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\ihead{Vertex algebras, lecture 3, scribe notes}
\ohead{\today}
\begin{document}

\title{Vertex algebras by Victor Kac. \textit{Lecture 3}: Fundamentals of formal distributions}
\author{Scribe notes by Darij Grinberg}
\date{version of \today}
\maketitle
\tableofcontents

\section*{Introduction}

These notes introduce some algebraic analogues of notions from classical
analysis such as Laurent series, differential operators, formal distributions
and the $\delta$-function. We believe that everything written below is
well-known (\textquotedblleft folklore\textquotedblright), but a good deal of
it is not easy to find in the literature. The notes are self-contained and
should be understandable for anyone who has prior experience with
(multivariate) formal power series, binomial coefficients and localization of
commutative rings.

Chapters \ref{chap.delta} and \ref{chap.lie} are scribe notes of a lecture in
Victor Kac's 18.276 class at MIT in Spring 2015.\footnote{The class is about
vertex algebras, but no vertex algebras appear in these notes.} Chapters
\ref{chap.poly} and \ref{chap.deri} are meant to set the stage and define the
requisite objects beforehand.

Please let me (\texttt{\href{darijgrinberg@gmail.com}{darijgrinberg@gmail.com}%
}) know of any mistakes!

\section{\label{chap.poly}Basics on polynomial-like objects}

\subsection{Polynomial-like objects}

Let us first define the objects which we are going to deal with. These objects
will be certain analogues of polynomials and power series. Some of them are
well-known, but some are apocryphal. We assume that the reader has some
familiarity with polynomials and formal power series; this will allow us to be
brief about them and also about the more baroque objects that we introduce
when they behave similarly to polynomials and formal power series.

In the following, $\mathbb{F}$ denotes a commutative ring\footnote{Here and in
the following, ``ring'' always means ``ring with $1$''.}, and $\mathbb{N}$
denotes $\left\{  0,1,2,\ldots\right\}  $. When $R$ is a ring (for instance,
$\mathbb{F}$), the word \textquotedblleft$R$-algebra\textquotedblright\ means
\textquotedblleft associative central $R$-algebra with $1$\textquotedblright,
unless the word \textquotedblleft algebra\textquotedblright\ is qualified by
additional adjectives (such as in \textquotedblleft
superalgebra\textquotedblright, \textquotedblleft Lie
algebra\textquotedblright\ or \textquotedblleft vertex
algebra\textquotedblright). We use the notation $\delta_{i,j}$ for $\left\{
\begin{array}
[c]{c}%
1,\text{ if }i=j;\\
0,\text{ if }i\neq j
\end{array}
\right.  $ whenever $i$ and $j$ are any two objects.

We shall work with the usual notion of $\mathbb{F}$-modules in the following,
but we shall keep in the back of our mind that all of our arguments can be
adapted (using the Koszul-Quillen sign convention) to the setting of
$\mathbb{F}$-supermodules (at the cost of sometimes having to require that $2$
is invertible in $\mathbb{F}$).

If $U$ is a $\mathbb{F}$-module and $z$ is a symbol, then we can consider the
following five $\mathbb{F}$-modules:

\begin{itemize}
\item The $\mathbb{F}$-module $U\left[  z\right]  $ of polynomials in $z$ with
coefficients in $U$. This $\mathbb{F}$-module consists of all families
$\left(  u_{i}\right)  _{i\in\mathbb{N}}\in U^{\mathbb{N}}$ such that all but
finitely many $i\in\mathbb{N}$ satisfy $u_{i}=0$. We write such families
$\left(  u_{i}\right)  _{i\in\mathbb{N}}$ in the forms $\sum_{i=0}^{\infty
}u_{i}z^{i}$ and $\sum_{i\in\mathbb{N}}u_{i}z^{i}$, and refer to them as
\textquotedblleft polynomials\textquotedblright\ (despite $U$ not necessarily
being a ring).

\item The $\mathbb{F}$-module $U\left[  z,z^{-1}\right]  $ of Laurent
polynomials in $z$ with coefficients in $U$. This $\mathbb{F}$-module consists
of all families $\left(  u_{i}\right)  _{i\in\mathbb{Z}}\in U^{\mathbb{Z}}$
such that all but finitely many $i\in\mathbb{Z}$ satisfy $u_{i}=0$. We write
such families $\left(  u_{i}\right)  _{i\in\mathbb{Z}}$ in the forms
$\sum_{i=-\infty}^{\infty}u_{i}z^{i}$ and $\sum_{i\in\mathbb{Z}}u_{i}z^{i}$,
and refer to them as \textquotedblleft Laurent polynomials\textquotedblright.

\item The $\mathbb{F}$-module $U\left[  \left[  z\right]  \right]  $ of formal
power series in $z$ with coefficients in $U$. This $\mathbb{F}$-module
consists of all families $\left(  u_{i}\right)  _{i\in\mathbb{N}}\in
U^{\mathbb{N}}$. We write such families $\left(  u_{i}\right)  _{i\in
\mathbb{N}}$ in the forms $\sum_{i=0}^{\infty}u_{i}z^{i}$ and $\sum
_{i\in\mathbb{N}}u_{i}z^{i}$, and refer to them as \textquotedblleft power
series\textquotedblright\ (or \textquotedblleft formal power
series\textquotedblright).

