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\ihead{Errata to ``The Theory of Witt Vectors''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{The Theory of Witt Vectors}

\textit{Joseph Rabinoff}

version of 7 May 2009 (posted on arXiv as
\href{http://arxiv.org/abs/1409.7445v1}{arXiv:1409.7445v1})

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

I will refer to the results appearing in the note \textquotedblleft The Theory
of Witt Vectors\textquotedblright\ by the numbers under which they appear in
this note (specifically, in its version of May 7th, 2009\footnote{This version
is exactly the version that is available on the arXiv under the identifier
\href{http://arxiv.org/abs/1409.7445v1}{arXiv:1409.7445v1}; it also is exactly
the version that used to be available at
\url{http://www.math.harvard.edu/~rabinoff/misc/witt.pdf}.}).

\setcounter{section}{9}

\section{Errata}

\begin{itemize}
\item \textbf{Page 2, proof of Lemma 1.4:} Replace \textquotedblleft%
$y^{p^{i}\left(  p-n\right)  }$\textquotedblright\ by \textquotedblleft%
$y^{p^{i-1}\left(  p-n\right)  }$\textquotedblright\ (in the displayed
equation, just after the binomial coefficient).

\item \textbf{Page 4:} \textquotedblleft Letting $X_{1},Y_{1},X_{2}%
,Y_{2},\ldots$ be indeterminates\textquotedblright\ should be
\textquotedblleft Letting $X_{0},X_{1},X_{1},Y_{1},X_{2},Y_{2},\ldots$ be
indeterminates\textquotedblright.

\item \textbf{Page 4, proof of Theorem 1.5:} Throughout the proof, the
indexing of the $S_{i}$ is wrong: Every \textquotedblleft$S_{p^{i}}%
$\textquotedblright\ should be replaced by \textquotedblleft$S_{i}%
$\textquotedblright; also, \textquotedblleft$S_{1}$\textquotedblright,
\textquotedblleft$S_{p}$\textquotedblright\ and \textquotedblleft$S_{p^{n}}%
$\textquotedblright\ should be \textquotedblleft$S_{0}$\textquotedblright,
\textquotedblleft$S_{1}$\textquotedblright\ and \textquotedblleft$S_{n}%
$\textquotedblright, respectively.

That said, of course, the notation you are using from \S 2 on is different.
Maybe it is worth pointing this out, that the $S_{k}$ introduced in \S 1 are
the $S_{p^{k}}$ (and not the $S_{k}$) in \S 2 and later.

\item \textbf{Page 4, proof of Theorem 1.5:} In the last displayed equation of
the proof, replace \textquotedblleft$\tau\left(  x_{1}\right)  +p\tau\left(
x_{1}\right)  $\textquotedblright\ by \textquotedblleft$\tau\left(
x_{0}\right)  +p\tau\left(  x_{1}\right)  $\textquotedblright. Similarly,
replace \textquotedblleft$\tau\left(  y_{1}\right)  +p\tau\left(
y_{1}\right)  $\textquotedblright\ by \textquotedblleft$\tau\left(
y_{0}\right)  +p\tau\left(  y_{1}\right)  $\textquotedblright, and replace
\textquotedblleft$\tau\left(  \widetilde{s}_{1}\right)  +p\tau\left(
\widetilde{s}_{1}\right)  $\textquotedblright\ by \textquotedblleft%
$\tau\left(  \widetilde{s}_{0}\right)  +p\tau\left(  \widetilde{s}_{1}\right)
$\textquotedblright.

\item \textbf{Page 7:} In the displayed equation \textquotedblleft%
$w_{n}\left(  \left(  S_{n}\right)  _{n\in P}\right)  =w_{n}\left(  \left(
X_{n}\right)  _{n\in P}\right)  +w_{n}\left(  \left(  Y_{n}\right)  _{n\in
P}\right)  $\textquotedblright, you are using the letter \textquotedblleft%
$n$\textquotedblright\ in two different meanings (as an index and as a
variable). I suggest replacing it by \textquotedblleft$w_{m}\left(  \left(
S_{n}\right)  _{n\in P}\right)  =w_{m}\left(  \left(  X_{n}\right)  _{n\in
P}\right)  +w_{m}\left(  \left(  Y_{n}\right)  _{n\in P}\right)
$\textquotedblright.

\item \textbf{Page 7, Example 2.12:} Replace \textquotedblleft$\sum_{i=0}%
^{n}p^{i}X_{i}^{p^{n-i}}$\textquotedblright\ by \textquotedblleft$\sum
_{i=0}^{n}p^{i}X_{p^{i}}^{p^{n-i}}$\textquotedblright.

\item \textbf{Page 8, Theorem 2.13:} \textquotedblleft
reprsentatives\textquotedblright\ $\rightarrow$ \textquotedblleft
representatives\textquotedblright.

\item \textbf{Page 8, proof of Theorem 2.13:} Replace \textquotedblleft%
$\widetilde{x}_{p^{i}}$\textquotedblright\ and \textquotedblleft%
$\widetilde{y}_{p^{i}}$\textquotedblright\ by \textquotedblleft$\widetilde{x}%
_{i}$\textquotedblright\ and \textquotedblleft$\widetilde{y}_{i}%
$\textquotedblright, respectively (in the third displayed equation of this proof).

\item \textbf{Page 10, proof of Lemma 3.2:} Replace \textquotedblleft$y_{n}%
\in\mathbf{Z}\left[  X_{1},\ldots,X_{n}\right]  $\textquotedblright\ by
\textquotedblleft$y_{n}\in\mathbf{Z}\left[  x_{1},\ldots,x_{n}\right]
$\textquotedblright.

\item \textbf{Page 10, proof of Lemma 3.2:} Replace \textquotedblleft for some
choice of $x_{n}\in A$\textquotedblright\ by \textquotedblleft for some choice
of $x_{n}\in B$\textquotedblright.

\item \textbf{Page 12, proof of Theorem 2.6:} When you say \textquotedblleft
The unicity of the ring structure on $W\left(  A\right)  $\textquotedblright,
it might be helpful to clarify that you are talking about the unicity of the
whole functor $W$, not of a single ring structure $W\left(  A\right)  $
considered in isolation. (I don't think the latter would be unique, without
the functoriality requirement.)

\item \textbf{Page 12, proof of Theorem 2.6:} I think your proof is missing a
part: the proof of the fact that every element of $W\left(  A\right)  $ has an
additive inverse (one of the ring axioms). To prove this, I would again argue
by functoriality (first constructing the additive inverse of the Witt vector
$X=\left(  X_{1},X_{2},X_{3},\ldots\right)  $ in $W\left(  \mathbf{Q}\left[
X_{1},X_{2},X_{3},\ldots\right]  \right)  $, then showing (using Lemma3.2)
that its Witt coordinates belong to $\mathbf{Z}\left[  X_{1},X_{2}%
,X_{3},\ldots\right]  $, then projecting it onto $W\left(  A\right)  $ for any
$A$).

\item \textbf{Page 12, \S 3:} I think it is worth explicitly stating (as a
proposition?) the fact that the bijection $x\mapsto f_{x}:W\left(  A\right)
\rightarrow\Lambda\left(  A\right)  $ introduced in Corollary 3.3 is a ring
homomorphism. This is easy to prove\footnote{It requires showing that
$f_{xy}=f_{x}\widehat{\cdot}f_{y}$ and $f_{x+y}=f_{x}\widehat{+}f_{y}$ for any
$x,y\in W\left(  A\right)  $ (where $\widehat{\cdot}$ and $\widehat{+}$ are
the multiplication and the addition in the ring $\Lambda\left(  A\right)  $),
and that $f_{1}=1-t$ and $f_{0}=1$. Proving that $f_{xy}=f_{x}\widehat{\cdot
}f_{y}$ for any $x,y\in W\left(  A\right)  $ is easy: it follows (by
functoriality) from $f_{XY}=f_{X}\widehat{\cdot}f_{Y}$ (where $X$ and $Y$ are
as in the proof of Theorem 2.6), which was proven in the proof of Theorem 2.6.
Similarly, the other properties can be shown.}, and used a few times in the
rest of your paper.

\item \textbf{Page 13, Proposition 4.5:} Replace \textquotedblleft$s_{n}=%
\begin{cases}
x_{n} & \text{if }x_{n}\neq0\\
y_{n} & \text{if }y_{n}\neq0
\end{cases}
$\textquotedblright\ by \textquotedblleft$s_{n}=%
\begin{cases}
x_{n}, & \text{if }y_{n}=0;\\
y_{n}, & \text{if }x_{n}=0
\end{cases}
$\textquotedblright. (Otherwise, you are making no claim about the case when
both $x_{n}$ and $y_{n}$ are $0$.)

\item \textbf{Page 13, proof of Proposition 4.5:} \textquotedblleft the
equality of polynomial equations\textquotedblright\ $\rightarrow$
\textquotedblleft the equality of polynomials\textquotedblright\ maybe?

\item \textbf{Page 14, \S 5:} \textquotedblleft Verchiebung\textquotedblright%
\ $\rightarrow$ \textquotedblleft Verschiebung\textquotedblright.

\item \textbf{Page 15, Lemma 5.2:} \textquotedblleft two natural
transformations\textquotedblright\ $\rightarrow$ \textquotedblleft two natural
transformations between group-valued functors\textquotedblright. (This is to
clarify that the natural transformations have to preserve the groups' addition.)

\item \textbf{Page 15, proof of Lemma 5.2:} \textquotedblleft Let
$A_{n}=A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $\textquotedblright\ should
be \textquotedblleft Let $A_{n}=K\left[  X_{1},X_{2},\ldots,X_{n}\right]
$\textquotedblright.

\item \textbf{Page 15, proof of Lemma 5.2:} The notation $A_{n}$ (for
$K\left[  X_{1},X_{2},\ldots,X_{n}\right]  $) conflicts with the notation
$A_{0}$ (for $K\left[  x\right]  $). I would suggest changing one of the two notations.

\item \textbf{Page 15, proof of Lemma 5.2:} \textquotedblleft
commutivity\textquotedblright\ $\rightarrow$ \textquotedblleft
commutativity\textquotedblright.

\item \textbf{Page 19, proof of Proposition 5.10:} Remove the sentence
\textquotedblleft By Proposition 5.9, we may assume that $n$ and $m$ are
prime\textquotedblright\ (nothing wrong about it, but it is unnecessary).
Replace \textquotedblleft Since $m$ and $n$ are distinct
primes\textquotedblright\ by \textquotedblleft Since $m$ and $n$ are
relatively prime\textquotedblright.

\item \textbf{Page 19, proof of Proposition 5.12:} The letter
\textquotedblleft$N$\textquotedblright\ should be replaced by a (operatorname,
or mathrm-shaped) \textquotedblleft$\operatorname*{N}$\textquotedblright%
\ several times (whenever it stands for the norm map).

\item \textbf{Page 19, proof of Proposition 5.12:} Replace \textquotedblleft
Now let $n>1$\textquotedblright\ by \textquotedblleft Now let $n\geq
1$\textquotedblright.

\item \textbf{Page 20, Theorem 5.14:} \textquotedblleft
reprsentatives\textquotedblright\ $\rightarrow$ \textquotedblleft
representatives\textquotedblright.

\item \textbf{Page 21, proof of Theorem 5.14:} Replace all three summation
signs by \textquotedblleft$\sum_{n=0}^{\infty}$\textquotedblright.

\item \textbf{Page 22, proof of Theorem 6.1:} Replace \textquotedblleft%
$\sigma_{m}\left(  y\right)  $\textquotedblright\ by \textquotedblleft%
$\sigma_{m}\left(  x\right)  $\textquotedblright\ (after \textquotedblleft so
by induction\textquotedblright).

\item \textbf{Page 22, proof of Theorem 6.1:} Replace \textquotedblleft%
$\sigma_{n}\left(  y\right)  =\sigma_{p_{i}}\left(  y\right)  \circ\sigma
_{m}\left(  y\right)  $\textquotedblright\ by \textquotedblleft$\sigma
_{n}\left(  x\right)  =\sigma_{p_{i}}\left(  \sigma_{m}\left(  x\right)
\right)  $\textquotedblright.

\item \textbf{Page 22, proof of Theorem 6.1:} Replace \textquotedblleft%
$\sigma_{n}\left(  y\right)  \equiv$\textquotedblright\ by \textquotedblleft%
$\sigma_{n}\left(  x\right)  \equiv$\textquotedblright\ (twice).

\item \textbf{Page 22, proof of Theorem 6.1:} Replace \textquotedblleft%
$\sigma_{n}\left(  y\right)  -$\textquotedblright\ by \textquotedblleft%
$\sigma_{n}\left(  x\right)  -$\textquotedblright.

\item \textbf{Page 24, \S 7:} Replace \textquotedblleft since is
difficult\textquotedblright\ by \textquotedblleft since it is
difficult\textquotedblright.

\item \textbf{Page 25, proof of Lemma 7.3:} In the first displayed equation of
this proof, add a \textquotedblleft$x$\textquotedblright\ addend on the left
hand side (the sum should start with $x$, not with $\dfrac{x^{p}}{p}$).

\item \textbf{Page 25, proof of Lemma 7.3:} Replace \textquotedblleft%
$x^{p^{n}}$ coefficient\textquotedblright\ by \textquotedblleft$x^{p^{n}}%
$-coefficient\textquotedblright.

\item \textbf{Page 25, proof of Lemma 7.3:} After \textquotedblleft$p\nmid
s$\textquotedblright, add \textquotedblleft and $s>1$\textquotedblright.

\item \textbf{Page 26, proof of Lemma 7.4:} Replace \textquotedblleft a $p$th
power\textquotedblright\ by \textquotedblleft a power of $p$\textquotedblright\ (twice).

\item \textbf{Page 26, Lemma 7.5:} Replace \textquotedblleft$\prod_{r\geq1}%
$\textquotedblright\ by \textquotedblleft$\prod_{r\geq0}$\textquotedblright.

\item \textbf{Page 26, Lemma 7.5:} I would suggest clarifying that the product
$\prod_{r\geq0}\operatorname*{hexp}\left(  x_{p^{r}}t^{p^{r}}\right)  $ is
taken in $A\left[  \left[  t\right]  \right]  $, not in $\Lambda\left(
A\right)  $.

\item \textbf{Page 26, proof of Lemma 7.5:} Replace every appearance of
\textquotedblleft$\prod_{r\geq1}$\textquotedblright, \textquotedblleft%
$\prod_{n\geq1}$\textquotedblright\ or \textquotedblleft$\prod_{m\geq1}%
$\textquotedblright\ by \textquotedblleft$\prod_{r\geq0}$\textquotedblright,
\textquotedblleft$\prod_{n\geq0}$\textquotedblright\ or \textquotedblleft%
$\prod_{m\geq0}$\textquotedblright, respectively.

\item \textbf{Page 27, Theorem 7.6:} Replace \textquotedblleft$\prod_{r\geq1}%
$\textquotedblright\ by \textquotedblleft$\prod_{r\geq0}$\textquotedblright.

\item \textbf{Page 27, Theorem 7.6:} I would suggest clarifying that the
product $\prod_{r\geq0}\operatorname*{hexp}\left(  x_{p^{r}}t^{p^{r}}\right)
$ is taken in $A\left[  \left[  t\right]  \right]  $, not in $\Lambda\left(
A\right)  $.

\item \textbf{Page 27:} In \textquotedblleft every element of
$\widehat{\Lambda}\left(  \mathcal{N}\right)  :=1+\mathcal{N}\left[  t\right]
$ can be written uniquely as a finite product $\prod\left(  1-a_{n}%
t^{n}\right)  $\textquotedblright, I would replace the product sign
\textquotedblleft$\prod$\textquotedblright\ by \textquotedblleft$\prod
_{n=0}^{\infty}$\textquotedblright\ (as the reader would otherwise expect
$\prod_{n=1}^{\infty}$ instead). (Alternatively, you can replace it by
\textquotedblleft$\prod_{n=1}^{\infty}$\textquotedblright, but then you'd need
to replace \textquotedblleft$\widehat{\Lambda}\left(  \mathcal{N}\right)
:=1+\mathcal{N}\left[  t\right]  $\textquotedblright\ by \textquotedblleft%
$\widehat{\Lambda}\left(  \mathcal{N}\right)  :=1+t\mathcal{N}\left[
t\right]  $\textquotedblright.)
\end{itemize}

\section{Addenda}

Let me now add some further properties of Witt vectors.

