topological base of qc open ``rational domains'' of $\Spa(A,A^+)$
and prove that the natural map
$$\Spa(A^{\wedge}, (A^+)^{\wedge})\to\Spa(A,A^+)$$
is a homeomorphism preserving the property of ``rational domain'' in {\em both} directions.
(This homeomorphism  property is  an essential technical device in the study of perfectoid spaces and 
``approximation'' arguments with them. It would be a mistake to only define the topological space $\Spa(A,A^+)$ under
a completeness hypothesis on $A$.)
\item Characterize when $\Spa(A,A^+)$ is empty, and likewise for its subset of analytic points.
\item For a non-archimedean field $k$ and a $k$-affinoid algebra $A$, 
discuss density of $\Sp(A)\subset\Spa(A,A^0)$ 
and how $\Spa(A,A^0)$ is related to the Berkovich space $M(A)$ and the Tate topos of ${\rm{Sp}}(A)$.

\item Why consider the case $A^+\neq A^0$? (This question is immediately suggested by (3) and occurs to everyone
who sees $\Spa(A,A^+)$ defined for the first time, especially if done without first going through a detailed
study of $\Spv(A)$ and $\Cont(A)$ culminating in Theorem \ref{Nullstellensatz}
as we have done following \cite{H1}.  Some simple
examples will illustrate why it would be a bad idea to insist on taking $A^+ = A^0$.)

\item Equip $\Spa(A,A^+)$ with a suitable structure presheaf (initially defined on the ``rational domains''
in (1)), and discuss when it is a sheaf.  Relate morphisms of  such ``structured locally ringed spaces''
to morphisms of pairs $(A, A^+)$.
\end{enumerate}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\subsection{Appendix A: The adic unit disc $\bD_C=\Spa(C\langle t\rangle, C^{0}\langle t\rangle)$}
%
%\begin{eg}
%We start discussing our first nontrivial example of adic space over a non-archimedean field $C$.
%We let $A:=C\langle t\rangle$, and $A^+=A^{0}=C^{0}\langle t\rangle$. We assume $C$ is complete and algebraically closed, so that the only irreducible polynomials over $C$ will be linear.
%
%We call $$\bD_C:=\Spa(A,A^+).$$
%
%We describe $\bD_C$ step by step, showing in particular that in the case, for example, when $C=\bC_p$, it turns out that we can find five different types of points, two of which are new to the rigid-analytic reader. 
%
%We call $|\cdot |$ the non-archimedean absolute value on $C$.
%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\paragraph{The classical points.}\ \\
%Pick $x\in C$ such that $|x|\leq 1$. Then we can consider the assignement $f\mapsto |f(x)|$ for all $f\in C\langle t\rangle$. This is indeed a valuation $$v_x: C\langle t\rangle\rightarrow\Gamma_x\subset\bR_{\geq 0}^{\times},$$ and if $f\in C^{0}$, $|f(x)|\leq 1$. Therefore, $x$ must indeed be a point of $\bD_C$.
%
%We compute the support of $v_{x}$. Evaluating at $x$ an element $f(t)\in C\langle t\rangle$ gives zero if and only if $t-x$ divides $f(t)$. This means that $(t-x)\subseteq\textup{supp}(v_{x})$, and since $(t-x)$ is a maximal ideal of $A$, we have equality. We call the points described in this paragraph ``classical'', as they correspond bijectively to the maximal ideals of $A$. Moreover, since we are assuming $C$ is algebraically closed, $A/\fm= C$, and these points are all rigid points in $\bD_C$.
%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\paragraph{Points on the limbs.}\ \\
%Let us now consider $x$ as before, and, in addition, $r\in\bR$ such that $0\leq r\leq 1$. We define
%$$v_{x,r}:C\langle t\rangle\to\bR^{\times}_{\geq 0}$$
%assigned by sending $f(t)=\sum_{m=0}^{\infty}a_m(t-x)^m$ to $$\sup_m\{|a_m|r^m\}.$$ Actually this supremum is a maximum, as $|a_m|\to 0$. 
%
%Again this assignment is a valuation on $C\langle t\rangle$, and $v_{x,r}(f)\le 1$ for all $f\in A^+$. Therefore, it is a point in $\bD_C$.
%
%We notice that, if we call $D(x,r)$ the closed disc in $C^{0}$ of center $x$ and radius $r$,
%$$\sup_m\{|a_m|r^m\}=\sup_{y\in D(x,r)}|f(y)|.$$
%Then $v_{x,r}$ actually {\em depends only on} $D(x,r)$. 