\item The $\mathbb{F}$-module $U\left(  \left(  z\right)  \right)  $ of
Laurent series in $z$ with coefficients in $U$. This $\mathbb{F}$-module
consists of all families $\left(  u_{i}\right)  _{i\in\mathbb{Z}}\in
U^{\mathbb{Z}}$ such that all but finitely many \textbf{negative}
$i\in\mathbb{Z}$ satisfy $u_{i}=0$. We write such families $\left(
u_{i}\right)  _{i\in\mathbb{Z}}$ in the forms $\sum_{i=-\infty}^{\infty}%
u_{i}z^{i}$ and $\sum_{i\in\mathbb{Z}}u_{i}z^{i}$, and refer to them as
\textquotedblleft Laurent series\textquotedblright.

\item The $\mathbb{F}$-module $U\left[  \left[  z,z^{-1}\right]  \right]  $ of
$U$-valued formal distributions. This $\mathbb{F}$-module consists of all
families $\left(  u_{i}\right)  _{i\in\mathbb{Z}}\in U^{\mathbb{Z}}$. We write
such families $\left(  u_{i}\right)  _{i\in\mathbb{Z}}$ in the forms
$\sum_{i=-\infty}^{\infty}u_{i}z^{i}$ and $\sum_{i\in\mathbb{Z}}u_{i}z^{i}$,
and refer to them as \textquotedblleft$U$-valued formal
distributions\textquotedblright\ (in analogy to the distributions of analysis,
although these formal distributions are much more elementary than the latter).
\end{itemize}

The $\mathbb{F}$-module structure on each of these five $\mathbb{F}$-modules
$U\left[  z\right]  $, $U\left[  z,z^{-1}\right]  $, $U\left[  \left[
z\right]  \right]  $, $U\left(  \left(  z\right)  \right)  $ and $U\left[
\left[  z,z^{-1}\right]  \right]  $ is defined to be componentwise (i.e., we
have $\left(  u_{i}\right)  _{i\in\mathbb{N}}+\left(  v_{i}\right)
_{i\in\mathbb{N}}=\left(  u_{i}+v_{i}\right)  _{i\in\mathbb{N}}$ and
$\lambda\cdot\left(  u_{i}\right)  _{i\in\mathbb{N}}=\left(  \lambda
u_{i}\right)  _{i\in\mathbb{N}}$ for $U\left[  z\right]  $, and similar rules
with $\mathbb{N}$ replaced by $\mathbb{Z}$ for the other four $\mathbb{F}%
$-modules). Of course, we have%
\begin{align*}
U  &  \subseteq U\left[  z\right]  \subseteq U\left[  z,z^{-1}\right]
\subseteq U\left(  \left(  z\right)  \right)  \subseteq U\left[  \left[
z,z^{-1}\right]  \right]  \ \ \ \ \ \ \ \ \ \ \text{and}\\
U\left[  z\right]   &  \subseteq U\left[  \left[  z\right]  \right]  \subseteq
U\left[  \left[  z,z^{-1}\right]  \right]  .
\end{align*}


We refer to the elements of any of the five $\mathbb{F}$-modules $U\left[
z\right]  $, $U\left[  z,z^{-1}\right]  $, $U\left[  \left[  z\right]
\right]  $, $U\left(  \left(  z\right)  \right)  $ and $U\left[  \left[
z,z^{-1}\right]  \right]  $ as \textquotedblleft polynomial-like
objects\textquotedblright. Given such an object -- say, $\sum_{i\in\mathbb{Z}%
}u_{i}z^{i}$ --, we shall refer to the elements $u_{i}$ of $U$ as its
\textit{coefficients}. More precisely, if $u=\sum_{i\in\mathbb{Z}}u_{i}z^{i}$,
then $u_{i}$ is called the $i$\textit{-th coefficient} of $u$ (or the
\textit{coefficient of }$z^{i}$ \textit{in }$u$).

Often, $U\left[  z,z^{-1}\right]  $ is denoted by $U\left[  z^{\pm1}\right]
$, and $U\left[  \left[  z,z^{-1}\right]  \right]  $ is denoted by $U\left[
\left[  z^{\pm1}\right]  \right]  $.

We shall use standard notations for elements of $U\left[  z\right]  $,
$U\left[  z,z^{-1}\right]  $, $U\left[  \left[  z\right]  \right]  $,
$U\left(  \left(  z\right)  \right)  $ and $U\left[  \left[  z,z^{-1}\right]
\right]  $. For instance, for given $u\in U$ and $m\in\mathbb{Z}$, we let
$uz^{m}$ denote the element $\left(  u\delta_{i,m}\right)  _{i\in\mathbb{Z}}$
of $U\left[  z,z^{-1}\right]  $ (that is, the Laurent polynomial whose $m$-th
coefficient is $u$ and whose all other coefficients are $0$). This is also an
element of $U\left[  z\right]  $ when $m\in\mathbb{N}$. When $U=\mathbb{F}$
and $m\in\mathbb{Z}$, we write $z^{m}$ for $1z^{m}$. We abbreviate $z^{1}$ by
$z$.