I shall use the notations of \S 2--\S 7 of your notes. In particular, the Witt
polynomials $w_{n}$ will be defined as in Definition 2.4 (not as in \S 1):
namely, by%
\[
w_{n}=\sum_{d\mid n}dX_{d}^{n/d}\in\mathbf{Z}\left[  \left\{  X_{d}:d\mid
n\right\}  \right]  .
\]
Thus, if $p$ is a prime and if $n$ is a nonnegative integer, then the
definition of $w_{p^{n}}$ yields
\begin{equation}
w_{p^{n}}=\sum_{d\mid p^{n}}dX_{d}^{p^{n}/d}=\sum_{i=0}^{n}p^{i}X_{p^{i}%
}^{p^{n-i}} \label{eq.wpn=}%
\end{equation}
(since the positive divisors of $p^{n}$ are $p^{0},p^{1},\ldots,p^{n}$).

Sam Raskin has pointed out to me that Theorem 6.6 can be generalized: namely,
the condition that $L$ be perfect can be dropped from the definition of a
$p$-ring. In other words, the following holds:

\begin{theorem}
\label{thm.6.6gen}Let $p$ be a prime. Let $R$ be a ring equipped with a
decreasing sequence $R=\mathfrak{a}_{0}\supset\mathfrak{a}_{1}\supset
\mathfrak{a}_{2}\supset\mathfrak{a}_{3}\supset\cdots$ of ideals such that
$\left(  \mathfrak{a}_{n}\cdot\mathfrak{a}_{m}\subset\mathfrak{a}_{n+m}\text{
for all }n\geq0\text{ and }m\geq0\right)  $. Equip $R$ with the topology
defined by this sequence of ideals. Assume that $R$ is Hausdorff and complete
for this topology. Assume that the residue ring $L=R/\mathfrak{a}_{1}$ has
$p\cdot1_{L}=0$. Let $\pi$ denote the canonical projection $R\rightarrow
R/\mathfrak{a}_{1}=L$.

Let $K$ be a perfect ring of characteristic $p$. Let $f:K\rightarrow L$ be a
ring homomorphism.

There exists a unique continuous ring homomorphism $\theta:W_{p}\left(
K\right)  \rightarrow R$ making the square%
\begin{equation}%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%W_p\left(K\right) \ar[r]^-{\theta} \ar[d]_{w_1} & R \ar[d]^{\pi} \\
%K \ar[r]^{f} & L
%}} }%
%BeginExpansion
\xymatrix{
W_p\left(K\right) \ar[r]^-{\theta} \ar[d]_{w_1} & R \ar[d]^{\pi} \\
K \ar[r]^{f} & L
}
%EndExpansion
\label{eq.thm.6.6gen.a.2}%
\end{equation}
commute.
\end{theorem}

Theorem \ref{thm.6.6gen} generalizes your Theorem 6.6.

The proof of Theorem \ref{thm.6.6gen} given below is (at least on a
superficial level) quite different from your proof of Theorem 6.6; it is
inspired by the proof of \cite[Lemma 4.4.1]{BriCon} (thanks again to Sam
Raskin for the reference).\footnote{As far as I recall, I devised this proof
based on discussions with Sam Raskin at MIT in 2016. Sam seems to have a
cleaner and shorter proof, which I unfortunately have never had the time to
fully comprehend. Sorry, Sam, for making a mess of your ideas!}

Before I prove Theorem \ref{thm.6.6gen}, let me derive various facts that will
be used in its proof and (some of which) are of independent interest:

\begin{proposition}
\label{prop.6.6.conty}Let $R$ be a ring equipped with a decreasing sequence
$R=\mathfrak{a}_{0}\supset\mathfrak{a}_{1}\supset\mathfrak{a}_{2}%
\supset\mathfrak{a}_{3}\supset\cdots$ of ideals such that $\left(
\mathfrak{a}_{n}\cdot\mathfrak{a}_{m}\subset\mathfrak{a}_{n+m}\text{ for all
}n\geq0\text{ and }m\geq0\right)  $. Equip $R$ with the topology defined by
this sequence of ideals. Assume that $R$ is Hausdorff for this topology.

\textbf{(a)} We have $\bigcap_{n\geq0}\mathfrak{a}_{n}=0$.

\textbf{(b)} If $u$ and $v$ are two elements of $R$ such that every $i\geq0$
satisfies $u\equiv v\operatorname{mod}\mathfrak{a}_{i}$, then $u=v$.

\textbf{(c)} For every $n\geq0$, let $\pi_{n}$ be the canonical projection
$R\rightarrow R/\mathfrak{a}_{n}$.

Let $A$ be a set, and let $\varphi:A\rightarrow R$ and $\psi:A\rightarrow R$
be two maps. Assume that $\pi_{n}\circ\varphi=\pi_{n}\circ\psi$ for every
$n\geq0$. Then, $\varphi=\psi$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.6.6.conty}.]Recall the definition of the
topology on $R$: A subset $S$ of $R$ is said to be open if and only if for
every $s\in S$, there exists an $n\geq0$ satisfying $s+\mathfrak{a}_{n}\subset
S$.

\textbf{(a)} Let $u\in\bigcap_{n\geq0}\mathfrak{a}_{n}$. We shall show that
$u=0$.

Let $P$ be an open subset of $R$ containing $u$. Let $Q$ be an open subset of
$R$ containing $0$. (We shall not use the fact that $Q$ is open.)

The subset $P$ of $R$ is open. In other words, for every $s\in P$, there
exists an $n\geq0$ satisfying $s+\mathfrak{a}_{n}\subset P$ (by the definition
of the topology on $R$). Applying this to $s=u$, we conclude that there exists
an $n\geq0$ satisfying $u+\mathfrak{a}_{n}\subset P$. Let $p$ be this $n$.
Thus, $p\geq0$ satisfies $u+\mathfrak{a}_{p}\subset P$.

On the other hand, $0\in Q$ (since $Q$ contains $0$).

But $u\in\bigcap_{n\geq0}\mathfrak{a}_{n}\subset\mathfrak{a}_{p}$ and thus
$u\equiv0\operatorname{mod}\mathfrak{a}_{p}$. Hence, $0\equiv
u\operatorname{mod}\mathfrak{a}_{p}$. In other words, $0\in u+\mathfrak{a}%
_{p}\subset P$. Combining this with $0\in Q$, we obtain $0\in P\cap Q$. Hence,
the set $P\cap Q$ contains at least one element (namely, $0$). Thus, $P\cap
Q\neq\varnothing$.

Now, forget that we fixed $P$ and $Q$. We thus have shown that if $P$ is an
open subset of $R$ containing $u$, and if $Q$ is an open subset of $R$
containing $0$, then $P\cap Q\neq\varnothing$. Since the topological space $R$
is Hausdorff, this shows that $u=0$.

Now, forget that we fixed $u$. We thus have proven that every $u\in
\bigcap_{n\geq0}\mathfrak{a}_{n}$ satisfies $u=0$. In other words,
$\bigcap_{n\geq0}\mathfrak{a}_{n}=0$. This proves Proposition
\ref{prop.6.6.conty} \textbf{(a)}.

\textbf{(b)} Let $u$ and $v$ be two elements of $R$ such that every $i\geq0$
satisfies $u\equiv v\operatorname{mod}\mathfrak{a}_{i}$. Hence, every $i\geq0$
satisfies $u-v\in\mathfrak{a}_{i}$ (since $u\equiv v\operatorname{mod}%
\mathfrak{a}_{i}$). In other words, $u-v\in\bigcap_{i\geq0}\mathfrak{a}%
_{i}=\bigcap_{n\geq0}\mathfrak{a}_{n}=0$ (by Proposition \ref{prop.6.6.conty}
\textbf{(a)}). In other words, $u-v=0$, so that $u=v$. This proves Proposition
\ref{prop.6.6.conty} \textbf{(b)}.

\textbf{(c)} Let $a\in A$. Let $n\geq0$. Then, $\pi_{n}\left(  \varphi\left(
a\right)  \right)  =\underbrace{\left(  \pi_{n}\circ\varphi\right)  }%
_{=\pi_{n}\circ\psi}\left(  a\right)  =\left(  \pi_{n}\circ\psi\right)
\left(  a\right)  =\pi_{n}\left(  \psi\left(  a\right)  \right)  $. Since
$\pi_{n}$ is the canonical projection $R\rightarrow R/\mathfrak{a}_{n}$, this
rewrites as $\varphi\left(  a\right)  \equiv\psi\left(  a\right)
\operatorname{mod}\mathfrak{a}_{n}$. In other words, $\varphi\left(  a\right)
-\psi\left(  a\right)  \in\mathfrak{a}_{n}$.

Now, forget that we fixed $n$. We thus have proven that $\varphi\left(
a\right)  -\psi\left(  a\right)  \in\mathfrak{a}_{n}$ for each $n\geq0$.
Hence, $\varphi\left(  a\right)  -\psi\left(  a\right)  \in\bigcap_{n\geq
0}\mathfrak{a}_{n}=0$ (by Proposition \ref{prop.6.6.conty} \textbf{(a)}), so
that $\varphi\left(  a\right)  -\psi\left(  a\right)  =0$ and thus
$\varphi\left(  a\right)  =\psi\left(  a\right)  $.

Now, forget that we fixed $a$. We have now shown that $\varphi\left(
a\right)  =\psi\left(  a\right)  $ for each $a\in A$. In other words,
$\varphi=\psi$. This proves Proposition \ref{prop.6.6.conty} \textbf{(c)}.
\end{proof}

\begin{proposition}
\label{prop.6.6.conty-riho}Let $R$ be a ring equipped with a decreasing
sequence $R=\mathfrak{a}_{0}\supset\mathfrak{a}_{1}\supset\mathfrak{a}%
_{2}\supset\mathfrak{a}_{3}\supset\cdots$ of ideals such that $\left(
\mathfrak{a}_{n}\cdot\mathfrak{a}_{m}\subset\mathfrak{a}_{n+m}\text{ for all
}n\geq0\text{ and }m\geq0\right)  $. Equip $R$ with the topology defined by
this sequence of ideals. Assume that $R$ is Hausdorff for this topology. For
every $n\geq0$, let $\pi_{n}$ be the canonical projection $R\rightarrow
R/\mathfrak{a}_{n}$.

Let $A$ be a ring. Let $\gamma:A\rightarrow R$ be a map. Assume that the map
$\pi_{n}\circ\gamma:A\rightarrow R/\mathfrak{a}_{n}$ is a ring homomorphism
for each $n\geq0$. Then, the map $\gamma$ is a ring homomorphism.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.6.6.conty-riho}.]We need to show the following
four claims:

\textit{Claim 1:} We have $\gamma\left(  0_{A}\right)  =0_{R}$.

\textit{Claim 2:} We have $\gamma\left(  a\right)  +\gamma\left(  b\right)
=\gamma\left(  a+b\right)  $ for any $a\in A$ and $b\in A$.

\textit{Claim 3:} We have $\gamma\left(  1_{A}\right)  =1_{R}$.

\textit{Claim 4:} We have $\gamma\left(  a\right)  \gamma\left(  b\right)
=\gamma\left(  ab\right)  $ for any $a\in A$ and $b\in A$.

We shall only prove Claim 4; all the other three claims can be proven in the
same vein.

\textit{Proof of Claim 4:} Let $a\in A$ and $b\in A$. Let $n$ be a nonnegative
integer. Then, the map $\pi_{n}\circ\gamma:A\rightarrow R/\mathfrak{a}_{n}$ is
a ring homomorphism (by our assumption). Thus, $\left(  \pi_{n}\circ
\gamma\right)  \left(  a\right)  \cdot\left(  \pi_{n}\circ\gamma\right)
\left(  b\right)  =\left(  \pi_{n}\circ\gamma\right)  \left(  ab\right)  $.

Now, recall that $\pi_{n}$ is the canonical projection $R\rightarrow
R/\mathfrak{a}_{n}$, and thus a ring homomorphism. Hence,%
\begin{align*}
\pi_{n}\left(  \gamma\left(  a\right)  \gamma\left(  b\right)  \right)   &
=\underbrace{\pi_{n}\left(  \gamma\left(  a\right)  \right)  }_{=\left(
\pi_{n}\circ\gamma\right)  \left(  a\right)  }\cdot\underbrace{\pi_{n}\left(
\gamma\left(  b\right)  \right)  }_{=\left(  \pi_{n}\circ\gamma\right)
\left(  b\right)  }=\left(  \pi_{n}\circ\gamma\right)  \left(  a\right)
\cdot\left(  \pi_{n}\circ\gamma\right)  \left(  b\right) \\
&  =\left(  \pi_{n}\circ\gamma\right)  \left(  ab\right)  =\pi_{n}\left(
\gamma\left(  ab\right)  \right)  .
\end{align*}
In other words, $\gamma\left(  a\right)  \gamma\left(  b\right)  \equiv
\gamma\left(  ab\right)  \operatorname{mod}\mathfrak{a}_{n}$ (since $\pi_{n}$
is the canonical projection $R\rightarrow R/\mathfrak{a}_{n}$). In other
words, $\gamma\left(  a\right)  \gamma\left(  b\right)  -\gamma\left(
ab\right)  \in\mathfrak{a}_{n}$.

Now, forget that we fixed $n$. We thus have shown that $\gamma\left(
a\right)  \gamma\left(  b\right)  -\gamma\left(  ab\right)  \in\mathfrak{a}%
_{n}$ for every $n\geq0$. Thus, $\gamma\left(  a\right)  \gamma\left(
b\right)  -\gamma\left(  ab\right)  \in\bigcap_{n\geq0}\mathfrak{a}_{n}=0$ (by
Proposition \ref{prop.6.6.conty} \textbf{(a)}). In other words, $\gamma\left(
a\right)  \gamma\left(  b\right)  -\gamma\left(  ab\right)  =0$, so that
$\gamma\left(  a\right)  \gamma\left(  b\right)  =\gamma\left(  ab\right)  $.
This proves Claim 4.

As we have said, the Claims 1, 2 and 3 can be shown by analogous arguments.
Thus, all four claims are proven, and the proof of Proposition
\ref{prop.6.6.conty-riho} is complete.
\end{proof}

\begin{proposition}
\label{prop.6.6.contin}Let $R$ be a ring equipped with a decreasing sequence
$R=\mathfrak{a}_{0}\supset\mathfrak{a}_{1}\supset\mathfrak{a}_{2}%
\supset\mathfrak{a}_{3}\supset\cdots$ of ideals such that $\left(
\mathfrak{a}_{n}\cdot\mathfrak{a}_{m}\subset\mathfrak{a}_{n+m}\text{ for all
}n\geq0\text{ and }m\geq0\right)  $. Equip $R$ with the topology defined by
this sequence of ideals. For every $n\geq0$, let $\pi_{n}$ be the canonical
projection $R\rightarrow R/\mathfrak{a}_{n}$.

Let $A$ be a topological space. Let $\varphi:A\rightarrow R$ be a map. Assume
that for each $n\geq0$, the map $\pi_{n}\circ\varphi:A\rightarrow
R/\mathfrak{a}_{n}$ is continuous (where $R/\mathfrak{a}_{n}$ is equipped with
the discrete topology). Then, the map $\varphi:A\rightarrow R$ is continuous.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.6.6.contin}.]Recall the definition of the
topology on $R$: A subset $S$ of $R$ is said to be open if and only if for
every $s\in S$, there exists an $n\geq0$ satisfying $s+\mathfrak{a}_{n}\subset
S$.