%
%For $r=0$, $v_{x,r}=v_0$, ie. it agrees with the classical point defined by $x$. 
%
%For $r=1$, $D(x,1)$ does not depend on $x$ at all! 
%
%In fact any $z\in C^{0}$ is such that $|z|\le 1$, and then $|x-z|\le\max(|z|,|x|)\leq 1$, meaning that $z\in D(x,1)$, which thus equals $C^{0}$.
%
%This already allows us to visualize a partial picture. We can draw the classical points as in the picture below, and from each one of them we can go further inside the disc, ultil we reach $v_{x,1}$, which is the same ``target'' for all limbs starting from classical points, since it does not depend on $x$. This ``central'' point is called {\em Gauss point}. 
%
%We distinguish between the case $r\in |C^{\times}|$ and $r\notin |C^{\times}|$, calling the points $v_{x,r}$ with $r\in |C^{\times}|$ {\em of type $(2)$}, and calling {\em of type $(3)$} the others.
%
%\begin{center}
%\includegraphics[width=0.65\textwidth]{Gauss}\ \\
%{Around the Gauss point.}
%\end{center}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\paragraph{Dead ends.}\ \\
%In the case in which $C$ is not ``spherically complete'', that is, there exists a sequence of closed discs
%$$D_1\supset D_2\supset ... $$
%such that $\bigcap_mD_m=\varnothing$ (eg., for $C=\bC_p$) then we can encounter the following valuation on $C\langle T\rangle$:
%
%$$f\mapsto \inf_m\sup_{y\in D_m}|f(y)|.$$
%
%It is readily seen that this determines a point in $\bD_C$.
%
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\paragraph{Points corresponding to rank $2$ valuations.}\ \\
%We let $x\in C^{0}$, and fix $r\in\bR$ such that $0<r\leq 1$. We define
%$$\Gamma_{<r}:=\bR^{\times}_{>0}\times\gamma^{\bZ}$$
%the abelian group endowed with the unique total order such that $s<\gamma<r$ for all $s<r$. 
%
%Then we can assign the map $v_{x,<r}: C\langle t\rangle\to\Gamma_{<r}\cup\{0\}$:
%
%$$f=\sum_{m=0}^{\infty}a_m(t-x)^m\longmapsto\max_m\{|a_m|\gamma^m\}.$$
%
%It turns out that this point only depends on the maximal ideal $\fm\subset C^{0}$. Analogously, we can define, for all $0<r<1$, $$\Gamma_{>r}:=\bR^{\times}_{>0}\times\gamma^{\bZ}$$
%endowed with the unique total order such that $s>\gamma>r$ for all $s>r$. 
%
%Then, $v_{x,>r}$ will be naturally assigned by
%$$f=\sum_{m=0}^{\infty}a_m(t-x)^m\longmapsto\max_m\{|a_m|\gamma^m\}.$$
%It turns out that this point only depends on $D(x,r)$, meaning that $v_{x,>r}=v_{y,>r}$ if and only if $v_{x,r}=v_{y,r}$. 
%
%In the case $r\notin |C^{\times}|$, then
%$$v_{x,>r}=v_{x,<r}=v_{x,r}$$
%and in the case $r\in |C^{\times}|$, this consists of one additional point at which a ray from $v_{x,r}$ starts, and to which $v_{x,r}$ specializes. Roughly, the points $v_{x,>r}$ correspond to rays towards the Gauss point, while the points $v_{x,<r}$ correspond to rays towards the classical points. These points are called {\em of type $(5)$}, and do not appear in the rigid-analytic setting, as they correspond to rank $2$ valuations. 
%
%We should also remarx that $v_{x,>1}$ is {\em outside} $\bD^C$, as
%$$v_{x,>1}(t-x)=(1\times \gamma)>(1\times \gamma^0)$$
%where $(1\times \gamma^0)$ is the unit element of $\bR^{\times}_{>0}\times\gamma^{\bZ}$.
%
%\begin{center}
%\includegraphics[width=0.65\textwidth]{Disc}\ \\
%{The adic unit disc over $\bC_p$.}
%\end{center}
%
%The above description simply says that, starting from the classical points, we find rays tending to the Gauss point, where the various points merge, depending on their distance. We also remark that the closure of the Gauss point is homeomorphic to $\bA^1_{\kappa}$, with the Gauss point as generic point. The closure of a point of type $(2)$ is, instead, homeomorphic to $\bP^1_{\kappa}$.