When $U$ has additional structure (such as a multiplication, or a module
structure over some $\mathbb{F}$-algebra), we can endow some of the five
$\mathbb{F}$-modules $U\left[  z\right]  $, $U\left[  z,z^{-1}\right]  $,
$U\left[  \left[  z\right]  \right]  $, $U\left(  \left(  z\right)  \right)  $
and $U\left[  \left[  z,z^{-1}\right]  \right]  $ with additional structure as
well. Here are some examples:

\begin{itemize}
\item If $U$ is a $\mathbb{F}$-algebra\footnote{This includes $U=\mathbb{F}$
as a particular case.}, then $U\left[  z\right]  $, $U\left[  z,z^{-1}\right]
$, $U\left[  \left[  z\right]  \right]  $ and $U\left(  \left(  z\right)
\right)  $ become $\mathbb{F}$-algebras, with multiplication given by the rule%
\begin{equation}
\left(  \sum_{i}u_{i}z^{i}\right)  \cdot\left(  \sum_{i}v_{i}z^{i}\right)
=\sum_{i}\left(  \sum_{j}u_{j}v_{i-j}\right)  z^{i}
\label{multiplication-rule}%
\end{equation}
(and unity defined to be $\sum_{i}\delta_{i,0}z^{i}$). Here, the sums range
over $\mathbb{N}$ in the case of $U\left[  z\right]  $, and over $\mathbb{Z}$
in the cases of $U\left[  z,z^{-1}\right]  $, $U\left[  \left[  z\right]
\right]  $ and $U\left(  \left(  z\right)  \right)  $. When $U$ is
commutative, then these four $\mathbb{F}$-algebras are commutative
$U$-algebras (with the action of $U$ being componentwise). The case of
$U=\mathbb{F}$ is the one most frequently encountered.

However, $U\left[  \left[  z,z^{-1}\right]  \right]  $ does not become a
$\mathbb{F}$-algebra in this way, not even for $U=\mathbb{F}$. In fact,
attempting to compute $\left(  \sum_{i\in\mathbb{Z}}z^{i}\right)  \cdot\left(
\sum_{i\in\mathbb{Z}}z^{i}\right)  $ according to (\ref{multiplication-rule})
would lead to the result $\sum_{i\in\mathbb{Z}}\left(  \sum_{j\in\mathbb{Z}%
}1\cdot1\right)  z^{i}$, which makes no sense (as the inner sum $\sum
_{j\in\mathbb{Z}}1\cdot1$ diverges in any meaningful topology). This is
probably the reason why you rarely see $U\left[  \left[  z,z^{-1}\right]
\right]  $ studied in literature; its elements cannot be
multiplied\footnote{This is similar to the lack of a reasonable notion of
product of distributions in analysis.}. However, it is at least possible to
multiply elements of $U\left[  \left[  z,z^{-1}\right]  \right]  $ with
Laurent polynomials (i.e., elements of $U\left[  z,z^{-1}\right]  $); this
multiplication again is defined according to (\ref{multiplication-rule}), and
it is well-defined because of the \textquotedblleft all but finitely many
$i\in\mathbb{N}$ satisfy $u_{i}=0$\textquotedblright\ condition in the
definition of $U\left[  z,z^{-1}\right]  $. This multiplication makes
$U\left[  \left[  z,z^{-1}\right]  \right]  $ into a $U\left[  z,z^{-1}%
\right]  $-module. This module usually contains torsion, however: we have%
\begin{align*}
\left(  1-z\right)  \cdot\left(  \sum_{i\in\mathbb{Z}}z^{i}\right)   &
=\sum_{i\in\mathbb{Z}}z^{i}-\sum_{i\in\mathbb{Z}}\underbrace{z\cdot z^{i}%
}_{=z^{i+1}}=\sum_{i\in\mathbb{Z}}z^{i}-\underbrace{\sum_{i\in\mathbb{Z}%
}z^{i+1}}_{=\sum_{i\in\mathbb{Z}}z^{i}}\\
&  =\sum_{i\in\mathbb{Z}}z^{i}-\sum_{i\in\mathbb{Z}}z^{i}=0.
\end{align*}


\item More generally, if $V$ is a $\mathbb{F}$-algebra and $U$ is a
$V$-module, then $U\left[  z\right]  $ becomes a $V\left[  z\right]  $-module,
$U\left[  z,z^{-1}\right]  $ becomes a $V\left[  z,z^{-1}\right]  $-module,
$U\left[  \left[  z\right]  \right]  $ becomes a $V\left[  \left[  z\right]
\right]  $-module, $U\left(  \left(  z\right)  \right)  $ becomes a $V\left(
\left(  z\right)  \right)  $-module, and $U\left[  \left[  z,z^{-1}\right]
\right]  $ becomes a $V\left[  z,z^{-1}\right]  $-module. Again, the actions
are given by (\ref{multiplication-rule}). This is particularly useful in the
case when $V=\mathbb{F}$. This particular case shows that whenever $U$ is an
$\mathbb{F}$-module, the $\mathbb{F}$-module $U\left[  z\right]  $ becomes an
$\mathbb{F}\left[  z\right]  $-module, $U\left[  z,z^{-1}\right]  $ becomes an
$\mathbb{F}\left[  z,z^{-1}\right]  $-module, $U\left[  \left[  z\right]
\right]  $ becomes an $\mathbb{F}\left[  \left[  z\right]  \right]  $-module,
$U\left(  \left(  z\right)  \right)  $ becomes an $\mathbb{F}\left(  \left(
z\right)  \right)  $-module, and $U\left[  \left[  z,z^{-1}\right]  \right]  $
becomes an $\mathbb{F}\left[  z,z^{-1}\right]  $-module. In particular, all
five $\mathbb{F}$-modules $U\left[  z\right]  $, $U\left[  z,z^{-1}\right]  $,
$U\left[  \left[  z\right]  \right]  $, $U\left(  \left(  z\right)  \right)  $
and $U\left[  \left[  z,z^{-1}\right]  \right]  $ thus become $\mathbb{F}%
\left[  z\right]  $-modules.