Let $U$ be an open subset of $R$. We shall show that the subset $\varphi
^{-1}\left(  U\right)  $ of $A$ is open.

Indeed, let $t\in\varphi^{-1}\left(  U\right)  $ be arbitrary. Thus,
$\varphi\left(  t\right)  \in U$.

The subset $U$ of $R$ is open. In other words, for every $s\in U$, there
exists an $n\geq0$ satisfying $s+\mathfrak{a}_{n}\subset U$ (by the definition
of the topology on $R$). Applying this to $s=\varphi\left(  t\right)  $, we
conclude that there exists an $n\geq0$ satisfying $\varphi\left(  t\right)
+\mathfrak{a}_{n}\subset U$. Consider this $n$.

Consider the set $R/\mathfrak{a}_{n}$ as a topological space, equipped with
the discrete topology. Then, each subset of $R/\mathfrak{a}_{n}$ is open. In
particular, the subset $\left\{  \pi_{n}\left(  \varphi\left(  t\right)
\right)  \right\}  $ of $R/\mathfrak{a}_{n}$ is open.

By our assumption, the map $\pi_{n}\circ\varphi:A\rightarrow R/\mathfrak{a}%
_{n}$ is continuous. Thus, for each open subset $Q$ of $R/\mathfrak{a}_{n}$,
the subset $\left(  \pi_{n}\circ\varphi\right)  ^{-1}\left(  Q\right)  $ of
$A$ is open. Applying this to $Q=\left\{  \pi_{n}\left(  \varphi\left(
t\right)  \right)  \right\}  $, we conclude that the subset $\left(  \pi
_{n}\circ\varphi\right)  ^{-1}\left(  \left\{  \pi_{n}\left(  \varphi\left(
t\right)  \right)  \right\}  \right)  $ of $A$ is open (since the subset
$\left\{  \pi_{n}\left(  \varphi\left(  t\right)  \right)  \right\}  $ of
$R/\mathfrak{a}_{n}$ is open). Denote this subset $\left(  \pi_{n}\circ
\varphi\right)  ^{-1}\left(  \left\{  \pi_{n}\left(  \varphi\left(  t\right)
\right)  \right\}  \right)  $ by $G$. Thus, $G=\left(  \pi_{n}\circ
\varphi\right)  ^{-1}\left(  \left\{  \pi_{n}\left(  \varphi\left(  t\right)
\right)  \right\}  \right)  $ is an open subset of $A$.

Furthermore, $G\subset\varphi^{-1}\left(  U\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $g\in G$. Thus, $g\in G=\left(  \pi
_{n}\circ\varphi\right)  ^{-1}\left(  \left\{  \pi_{n}\left(  \varphi\left(
t\right)  \right)  \right\}  \right)  $. Hence, $\left(  \pi_{n}\circ
\varphi\right)  \left(  g\right)  \in\left\{  \pi_{n}\left(  \varphi\left(
t\right)  \right)  \right\}  $, so that $\left(  \pi_{n}\circ\varphi\right)
\left(  g\right)  =\pi_{n}\left(  \varphi\left(  t\right)  \right)  $. Hence,
$\pi_{n}\left(  \varphi\left(  t\right)  \right)  =\left(  \pi_{n}\circ
\varphi\right)  \left(  g\right)  =\pi_{n}\left(  \varphi\left(  g\right)
\right)  $. In other words, $\varphi\left(  t\right)  \equiv\varphi\left(
g\right)  \operatorname{mod}\mathfrak{a}_{n}$ (since $\pi_{n}$ is the
canonical projection $R\rightarrow R/\mathfrak{a}_{n}$). Thus, $\varphi\left(
g\right)  \in\varphi\left(  t\right)  +\mathfrak{a}_{n}\subset U$. Hence,
$g\in\varphi^{-1}\left(  U\right)  $.
\par
Now, forget that we fixed $g$. We thus have proven that $g\in\varphi
^{-1}\left(  U\right)  $ for each $g\in G$. In other words, $G\subset
\varphi^{-1}\left(  U\right)  $.} and $t\in G$%
\ \ \ \ \footnote{\textit{Proof.} We have $\left(  \pi_{n}\circ\varphi\right)
\left(  t\right)  =\pi_{n}\left(  \varphi\left(  t\right)  \right)
\in\left\{  \pi_{n}\left(  \varphi\left(  t\right)  \right)  \right\}  $ and
thus $t\in\left(  \pi_{n}\circ\varphi\right)  ^{-1}\left(  \left\{  \pi
_{n}\left(  \varphi\left(  t\right)  \right)  \right\}  \right)  =G$.}. Hence,
$G$ is an open neighborhood of $t$ (since $G$ is open and satisfies $t\in G$)
and is a subset of $\varphi^{-1}\left(  U\right)  $ (since $G\subset
\varphi^{-1}\left(  U\right)  $). We thus have shown that there exists some
open neighborhood of $t$ that is a subset of $\varphi^{-1}\left(  U\right)  $
(namely, $G$).

Now, forget that we fixed $t$. We have now proven that for each $t\in
\varphi^{-1}\left(  U\right)  $, there exists some open neighborhood of $t$
that is a subset of $\varphi^{-1}\left(  U\right)  $. This means that the
subset $\varphi^{-1}\left(  U\right)  $ of $A$ is open (by one of the criteria
for openness).

Now, forget that we fixed $U$. We thus have shown that if $U$ is an open
subset of $R$, then the subset $\varphi^{-1}\left(  U\right)  $ of $A$ is
open. In other words, the map $\varphi:A\rightarrow R$ is continuous. This
proves Proposition \ref{prop.6.6.contin}.
\end{proof}

\begin{proposition}
\label{prop.6.6.ideals}Let $p$ be a positive integer. Let $R$ be a ring
equipped with a decreasing sequence $R=\mathfrak{a}_{0}\supset\mathfrak{a}%
_{1}\supset\mathfrak{a}_{2}\supset\mathfrak{a}_{3}\supset\cdots$ of ideals
such that
\begin{equation}
\left(  \mathfrak{a}_{n}\cdot\mathfrak{a}_{m}\subset\mathfrak{a}_{n+m}\text{
for all }n\geq0\text{ and }m\geq0\right)  . \label{eq.prop.6.6.ideals.anam}%
\end{equation}
Assume that the residue ring $L=R/\mathfrak{a}_{1}$ has $p\cdot1_{L}=0$.

\textbf{(a)} We have $\mathfrak{a}_{1}^{n}\subset\mathfrak{a}_{n}$ for every
$n\geq0$.

\textbf{(b)} We have $p\cdot1_{R}\in\mathfrak{a}_{1}$ and $pR\subset
\mathfrak{a}_{1}$.

\textbf{(c)} We have $p^{n}\cdot1_{R}\in\mathfrak{a}_{n}$ for every $n\geq0$.

\textbf{(d)} If $x$ and $y$ are two elements of $R$ satisfying $x\equiv
y\operatorname{mod}\mathfrak{a}_{1}$, then%
\[
x^{p^{i}}\equiv y^{p^{i}}\operatorname{mod}\mathfrak{a}_{i+1}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\geq0.
\]


\textbf{(e)} Every $u\in\mathfrak{a}_{1}$, every $n\geq0$ and every
$i\in\left\{  0,1,\ldots,n\right\}  $ satisfy $p^{i}u^{p^{n-i}}\in
\mathfrak{a}_{n}$.

\textbf{(f)} If $a$ and $b$ are two elements of $R$ and $i$ is a positive
integer satisfying $a\equiv b\operatorname{mod}\mathfrak{a}_{i}$, then%
\[
a^{p}\equiv b^{p}\operatorname{mod}\mathfrak{a}_{i+1}.
\]


\textbf{(g)} We have $p\mathfrak{a}_{n-1}\subset\mathfrak{a}_{n}$ for every
positive integer $n$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.6.6.ideals}.]\textbf{(a)} Using
(\ref{eq.prop.6.6.ideals.anam}), we can easily prove Proposition
\ref{prop.6.6.ideals} \textbf{(a)} (by induction on $n$). (The induction base
follows from $\mathfrak{a}_{1}^{0}\subset R=\mathfrak{a}_{0}$.)

\textbf{(b)} We have $p\cdot1_{R}\in\mathfrak{a}_{1}$ (since the projection of
$p\cdot1_{R}$ onto $R/\mathfrak{a}_{1}=L$ is $p\cdot1_{L}=0$) and thus
$pR=\underbrace{\left(  p\cdot1_{R}\right)  }_{\in\mathfrak{a}_{1}}%
R\subset\mathfrak{a}_{1}R\subset\mathfrak{a}_{1}$ (since $\mathfrak{a}_{1}$ is
an ideal of $R$). This proves Proposition \ref{prop.6.6.ideals} \textbf{(b)}.

\textbf{(c)} We have $p^{n}\cdot1_{R}=\left(  \underbrace{p\cdot1_{R}}%
_{\in\mathfrak{a}_{1}}\right)  ^{n}\in\mathfrak{a}_{1}^{n}\subset
\mathfrak{a}_{n}$ (by Proposition \ref{prop.6.6.ideals} \textbf{(a)}) for
every $n\geq0$. This proves Proposition \ref{prop.6.6.ideals} \textbf{(c)}.

\textbf{(g)} Let $n$ be a positive integer. Proposition \ref{prop.6.6.ideals}
\textbf{(b)} yields $pR\subset\mathfrak{a}_{1}$. Now,
$p\underbrace{\mathfrak{a}_{n-1}}_{\subset R\mathfrak{a}_{n-1}}\subset
\underbrace{pR}_{\subset\mathfrak{a}_{1}}\mathfrak{a}_{n-1}\subset
\mathfrak{a}_{1}\cdot\mathfrak{a}_{n-1}\subset\mathfrak{a}_{n}$ (by
(\ref{eq.prop.6.6.ideals.anam}), applied to $1$ and $n-1$ instead of $n$ and
$m$). This proves Proposition \ref{prop.6.6.ideals} \textbf{(g)}.

\textbf{(f)} Let $a$ and $b$ be two elements of $R$, and let $i$ be a positive
integer satisfying $a\equiv b\operatorname{mod}\mathfrak{a}_{i}$.

Since $i$ is a positive integer, we have $i\geq1$. Thus, $\mathfrak{a}%
_{i}\subset\mathfrak{a}_{1}$ (since $\mathfrak{a}_{0}\supset\mathfrak{a}%
_{1}\supset\mathfrak{a}_{2}\supset\mathfrak{a}_{3}\supset\cdots$).

Proposition \ref{prop.6.6.ideals} \textbf{(b)} yields $p\cdot1_{R}%
\in\mathfrak{a}_{1}$ and $pR\subset\mathfrak{a}_{1}$.

It is well-known that every $u\in R$ and $v\in R$ and $g\geq0$ satisfy%
\begin{equation}
u^{g}-v^{g}=\left(  u-v\right)  \left(  \sum_{k=0}^{g-1}u^{k}v^{g-1-k}\right)
. \label{pf.thm.6.6gen.xpi.pf.0}%
\end{equation}


But $a\equiv b\operatorname{mod}\mathfrak{a}_{i}$. Hence, $a-b\in
\mathfrak{a}_{i}\subset\mathfrak{a}_{1}$, so that $a\equiv b\operatorname{mod}%
\mathfrak{a}_{1}$. Also,%
\begin{align*}
\sum_{k=0}^{p-1}\underbrace{a^{k}}_{\substack{\equiv b^{k}\operatorname{mod}%
\mathfrak{a}_{1}\\\text{(since }a\equiv b\operatorname{mod}\mathfrak{a}%
_{1}\text{)}}}b^{p-1-k}  &  \equiv\sum_{k=0}^{p-1}\underbrace{b^{k}b^{p-1-k}%
}_{=b^{p-1}}=\sum_{k=0}^{p-1}b^{p-1}=pb^{p-1}\\
&  =\underbrace{p\cdot1_{R}}_{\substack{\equiv0\operatorname{mod}%
\mathfrak{a}_{1}\\\text{(since }p\cdot1_{R}\in\mathfrak{a}_{1}\text{)}%
}}b^{p-1}\equiv0\operatorname{mod}\mathfrak{a}_{1};
\end{align*}
thus, $\sum_{k=0}^{p-1}a^{k}b^{p-1-k}\in\mathfrak{a}_{1}$. Now,
(\ref{pf.thm.6.6gen.xpi.pf.0}) (applied to $a=u$, $b=v$ and $g=p$) yields%
\[
a^{p}-b^{p}=\underbrace{\left(  a-b\right)  }_{\in\mathfrak{a}_{i}%
}\underbrace{\left(  \sum_{k=0}^{p-1}a^{k}b^{p-1-k}\right)  }_{\in
\mathfrak{a}_{1}}\in\mathfrak{a}_{i}\mathfrak{a}_{1}\subset\mathfrak{a}_{i+1}%
\]
(by (\ref{eq.prop.6.6.ideals.anam}), applied to $n=i$ and $m=1$). In other
words, $a^{p}\equiv b^{p}\operatorname{mod}\mathfrak{a}_{i+1}$. This proves
Proposition \ref{prop.6.6.ideals} \textbf{(f)}.

\textbf{(d)} Proposition \ref{prop.6.6.ideals} \textbf{(d)} can be proven by
induction over $i$, similarly to the proof of Lemma 1.4. However, let me
sketch a slightly different proof, for the sake of diversity. We proceed by
induction over $i$. The case of $i=0$ is obvious. The induction step proceeds
as follows: Let $i\geq1$, and assume that $x^{p^{i-1}}\equiv y^{p^{i-1}%
}\operatorname{mod}\mathfrak{a}_{i}$. We must show that $x^{p^{i}}\equiv
y^{p^{i}}\operatorname{mod}\mathfrak{a}_{i+1}$.

Set $a=x^{p^{i-1}}$ and $b=y^{p^{i-1}}$. Then, $a=x^{p^{i-1}}\equiv
y^{p^{i-1}}=b\operatorname{mod}\mathfrak{a}_{i}$. Hence, Proposition
\ref{prop.6.6.ideals} \textbf{(f)} yields $a^{p}\equiv b^{p}\operatorname{mod}%
\mathfrak{a}_{i+1}$. Since $a=x^{p^{i-1}}$, we have $a^{p}=\left(  x^{p^{i-1}%
}\right)  ^{p}=x^{p^{i}}$. Similarly, $b^{p}=y^{p^{i}}$. Thus, $x^{p^{i}%
}=a^{p}\equiv b^{p}=y^{p^{i}}\operatorname{mod}\mathfrak{a}_{i+1}$. This
completes the induction step; thus, Proposition \ref{prop.6.6.ideals}
\textbf{(d)} is proven.

