\input{preamble.tex}
\unittitle{Master theorems for slope factorization}

I have used several constructions called ``slope factorizations'' so far.
In this unit, I give a general framework that includes all of them.

Throughout this unit only, 
rings are unital but \emph{not} necessarily commutative.

\section{The master theorems}

For $R$ a ring, we define a
\emph{norm} on $R$ as in the case of a field, i.e., it is a function
$|\cdot|: R \to \RR_{\geq 0}$ 
satisfying the following conditions.
\begin{enumerate}
\item[(a)]
For $r \in R$, $|r| =0$ if and only if $r=0$.
\item[(b)]
For $r,s \in R$, $|r+s| \leq \max\{|r|, |s|\}$.
\item[(c)]
For $r,s \in R$, $|rs| = |r| |s|$.
\end{enumerate}

A \emph{plus-minus decomposition} of $R$ is a direct sum decomposition
$R_- \oplus R_0 \oplus R_+$ of the additive group of $R$ such that
\begin{align*}
&1 \in R_0, \quad R_0 \cdot R_0 \subseteq R_0, \\
&(R_0 \oplus R_-) \cdot R_- \cdot (R_0 \oplus R_-) \subseteq R_-, \\
&(R_0 \oplus R_+) \cdot R_+ \cdot (R_0 \oplus R_+) \subseteq R_+.
\end{align*}
For $r \in R$, write $f_-(r), f_0(r), f_+(r)$ for the components of $r$
under the decomposition.

\begin{theorem} \label{T:master1}
Let $R$ be a ring
complete under the norm $|\cdot|$,
equipped with a plus-minus decomposition.
Then for any $r \in R$ such that 
\[
f_0(r) \in R_0^\times,
\qquad
|f_+(r)| < |f_0(r)|, \qquad
|f_-(r)|< |f_0(r)|,
\]
there exists a unique 
factorization $r = s_+ s_-$ with
\[
s_+ \in R_0 \oplus R_+,
\qquad
s_- -1 \in R_-, \qquad
|s_+| \leq |f_0(r)|, \qquad
|s_- -1| < |f_0(r)|.
\]
\end{theorem}
\begin{proof}
By multiplying on the left by the inverse of $f_0(r)$, we may reduce to the case
where $f_0(r) = 1$, $|f_+(r)| < 1$, $|f_-(r)| < 1$. 
Put $\eta = \max\{|f_+(r)|, |f_-(r)|\}$.

We first check existence. Put $s_{+,0} = 1 + f_+(r)$, $s_{-,0} = 1 + f_-(r)$.
Given $s_{+,l}, s_{-,l}$ such that
$s_{+,l} \in R_0 \oplus R_+$, $|s_{+,l}-1| \leq \eta$,
$s_{-,l} -1 \in R_-$, $|s_{+,l}-1| \leq \eta$,
put $\delta_l = r - s_{+,l} s_{-,l}$.
Then set
\[
s_{+,l+1} = s_+ + f_0(\delta_l) + f_+(\delta_l), \qquad
s_{-,l+1} = s_- + f_-(\delta_l).
\]
We then see that 
\begin{align*}
|\delta_{l+1}| 
&= |(f_0(\delta_l) + f_+(\delta_l))(1 - s_{-,l}) 
+ (1- s_{+,l})f_-(\delta_l)| \\
&\leq \eta |\delta_l|.
\end{align*}
Thus the sequences $s_{+,l}, s_{-,l}$ converge to limits
$s_+, s_-$ with the desired properties.

We next check uniqueness; again, we may assume $f_0(r) = 1$.
If $s'_+ s'_-$ is another factorization of the desired form,
put $\rho = \max\{|s_+ - s'_+|, |s_- - s'_-|\}$.
Put 
\begin{align*}
x &= (s_+ - s'_+) s_- \\
&= s'_+ (s'_- - s_-) \\
&= (s_+ - s'_+) + (s_+ - s'_+)(s_- - 1) \\
&= (s'_- - s_-) + (s'_+-1) (s'_- - s_-).
\end{align*}
The third expression yields $|f_-(x)| \leq \rho \eta$;
the fourth expression yields $|f_0(x)|, |f_+(x)| \leq \rho \eta$.
However, the first two expressions together force $|x| = \eta$,
which is a contradiction unless $\rho=0$.
\end{proof}

Theorem~\ref{T:master1} works well for rings of power series but not
for rings of polynomials; for that, we must add some extra structure.
Let $R$ be a ring equipped with
a plus-minus decomposition. A \emph{plus-minus grading} on $R$ is
a pair of functions 
\[
\deg_*: R_* \to \ZZ_{\geq 0} \qquad (* \in \{+,-\})
\]
satisfying the following conditions.
\begin{enumerate}
\item[(a)]
For $* \in \{+, -\}$ and $r_1, r_2 \in R_*$,
$\deg_*(r_1 + r_2) \leq \max\{\deg_*(r_1), \deg_*(r_2)\}$.
\item[(b)]
For $* \in \{+, -\}$ and $r_1, r_2 \in R$,
$\deg_*(f_*(r_1 r_2)) \leq \deg_*(f_*(r_1)) + \deg_*(f_*(r_2))$.
\end{enumerate}

