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\textbf{
18.727, Topics in Algebraic Geometry (rigid analytic geometry)\\
Kiran S. Kedlaya, fall 2004 \\
More on affinoid algebras (amended 4 Oct 04)
}
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\subsection*{Addenda on the spectral seminorm}

A norm on a Banach algebra $A$ is \emph{power-multiplicative} if
$\|f^n\| = \|f\|^n$ for any $f \in A$ and any positive integer $n$.
Our proof that the Gauss norm on $T_n$ has a topological characterization
adapts to show that for any affinoid algebra $A$, there is at most
one power-multiplicative Banach norm on $A$. We now know that if a
power-multiplicative norm exists, it must be the spectral seminorm;
hence such a norm exists if and only if $A$ is reduced.

In fact, the spectral seminorm is ``minimal'' in the following sense
[BGR, Corollary 3.8.2/2].
\begin{prop}
Let $A$ be an affinoid algebra equipped with a Banach
norm $\|\cdot\|$. Then
for all $f \in A$, $\|f\|_{\spec} \leq \|f\|$. In particular,
$|f(x)| \leq \|f\|$ for any $x \in \Max A$ (we already showed this if
$\|\cdot\|$ is a quotient norm from a Tate algebra).
\end{prop}
\begin{proof}
Apply the formula
\[
\|f\|_{\spec} = \lim_{n \to \infty} \|f^n\|^{1/n}
\]
and note that $\|f^n\| \leq \|f\|^n$ because $\|\cdot\|$ is a Banach
algebra norm.
\end{proof}

This yields the following 
characterization of nilpotent elements, in the vein of our characterization
of power-bounded elements [BGR, Proposition~6.2.3/2].
\begin{prop}
For $A$ an affinoid algebra and $f \in A$, the following 
statements are equivalent:
\begin{enumerate}
\item[(a)]
$f$ is topologically nilpotent (i.e., $\{f^n\}$ is a null sequence in $A$);
\item[(b)]
$|f(x)| < 1$ for all $x \in \Max A$;
\item[(c)]
$\|f\|_{\spec} < 1$.
\end{enumerate}
\end{prop}
\begin{proof}
The equivalence of (b) and (c) follows from the maximum modulus principle,
and (a) implies (c) by the previous proposition. Given (c), choose
$c \in K$ and $m \in \NN$ such that $|c| > 1$ but $\|cf^m\|_{\spec} \leq 1$.
Then $cf^m$ is power-bounded (from last time), 
so $c^{-1}(cf^m) = f^m$ is topologically nilpotent, as then is $f$.
Thus (c) implies (a), and we are done.
\end{proof}

\subsection*{Spectral norms are Banach norms}

We now know that the spectral seminorm on a reduced affinoid algebra
is a norm. However, more than that is true: it is a Banach norm.
(This proof is from [FvdP, Theorem~3.4.9]; the proof in
[BGR, Theorem 6.2.4/1] is a bit more intricate.)

We say a bounded map $f: A \to B$ of Banach spaces is \emph{strict} if
the bijection between $A/\ker(f)$ with its quotient topology and
$\image(f)$ with its subspace topology is an isomorphism of normed spaces.
Since $\ker(f)$ is closed, by the open mapping theorem this is
equivalent to saying that $\image(f)$ is closed in $B$. 

Note that if 
$A$ and $B$ are affinoid algebras and $f$ is finite, it is automatically
strict, because $f(A)$ is an $A$-submodule of $B$, and any submodule
of a finitely generated module over an affinoid algebra is closed
(see lemma from ``$p$-adic functional analysis 2'').
In fact, if $A$ is an affinoid algebra and $M_1, M_2$ are finitely
generated modules over $A$, then every $A$-algebra homomorphism
$M_1 \to M_2$ is strict.

On the other hand, here's an
example of a bounded non-strict map: the homomorphism
$T_1 \to T_1$ sending $x_1$ to $cx_1$, where $c \in K$ satisfies $|c| < 1$.

\begin{lemma}
Let $A \hookrightarrow B$ be a strict inclusion of affinoid algebras. Then
the spectral seminorm on $B$ restricts to the spectral seminorm on $A$.
\end{lemma}
\begin{proof}
Choose a Banach norm $\|\cdot\|$ on $B$; it then restricts to a Banach
norm on $A$ by strictness (and the fact that any two Banach norms on
an affinoid algebra are equivalent), and applying the formula
$\|f\|_{\spec} = \lim_{n \to \infty} \|f^n\|^{1/n}$ gives us the claim.
\end{proof}

\begin{theorem}
Let $A$ be a reduced affinoid algebra. Then $A$ is complete under
$\|\cdot\|_{\spec}$. In particular, every Banach algebra norm on
$A$ is equivalent to the spectral norm.
\end{theorem}
\begin{proof}
The last sentence will follow from what we showed earlier: any two
Banach algebra norms on an affinoid algebra are equivalent. So we focus
on showing that $A$ is complete.