\textbf{(e)} Let $u\in\mathfrak{a}_{1}$, $n\geq0$ and $i\in\left\{
0,1,\ldots,n\right\}  $. We have $u\in\mathfrak{a}_{1}$, so that
$u\equiv0\operatorname{mod}\mathfrak{a}_{1}$. Thus, Proposition
\ref{prop.6.6.ideals} \textbf{(d)} (applied to $u$, $0$ and $n-i$ instead of
$x$, $y$ and $i$) yields $u^{p^{n-i}}\equiv0^{p^{n-i}}=0\operatorname{mod}%
\mathfrak{a}_{\left(  n-i\right)  +1}$ (since $p^{n-i}$ is a positive
integer). Thus, $u^{p^{n-i}}\in\mathfrak{a}_{\left(  n-i\right)  +1}%
\subset\mathfrak{a}_{n-i}$. Now,%
\begin{align*}
p^{i}\underbrace{u^{p^{n-i}}}_{\in\mathfrak{a}_{n-i}\subset R\mathfrak{a}%
_{n-i}}  &  \in\underbrace{p^{i}R}_{=p^{i}\cdot1_{R}R}\mathfrak{a}%
_{n-i}=\underbrace{p^{i}\cdot1_{R}}_{\substack{\in\mathfrak{a}_{i}\\\text{(by
Proposition \ref{prop.6.6.ideals} \textbf{(c)},}\\\text{applied to }i\text{
instead of }n\text{)}}}R\mathfrak{a}_{n-i}\\
&  \subset\mathfrak{a}_{i}\underbrace{R\mathfrak{a}_{n-i}}_{\substack{\subset
\mathfrak{a}_{n-i}\\\text{(since }\mathfrak{a}_{n-i}\text{ is an}\\\text{ideal
of }R\text{)}}}\subset\mathfrak{a}_{i}\mathfrak{a}_{n-i}\subset\mathfrak{a}%
_{n}%
\end{align*}
(by (\ref{eq.prop.6.6.ideals.anam}), applied to $i$ and $n-i$ instead of $n$
and $m$). This proves Proposition \ref{prop.6.6.ideals} \textbf{(e)}.
\end{proof}

The following proposition is just the fundamental theorem on homomorphisms (in
a slight disguise):

\begin{proposition}
\label{prop.6.6.hs}Let $A$, $B$ and $C$ be three rings. Let $\varphi
:A\rightarrow B$ be a surjective ring homomorphism. Let $\psi:A\rightarrow C$
be a ring homomorphism such that $\psi\left(  \operatorname*{Ker}%
\varphi\right)  =0$. Then, there exists a unique ring homomorphism
$\zeta:B\rightarrow C$ such that $\zeta\circ\varphi=\psi$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.6.6.hs}.]The surjective ring homomorphism
$\varphi:A\rightarrow B$ induces a canonical ring isomorphism $B\cong
A/\operatorname*{Ker}\varphi$. Thus, the ring $B$ and the ring homomorphism
$\varphi:A\rightarrow B$ satisfy the universal property of the quotient
$A/\operatorname*{Ker}\varphi$. But this is precisely the statement of
Proposition \ref{prop.6.6.hs}.
\end{proof}

\begin{proposition}
\label{prop.6.6.wn}Let $p$ be a prime. Let $R$ be a ring equipped with a
decreasing sequence $R=\mathfrak{a}_{0}\supset\mathfrak{a}_{1}\supset
\mathfrak{a}_{2}\supset\mathfrak{a}_{3}\supset\cdots$ of ideals such that
$\left(  \mathfrak{a}_{n}\cdot\mathfrak{a}_{m}\subset\mathfrak{a}_{n+m}\text{
for all }n\geq0\text{ and }m\geq0\right)  $. Assume that the residue ring
$L=R/\mathfrak{a}_{1}$ has $p\cdot1_{L}=0$. Let $\pi$ denote the canonical
projection $R\rightarrow R/\mathfrak{a}_{1}=L$. For every $n\geq0$, let
$\pi_{n}$ be the canonical projection $R\rightarrow R/\mathfrak{a}_{n}$.

Let $n\geq0$. There exists a unique ring homomorphism $\widetilde{w}_{p^{n}%
}:W_{p}\left(  L\right)  \rightarrow R/\mathfrak{a}_{n}$ such that the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{x}{\xymatrix@C=5pc{
%W_p\left(R\right) \ar[r]^-{w_{p^n}} \ar[d]_{W_p\left(\pi\right)}
%& R \ar[d]^{\pi_n} \\
%W_p\left(L\right) \ar[r]_-{\widetilde{w}_{p^n}} & R / \mathfrak{a}_n
%}} }%
%BeginExpansion
\xymatrix@C=5pc{
W_p\left(R\right) \ar[r]^-{w_{p^n}} \ar[d]_{W_p\left(\pi\right)}
& R \ar[d]^{\pi_n} \\
W_p\left(L\right) \ar[r]_-{\widetilde{w}_{p^n}} & R / \mathfrak{a}_n
}
%EndExpansion
\label{eq.prop.6.6.wn.diag}%
\end{equation}
commutes.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.6.6.wn}.]The map $\pi:R\rightarrow L$ is
surjective (being a projection). Hence, the map $W_{p}\left(  \pi\right)
:W_{p}\left(  R\right)  \rightarrow W_{p}\left(  L\right)  $ is also
surjective\footnote{\textit{Proof.} Recall that $W_{p}\left(  R\right)
=R^{P_{p}}$ (as sets) and $W_{p}\left(  L\right)  =L^{P_{p}}$ (as sets). The
map $W_{p}\left(  \pi\right)  $ is defined by%
\[
W_{p}\left(  \pi\right)  \left(  \left(  a_{n}\right)  _{n\in P_{p}}\right)
=\left(  \pi\left(  a_{n}\right)  \right)  _{n\in P_{p}}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  a_{n}\right)  _{n\in P_{p}}\in
W_{p}\left(  R\right)  .
\]
In other words, the map $W_{p}\left(  \pi\right)  :W_{p}\left(  R\right)
\rightarrow W_{p}\left(  L\right)  $ is identical with the map $\prod_{q\in
P_{p}}\pi:R^{P_{p}}\rightarrow L^{P_{p}}$. But the latter map is clearly
surjective (since the map $\pi$ is surjective). Hence, the former map is
surjective. Qed.}. Furthermore, this map $W_{p}\left(  \pi\right)  $ is a ring
homomorphism (since $\pi$ is a ring homomorphism).

Now, let $x\in\operatorname*{Ker}\left(  W_{p}\left(  \pi\right)  \right)  $.
We are going to show that $w_{p^{n}}\left(  x\right)  \in\mathfrak{a}_{n}$.

Indeed, the definition of $W_{p}\left(  \pi\right)  $ yields%
\[
\left(  W_{p}\left(  \pi\right)  \right)  \left(  x\right)  =\left(
\pi\left(  x_{m}\right)  \right)  _{m\in P_{p}}.
\]
Hence, $\left(  \pi\left(  x_{m}\right)  \right)  _{m\in P_{p}}=\left(
W_{p}\left(  \pi\right)  \right)  \left(  x\right)  =0$ (since $x\in
\operatorname*{Ker}\left(  W_{p}\left(  \pi\right)  \right)  $). In other
words, $\pi\left(  x_{m}\right)  =0$ for every $m\in P_{p}$. In other words,
\begin{equation}
x_{m}\in\mathfrak{a}_{1}\ \ \ \ \ \ \ \ \ \ \text{for every }m\in P_{p}
\label{pf.prop.6.6.wn.1}%
\end{equation}
(because $\pi$ is the canonical projection $R\rightarrow R/\mathfrak{a}_{1}$,
and therefore the equality $\pi\left(  x_{m}\right)  =0$ means that $x_{m}%
\in\mathfrak{a}_{1}$).

Every $i\in\left\{  0,1,\ldots,n\right\}  $ satisfies $p^{i}\in P_{p}$ and
thus $x_{p^{i}}\in\mathfrak{a}_{1}$ (by (\ref{pf.prop.6.6.wn.1}), applied to
$m=p^{i}$), so that
\begin{equation}
p^{i}x_{p^{i}}^{p^{n-i}}\in\mathfrak{a}_{n} \label{pf.prop.6.6.wn.2}%
\end{equation}
(by Proposition \ref{prop.6.6.ideals} \textbf{(e)}, applied to $u=x_{p^{i}}$).

From (\ref{eq.wpn=}), we obtain%
\[
w_{p^{n}}\left(  x\right)  =\sum_{i=0}^{n}\underbrace{p^{i}x_{p^{i}}^{p^{n-i}%
}}_{\substack{\in\mathfrak{a}_{n}\\\text{(by (\ref{pf.prop.6.6.wn.2}))}}%
}\in\sum_{i=0}^{n}\mathfrak{a}_{n}\subset\mathfrak{a}_{n}.
\]
Thus, $\pi_{n}\left(  w_{p^{n}}\left(  x\right)  \right)  =0$ (since $\pi_{n}$
is the canonical projection $R\rightarrow R/\mathfrak{a}_{n}$). Hence,
$\left(  \pi_{n}\circ w_{p^{n}}\right)  \left(  x\right)  =\pi_{n}\left(
w_{p^{n}}\left(  x\right)  \right)  =0$.

Now, forget that we fixed $x$. We thus have shown that
\[
\left(  \pi_{n}\circ w_{p^{n}}\right)  \left(  x\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every }x\in\operatorname*{Ker}\left(
W_{p}\left(  \pi\right)  \right)  .
\]
In other words,%
\[
\left(  \pi_{n}\circ w_{p^{n}}\right)  \left(  \operatorname*{Ker}\left(
W_{p}\left(  \pi\right)  \right)  \right)  =0.
\]
On the other hand, $w_{p^{n}}:W_{p}\left(  R\right)  \rightarrow R$ is a ring
homomorphism (by Theorem 2.6 \textbf{(ii)}, applied to $R$, $P_{p}$ and
$p^{n}$ instead of $A$, $P$ and $n$). The map $\pi_{n}:R\rightarrow
R/\mathfrak{a}_{n}$ also is a ring homomorphism (being the canonical
projection). Hence, the map $\pi_{n}\circ w_{p^{n}}:W_{p}\left(  R\right)
\rightarrow R/\mathfrak{a}_{n}$ is a ring homomorphism (being the composition
of two ring homomorphisms).

Thus, Proposition \ref{prop.6.6.hs} (applied to $A=W_{p}\left(  R\right)  $,
$B=W_{p}\left(  L\right)  $, $C=R/\mathfrak{a}_{n}$, $\varphi=W_{p}\left(
\pi\right)  $ and $\psi=\pi_{n}\circ w_{p^{n}}$) yields that there exists a
unique ring homomorphism $\zeta:W_{p}\left(  L\right)  \rightarrow
R/\mathfrak{a}_{n}$ such that $\zeta\circ W_{p}\left(  \pi\right)  =\pi
_{n}\circ w_{p^{n}}$. Renaming $\zeta$ as $\widetilde{w}_{p^{n}}$ in this
statement, we obtain the following: There exists a unique ring homomorphism
$\widetilde{w}_{p^{n}}:W_{p}\left(  L\right)  \rightarrow R/\mathfrak{a}_{n}$
such that $\widetilde{w}_{p^{n}}\circ W_{p}\left(  \pi\right)  =\pi_{n}\circ
w_{p^{n}}$. In other words, there exists a unique ring homomorphism
$\widetilde{w}_{p^{n}}:W_{p}\left(  L\right)  \rightarrow R/\mathfrak{a}_{n}$
such that the diagram (\ref{eq.prop.6.6.wn.diag}) commutes. This proves
Proposition \ref{prop.6.6.wn}.
\end{proof}

\begin{definition}
\label{def.multiplicative}Let $K$ and $R$ be two rings. A map $\varphi
:K\rightarrow R$ is said to be \textit{multiplicative} if and only if it
satisfies $\varphi\left(  1\right)  =1$ and $\left(  \varphi\left(  ab\right)
=\varphi\left(  a\right)  \varphi\left(  b\right)  \text{ for every }a\in
K\text{ and }b\in K\right)  $.
\end{definition}

\begin{proposition}
\label{prop.multiplicative.power}Let $K$ and $R$ be two rings. Let
$\varphi:K\rightarrow R$ be a multiplicative map. Let $v\in K$ and $i\geq0$.
Then, $\varphi\left(  v^{i}\right)  =\left(  \varphi\left(  v\right)  \right)
^{i}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.multiplicative.power}.]This follows by
straightforward induction on $i$.
\end{proof}

\begin{proposition}
\label{prop.6.6.multuniq}Let $p$ be a prime. Let $R$ be a ring equipped with a
decreasing sequence $R=\mathfrak{a}_{0}\supset\mathfrak{a}_{1}\supset
\mathfrak{a}_{2}\supset\mathfrak{a}_{3}\supset\cdots$ of ideals such that
$\left(  \mathfrak{a}_{n}\cdot\mathfrak{a}_{m}\subset\mathfrak{a}_{n+m}\text{
for all }n\geq0\text{ and }m\geq0\right)  $. Equip $R$ with the topology
defined by this sequence of ideals. Assume that $R$ is Hausdorff for this topology.

Assume that the residue ring $L=R/\mathfrak{a}_{1}$ has $p\cdot1_{L}=0$. Let
$\pi$ denote the canonical projection $R\rightarrow R/\mathfrak{a}_{1}=L$.

Let $K$ be a perfect ring of characteristic $p$. Let $f:K\rightarrow L$ be any
map. Then, there exists at most one multiplicative map $r:K\rightarrow R$
satisfying
\begin{equation}
\left(  \pi\left(  r\left(  a\right)  \right)  =f\left(  a\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in K\right)  .
\label{eq.prop.6.6.multuniq.ass}%
\end{equation}

\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.6.6.multuniq}.]We need to show that if $r_{1}$
and $r_{2}$ are two multiplicative maps $r:K\rightarrow R$ satisfying
(\ref{eq.prop.6.6.multuniq.ass}), then $r_{1}=r_{2}$.

So let $r_{1}$ and $r_{2}$ be two multiplicative maps $r:K\rightarrow R$
satisfying (\ref{eq.prop.6.6.multuniq.ass}). We must show that $r_{1}=r_{2}$.

Let $u\in K$. Let $n\geq0$. Then, the element $u^{1/p^{n}}$ of $K$ is
well-defined (since $K$ is a perfect ring of characteristic $p$). Moreover,
$r_{1}$ is multiplicative; thus, Proposition \ref{prop.multiplicative.power}
(applied to $\varphi=r_{1}$, $v=u^{1/p^{n}}$ and $i=p^{n}$) yields
$r_{1}\left(  \left(  u^{1/p^{n}}\right)  ^{p^{n}}\right)  =\left(
r_{1}\left(  u^{1/p^{n}}\right)  \right)  ^{p^{n}}$. Since $\left(
u^{1/p^{n}}\right)  ^{p^{n}}=u$, this rewrites as $r_{1}\left(  u\right)
=\left(  r_{1}\left(  u^{1/p^{n}}\right)  \right)  ^{p^{n}}$. Similarly,
$r_{2}\left(  u\right)  =\left(  r_{2}\left(  u^{1/p^{n}}\right)  \right)
^{p^{n}}$.

But the map $r_{1}$ satisfies (\ref{eq.prop.6.6.multuniq.ass}). Hence,
(\ref{eq.prop.6.6.multuniq.ass}) (applied to $r=r_{1}$ and $a=u^{1/p^{n}}$)
yields $\pi\left(  r_{1}\left(  u^{1/p^{n}}\right)  \right)  =f\left(
u^{1/p^{n}}\right)  $. Similarly, $\pi\left(  r_{2}\left(  u^{1/p^{n}}\right)
\right)  =f\left(  u^{1/p^{n}}\right)  $. Hence, $\pi\left(  r_{1}\left(
u^{1/p^{n}}\right)  \right)  =f\left(  u^{1/p^{n}}\right)  =\pi\left(
r_{2}\left(  u^{1/p^{n}}\right)  \right)  $. In other words, $r_{1}\left(
u^{1/p^{n}}\right)  \equiv r_{2}\left(  u^{1/p^{n}}\right)  \operatorname{mod}%
\mathfrak{a}_{1}$ (since $\pi$ is the canonical projection $R\rightarrow
R/\mathfrak{a}_{1}$). Hence, Proposition \ref{prop.6.6.ideals} \textbf{(d)}
(applied to $x=r_{1}\left(  u^{1/p^{n}}\right)  $, $y=r_{2}\left(  u^{1/p^{n}%
}\right)  $ and $i=n$) yields $\left(  r_{1}\left(  u^{1/p^{n}}\right)
\right)  ^{p^{n}}\equiv\left(  r_{2}\left(  u^{1/p^{n}}\right)  \right)
^{p^{n}}\operatorname{mod}\mathfrak{a}_{n+1}$. Thus,%
\[
r_{1}\left(  u\right)  =\left(  r_{1}\left(  u^{1/p^{n}}\right)  \right)
^{p^{n}}\equiv\left(  r_{2}\left(  u^{1/p^{n}}\right)  \right)  ^{p^{n}}%
=r_{2}\left(  u\right)  \operatorname{mod}\mathfrak{a}_{n+1}.
\]
Thus, $r_{1}\left(  u\right)  -r_{2}\left(  u\right)  \in\mathfrak{a}%
_{n+1}\subset\mathfrak{a}_{n}$.