\item The product of an element of $\mathbb{F}\left[  z,z^{-1}\right]  $ with
an element of $U\left[  \left[  z,z^{-1}\right]  \right]  $ is well-defined
and belongs to $U\left[  \left[  z,z^{-1}\right]  \right]  $ (since $U\left[
\left[  z,z^{-1}\right]  \right]  $ is an $\mathbb{F}\left[  z,z^{-1}\right]
$-module). The product of an element of $\mathbb{F}\left[  \left[
z,z^{-1}\right]  \right]  $ with an element of $U\left[  z,z^{-1}\right]  $ is
also well-defined, and also belongs to $U\left[  \left[  z,z^{-1}\right]
\right]  $; but this notion of product makes neither $U\left[  z,z^{-1}%
\right]  $ nor $U\left[  \left[  z,z^{-1}\right]  \right]  $ into an
$\mathbb{F}\left[  \left[  z,z^{-1}\right]  \right]  $-module (yet it is
useful nevertheless).

\item We defined $\mathbb{F}$-algebra structures on $U\left[  z\right]  $,
$U\left[  z,z^{-1}\right]  $, $U\left[  \left[  z\right]  \right]  $ and
$U\left(  \left(  z\right)  \right)  $ for any $\mathbb{F}$-algebra $U$. Here,
we only used the associativity of $U$ to ensure that these new $\mathbb{F}%
$-algebra structures are associative, and we only used the unity of $U$ to
construct a unity for these new $\mathbb{F}$-algebra structures. Thus, in the
same way, we can obtain nonassociative nonunital $\mathbb{F}$-algebra
structures on $U\left[  z\right]  $, $U\left[  z,z^{-1}\right]  $, $U\left[
\left[  z\right]  \right]  $ and $U\left(  \left(  z\right)  \right)  $
whenever $U$ is a nonassociative nonunital $\mathbb{F}$-algebra. In
particular, this construction works for Lie algebras: If $\mathfrak{g}$ is a
$\mathbb{F}$-Lie algebra, then $\mathfrak{g}\left[  z\right]  $,
$\mathfrak{g}\left[  z,z^{-1}\right]  $, $\mathfrak{g}\left[  \left[
z\right]  \right]  $ and $\mathfrak{g}\left(  \left(  z\right)  \right)  $
become $\mathbb{F}$-Lie algebras, with Lie bracket defined by the rule%
\[
\left[  \sum_{i}u_{i}z^{i},\sum_{i}v_{i}z^{i}\right]  =\sum_{i}\left(
\sum_{j}\left[  u_{j},v_{i-j}\right]  \right)  z^{i}.
\]
\footnote{\textbf{Exercise:} Check that $\mathfrak{g}\left[  z\right]  $,
$\mathfrak{g}\left[  z,z^{-1}\right]  $, $\mathfrak{g}\left[  \left[
z\right]  \right]  $ and $\mathfrak{g}\left(  \left(  z\right)  \right)  $
actually become $\mathbb{F}$-Lie algebras using this definition. (This is not
completely obvious, because the $\left[  a,a\right]  =0$ axiom does not
directly get inherited from $\mathfrak{g}$. But it is still easy to check.)}

\item Given any $\mathbb{Z}$-grading on an $\mathbb{F}$-module $U$ and any
integer $d$, we can define a $\mathbb{Z}$-grading on $U\left[  z\right]  $ by
giving each $uz^{i}$ (for $u\in U$ homogeneous and $i\in\mathbb{N}$) the
degree $\deg u+id$. Similarly, we can define a $\mathbb{Z}$-grading on
$U\left[  z,z^{-1}\right]  $ (but not on $U\left[  \left[  z\right]  \right]
$, $U\left(  \left(  z\right)  \right)  $ or $U\left[  \left[  z,z^{-1}%
\right]  \right]  $, unless we content ourselves with an \textquotedblleft
almost-grading\textquotedblright\ \footnote{By an ``almost-grading'' of a
topological $\mathbb{F}$-module $P$, I mean a family $\left(  P_{n}\right)
_{n\in\mathbb{Z}}$ of $\mathbb{F}$-submodules $P_{n} \subseteq P$ such that
the internal direct sum $\bigoplus\limits_{n\in\mathbb{Z}} P_{n}$ is
well-defined and is dense in $P$. Such almost-gradings exist on $U\left[
\left[  z\right]  \right]  $, $U\left(  \left(  z\right)  \right)  $ and
$U\left[  \left[  z,z^{-1}\right]  \right]  $.}). These gradings turn
$U\left[  z\right]  $ and $U\left[  z,z^{-1}\right]  $ into graded
$\mathbb{F}$-algebras. When the $\mathbb{Z}$-grading on $U$ is trivial (i.e.,
everything in $U$ is homogeneous of degree $0$) and $d=0$, these gradings are
the \textquotedblleft grading by degree\textquotedblright\ (i.e., each
$uz^{i}$ has degree $i$).