We first reduce to the case where $A$ is an integral domain. Let
$\gothp_1, \dots, \gothp_m$ be the minimal primes of $A$. Choose a 
Banach norm $\|\cdot\|_A$ on $A$, put $A_i = A/\gothp_i$ (which is 
an integral domain), and
equip each $A_i$ with the quotient norm induced by $\|\cdot\|_A$.
Then $A_1 \oplus \cdots \oplus A_m$ becomes a finitely generated
Banach module over $A$ under the max norm
\[
(a_1, \dots, a_m) = \max_i \{\|a_i\|\}.
\]
Let $\iota: A \to A_1 \oplus \cdots \oplus A_m$ be the canonical injection;
then $\iota(A)$ is an $A$-submodule of $A_1 \oplus \cdots \oplus A_m$.
By the lemma from ``$p$-adic functional analysis 2'', $\iota(A)$
is closed in $A_1 \oplus \cdots \oplus A_m$, and $\iota$ is an isomorphism
onto its image by the open mapping theorem. The map $\iota$
is isometric for the spectral norms (because $\Max A = \cup_i \Max A_i$), 
so proving that the spectral
norm on each $A_i$ is equivalent to the quotient norm proves that
the spectral norm on $A$ is equivalent to $\|\cdot\|_A$.

From now on, we assume $A$ is an integral domain.
If $B$ is a reduced affinoid algebra containing $A$ and $A \hookrightarrow
B$ is strict (e.g., if $A \hookrightarrow B$ is finite), showing that $B$
is complete under its spectral seminorm implies that $A$ is complete
under its spectral seminorm, thanks to the previous lemma.
In particular, we can write $A$ as a finite integral extension of $T_d$
for some $d \geq 0$, and take $B$ to be the integral closure of $T_d$
in the normal closure of $\Frac A$ over $\Frac T_d$.
This lets us break the problem into two steps.
\begin{enumerate}
\item[(a)] Show that if $A$ is finite over $T_d$ and $\Frac A$
is purely inseparable over $\Frac T_d$, then the spectral norm on $A$
is complete.
\item[(b)] Show that if $B$ is finite over $A$, $\Frac B$
is Galois over $\Frac A$, and the spectral norm on $A$ is fully multiplicative
and complete, then the spectral norm on $B$ is complete.
\end{enumerate}
So we work on these two steps separately.
\begin{enumerate}
\item[(a)]
There is nothing to show unless the characteristic of $K$ is $p>0$.
Note that for any affinoid algebra $A$ over $K$ and any field extension
$K'$ containing $K$ and complete for a norm extending that on $K$,
the inclusion of $A$ into the completed tensor product $A \widehat{\otimes}_K
K'$ is strict. In particular, we may reduce to the case where
$K$ is algebraically closed. In that case,
\[
A \subseteq K \langle x_1^{1/p^m}, \dots, x_d^{1/p^m} \rangle
\]
for some $m$,
so it's enough to check the completeness of 
$K \langle x_1^{1/p^m}, \dots, x_d^{1/p^m} \rangle$ under its spectral
norm. But again, this is true because that spectral norm is the Gauss norm.
\item[(b)]
Put $G = \Gal(\Frac B/\Frac A)$, and put
\[
\Trace(f) = \sum_{g \in G} f^g;
\]
this gives a map from $\Frac B$ to $\Frac A$ such that
$\|\Trace(f)\|_{A,\spec} \leq \|f\|_{B,\spec}$. (This trace is the same
as the trace of multiplication by $f$ as 
a $\Frac A$-linear transformation on $\Frac B$.) From basic algebra,
we know that the $\Frac A$-linear pairing
\[
(x,y) \mapsto \Trace(xy)
\]
on $\Frac B$ is nondegenerate.