Now, forget that we fixed $n$. We thus have shown that $r_{1}\left(  u\right)
-r_{2}\left(  u\right)  \in\mathfrak{a}_{n}$ for every $n\geq0$. In other
words, $r_{1}\left(  u\right)  -r_{2}\left(  u\right)  \in\bigcap_{n\geq
0}\mathfrak{a}_{n}=0$ (by Proposition \ref{prop.6.6.conty} \textbf{(a)}). In
other words, $r_{1}\left(  u\right)  -r_{2}\left(  u\right)  =0$, so that
$r_{1}\left(  u\right)  =r_{2}\left(  u\right)  $.

Now, forget that we fixed $u$. We thus have shown that $r_{1}\left(  u\right)
=r_{2}\left(  u\right)  $ for every $u\in K$. In other words, $r_{1}=r_{2}$.
This completes the proof of Proposition \ref{prop.6.6.multuniq}.
\end{proof}

\begin{theorem}
\label{thm.6.6teich}Let $p$ be a prime. Let $R$ be a ring equipped with a
decreasing sequence $R=\mathfrak{a}_{0}\supset\mathfrak{a}_{1}\supset
\mathfrak{a}_{2}\supset\mathfrak{a}_{3}\supset\cdots$ of ideals such that
$\left(  \mathfrak{a}_{n}\cdot\mathfrak{a}_{m}\subset\mathfrak{a}_{n+m}\text{
for all }n\geq0\text{ and }m\geq0\right)  $. Equip $R$ with the topology
defined by this sequence of ideals. Assume that $R$ is Hausdorff and complete
for this topology. Assume that the residue ring $L=R/\mathfrak{a}_{1}$ has
$p\cdot1_{L}=0$. Let $\pi$ denote the canonical projection $R\rightarrow
R/\mathfrak{a}_{1}=L$.

Let $K$ be a perfect ring of characteristic $p$. Let $f:K\rightarrow L$ be a
ring homomorphism.

There exists a unique multiplicative map $r:K\rightarrow R$ satisfying%
\begin{equation}
\left(  \pi\left(  r\left(  a\right)  \right)  =f\left(  a\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in K\right)  .
\label{eq.thm.6.6gen.a.1}%
\end{equation}

\end{theorem}

Theorem \ref{thm.6.6teich} is a generalization of the existence of a
Teichm\"{u}ller system of representatives for a perfect ring\footnote{More
precisely: If we apply it to $L=K$ and $f=\operatorname*{id}\nolimits_{K}$,
where $R$ is a $p$-ring with residue ring $K$, then we recover the classical
result that there exists a unique Teichm\"{u}ller system of representatives
$K\rightarrow R$.}.

\begin{proof}
[Proof of Theorem \ref{thm.6.6teich}.]Proposition \ref{prop.6.6.ideals}
\textbf{(b)} shows that $p\cdot1_{R}\in\mathfrak{a}_{1}$ and $pR\subset
\mathfrak{a}_{1}$.

The map $\pi$ is surjective (since it is the canonical projection
$R\rightarrow R/\mathfrak{a}_{1}$).

Proposition \ref{prop.6.6.multuniq} shows that there exists \textbf{at most
one} multiplicative map $r:K\rightarrow R$ satisfying (\ref{eq.thm.6.6gen.a.1}%
). It thus remains to prove that there exists \textbf{at least one} such map.
In other words, it remains to construct such a map.

Let $a\in K$. Then, the element $a^{1/p^{i}}$ of $K$ is defined for every
$i\in\left\{  0,1,2,\ldots\right\}  $ (since $K$ is a perfect ring of
characteristic $p$).

For every $i\in\left\{  0,1,2,\ldots\right\}  $, choose some $g_{i}\in R$ such
that $\pi\left(  g_{i}\right)  =f\left(  a^{1/p^{i}}\right)  $. (This is
well-defined, since $\pi$ is surjective.) Thus, a sequence $\left(
g_{0},g_{1},g_{2},\ldots\right)  \in R^{\infty}$ of elements of $R$ is
defined. We have%
\begin{equation}
g_{i}^{p^{i}}\equiv g_{i+1}^{p^{i+1}}\operatorname{mod}\mathfrak{a}%
_{i+1}\ \ \ \ \ \ \ \ \ \ \text{for every }i\geq0.
\label{pf.thm.6.6gen.g.converges}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.6.6gen.g.converges}):} Let $i\geq0$.
The definition of $g_{i}$ yields $\pi\left(  g_{i}\right)  =f\left(
a^{1/p^{i}}\right)  $. The definition of $g_{i+1}$ yields $\pi\left(
g_{i+1}\right)  =f\left(  a^{1/p^{i+1}}\right)  $. Hence,%
\begin{align*}
\pi\left(  g_{i+1}^{p}\right)   &  =\left(  \underbrace{\pi\left(
g_{i+1}\right)  }_{=f\left(  a^{1/p^{i+1}}\right)  }\right)  ^{p}=\left(
f\left(  a^{1/p^{i+1}}\right)  \right)  ^{p}=f\left(  \underbrace{\left(
a^{1/p^{i+1}}\right)  ^{p}}_{=a^{1/p^{i}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is a ring homomorphism}%
\right) \\
&  =f\left(  a^{1/p^{i}}\right)  =\pi\left(  g_{i}\right)  .
\end{align*}
In other words, $g_{i+1}^{p}\equiv g_{i}\operatorname{mod}\mathfrak{a}_{1}$.
Hence, Proposition \ref{prop.6.6.ideals} \textbf{(d)} (applied to
$x=g_{i+1}^{p}$ and $y=g_{i}$) yields $\left(  g_{i+1}^{p}\right)  ^{p^{i}%
}\equiv g_{i}^{p^{i}}\operatorname{mod}\mathfrak{a}_{i+1}$. Now,
$g_{i+1}^{p^{i+1}}=\left(  g_{i+1}^{p}\right)  ^{p^{i}}\equiv g_{i}^{p^{i}%
}\operatorname{mod}\mathfrak{a}_{i+1}$. This proves
(\ref{pf.thm.6.6gen.g.converges}).} Thus, the sequence $\left(  g_{0}^{p^{0}%
},g_{1}^{p^{1}},g_{2}^{p^{2}},\ldots\right)  $ is Cauchy with respect to the
topology on $R$. Hence, the limit $\lim\limits_{n\rightarrow\infty}%
g_{n}^{p^{n}}$ is defined. Denote this limit by $g^{\prime}$. Thus,
$g^{\prime}=\lim\limits_{n\rightarrow\infty}g_{n}^{p^{n}}$. From
(\ref{pf.thm.6.6gen.g.converges}), we obtain%
\begin{equation}
g^{\prime}\equiv g_{i}^{p^{i}}\operatorname{mod}\mathfrak{a}_{i+1}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\geq0. \label{pf.thm.6.6gen.g.g'.1}%
\end{equation}


We notice the following fact: If $u\in R$ and $i\in\left\{  0,1,2,\ldots
\right\}  $ are such that $\pi\left(  u\right)  =f\left(  a^{1/p^{i}}\right)
$, then%
\begin{equation}
g^{\prime}\equiv u^{p^{i}}\operatorname{mod}\mathfrak{a}_{i+1}.
\label{pf.thm.6.6gen.g.g'.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.6.6gen.g.g'.2}):} Let $u\in R$ and
$i\in\left\{  0,1,2,\ldots\right\}  $ be such that $\pi\left(  u\right)
=f\left(  a^{1/p^{i}}\right)  $. From $\pi\left(  u\right)  =f\left(
a^{1/p^{i}}\right)  =\pi\left(  g_{i}\right)  $, we obtain $u\equiv
g_{i}\operatorname{mod}\mathfrak{a}_{1}$. Hence, Proposition
\ref{prop.6.6.ideals} \textbf{(d)} (applied to $x=u$ and $y=g_{i}$) yields
$u^{p^{i}}\equiv g_{i}^{p^{i}}\operatorname{mod}\mathfrak{a}_{i+1}$. Now,
$u^{p^{i}}\equiv g_{i}^{p^{i}}\equiv g^{\prime}\operatorname{mod}%
\mathfrak{a}_{i+1}$ (by (\ref{pf.thm.6.6gen.g.g'.1})). This proves
(\ref{pf.thm.6.6gen.g.g'.2}).}

The value of $g^{\prime}$ does not depend on the choice of the elements
$g_{i}$\ \ \ \ \footnote{\textit{Proof.} We need to show that if
$g_{1}^{\prime}$ and $g_{2}^{\prime}$ are two possible values of $g^{\prime}$,
then $g_{1}^{\prime}=g_{2}^{\prime}$.
\par
Indeed, let $g_{1}^{\prime}$ and $g_{2}^{\prime}$ be two possible values of
$g^{\prime}$. Let $i\in\left\{  0,1,2,\ldots\right\}  $. Pick any $u\in R$
satisfying $\pi\left(  u\right)  =f\left(  a^{1/p^{i}}\right)  $ (such a $u$
exists, since $\pi$ is surjective). Then, (\ref{pf.thm.6.6gen.g.g'.2})
(applied to $g_{1}^{\prime}$ instead of $g^{\prime}$) yields $g_{1}^{\prime
}\equiv u^{p^{i}}\operatorname{mod}\mathfrak{a}_{i+1}$ (since $g_{1}^{\prime}$
is one possible value of $g^{\prime}$). Similarly, $g_{2}^{\prime}\equiv
u^{p^{i}}\operatorname{mod}\mathfrak{a}_{i+1}$. Thus, $g_{1}^{\prime}\equiv
u^{p^{i}}\equiv g_{2}^{\prime}\operatorname{mod}\mathfrak{a}_{i+1}$. In other
words, $g_{1}^{\prime}-g_{2}^{\prime}\in\mathfrak{a}_{i+1}\subset
\mathfrak{a}_{i}$, so that $g_{1}^{\prime}\equiv g_{2}^{\prime}%
\operatorname{mod}\mathfrak{a}_{i}$.
\par
Now, forget that we fixed $i$. We thus have proven that $g_{1}^{\prime}\equiv
g_{2}^{\prime}\operatorname{mod}\mathfrak{a}_{i}$ for each $i\in\left\{
0,1,2,\ldots\right\}  $. Hence, $g_{1}^{\prime}=g_{2}^{\prime}$ (by
Proposition \ref{prop.6.6.conty} \textbf{(b)}, applied to $u=g_{1}^{\prime}$
and $v=g_{2}^{\prime}$), qed.}. We denote this value of $g^{\prime}$ by
$r\left(  a\right)  $ (to stress its dependence on $a$). It has the property
that if $u\in R$ and $i\in\left\{  0,1,2,\ldots\right\}  $ are such that
$\pi\left(  u\right)  =f\left(  a^{1/p^{i}}\right)  $, then%
\begin{equation}
r\left(  a\right)  \equiv u^{p^{i}}\operatorname{mod}\mathfrak{a}_{i+1}
\label{pf.thm.6.6gen.g.g(a).2}%
\end{equation}
(indeed, this is just a restatement of (\ref{pf.thm.6.6gen.g.g'.2}) using the
new notation $r\left(  a\right)  $ for $g^{\prime}$).

Now, forget that we fixed $a$. We thus have defined an element $r\left(
a\right)  \in R$ for each $a\in K$, and noticed that this element satisfies
(\ref{pf.thm.6.6gen.g.g(a).2}). Thus, we have defined a map $r:K\rightarrow R$
such that every $a\in K$ satisfies (\ref{pf.thm.6.6gen.g.g(a).2}). In order to
prove Theorem \ref{thm.6.6teich}, it remains to prove that this map $r$ is
multiplicative and satisfies (\ref{eq.thm.6.6gen.a.1}).

\textit{Proof that }$r$ \textit{is multiplicative:} We have $r\left(
1\right)  =1$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{
0,1,2,\ldots\right\}  $. Then, $f\left(  \underbrace{1^{1/p^{i}}}_{=1}\right)
=f\left(  1\right)  =1$ (since $f$ is a ring homomorphism), and thus
$\pi\left(  1\right)  =1=f\left(  1^{1/p^{i}}\right)  $. Hence,
(\ref{pf.thm.6.6gen.g.g(a).2}) (applied to $a=1$ and $u=1$) yields $r\left(
1\right)  \equiv1^{p^{i}}=1\operatorname{mod}\mathfrak{a}_{i+1}$. Thus,
$r\left(  1\right)  -1\in\mathfrak{a}_{i+1}\subset\mathfrak{a}_{i}$, so that
$r\left(  1\right)  \equiv1\operatorname{mod}\mathfrak{a}_{i}$.
\par
Now, forget that we fixed $i$. We thus have shown that $r\left(  1\right)
\equiv1\operatorname{mod}\mathfrak{a}_{i}$ for every $i\in\left\{
0,1,2,\ldots\right\}  $. Hence, $r\left(  1\right)  =1$ (by Proposition
\ref{prop.6.6.conty} \textbf{(b)}, applied to $u=r\left(  1\right)  $ and
$v=1$), qed.}. Also, $r\left(  ab\right)  =r\left(  a\right)  r\left(
b\right)  $ for every $a\in K$ and $b\in K$\ \ \ \ \footnote{\textit{Proof.}
Let $a\in K$ and $b\in K$. Let $i\in\left\{  0,1,2,\ldots\right\}  $.
\par
Pick some $u\in R$ such that $\pi\left(  u\right)  =f\left(  a^{1/p^{i}%
}\right)  $. (Such a $u$ exists, since $\pi$ is surjective.) Hence,
(\ref{pf.thm.6.6gen.g.g(a).2}) yields $r\left(  a\right)  \equiv u^{p^{i}%
}\operatorname{mod}\mathfrak{a}_{i+1}$.
\par
Pick some $v\in R$ such that $\pi\left(  v\right)  =f\left(  b^{1/p^{i}%
}\right)  $. (Such a $v$ exists, since $\pi$ is surjective.) Hence,
(\ref{pf.thm.6.6gen.g.g(a).2}) (applied to $b$ and $v$ instead of $a$ and $u$)
yields $r\left(  b\right)  \equiv v^{p^{i}}\operatorname{mod}\mathfrak{a}%
_{i+1}$.
\par
Now,%
\begin{align*}
\pi\left(  uv\right)   &  =\underbrace{\pi\left(  u\right)  }_{=f\left(
a^{1/p^{i}}\right)  }\underbrace{\pi\left(  v\right)  }_{=f\left(  b^{1/p^{i}%
}\right)  }=f\left(  a^{1/p^{i}}\right)  f\left(  b^{1/p^{i}}\right) \\
&  =f\left(  \underbrace{a^{1/p^{i}}b^{1/p^{i}}}_{=\left(  ab\right)
^{1/p^{i}}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is a ring
homomorphism}\right) \\
&  =f\left(  \left(  ab\right)  ^{1/p^{i}}\right)  .
\end{align*}
Thus, (\ref{pf.thm.6.6gen.g.g(a).2}) (applied to $ab$ and $uv$ instead of $a$
and $u$) yields $r\left(  ab\right)  \equiv\left(  uv\right)  ^{p^{i}%
}=\underbrace{u^{p^{i}}}_{\equiv r\left(  a\right)  \operatorname{mod}%
\mathfrak{a}_{i+1}}\underbrace{v^{p^{i}}}_{\equiv r\left(  b\right)
\operatorname{mod}\mathfrak{a}_{i+1}}\equiv r\left(  a\right)  r\left(
b\right)  \operatorname{mod}\mathfrak{a}_{i+1}$. Thus, $r\left(  ab\right)
-r\left(  a\right)  r\left(  b\right)  \in\mathfrak{a}_{i+1}\subset
\mathfrak{a}_{i}$, so that $r\left(  ab\right)  \equiv r\left(  a\right)
r\left(  b\right)  \operatorname{mod}\mathfrak{a}_{i}$.
\par
Now, forget that we fixed $i$. We thus have shown that $r\left(  ab\right)
\equiv r\left(  a\right)  r\left(  b\right)  \operatorname{mod}\mathfrak{a}%
_{i}$ for all $i\in\left\{  0,1,2,\ldots\right\}  $. Thus, $r\left(
ab\right)  =r\left(  a\right)  r\left(  b\right)  $ (by Proposition
\ref{prop.6.6.conty} \textbf{(b)}, applied to $u=r\left(  ab\right)  $ and
$v=r\left(  a\right)  r\left(  b\right)  $), qed.}. Hence, the map $r$ is multiplicative.