\item The $\mathbb{F}$-modules $U\left[  z\right]  $, $U\left[  z,z^{-1}%
\right]  $, $U\left[  \left[  z\right]  \right]  $, $U\left(  \left(
z\right)  \right)  $ and $U\left[  \left[  z,z^{-1}\right]  \right]  $ are
automatically endowed with topologies, which are defined as follows: Endow
$U^{\mathbb{Z}}$ (a direct product of infinitely many copies of $U$) with the
direct-product topology (where each copy of $U$ is given the discrete
topology), and pull back this topology onto $U\left[  \left[  z,z^{-1}\right]
\right]  $ via the $\mathbb{F}$-module isomorphism $U\left[  \left[
z,z^{-1}\right]  \right]  \rightarrow U^{\mathbb{Z}},\ \sum_{i\in\mathbb{Z}%
}u_{i}z^{i}\mapsto\left(  u_{i}\right)  _{i\in\mathbb{Z}}$. This defines a
topology on $U\left[  \left[  z,z^{-1}\right]  \right]  $. Topologies on the
four $\mathbb{F}$-modules $U\left[  z\right]  $, $U\left[  z,z^{-1}\right]  $,
$U\left[  \left[  z\right]  \right]  $ and $U\left(  \left(  z\right)
\right)  $ are defined by restricting this topology (since these four
$\mathbb{F}$-modules are $\mathbb{F}$-submodules of $U\left[  \left[
z,z^{-1}\right]  \right]  $). These topologies are called \textquotedblleft
topologies of coefficientwise convergence\textquotedblright, due to the
following fact (which we state for $U\left[  \left[  z,z^{-1}\right]  \right]
$ as an example): A sequence $\left(  u_{\left(  n\right)  }\right)
_{u\in\mathbb{N}}$ of formal distributions $u_{\left(  n\right)  }\in U\left[
\left[  z,z^{-1}\right]  \right]  $ converges to a formal distribution $u\in
U\left[  \left[  z,z^{-1}\right]  \right]  $ with respect to this topology if
and only if for every $i\in\mathbb{Z}$, we have%
\begin{align*}
\left(  \text{the }i\text{-th coefficient of }u_{\left(  n\right)  }\right)
&  =\left(  \text{the }i\text{-th coefficient of }u\right) \\
&  \ \ \ \ \ \ \ \ \ \ \text{for all sufficiently high }n\in\mathbb{N}.
\end{align*}


The topologies that we introduced are Hausdorff and respect the $\mathbb{F}%
$-module structures (i.e., they turn our $\mathbb{F}$-modules into topological
$\mathbb{F}$-modules). They also respect multiplication when $U$ is an
$\mathbb{F}$-algebra\footnote{Here is what we mean by this: If $U$ is an
$\mathbb{F}$-algebra, then the multiplication maps of the $\mathbb{F}%
$-algebras $U\left[  z\right]  $, $U\left[  z,z^{-1}\right]  $, $U\left[
\left[  z\right]  \right]  $ and $U\left(  \left(  z\right)  \right)  $ are
continuous with respect to these topologies, and so are the maps $U\left[
z,z^{-1}\right]  \times U\left[  \left[  z,z^{-1}\right]  \right]  \rightarrow
U\left[  \left[  z,z^{-1}\right]  \right]  $ and $U\left[  \left[
z,z^{-1}\right]  \right]  \times U\left[  z,z^{-1}\right]  \rightarrow
U\left[  \left[  z,z^{-1}\right]  \right]  $ that send every $\left(
f,g\right)  $ to $fg$.}. Furthermore, the set $U\left[  z\right]  $ is dense
$U\left[  \left[  z\right]  \right]  $, and the set $U\left[  z,z^{-1}\right]
$ is dense in each of $U\left(  \left(  z\right)  \right)  $ and $U\left[
\left[  z,z^{-1}\right]  \right]  $. This ensures that, when we are proving
certain kinds of identities in $U\left[  \left[  z\right]  \right]  $,
$U\left(  \left(  z\right)  \right)  $ or $U\left[  \left[  z,z^{-1}\right]
\right]  $ (namely, the kind where both sides depend continuously on the
inputs), we can WLOG assume that the inputs are polynomials (for $U\left[
\left[  z\right]  \right]  $) resp. Laurent polynomials (for $U\left(  \left(
z\right)  \right)  $ and for $U\left[  \left[  z,z^{-1}\right]  \right]  $).
\end{itemize}

\subsection{\label{sect.polynomials.multivar}Polynomial-like objects in
multiple variables}