Choose $e_1, \dots, e_n \in B$ which form a basis for $\Frac B$ over $\Frac A$,
and let $e_1^*, \dots, e_n^*$ be the dual basis for the trace pairing.
We show that $Ae_1 + \cdots + Ae_n$ is a Banach module under the
spectral norm, by showing that the spectral norm is equivalent to the
maximum norm
\[
\|f_1 e_1 + \cdots + f_1 e_n\|_{B, \spec}
= \max_i \{\|f_i\|_{A,\spec}\}.
\]
Choose $a_0 \in A$ such that $a_0 e_j^* \in \gotho_B^{\spec}$ for
$j=1, \dots, n$; now given $f_1 e_1 + \cdots + f_n e_n \in
Ae_1 + \cdots + Ae_n$, we have
\begin{align*}
a_0 f_j &= \Trace(a_0 e_j^* \sum_i f_i e_i) \\
\|\Trace(a_0 e_j^* \sum_i f_i e_i)\|_{A,\spec}
&\leq \|a_0 e_j^* \sum_i f_i e_i\|_{B,\spec} \\\
&\leq \|\sum_i f_i e_i \|_{B,\spec}.
\end{align*}
Therefore
\[
\|\sum_i f_i e_i\|_{B,\spec} \geq \|a_0\|_{A,\spec} \max_i
\{\|f_i\|_{A,\spec} \}.
\]
Since we also have
\[
\|\sum_i f_i e_i\|_{B,\spec}
\leq \max_i \{\|f_i\|_{A,\spec}\}
\max_i \{\|e_i\|_{B,\spec}\},
\]
the spectral norm restricted to $Ae_1 + \cdots + Ae_n$ is
equivalent to the maximum norm, and so is a Banach norm.

For some $a \in A$, $aB \subseteq Ae_1 + \cdots + Ae_n$;
since the spectral norm on $A$ is multiplicative, the spectral norm
on $B$ is thus complete.
\end{enumerate}
\end{proof}

Note that this theorem can also be interpreted as follows: a sequence
of elements of $A$ converges to zero under some (any) Banach algebra norm if 
and only if it converges uniformly to zero on $\Max A$.


\subsection*{The reduction of an affinoid algebra}

In case you are wondering when the spectral seminorm is not just a norm
but is actually fully multiplicative (like the Gauss norm), here is your
answer. Recall that for $A$ an affinoid algebra, we defined
\[
\gotho_A^{\spec} = \{f \in A: \|f\|_{\spec} \leq 1\}.
\]
Now define
\[
\gothm_A^{\spec} = \{f \in A: \|f\|_{\spec} < 1\}
\]
and $\overline{A}^{\spec} = \gotho_A^{\spec}/\gothm_A^{\spec}$;
we call the latter the \emph{reduction} of $A$. Then we have
the following [BGR, Proposition~6.2.3/5].

\begin{prop}
The spectral seminorm is a fully multiplicative norm if and only if
$A$ is reduced and $\overline{A}$ is an integral domain.
\end{prop}
Note that $A$ being an integral domain is not enough; see exercises.
\begin{proof}
We already know that the spectral seminorm is a norm if and only if
$A$ is reduced; also, if $A$ is reduced and
the spectral norm is fully multiplicative,
then the product of elements of spectral norm 1 again has spectral
norm 1, so $\overline{A}$ is an integral domain. Conversely, suppose
$A$ is reduced and $\overline{A}$ is an integral domain.
Given $f,g \in A$ nonzero, 
there exists an integer $n$ such that $\|f\|_{\spec}^n$
and $\|g\|_{\spec}^n$ belong to $|K^*|$, by the maximum modulus principle.
(Namely, the spectral seminorm is always the norm of the evaluation of $f$
at some point whose residue field is finite over $K$.)
Choose $c,d \in K^*$ with $c \|f\|_{\spec}^n = d\|g\|_{\spec}^n = 1$.
Then the product of the images of $cf^n$ and $dg^n$ in $\overline{A}$
must be nonzero because $\overline{A}$ is an integral domain; that is,
$\|cf^n dg^n\|_{\spec} = 1$. Hence
\[
1 = \|cf^n dg^n\|_{\spec} = c \|f\|_{\spec}^n \cdot d \|g\|_{\spec}^n
\]
and (by power-multiplicativity of the spectral seminorm)
it follows that $\|fg\|_{\spec} = \|f\|_{\spec} \cdot \|g\|_{\spec}$.
\end{proof}

\subsection*{Exercises}

\begin{enumerate}
\item 
Give an explicit example of an affinoid algebra $A$ which is an integral
domain, but whose spectral seminorm is not fully multiplicative. (Hint:
consider power series in $x$ and $x^{-1}$ which converge on a suitable
annulus; your motivation should functions on a complex annulus which
have their suprema on opposite boundary components. 
We'll look more at this geometric situation in the next few lectures.)
\item
Let $A$ be an affinoid algebra, and suppose $f \in A$ satisfies
$\|f\|_{A,\spec} = 1$. Prove that the natural map from $A$
to $B = A\langle x \rangle/(fx-1)$ is strict. (Such a $B$ is
called a \emph{strict localization} of $A$.)
\end{enumerate}


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