\textit{Proof that }$r$ \textit{satisfies (\ref{eq.thm.6.6gen.a.1}):} Let
$a\in K$. Pick some $u\in R$ such that $\pi\left(  u\right)  =f\left(
a\right)  $. (Such a $u$ exists, since $\pi$ is surjective.) Now, $\pi\left(
u\right)  =f\left(  \underbrace{a}_{=a^{1}=a^{1/p^{0}}}\right)  =f\left(
a^{1/p^{0}}\right)  $. Hence, (\ref{pf.thm.6.6gen.g.g(a).2}) (applied to
$i=0$) yields $r\left(  a\right)  \equiv u^{p^{0}}=u^{1}=u\operatorname{mod}%
\mathfrak{a}_{1}$. Hence, $\pi\left(  r\left(  a\right)  \right)  =\pi\left(
u\right)  =f\left(  a\right)  $. Thus, (\ref{eq.thm.6.6gen.a.1}) is proven.

So we have defined a map $r:K\rightarrow R$, and showed that this map $r$ is
multiplicative and satisfies (\ref{eq.thm.6.6gen.a.1}). Thus, we have
constructed the map $r$ whose existence was alleged in Theorem
\ref{thm.6.6teich}. This proves Theorem \ref{thm.6.6teich}.
\end{proof}

\begin{lemma}
\label{lem.WpK.p-th-mod-p.1}Let $p$ be a prime. Let $K$ be a perfect ring of
characteristic $p$. Let $x\in W_{p}\left(  K\right)  $. Then, there exist two
elements $y$ and $z$ of $W_{p}\left(  K\right)  $ such that $x=y^{p}+pz$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.WpK.p-th-mod-p.1}.]The ring $K$ is a perfect ring of
characteristic $p$. Hence, an element $x_{m}^{1/p}$ of $K$ is well-defined for
every $m\in P_{p}$. Now, define $a\in W_{p}\left(  K\right)  $ by $a=\left(
x_{m}^{1/p}\right)  _{m\in P_{p}}$. Then, $a_{m}=x_{m}^{1/p}$ for every $m\in
P_{p}$. In other words, $a_{m}^{p}=x_{m}$ for every $m\in P_{p}$.

Let $b=F_{p}\left(  a\right)  $. Then, Proposition 5.12 (applied to $P_{p}$,
$K$, $a$ and $b$ instead of $P$, $A$, $x$ and $y$) yields the following:

\begin{enumerate}
\item We have $b_{m}\equiv a_{m}^{p}\operatorname{mod}pK$ for every $m\in
P_{p}$.

\item We have $F_{p}\left(  a\right)  \equiv a^{p}\operatorname{mod}%
pW_{p}\left(  K\right)  $.
\end{enumerate}

We have $F_{p}\left(  a\right)  \equiv a^{p}\operatorname{mod}pW_{p}\left(
K\right)  $. In other words, there exists some $c\in W_{p}\left(  K\right)  $
such that $F_{p}\left(  a\right)  =a^{p}+pc$. Consider this $c$.

We have $pK=0$ (since $K$ has characteristic $p$). Now, for every $m\in P_{p}%
$, we have $b_{m}\equiv a_{m}^{p}\operatorname{mod}pK$, thus $b_{m}=a_{m}^{p}$
(since $pK=0$), hence $b_{m}=a_{m}^{p}=x_{m}$ (since $a_{m}^{p}=x_{m}$).
Therefore, $b=x$. Comparing this with $b=F_{p}\left(  a\right)  $, we obtain
$F_{p}\left(  a\right)  =x$. Thus, $x=F_{p}\left(  a\right)  =a^{p}+pc$.
Hence, there exist two elements $y$ and $z$ of $W_{p}\left(  K\right)  $ such
that $x=y^{p}+pz$ (namely, $y=a$ and $z=c$). This proves Lemma
\ref{lem.WpK.p-th-mod-p.1}.
\end{proof}

\begin{noncompile}
The following lemma was used in a previous version of a proof:

\begin{lemma}
\label{lem.WpK.p-th-mod-p.2}Let $p$ be a prime. Let $K$ be a perfect ring of
characteristic $p$. Let $x\in W_{p}\left(  K\right)  $. Let $n\geq0$. Then,
there exists an $\left(  y_{0},y_{1},\ldots,y_{n-1}\right)  \in\left(
W_{p}\left(  K\right)  \right)  ^{n}$ such that $x\equiv\sum_{i=0}^{n-1}%
p^{i}y_{i}^{p}\operatorname{mod}p^{n}W_{p}\left(  K\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.WpK.p-th-mod-p.2}.]We shall prove Lemma
\ref{lem.WpK.p-th-mod-p.2} by induction over $n$.

\textit{Induction base:} The base case (i.e., the case of $n=0$) of our
induction is trivial (because a congruence modulo $p^{0}W_{p}\left(  K\right)
$ is always satisfied).

\textit{Induction step:} Let $N$ be a positive integer. Assume that Lemma
\ref{lem.WpK.p-th-mod-p.2} holds for $n=N-1$. We must now show that Lemma
\ref{lem.WpK.p-th-mod-p.2} holds for $n=N$.

Let $x\in W_{p}\left(  K\right)  $. Lemma \ref{lem.WpK.p-th-mod-p.1} shows
that there exist two elements $y$ and $z$ of $W_{p}\left(  K\right)  $ such
that $x=y^{p}+pz$. Denote these $y$ and $z$ by $a$ and $b$. Thus, $a$ and $b$
are two elements of $W_{p}\left(  K\right)  $ such that $x=a^{p}+pb$.

We have assumed that Lemma \ref{lem.WpK.p-th-mod-p.2} holds for $n=N-1$.
Hence, Lemma \ref{lem.WpK.p-th-mod-p.2} (applied to $b$ and $N-1$ instead of
$x$ and $n$) yields that there exists an $\left(  y_{0},y_{1},\ldots
,y_{N-2}\right)  \in\left(  W_{p}\left(  K\right)  \right)  ^{N-2}$ such that
$b\equiv\sum_{i=0}^{N-2}p^{i}y_{i}^{p}\operatorname{mod}p^{N-1}W_{p}\left(
K\right)  $. Denote this $\left(  y_{0},y_{1},\ldots,y_{N-2}\right)  $ by
$\left(  c_{0},c_{1},\ldots,c_{N-2}\right)  $. Thus, $\left(  c_{0}%
,c_{1},\ldots,c_{N-2}\right)  $ is an element of $\left(  W_{p}\left(
K\right)  \right)  ^{N-2}$ such that $b\equiv\sum_{i=0}^{N-2}p^{i}c_{i}%
^{p}\operatorname{mod}p^{N-1}W_{p}\left(  K\right)  $.

Multiplying the congruence $b\equiv\sum_{i=0}^{N-2}p^{i}c_{i}^{p}%
\operatorname{mod}p^{N-1}W_{p}\left(  K\right)  $ by $p$, we obtain
\begin{align*}
pb  &  \equiv p\sum_{i=0}^{N-2}p^{i}c_{i}^{p}=\sum_{i=0}^{N-2}p^{i+1}c_{i}%
^{p}=\sum_{i=1}^{N-1}p^{i}c_{i-1}^{p}\operatorname{mod}p^{N}W_{p}\left(
K\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }i-1\text{ for
}i\text{ in the sum}\right)  .
\end{align*}
Now,%
\[
x=a^{p}+\underbrace{pb}_{\equiv\sum_{i=1}^{N-1}p^{i}c_{i-1}^{p}%
\operatorname{mod}p^{N}W_{p}\left(  K\right)  }\equiv a^{p}+\sum_{i=1}%
^{N-1}p^{i}c_{i-1}^{p}\operatorname{mod}p^{N}W_{p}\left(  K\right)  .
\]
In other words, if we define an $\left(  y_{0},y_{1},\ldots,y_{N-1}\right)
\in\left(  W_{p}\left(  K\right)  \right)  ^{N}$ by $\left(  y_{0}%
,y_{1},\ldots,y_{N-1}\right)  =\left(  a,c_{0},c_{1},\ldots,c_{N-2}\right)  $,
then this $\left(  y_{0},y_{1},\ldots,y_{N-1}\right)  $ satisfies $x\equiv
\sum_{i=0}^{N-1}p^{i}y_{i}^{p}\operatorname{mod}p^{N}W_{p}\left(  K\right)  $.
Hence, there exists an $\left(  y_{0},y_{1},\ldots,y_{N-1}\right)  \in\left(
W_{p}\left(  K\right)  \right)  ^{N}$ such that $x\equiv\sum_{i=0}^{N-1}%
p^{i}y_{i}^{p}\operatorname{mod}p^{N}W_{p}\left(  K\right)  $ (namely,
$\left(  y_{0},y_{1},\ldots,y_{N-1}\right)  =\left(  a,c_{0},c_{1}%
,\ldots,c_{N-2}\right)  $).

Now, forget that we fixed $x$. We thus have proven that for every $x\in
W_{p}\left(  K\right)  $, there exists an $\left(  y_{0},y_{1},\ldots
,y_{N-1}\right)  \in\left(  W_{p}\left(  K\right)  \right)  ^{N}$ such that
$x\equiv\sum_{i=0}^{N-1}p^{i}y_{i}^{p}\operatorname{mod}p^{N}W_{p}\left(
K\right)  $. In other words, Lemma \ref{lem.WpK.p-th-mod-p.2} holds for $n=N$.
This completes the induction step. Thus, Lemma \ref{lem.WpK.p-th-mod-p.2} is
proven by induction.
\end{proof}
\end{noncompile}

\begin{proposition}
\label{prop.6.6.Funiq}Let $p$ be a prime. Let $R$ be a ring equipped with a
decreasing sequence $R=\mathfrak{a}_{0}\supset\mathfrak{a}_{1}\supset
\mathfrak{a}_{2}\supset\mathfrak{a}_{3}\supset\cdots$ of ideals such that
$\left(  \mathfrak{a}_{n}\cdot\mathfrak{a}_{m}\subset\mathfrak{a}_{n+m}\text{
for all }n\geq0\text{ and }m\geq0\right)  $. Equip $R$ with the topology
defined by this sequence of ideals. Assume that $R$ is Hausdorff for this topology.

Assume that the residue ring $L=R/\mathfrak{a}_{1}$ has $p\cdot1_{L}=0$. Let
$\pi$ denote the canonical projection $R\rightarrow R/\mathfrak{a}_{1}=L$.

Let $K$ be a perfect ring of characteristic $p$. Let $G:W_{p}\left(  K\right)
\rightarrow L$ be any map.

Then, there exists at most one ring homomorphism $\theta:W_{p}\left(
K\right)  \rightarrow R$ such that $\pi\circ\theta=G$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.6.6.Funiq}.]We need to show that if
$\theta_{1}$ and $\theta_{2}$ are two ring homomorphisms $\theta:W_{p}\left(
K\right)  \rightarrow R$ such that $\pi\circ\theta=G$, then $\theta_{1}%
=\theta_{2}$.

So let $\theta_{1}$ and $\theta_{2}$ be two ring homomorphisms $\theta
:W_{p}\left(  K\right)  \rightarrow R$ such that $\pi\circ\theta=G$. We must
show that $\theta_{1}=\theta_{2}$.

Define the maps $\pi_{n}$ for all $n\geq0$ as in Proposition
\ref{prop.6.6.conty} \textbf{(c)}.

We first observe that
\begin{equation}
\pi_{1}\circ\theta_{1}=\pi_{1}\circ\theta_{2}.
\label{pf.prop.6.6.Funiq.main.pf.cong1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.6.6.Funiq.main.pf.cong1}):} Both maps
$\pi_{1}$ and $\pi$ are defined as the canonical projection $R\rightarrow
R/\mathfrak{a}_{1}$. Hence, these two maps are equal. In other words, $\pi
_{1}=\pi$.
\par
We know that $\theta_{1}$ is a ring homomorphism $\theta:W_{p}\left(
K\right)  \rightarrow R$ such that $\pi\circ\theta=G$. Hence, $\pi\circ
\theta_{1}=G$. The same argument (applied to $\theta_{2}$ instead of
$\theta_{1}$) shows that $\pi\circ\theta_{2}=G$. Hence, $\pi\circ\theta
_{1}=G=\pi\circ\theta_{2}$. In light of $\pi_{1}=\pi$, this rewrites as
$\pi_{1}\circ\theta_{1}=\pi_{1}\circ\theta_{2}$. This proves
(\ref{pf.prop.6.6.Funiq.main.pf.cong1}).}

We shall show that
\begin{equation}
\pi_{n}\circ\theta_{1}=\pi_{n}\circ\theta_{2}\ \ \ \ \ \ \ \ \ \ \text{for
every }n\geq0. \label{pf.prop.6.6.Funiq.main}%
\end{equation}


\textit{Proof of (\ref{pf.prop.6.6.Funiq.main}):} We shall prove
(\ref{pf.prop.6.6.Funiq.main}) by induction on $n$:

\textit{Induction base:} The map $\pi_{0}$ is the canonical projection
$R\rightarrow R/\mathfrak{a}_{0}=0$ (since $\mathfrak{a}_{0}=R$). Hence, any
two maps become equal when composed with $\pi_{0}$. In particular, we thus
have $\pi_{0}\circ\theta_{1}=\pi_{0}\circ\theta_{2}$. In other words,
(\ref{pf.prop.6.6.Funiq.main}) holds for $n=0$. This completes the induction base.

\textit{Induction step:} Let $N$ be a positive integer. Assume that
(\ref{pf.prop.6.6.Funiq.main}) holds for $n=N-1$. We must show that
(\ref{pf.prop.6.6.Funiq.main}) holds for $n=N$. In other words, we need to
prove that $\pi_{N}\circ\theta_{1}=\pi_{N}\circ\theta_{2}$.

If $N=1$, then this follows immediately from
(\ref{pf.prop.6.6.Funiq.main.pf.cong1}). Hence, we WLOG assume that $N\neq1$.
Hence, $N\geq2$ (since $N$ is a positive integer). Thus, $N-1$ is a positive integer.