We have so far studied the case of a single variable $z$; but it is possible
to define similar structures in multiple variables. Let us briefly sketch how
they are defined. Let $U$ be a $\mathbb{F}$-module, and let $\left(
x_{j}\right)  _{j\in J}$ be a family of symbols. We briefly denote the family
$\left(  x_{j}\right)  _{j\in J}$ as $\mathbf{x}$.

\begin{itemize}
\item The $\mathbb{F}$-module $U\left[  \mathbf{x}\right]  =U\left[
x_{j}\ \mid\ j\in J\right]  $ consists of all families $\left(  u_{\mathbf{i}%
}\right)  _{\mathbf{i}\in\mathbb{N}^{J}}\in U^{\mathbb{N}^{J}}$ such that all
but finitely many $\mathbf{i}\in\mathbb{N}^{J}$ satisfy $u_{\mathbf{i}}=0$. We
write such families $\left(  u_{\mathbf{i}}\right)  _{\mathbf{i}\in
\mathbb{N}^{J}}$ in the form $\sum_{\mathbf{i}\in\mathbb{N}^{J}}u_{\mathbf{i}%
}\mathbf{x}^{\mathbf{i}}$, and refer to them as \textquotedblleft
polynomials\textquotedblright\ in the $x_{j}$ with coefficients in $U$.

\item The $\mathbb{F}$-module $U\left[  \mathbf{x},\mathbf{x}^{-1}\right]
=U\left[  x_{j},x_{j}^{-1}\ \mid\ j\in J\right]  $ consists of all families
$\left(  u_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbb{Z}^{J}}\in
U^{\mathbb{Z}^{J}}$ such that all but finitely many $\mathbf{i}\in
\mathbb{Z}^{J}$ satisfy $u_{\mathbf{i}}=0$. We write such families $\left(
u_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbb{Z}^{J}}$ in the form
$\sum_{\mathbf{i}\in\mathbb{Z}^{J}}u_{\mathbf{i}}\mathbf{x}^{\mathbf{i}}$, and
refer to them as \textquotedblleft Laurent polynomials\textquotedblright\ in
the $x_{j}$ with coefficients in $U$.

\item The $\mathbb{F}$-module $U\left[  \left[  \mathbf{x}\right]  \right]
=U\left[  \left[  x_{j}\ \mid\ j\in J\right]  \right]  $ consists of all
families $\left(  u_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbb{N}^{J}}\in
U^{\mathbb{N}^{J}}$. We write such families $\left(  u_{\mathbf{i}}\right)
_{\mathbf{i}\in\mathbb{N}^{J}}$ in the form $\sum_{\mathbf{i}\in\mathbb{N}%
^{J}}u_{\mathbf{i}}\mathbf{x}^{\mathbf{i}}$, and refer to them as
\textquotedblleft formal power series\textquotedblright\ in the $x_{j}$ with
coefficients in $U$.

\item The $\mathbb{F}$-module $U\left(  \left(  \mathbf{x}\right)  \right)
=U\left(  \left(  x_{j}\ \mid\ j\in J\right)  \right)  $ consists of all
families $\left(  u_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbb{Z}^{J}}\in
U^{\mathbb{Z}^{J}}$ for which there exists an $N\in\mathbb{Z}$ such that all
$\mathbf{i}\in\mathbb{Z}^{J}\setminus\left\{  N,N+1,N+2,\ldots\right\}  ^{J}$
satisfy $u_{\mathbf{i}}=0$. We write such families $\left(  u_{\mathbf{i}%
}\right)  _{\mathbf{i}\in\mathbb{Z}^{J}}$ in the form $\sum_{\mathbf{i}%
\in\mathbb{Z}^{J}}u_{\mathbf{i}}\mathbf{x}^{\mathbf{i}}$, and refer to them as
\textquotedblleft Laurent series\textquotedblright\ in the $x_{j}$ with
coefficients in $U$. (This is the only definition you might not have
immediately guessed.)

\item The $\mathbb{F}$-module $U\left[  \left[  \mathbf{x},\mathbf{x}%
^{-1}\right]  \right]  =U\left[  \left[  x_{j},x_{j}^{-1}\ \mid\ j\in
J\right]  \right]  $ consists of all families $\left(  u_{\mathbf{i}}\right)
_{\mathbf{i}\in\mathbb{Z}^{J}}\in U^{\mathbb{Z}^{J}}$. We write such families
$\left(  u_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbb{Z}^{J}}$ in the form
$\sum_{\mathbf{i}\in\mathbb{Z}^{J}}u_{\mathbf{i}}\mathbf{x}^{\mathbf{i}}$, and
refer to them as \textquotedblleft$U$-valued formal
distributions\textquotedblright\ in the $x_{j}$ with coefficients in $U$.
\end{itemize}