\begin{theorem}
Let $R$ be a ring equipped with a norm $|\cdot|$,
a plus-minus decomposition, and a plus-minus grading.
Assume that for $n \in \ZZ_{\geq0}$ and $* \in \{+,-\}$,
the set $\{r_* \in R_*: \deg_*(r_*) \leq n\}$ is complete under $|\cdot|$.
Then for any $r \in R$ such that 
\[
f_0(r) \in R_0^\times,
\qquad
|f_+(r)| < |f_0(r)|, \qquad
|f_-(r)|< |f_0(r)|,
\]
there exists a unique 
factorization $r = s_+ s_-$ with
\[
s_+ \in R_0 \oplus R_+,
\qquad
s_- -1 \in R_-, \qquad
|s_+| \leq |f_0(r)|, \qquad
|s_- -1| < |f_0(r)|.
\]
\end{theorem}
\begin{proof}
The proof is the same as that of Theorem~\ref{T:master1},
except that we must note that for each $l$,
$\deg_*(s_{*,l}) \leq \deg_*(f_*(r))$.
This again allows us to form a limit $s_*$ of $s_{*,l}$, and to
proceed as before.
\end{proof}

\section{An even stronger theorem}

Both of the previously stated theorems may be deduced from the following
theorem.
\begin{theorem} \label{T:master3}
Let $R$ be a ring equipped with a nonarchimedean norm.
Suppose the nonzero elements $a,b,c \in R$ and the additive subgroups $U,V,W 
\subseteq R$ satisfy the following conditions.
\begin{enumerate}
\item[(a)]
The spaces $U,V$ are complete under the norm, and $UV \subseteq W$.
\item[(b)]
The map $f(u,v) = au + bv$ is a surjection of $U \times V$ onto $W$.
\item[(c)]
There exists $\lambda > 0$ such that 
\[
|f(u,v)| \geq \lambda \max\{|a||v|, |b||u|\} \qquad (u \in U, v \in V).
\]
\item[(d)]
We have $ab - c \in W$ and
\[
|ab-c| < \lambda^2 |c|.
\]
\end{enumerate}
Then there exists a unique pair $x \in U, y \in V$ such that
\[
c = (a+x)(b+y), \qquad |x| < \lambda |a|, \qquad |y| < \lambda |b|.
\]
For this $x,y$, we also have
\[
|x| \leq \lambda^{-1} |ab-c| |b|^{-1}, \qquad
|y| \leq \lambda^{-1} |ab-c| |a|^{-1}.
\]
\end{theorem}
\begin{proof}
We define a norm on $U \times V$ by setting
\[
|(u,v)| = \max\{|a||v|, |b||u|\}.
\]
so that (c) implies
\[
\lambda |(u,v)| \leq |f(u,v)| \leq |(u,v)|.
\]
In particular, $\lambda \leq 1$, so $|ab-c| < |ab| = |c|$.

Since $a,b$ are nonzero, (c) implies that $f$ is injective. By (b),
$f$ is in fact a bijective group homomorphism between $U \times V$
and $W$. It follows that for all $w \in W$,
\[
|f^{-1}(w)| \leq \lambda^{-1} |w|.
\]
By (d), we may choose $\lambda \in (0, \lambda)$ with 
$|ab-c| \leq \lambda \mu |c|$. Define
\[
B_\mu = \{(u,v) \in U \times V: |(u,v)| \leq \mu |c|\}.
\]
For $(u,v) \in B_\mu$, we have
\[
|a||v| \leq |(u,v)| \leq \mu|c| = \mu|a||b|,
\]
so $|v| \leq \mu|b|$. Similarly $|u| \leq \mu|a|$. As a result,
\begin{align*}
|f^{-1}(c - ab - uv)| &\leq \lambda^{-1} |c-ab-uv| \\
&\leq \lambda^{-1} \max\{|c-ab|, |uv|\} \\
&\leq \lambda^{-1} \max\{\lambda \mu |c|, \mu^2 |a||b|\} \\
&= \mu |c|.
\end{align*}
Consequently, the map $g(u,v) = f^{-1}(c-ab-uv)$ carries $B_\mu$
into itself.

We next show that $g$ is contractive. For $(u,v), (t,s) \in B_\mu$,
\begin{align*}
|g(u,v) - g(t,s)| &\leq |f^{-1}(ts-uv)| \\
&\leq \lambda^{-1} |ts-uv| \\
&\leq \lambda^{-1} |t(s-v) + (t-u)v| \\
&\leq \lambda^{-1} \max\{\mu|a||s-v|, \mu|t-u||b|\} \\
&\leq \lambda^{-1} \mu |(u-t,v-s)| \\
&= \lambda^{-1} \mu |(u-t) - (v-s)|,
\end{align*}
which has the desired effect because $\lambda^{-1} \mu < 1$.

Since $g$ is contractive on $B_\mu$, and $U \times V$ is complete,
there is a unique $(x,y) \in U \times V$ fixed by $g$. That is,
\[
ax + by = f(x,y) = f(g(x,y)) = c-ab - xy
\]
and so
\[
c = (a+x)(b+y).
\]
Moreover, there is a unique such $(x,y)$ in the union of all of the
$B_\mu$, and that element belongs to the intersection of all of the
$B_\mu$.
\end{proof}

\notes

Theorem~\ref{T:master3} and its proof are due to Christol 
[Chr83, Proposition~1.5.1].

\end{document}