We have assumed that (\ref{pf.prop.6.6.Funiq.main}) holds for $n=N-1$. In
other words, $\pi_{N-1}\circ\theta_{1}=\pi_{N-1}\circ\theta_{2}$. For every
$a\in W_{p}\left(  K\right)  $, we have%
\begin{equation}
\theta_{1}\left(  a\right)  \equiv\theta_{2}\left(  a\right)
\operatorname{mod}\mathfrak{a}_{N-1} \label{pf.prop.6.6.Funiq.main.pf.cong2a}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.6.6.Funiq.main.pf.cong2a}):} Let
$a\in W_{p}\left(  K\right)  $. Then,%
\[
\pi_{N-1}\left(  \theta_{1}\left(  a\right)  \right)  =\underbrace{\left(
\pi_{N-1}\circ\theta_{1}\right)  }_{=\pi_{N-1}\circ\theta_{2}}\left(
a\right)  =\left(  \pi_{N-1}\circ\theta_{2}\right)  \left(  a\right)
=\pi_{N-1}\left(  \theta_{2}\left(  a\right)  \right)  .
\]
In other words, $\theta_{1}\left(  a\right)  \equiv\theta_{2}\left(  a\right)
\operatorname{mod}\mathfrak{a}_{N-1}$ (since $\pi_{N-1}$ is the canonical
projection $R\rightarrow R/\mathfrak{a}_{N-1}$). This proves
(\ref{pf.prop.6.6.Funiq.main.pf.cong2a}).} and%
\begin{equation}
\left(  \theta_{1}\left(  a\right)  \right)  ^{p}\equiv\left(  \theta
_{2}\left(  a\right)  \right)  ^{p}\operatorname{mod}\mathfrak{a}_{N}
\label{pf.prop.6.6.Funiq.main.pf.cong2b}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.6.6.Funiq.main.pf.cong2b}):} Let
$a\in W_{p}\left(  K\right)  $. From (\ref{pf.prop.6.6.Funiq.main.pf.cong2a}),
we have $\theta_{1}\left(  a\right)  \equiv\theta_{2}\left(  a\right)
\operatorname{mod}\mathfrak{a}_{N-1}$. Thus, Proposition \ref{prop.6.6.ideals}
\textbf{(f)} (applied to $\theta_{1}\left(  a\right)  $, $\theta_{2}\left(
a\right)  $ and $N-1$ instead of $a$, $b$ and $i$) yields $\left(  \theta
_{1}\left(  a\right)  \right)  ^{p}\equiv\left(  \theta_{2}\left(  a\right)
\right)  ^{p}\operatorname{mod}\mathfrak{a}_{\left(  N-1\right)  +1}$. In
other words, $\left(  \theta_{1}\left(  a\right)  \right)  ^{p}\equiv\left(
\theta_{2}\left(  a\right)  \right)  ^{p}\operatorname{mod}\mathfrak{a}_{N}$.
This proves (\ref{pf.prop.6.6.Funiq.main.pf.cong2b}).} and%
\begin{equation}
p\theta_{1}\left(  a\right)  \equiv p\theta_{2}\left(  a\right)
\operatorname{mod}\mathfrak{a}_{N} \label{pf.prop.6.6.Funiq.main.pf.cong2c}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.6.6.Funiq.main.pf.cong2c}):} Let
$a\in W_{p}\left(  K\right)  $. From (\ref{pf.prop.6.6.Funiq.main.pf.cong2a}),
we have $\theta_{1}\left(  a\right)  \equiv\theta_{2}\left(  a\right)
\operatorname{mod}\mathfrak{a}_{N-1}$. In other words, $\theta_{1}\left(
a\right)  -\theta_{2}\left(  a\right)  \in\mathfrak{a}_{N-1}$. Now,%
\[
p\theta_{1}\left(  a\right)  -p\theta_{2}\left(  a\right)
=p\underbrace{\left(  \theta_{1}\left(  a\right)  -\theta_{2}\left(  a\right)
\right)  }_{\in\mathfrak{a}_{N-1}}\in p\mathfrak{a}_{N-1}\subset
\mathfrak{a}_{N}%
\]
(by Proposition \ref{prop.6.6.ideals} \textbf{(g)}, applied to $n=N$). Thus,
$p\theta_{1}\left(  a\right)  \equiv p\theta_{2}\left(  a\right)
\operatorname{mod}\mathfrak{a}_{N}$. This proves
(\ref{pf.prop.6.6.Funiq.main.pf.cong2c}).}.

Proposition \ref{prop.6.6.ideals} \textbf{(b)} yields $pR\subset
\mathfrak{a}_{1}$.

Let $x\in W_{p}\left(  K\right)  $. Lemma \ref{lem.WpK.p-th-mod-p.1} shows
that there exist two elements $y$ and $z$ of $W_{p}\left(  K\right)  $ such
that $x=y^{p}+pz$. Consider these $y$ and $z$. Applying the map $\theta_{1}$
to the equality $x=y^{p}+pz$, we find%
\begin{equation}
\theta_{1}\left(  x\right)  =\theta_{1}\left(  y^{p}+pz\right)  =\left(
\theta_{1}\left(  y\right)  \right)  ^{p}+p\theta_{1}\left(  z\right)
\label{pf.prop.6.6.Funiq.main.pf.F1x=}%
\end{equation}
(since $\theta_{1}$ is a ring homomorphism). The same argument (applied to
$\theta_{2}$ instead of $\theta_{1}$) shows that%
\begin{equation}
\theta_{2}\left(  x\right)  =\left(  \theta_{2}\left(  y\right)  \right)
^{p}+p\theta_{2}\left(  z\right)  . \label{pf.prop.6.6.Funiq.main.pf.F2x=}%
\end{equation}
Now, (\ref{pf.prop.6.6.Funiq.main.pf.F1x=}) becomes%
\begin{align*}
\theta_{1}\left(  x\right)   &  =\underbrace{\left(  \theta_{1}\left(
y\right)  \right)  ^{p}}_{\substack{\equiv\left(  \theta_{2}\left(  y\right)
\right)  ^{p}\operatorname{mod}\mathfrak{a}_{N}\\\text{(by
(\ref{pf.prop.6.6.Funiq.main.pf.cong2b}), applied to }a=y\text{)}%
}}+\underbrace{p\theta_{1}\left(  z\right)  }_{\substack{\equiv p\theta
_{2}\left(  z\right)  \operatorname{mod}\mathfrak{a}_{N}\\\text{(by
(\ref{pf.prop.6.6.Funiq.main.pf.cong2c}), applied to }a=z\text{)}}}\\
&  \equiv\left(  \theta_{2}\left(  y\right)  \right)  ^{p}+p\theta_{2}\left(
z\right)  =\theta_{2}\left(  x\right)  \operatorname{mod}\mathfrak{a}_{N}%
\end{align*}
(by (\ref{pf.prop.6.6.Funiq.main.pf.F2x=})). In other words, $\pi_{N}\left(
\theta_{1}\left(  x\right)  \right)  =\pi_{N}\left(  \theta_{2}\left(
x\right)  \right)  $. Thus, $\left(  \pi_{N}\circ\theta_{1}\right)  \left(
x\right)  =\pi_{N}\left(  \theta_{1}\left(  x\right)  \right)  =\pi_{N}\left(
\theta_{2}\left(  x\right)  \right)  =\left(  \pi_{N}\circ\theta_{2}\right)
\left(  x\right)  $.

Now, forget that we fixed $x$. We thus have proven that $\left(  \pi_{N}%
\circ\theta_{1}\right)  \left(  x\right)  =\left(  \pi_{N}\circ\theta
_{2}\right)  \left(  x\right)  $ for every $x\in W_{p}\left(  K\right)  $. In
other words, $\pi_{N}\circ\theta_{1}=\pi_{N}\circ\theta_{2}$. This completes
the induction step. The induction proof of (\ref{pf.prop.6.6.Funiq.main}) is
thus complete.

Proposition \ref{prop.6.6.conty} \textbf{(c)} (applied to $A=W_{p}\left(
K\right)  $, $\varphi=\theta_{1}$ and $\psi=\theta_{2}$) thus yields
$\theta_{1}=\theta_{2}$. This completes the proof of Proposition
\ref{prop.6.6.Funiq}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.6.6gen}.]Proposition \ref{prop.6.6.ideals}
\textbf{(b)} shows that $p\cdot1_{R}\in\mathfrak{a}_{1}$ and $pR\subset
\mathfrak{a}_{1}$.

The map $\pi$ is surjective (since it is the canonical projection
$R\rightarrow R/\mathfrak{a}_{1}$).

Theorem \ref{thm.6.6teich} shows that there exists a unique multiplicative map
$r:K\rightarrow R$ satisfying (\ref{eq.thm.6.6gen.a.1}). Consider this $r$.

The ring $K$ is perfect of characteristic $p$. Hence, an element $v^{1/p^{n}}$
of $K$ is well-defined for each $v\in K$ and each $n\geq0$. In particular, an
element $x_{p^{n}}^{1/p^{n}}$ of $K$ is well-defined for each $x\in
W_{p}\left(  K\right)  $ and each $n\geq0$.

Define a map $\Theta:W_{p}\left(  K\right)  \rightarrow R$ by%
\[
\left(  \Theta\left(  x\right)  =\sum_{n\geq0}r\left(  x_{p^{n}}^{1/p^{n}%
}\right)  p^{n}\ \ \ \ \ \ \ \ \ \ \text{for every }x\in W_{p}\left(
K\right)  \right)  .
\]
This is well-defined (because the infinite sum $\sum_{n\geq0}r\left(
x_{p^{n}}^{1/p^{n}}\right)  p^{n}$ converges\footnote{since $r\left(
x_{p^{n}}^{1/p^{n}}\right)  p^{n}=r\left(  x_{p^{n}}^{1/p^{n}}\right)
\underbrace{p^{n}\cdot1_{R}}_{\substack{\in\mathfrak{a}_{n}\\\text{(by
Proposition \ref{prop.6.6.ideals} \textbf{(c)})}}}\in\mathfrak{a}_{n}$ for
every $n\geq0$}). Moreover, the square (\ref{eq.thm.6.6gen.a.2}) commutes for
$\theta=\Theta$.\ \ \ \ \footnote{\textit{Proof.} Let $x\in W_{p}\left(
K\right)  $. Every $n\geq1$ satisfies%
\[
r\left(  x_{p^{n}}^{1/p^{n}}\right)  p^{n}=\underbrace{p^{n}}_{=pp^{n-1}%
}r\left(  x_{p^{n}}^{1/p^{n}}\right)  =p\underbrace{p^{n-1}r\left(  x_{p^{n}%
}^{1/p^{n}}\right)  }_{\in R}\in pR\subset\mathfrak{a}_{1}%
\]
and thus%
\begin{equation}
\pi\left(  r\left(  x_{p^{n}}^{1/p^{n}}\right)  p^{n}\right)  =0
\label{pf.thm.6.6gen.fn1.pir=0}%
\end{equation}
(since $\pi$ is the canonical projection $R\rightarrow R/\mathfrak{a}_{1}$).
Now,%
\begin{align*}
\left(  \pi\circ\Theta\right)  \left(  x\right)   &  =\pi\left(
\underbrace{\Theta\left(  x\right)  }_{=\sum_{n\geq0}r\left(  x_{p^{n}%
}^{1/p^{n}}\right)  p^{n}}\right)  =\pi\left(  \sum_{n\geq0}r\left(  x_{p^{n}%
}^{1/p^{n}}\right)  p^{n}\right)  =\sum_{n\geq0}\pi\left(  r\left(  x_{p^{n}%
}^{1/p^{n}}\right)  p^{n}\right) \\
&  =\pi\left(  \underbrace{r\left(  x_{p^{0}}^{1/p^{0}}\right)  p^{0}%
}_{=r\left(  x_{1}\right)  1=r\left(  x_{1}\right)  }\right)  +\sum_{n\geq
1}\underbrace{\pi\left(  r\left(  x_{p^{n}}^{1/p^{n}}\right)  p^{n}\right)
}_{\substack{=0\\\text{(by (\ref{pf.thm.6.6gen.fn1.pir=0}))}}}\\
&  =\pi\left(  r\left(  x_{1}\right)  \right)  +\underbrace{\sum_{n\geq1}%
0}_{=0}=\pi\left(  r\left(  x_{1}\right)  \right)  =f\left(  x_{1}\right)
\end{align*}
(by (\ref{eq.thm.6.6gen.a.1}), applied to $a=x_{1}$). Comparing this with%
\[
\left(  f\circ w_{1}\right)  \left(  x\right)  =f\left(  \underbrace{w_{1}%
\left(  x\right)  }_{=x_{1}}\right)  =f\left(  x_{1}\right)  ,
\]
we obtain $\left(  \pi\circ\Theta\right)  \left(  x\right)  =\left(  f\circ
w_{1}\right)  \left(  x\right)  $. Since we have proven this for every $x\in
W_{p}\left(  K\right)  $, we thus conclude that $\pi\circ\Theta=f\circ w_{1}$.
In other words, $\pi\circ\theta=f\circ w_{1}$ for $\theta=\Theta$. In other
words, the square (\ref{eq.thm.6.6gen.a.2}) commutes for $\theta=\Theta$.}

Notice that every $x\in W_{p}\left(  K\right)  $ satisfies%
\begin{equation}
\Theta\left(  x\right)  \equiv\sum_{k=0}^{i}r\left(  x_{p^{k}}^{1/p^{k}%
}\right)  p^{k}\operatorname{mod}\mathfrak{a}_{i+1}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\geq-1. \label{pf.thm.6.6gen.b.Fmod}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.6.6gen.b.Fmod}):} Let $x\in
W_{p}\left(  K\right)  $ and $i\geq-1$. Then, Proposition
\ref{prop.6.6.ideals} \textbf{(c)} yields that $p^{n}\cdot1_{R}\in
\mathfrak{a}_{n}$ for every $n\geq0$. Now,
\begin{align*}
\Theta\left(  x\right)   &  =\sum_{n\geq0}r\left(  x_{p^{n}}^{1/p^{n}}\right)
p^{n}=\sum_{n=0}^{i}r\left(  x_{p^{n}}^{1/p^{n}}\right)  p^{n}+\sum_{n\geq
i+1}\underbrace{r\left(  x_{p^{n}}^{1/p^{n}}\right)  p^{n}}_{=r\left(
x_{p^{n}}^{1/p^{n}}\right)  p^{n}\cdot1_{R}}\\
&  =\sum_{n=0}^{i}r\left(  x_{p^{n}}^{1/p^{n}}\right)  p^{n}+\sum_{n\geq
i+1}r\left(  x_{p^{n}}^{1/p^{n}}\right)  \underbrace{p^{n}\cdot1_{R}%
}_{\substack{\in\mathfrak{a}_{n}\subset\mathfrak{a}_{i+1}\\\text{(since }n\geq
i+1\text{)}}}\in\sum_{n=0}^{i}r\left(  x_{p^{n}}^{1/p^{n}}\right)
p^{n}+\underbrace{\sum_{n\geq i+1}r\left(  x_{p^{n}}^{1/p^{n}}\right)
\mathfrak{a}_{i+1}}_{\subset\mathfrak{a}_{i+1}}\\
&  \subset\sum_{n=0}^{i}r\left(  x_{p^{n}}^{1/p^{n}}\right)  p^{n}%
+\mathfrak{a}_{i+1}.
\end{align*}
In other words, $\Theta\left(  x\right)  \equiv\sum_{n=0}^{i}r\left(
x_{p^{n}}^{1/p^{n}}\right)  p^{n}\operatorname{mod}\mathfrak{a}_{i+1}$.
Renaming the summation index $n$ as $k$ in this congruence, we obtain
$\Theta\left(  x\right)  \equiv\sum_{k=0}^{i}r\left(  x_{p^{k}}^{1/p^{k}%
}\right)  p^{k}\operatorname{mod}\mathfrak{a}_{i+1}$, qed.}