Again, the first four of these five $\mathbb{F}$-modules become $\mathbb{F}%
$-algebras when $U$ itself is a $\mathbb{F}$-algebra. The multiplication rule,
like (\ref{multiplication-rule}), is \textquotedblleft what you would
expect\textquotedblright\ if you are told that $\mathbf{x}^{\mathbf{i}}$
stands for the monomial $\prod_{j\in J}x_{j}^{i_{j}}$ (where $\mathbf{i}%
=\left(  i_{j}\right)  _{j\in J}$). Explicitly:%
\[
\left(  \sum_{\mathbf{i}}u_{\mathbf{i}}\mathbf{x}^{\mathbf{i}}\right)
\cdot\left(  \sum_{\mathbf{i}}v_{\mathbf{i}}\mathbf{x}^{\mathbf{i}}\right)
=\sum_{\mathbf{i}}\left(  \sum_{\mathbf{j}}u_{\mathbf{j}}v_{\mathbf{i}%
-\mathbf{j}}\right)  \mathbf{x}^{\mathbf{i}},
\]
where the subtraction on $\mathbb{N}^{J}$ and on $\mathbb{Z}^{J}$ is
componentwise. Again, the same formula can be used to make $U\left[  \left[
\mathbf{x},\mathbf{x}^{-1}\right]  \right]  $ into a $U\left[  \mathbf{x}%
,\mathbf{x}^{-1}\right]  $-module.

Again, $U\left[  \mathbf{x},\mathbf{x}^{-1}\right]  $ and $U\left[  \left[
\mathbf{x},\mathbf{x}^{-1}\right]  \right]  $ are often called $U\left[
\mathbf{x}^{\pm1}\right]  $ and $U\left[  \left[  \mathbf{x}^{\pm1}\right]
\right]  $, respectively.

We will often be working with polynomials and power series (and similar
objects) in two variables. For instance, $U\left[  z,w\right]  $ (with $z$ and
$w$ being two symbols) means $U\left[  \mathbf{x}\right]  $, where
$\mathbf{x}$ is the two-element family $\left(  z,w\right)  $. Similarly,
$U\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $ means $U\left[  \left[
\mathbf{x},\mathbf{x}^{-1}\right]  \right]  $ for the same family $\mathbf{x}%
$. Explicitly,%
\[
U\left[  z,w\right]  =\left\{  \sum_{\left(  i,j\right)  \in\mathbb{N}^{2}%
}u_{\left(  i,j\right)  }z^{i}w^{j}\ \mid\ \text{all but finitely many
}\left(  i,j\right)  \in\mathbb{N}^{2}\text{ satisfy }u_{\left(  i,j\right)
}=0\right\}
\]
and%
\[
U\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  =\left\{  \sum_{\left(
i,j\right)  \in\mathbb{Z}^{2}}u_{\left(  i,j\right)  }z^{i}w^{j}\right\}
\]
(where the $u_{\left(  i,j\right)  }$ are supposed to live in $U$ both times).
Similarly,
\begin{align*}
&  U\left[  z,z^{-1},w,w^{-1}\right] \\
&  =\left\{  \sum_{\left(  i,j\right)  \in\mathbb{Z}^{2}}u_{\left(
i,j\right)  }z^{i}w^{j}\ \mid\ \text{all but finitely many }\left(
i,j\right)  \in\mathbb{Z}^{2}\text{ satisfy }u_{\left(  i,j\right)
}=0\right\}  ,
\end{align*}%
\[
U\left[  \left[  z,w\right]  \right]  =\left\{  \sum_{\left(  i,j\right)
\in\mathbb{N}^{2}}u_{\left(  i,j\right)  }z^{i}w^{j}\right\}
\]
and
\begin{align}
&  U\left(  \left(  z,w\right)  \right) \nonumber\\
&  =\left\{  \sum_{\left(  i,j\right)  \in\mathbb{Z}^{2}}u_{\left(
i,j\right)  }z^{i}w^{j}\ \mid\ \text{there exists an }N\in\mathbb{Z}\text{
such that}\right. \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.
\phantom{\sum_{\left( i,j\right) \in\mathbb{Z}^{2}}}\text{all }\left(
i,j\right)  \in\mathbb{Z}^{2}\setminus\left\{  N,N+1,N+2,\ldots\right\}
^{2}\text{ satisfy }u_{\left(  i,j\right)  }=0\right\}  .
\label{eq.polynomials.multivar.laurent}%
\end{align}


There are obvious isomorphisms $U\left[  z,w\right]  \cong\left(  U\left[
z\right]  \right)  \left[  w\right]  \cong\left(  U\left[  w\right]  \right)
\left[  z\right]  $ which preserve the $\mathbb{F}$-module structure and, if
$U$ is an $\mathbb{F}$-algebra, also the $\mathbb{F}$-algebra structure.
Similarly, there are obvious isomorphisms $U\left[  \left[  z,w\right]
\right]  \cong\left(  U\left[  \left[  z\right]  \right]  \right)  \left[
\left[  w\right]  \right]  \cong\left(  U\left[  \left[  w\right]  \right]
\right)  \left[  \left[  z\right]  \right]  $ which also preserve said
structures.\footnote{\textbf{Caveat:} These isomorphisms do not preserve the
topology! The sequence $\left(  \left(  z+w\right)  ^{n}\right)
_{n\in\mathbb{N}}$ converges to $0$ in the topology on $U\left[  z,w\right]
$, but not in the topology on $\left(  U\left[  z\right]  \right)  \left[
w\right]  $ (unless the latter topology is defined more subtly).} Similar
isomorphisms exist for Laurent polynomials and for formal distributions, but
not for Laurent series.