%\end{eg}
%
%
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\subsection{Appendix B: an interesting locus in $\Spa(A,A^+)$}
%
%Let $A$ be a commutative ring, and call $X:=\Spv(A)$ and $X_{\rm triv}$ the subspace of trivial valuations on $A$. The continuous map
%$$\varphi: X\to\Spec(A)$$
%sending $v\in X$ to its support, restricts to a homeomorphism between $X_{\rm triv}$ and $\Spec(A)$. Indeed, given $\fp\in\Spec(A)$, we define
%$$v_{\fp}(a):=\begin{cases} 1 & \textup{if}\ a\notin\fp\\
%						0 & \textup{if}\ a\in\fp
%\end{cases}$$
%and $\fp\mapsto v_{\fp}$ is an inverse map to $\varphi$. Given $x,s\in A$
%$$X_{\rm triv}\cap X(x/s)=\{v\in X_0\mid v(x)\le v(s)\neq 0\}=\{v\in X_{\rm triv}\mid v(s)\neq 0\}$$
%is sent by $\varphi$ onto $\{\fp\in\Spec(A)\mid s\notin\fp\}=D(s)$, which proves $\varphi$ is bijective and continuous $X_{\rm triv}\to\Spec(A)$, with continuous inverse.
%
%Now let $A$ be a Noetherian adic ring, with ideal of definition $I$. We can consider the closed locus $V(I)$ in $\Spec(A)$, which is $\Spf(A)$, and we wonder what subspace $Z_{0}$ of $X$ is such that $\varphi$ restricts to a homeomorphism $Z_0\simeq\Spf(A)$. 
%
%Given that $X_{\rm triv}$ is homeomorphic to $\Spec(A)$ as a subspace of $X$, such $Z_0$ shall be found to be a subspace of $X_{\rm triv}$. Indeed, $Z_0$ is just the locus of those valuations in $X$ which are trivial and with open support, so let us call it $Z_{\rm triv}$.
%
%All points in $Z_{\rm triv}$ are continuous valuations on $A$ because they have open support. Since they have no proper horizontal specialization, $$Z_{\rm triv}\subseteq\Spv(A,A)\cap\Cont(A)\cap\varphi^{-1}(\Spf(A))\subseteq\Spv(A,A)\cap\Cont(A).$$
%
%Now let us consider the adic spectrum $\Spa(A,A)$. $Z_{\rm triv}$ certainly lives in it, and we claim that in fact
%$$Z_{\rm triv}=\Spa(A,A)\cap\Spv(A,A)$$
%
%We can show
%$$X_{\rm triv}=\{v\in\Spv(A)\mid v(A)=v(A)_{\le 1}\}\cap\Spv(A,A).$$
%This amounts indeed to saying that as soon as a valuation $v$ on $A$ has no proper horizontal specializations, and satisfies $v(A)_{\le 1}=v(A)$, then it is trivial, and is immediate from the fact that $c\Gamma_v$ is the minimal convex subgroup of $\Gamma_v$ containing $v(A)_{\ge 1}=v(A)_{=1}$, implying $\Gamma_v=1$.
%
%We conclude that $X_{\rm triv}\cap\Spa(A,A)$, which is in turn $\Spv(A,A)\cap\Spa(A,A)$, is actually
%$$\Spa(A,A)\cap\Spv(A,A)\cap\varphi^{-1}(\Spf(A)).$$
%
%A consequence of the above observations is that we have a retraction
%$$\rho:\Spa(A,A)\twoheadrightarrow Z_{\rm triv}$$
%given by $\rho(v):=v_{\vert_{c\Gamma_v}}$. 
%
%In the last lecture, we shall equip $\Spa(A,A)$ with a presheaf of rings which will make it into a locally ringed space. In fact such presheaf, thanks to the adic and Noetherian assumptions on our $A$, will be a sheaf.
%
%We shall therefore equip $Z_{\rm triv}$ with the sheaf of rings $\rho_*\cO_{\Spa(A,A)}$. We are, in the end, given assignments
%$$(\Spf(A),\cO_{\Spf(A)})\mapsto (\Spa(A,A),\cO_{\Spa(A,A)})\mapsto (Z_{\rm triv}, \rho_*\cO_{\Spa(A,A)}),$$
%which are functorial, and the latter is seen to be a quasi-inverse functor to the former functor, a result which lies at the heart of Huber's theorem saying that the category of locally Noetherian formal schemes embeds as a full subcategory into the category of adic spaces, and will be discussed in the last set of notes.
%



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\small
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