For every $n\geq0$, let $\pi_{n}$ be the canonical projection $R\rightarrow
R/\mathfrak{a}_{n}$. Then, every $x\in W_{p}\left(  K\right)  $ and $n\geq0$
satisfy%
\begin{equation}
\left(  \pi_{n}\circ\Theta\right)  \left(  x\right)  =\pi_{n}\left(
\sum_{i=0}^{n}p^{i}\left(  r\left(  x_{p^{i}}^{1/p^{n}}\right)  \right)
^{p^{n-i}}\right)  . \label{pf.thm.6.6gen.b.Fmod-pi}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.6.6gen.b.Fmod-pi}):} Let $x\in
W_{p}\left(  K\right)  $ and $n\geq0$. Applying (\ref{pf.thm.6.6gen.b.Fmod})
to $i=n$, we obtain%
\[
\Theta\left(  x\right)  \equiv\sum_{k=0}^{n}r\left(  x_{p^{k}}^{1/p^{k}%
}\right)  p^{k}=\sum_{k=0}^{n}p^{k}r\left(  x_{p^{k}}^{1/p^{k}}\right)
=\sum_{i=0}^{n}p^{i}r\left(  x_{p^{i}}^{1/p^{i}}\right)  \operatorname{mod}%
\mathfrak{a}_{n+1}.
\]
In other words, $\Theta\left(  x\right)  -\sum_{i=0}^{n}p^{i}r\left(
x_{p^{i}}^{1/p^{i}}\right)  \in\mathfrak{a}_{n+1}\subset\mathfrak{a}_{n}$, so
that%
\begin{equation}
\Theta\left(  x\right)  \equiv\sum_{i=0}^{n}p^{i}r\left(  x_{p^{i}}^{1/p^{i}%
}\right)  \operatorname{mod}\mathfrak{a}_{n}.
\label{pf.thm.6.6gen.b.Fmod-pi.pf.1}%
\end{equation}
\par
Now, fix $i\in\left\{  0,1,\ldots,n\right\}  $. Recall that the map $r$ is
multiplicative. Thus, Proposition \ref{prop.multiplicative.power} (applied to
$r$, $x_{p^{i}}^{1/p^{n}}$ and $p^{n-i}$ instead of $\varphi$, $v$ and $i$)
yields $r\left(  \left(  x_{p^{i}}^{1/p^{n}}\right)  ^{p^{n-i}}\right)
=\left(  r\left(  x_{p^{i}}^{1/p^{n}}\right)  \right)  ^{p^{n-i}}$. Hence,%
\begin{equation}
\left(  r\left(  x_{p^{i}}^{1/p^{n}}\right)  \right)  ^{p^{n-i}}=r\left(
\underbrace{\left(  x_{p^{i}}^{1/p^{n}}\right)  ^{p^{n-i}}}_{=x_{p^{i}%
}^{p^{n-i}/p^{n}}=x_{p^{i}}^{1/p^{i}}}\right)  =r\left(  x_{p^{i}}^{1/p^{i}%
}\right)  . \label{pf.thm.6.6gen.b.Fmod-pi.pf.2}%
\end{equation}
\par
Now, forget that we fixed $i$. Thus, we have shown that
(\ref{pf.thm.6.6gen.b.Fmod-pi.pf.2}) holds for each $i\in\left\{
0,1,\ldots,n\right\}  $. Now, (\ref{pf.thm.6.6gen.b.Fmod-pi.pf.1}) becomes%
\[
\Theta\left(  x\right)  \equiv\sum_{i=0}^{n}p^{i}\underbrace{r\left(
x_{p^{i}}^{1/p^{i}}\right)  }_{\substack{=\left(  r\left(  x_{p^{i}}^{1/p^{n}%
}\right)  \right)  ^{p^{n-i}}\\\text{(by (\ref{pf.thm.6.6gen.b.Fmod-pi.pf.2}%
))}}}=\sum_{i=0}^{n}p^{i}\left(  r\left(  x_{p^{i}}^{1/p^{n}}\right)  \right)
^{p^{n-i}}\operatorname{mod}\mathfrak{a}_{n}.
\]
In other words, $\pi_{n}\left(  \Theta\left(  x\right)  \right)  =\pi
_{n}\left(  \sum_{i=0}^{n}p^{i}\left(  r\left(  x_{p^{i}}^{1/p^{n}}\right)
\right)  ^{p^{n-i}}\right)  $ (since $\pi_{n}$ is the canonical projection
$R\rightarrow R/\mathfrak{a}_{n}$). Thus,%
\[
\left(  \pi_{n}\circ\Theta\right)  \left(  x\right)  =\pi_{n}\left(
\Theta\left(  x\right)  \right)  =\pi_{n}\left(  \sum_{i=0}^{n}p^{i}\left(
r\left(  x_{p^{i}}^{1/p^{n}}\right)  \right)  ^{p^{n-i}}\right)  ,
\]
and so (\ref{pf.thm.6.6gen.b.Fmod-pi}) is proven.} Hence, for every $n\geq0$,
the map $\pi_{n}\circ\Theta:W_{p}\left(  K\right)  \rightarrow R/\mathfrak{a}%
_{n}$ is continuous (where $R/\mathfrak{a}_{n}$ is equipped with the discrete
topology)\footnote{\textit{Proof.} Let $n\geq0$. Then, for every $x\in
W_{p}\left(  K\right)  $, the formula (\ref{pf.thm.6.6gen.b.Fmod-pi})
expresses $\left(  \pi_{n}\circ\Theta\right)  \left(  x\right)  $ through the
first $N+1$ coordinates $x_{p^{0}},x_{p^{1}},\ldots,x_{p^{N}}$ of $x$. Hence,
$\left(  \pi_{n}\circ\Theta\right)  \left(  x\right)  $ depends only on the
first $N+1$ coordinates $x_{p^{0}},x_{p^{1}},\ldots,x_{p^{N}}$ of $x$. Thus,
the map $\pi_{n}\circ\Theta:W_{p}\left(  K\right)  \rightarrow R/\mathfrak{a}%
_{n}$ is continuous (by the definition of the topology on $W_{p}\left(
K\right)  $). Qed.}. Therefore, Proposition \ref{prop.6.6.contin} (applied to
$A=W_{p}\left(  K\right)  $ and $\varphi=\Theta$) shows that the map
$\Theta:W_{p}\left(  K\right)  \rightarrow R$ is continuous.

Now, fix a nonnegative integer $n$.

Let $\omega$ denote the map $K\rightarrow K,\ u\mapsto u^{p}$. Thus, $\omega$
is the Frobenius homomorphism of the ring $K$, and is a ring homomorphism
(since the ring $K$ has characteristic $p$) and invertible (since the ring $K$
is perfect). Hence, its inverse map $\omega^{-1}$ is also a ring homomorphism.
Thus, the map $\omega^{-n}$ is also a ring homomorphism. It is easy to see
that%
\begin{equation}
\omega^{-i}\left(  v\right)  =v^{1/p^{i}}\ \ \ \ \ \ \ \ \ \ \text{for every
}v\in K\text{ and }i\geq0. \label{pf.thm.6.6gen.b.omega-i}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.6.6gen.b.omega-i}):} We have
$\omega\left(  v\right)  =v^{p}$ for every $v\in K$. Thus, $\omega^{-1}\left(
v\right)  =v^{1/p}$ for every $v\in K$. Thus, we can easily see (by a
straightforward induction over $i$) that $\omega^{-i}\left(  v\right)
=v^{1/p^{i}}$ for every $v\in K$ and $i\geq0$. Thus,
(\ref{pf.thm.6.6gen.b.omega-i}) is proven.} We have%
\begin{equation}
\left(  f\circ\omega^{-n}\right)  \left(  v\right)  =\pi\left(  r\left(
v^{1/p^{n}}\right)  \right)  \label{pf.thm.6.6gen.b.4}%
\end{equation}
for each $v\in K$.\ \ \ \ \footnote{\textit{Proof of (\ref{pf.thm.6.6gen.b.4}%
):} Let $v\in K$. Then,%
\[
\left(  f\circ\omega^{-n}\right)  \left(  v\right)  =f\left(
\underbrace{\omega^{-n}\left(  v\right)  }_{\substack{=v^{1/p^{n}}\\\text{(by
(\ref{pf.thm.6.6gen.b.omega-i}), applied to }i=n\text{)}}}\right)  =f\left(
v^{1/p^{n}}\right)  =\pi\left(  r\left(  v^{1/p^{n}}\right)  \right)
\]
(because (\ref{eq.thm.6.6gen.a.1}) (applied to $a=v^{1/p^{n}}$) yields
$\pi\left(  r\left(  v^{1/p^{n}}\right)  \right)  =f\left(  v^{1/p^{n}%
}\right)  $). This proves (\ref{pf.thm.6.6gen.b.4}).}

Proposition \ref{prop.6.6.wn} shows that there exists a unique ring
homomorphism $\widetilde{w}_{p^{n}}:W_{p}\left(  L\right)  \rightarrow
R/\mathfrak{a}_{n}$ such that the diagram (\ref{eq.prop.6.6.wn.diag})
commutes. Consider this $\widetilde{w}_{p^{n}}$.

But $f\circ\omega^{-n}:K\rightarrow L$ is a ring homomorphism (since both $f$
and $\omega^{-n}$ are ring homomorphisms). Thus, $W_{p}\left(  f\circ
\omega^{-n}\right)  :W_{p}\left(  K\right)  \rightarrow W_{p}\left(  L\right)
$ is a ring homomorphism as well. Now, we shall show that%
\begin{equation}
\widetilde{w}_{p^{n}}\circ W_{p}\left(  f\circ\omega^{-n}\right)  =\pi
_{n}\circ\Theta. \label{pf.thm.6.6gen.b.main}%
\end{equation}


\textit{Proof of (\ref{pf.thm.6.6gen.b.main}):} Let $x\in W_{p}\left(
K\right)  $. The definition of $W_{p}\left(  f\circ\omega^{-n}\right)  $
yields%
\begin{equation}
\left(  W_{p}\left(  f\circ\omega^{-n}\right)  \right)  \left(  x\right)
=\left(  \underbrace{\left(  f\circ\omega^{-n}\right)  \left(  x_{m}\right)
}_{\substack{=\pi\left(  r\left(  x_{m}^{1/p^{n}}\right)  \right)  \\\text{(by
(\ref{pf.thm.6.6gen.b.4}), applied}\\\text{to }v=x_{m}\text{)}}}\right)
_{m\in P_{p}}=\left(  \pi\left(  r\left(  x_{m}^{1/p^{n}}\right)  \right)
\right)  _{m\in P_{p}}. \label{pf.thm.6.6gen.b.main.pf.1}%
\end{equation}


On the other hand, define $y\in W_{p}\left(  R\right)  $ by $y=\left(
r\left(  x_{m}^{1/p^{n}}\right)  \right)  _{m\in P_{p}}$. Then, the definition
of $W_{p}\left(  \pi\right)  $ yields%
\[
\left(  W_{p}\left(  \pi\right)  \right)  \left(  y\right)  =\left(
\pi\left(  r\left(  x_{m}^{1/p^{n}}\right)  \right)  \right)  _{m\in P_{p}%
}=\left(  W_{p}\left(  f\circ\omega^{-n}\right)  \right)  \left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.6.6gen.b.main.pf.1}%
)}\right)  .
\]
Applying the map $\widetilde{w}_{p^{n}}:W_{p}\left(  L\right)  \rightarrow
R/\mathfrak{a}_{n}$ to both sides of this equality, we obtain%
\[
\widetilde{w}_{p^{n}}\left(  \left(  W_{p}\left(  \pi\right)  \right)  \left(
y\right)  \right)  =\widetilde{w}_{p^{n}}\left(  \left(  W_{p}\left(
f\circ\omega^{-n}\right)  \right)  \left(  x\right)  \right)  =\left(
\widetilde{w}_{p^{n}}\circ W_{p}\left(  f\circ\omega^{-n}\right)  \right)
\left(  x\right)  .
\]
Thus,%
\begin{align}
\left(  \widetilde{w}_{p^{n}}\circ W_{p}\left(  f\circ\omega^{-n}\right)
\right)  \left(  x\right)   &  =\widetilde{w}_{p^{n}}\left(  \left(
W_{p}\left(  \pi\right)  \right)  \left(  y\right)  \right)
=\underbrace{\left(  \widetilde{w}_{p^{n}}\circ W_{p}\left(  \pi\right)
\right)  }_{\substack{=\pi_{n}\circ w_{p^{n}}\\\text{(since the diagram
(\ref{eq.prop.6.6.wn.diag})}\\\text{commutes)}}}\left(  y\right) \nonumber\\
&  =\left(  \pi_{n}\circ w_{p^{n}}\right)  \left(  y\right)  =\pi_{n}\left(
w_{p^{n}}\left(  y\right)  \right)  . \label{pf.thm.6.6gen.b.main.pf.4}%
\end{align}


On the other hand, (\ref{eq.wpn=}) yields%
\[
w_{p^{n}}\left(  y\right)  =\sum_{i=0}^{n}p^{i}\left(  r\left(  x_{p^{i}%
}^{1/p^{n}}\right)  \right)  ^{p^{n-i}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}y=\left(  r\left(  x_{m}^{1/p^{n}}\right)  \right)  _{m\in P_{p}}\right)  .
\]
Applying the map $\pi_{n}$ to both sides of this equality, we obtain%
\[
\pi_{n}\left(  w_{p^{n}}\left(  y\right)  \right)  =\pi_{n}\left(  \sum
_{i=0}^{n}p^{i}\left(  r\left(  x_{p^{i}}^{1/p^{n}}\right)  \right)
^{p^{n-i}}\right)  =\left(  \pi_{n}\circ\Theta\right)  \left(  x\right)
\]
(by (\ref{pf.thm.6.6gen.b.Fmod-pi})). Hence, (\ref{pf.thm.6.6gen.b.main.pf.4})
becomes%
\begin{equation}
\left(  \widetilde{w}_{p^{n}}\circ W_{p}\left(  f\circ\omega^{-n}\right)
\right)  \left(  x\right)  =\pi_{n}\left(  w_{p^{n}}\left(  y\right)  \right)
=\left(  \pi_{n}\circ\Theta\right)  \left(  x\right)  .
\label{pf.thm.6.6gen.b.main.pf.8}%
\end{equation}


Now, forget that we fixed $x$. We thus have proven
(\ref{pf.thm.6.6gen.b.main.pf.8}) for each $x\in W_{p}\left(  K\right)  $. In
other words, we have $\widetilde{w}_{p^{n}}\circ W_{p}\left(  f\circ
\omega^{-n}\right)  =\pi_{n}\circ\Theta$. Thus, (\ref{pf.thm.6.6gen.b.main})
is proven.

Now, the map $\widetilde{w}_{p^{n}}\circ W_{p}\left(  f\circ\omega
^{-n}\right)  $ is a ring homomorphism (since it is the composition of the two
ring homomorphisms $\widetilde{w}_{p^{n}}$ and $W_{p}\left(  f\circ\omega
^{-n}\right)  $). In view of (\ref{pf.thm.6.6gen.b.main}), this rewrites as
follows: The map $\pi_{n}\circ\Theta$ is a ring homomorphism.

Now, forget that we fixed $n$. We thus have shown that the map $\pi_{n}%
\circ\Theta$ is a ring homomorphism for each $n\geq0$. Hence, Proposition
\ref{prop.6.6.conty-riho} (applied to $A=W_{p}\left(  K\right)  $ and
$\gamma=\Theta$) shows that $\Theta:W_{p}\left(  K\right)  \rightarrow R$ is a
ring homomorphism. Hence, there exists a continuous ring homomorphism
$\theta:W_{p}\left(  K\right)  \rightarrow R$ making the square
(\ref{eq.thm.6.6gen.a.2}) commute (namely, $\theta=\Theta$).

It thus only remains to show that there exists \textbf{at most} one such
homomorphism. It will clearly be enough to prove that there exists at most one
ring homomorphism $\theta:W_{p}\left(  K\right)  \rightarrow R$ making the
square (\ref{eq.thm.6.6gen.a.2}) commute. In other words, it will be enough to
prove that there exists at most one ring homomorphism $\theta:W_{p}\left(
K\right)  \rightarrow R$ such that $\pi\circ\theta=f\circ w_{1}$. But this
follows from Proposition \ref{prop.6.6.Funiq} (applied to $G=f\circ w_{1}$).
Thus, the proof of Theorem \ref{thm.6.6gen} is complete.
\end{proof}

\begin{thebibliography}{99999999}                                                                                         %


\bibitem[BriCon09]{BriCon}Olivier Brinon, Brian Conrad, \textit{CMI Summer
School notes on }$p$\textit{-adic Hodge theory (preliminary version)}, 24 June
2009.\newline\url{http://math.stanford.edu/~conrad/papers/notes.pdf}
\end{thebibliography}


\end{document}