Sometimes we will also study \textquotedblleft intermediate\textquotedblright%
\ $\mathbb{F}$-modules such as $U\left[  \left[  z,z^{-1},w\right]  \right]  $
(intermediate between $U\left[  \left[  z,z^{-1}\right]  \right]  $ and
$U\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $); as you would expect,
it is defined by%
\[
U\left[  \left[  z,z^{-1},w\right]  \right]  =\left\{  \sum_{\left(
i,j\right)  \in\mathbb{Z}\times\mathbb{N}}u_{\left(  i,j\right)  }z^{i}%
w^{j}\right\}  .
\]
(It can also be viewed as $\left(  U\left[  \left[  z,z^{-1}\right]  \right]
\right)  \left[  \left[  w\right]  \right]  $, or as $\left(  U\left[  \left[
w\right]  \right]  \right)  \left[  \left[  z,z^{-1}\right]  \right]  $, both
times by canonical isomorphism.) Some authors write $U\left[  \left[  z^{\pm
1},w\right]  \right]  $ for $U\left[  \left[  z,z^{-1},w\right]  \right]  $.

As we said, $U\left[  \left[  z,z^{-1}\right]  \right]  $ is usually not a
$\mathbb{F}$-algebra even if $U$ is a $\mathbb{F}$-algebra. However, products
of \textquotedblleft formal distributions\textquotedblright\ can be
well-defined in various specific cases for particular reasons. Most
importantly, two formal distributions in distinct variables can be multiplied:
for example, if $U$ is a $\mathbb{F}$-algebra, then we can multiply
$\sum_{i\in\mathbb{Z}}u_{i}z^{i}\in U\left[  \left[  z,z^{-1}\right]  \right]
$ with $\sum_{i\in\mathbb{Z}}v_{i}w^{i}\in U\left[  \left[  w,w^{-1}\right]
\right]  $ to obtain%
\[
\left(  \sum_{i\in\mathbb{Z}}u_{i}z^{i}\right)  \cdot\left(  \sum
_{i\in\mathbb{Z}}v_{i}w^{i}\right)  =\sum_{\left(  i,j\right)  \in
\mathbb{Z}^{2}}u_{i}v_{j}z^{i}w^{j}\in U\left[  \left[  z,z^{-1}%
,w,w^{-1}\right]  \right]  .
\]


Later (in the proof of Proposition \ref{exe.3.2} \textbf{(a)}) we will
encounter a different case in which two formal distributions can be
multiplied\footnote{Namely, we will see a situation in which one power series
belongs to $U\left[  \left[  z,z^{-1}\right]  \right]  $, while the other
belongs to $\left(  U\left[  z,z^{-1}\right]  \right)  \left[  \left[
w,w^{-1}\right]  \right]  $ (which is embedded in $U\left[  \left[
z,z^{-1},w,w^{-1}\right]  \right]  $ via the identification of $U\left[
\left[  z,z^{-1},w,w^{-1}\right]  \right]  $ with $\left(  U\left[  \left[
z,z^{-1}\right]  \right]  \right)  \left[  \left[  w,w^{-1}\right]  \right]
$). The reader can come up with some even more general situations in which
multiplication is possible.}.

\subsection{Substitution}

In many cases, it is also possible to substitute things for variables into
polynomial-like objects, although, the less \textquotedblleft
tame\textquotedblright\ the objects are, the more we need to require of what
we substitute into them. Here are some examples:

\begin{itemize}
\item If $U$ is a commutative $\mathbb{F}$-algebra, and if $f\in U\left[
z\right]  $, then any element of a $U$-algebra can be substituted for $z$ in
$f$.

\item If $U$ is a commutative $\mathbb{F}$-algebra, and if $f\in U\left[
z,z^{-1}\right]  $, then any invertible element of a $U$-algebra can be
substituted for $z$ in $f$.

\item If $U$ is a commutative $\mathbb{F}$-algebra, and if $f\in U\left[
\left[  z\right]  \right]  $, and if $I$ is an ideal of a topological
$U$-algebra $P$ such that the $I$-adic topology on $P$ is complete, then any
element of $I$ can be substituted for $z$ in $f$. In particular, this shows
that any nilpotent element of a $U$-algebra can be substituted for $z$ in $f$,
and also that any power series with zero constant term can be substituted for
$z$ in $f$.

\item If $U$ is a commutative $\mathbb{F}$-algebra, and if $f\in U\left(
\left(  z\right)  \right)  $, then we can substitute any positive power of $z$
or of any other formal indeterminate for $z$ in $f$.

\item If $U$ is a commutative $\mathbb{F}$-algebra, and if $f\in U\left[
\left[  z,z^{-1}\right]  \right]  $, then we can substitute any nonzero power
of $z$ or of any other formal indeterminate for $z$ in $f$.
\end{itemize}

As usual, the result of substituting any object $p$ for $z$ in $f$ will be
denoted by $f\left(  p\right)  $.

In the case of multiple variables, some of these rules need additional
conditions. For example, we can only substitute \textit{commuting} elements of
a $U$-algebra for the variables in a polynomial. We can substitute two
distinct indeterminates for $z$ and $w$ in an element $f\in U\left[  \left[
z,z^{-1},w,w^{-1}\right]  \right]  $, but not two equal indeterminates unless
$f$ has special properties. In particular, if $f\in U\left[  \left[
z,z^{-1},w,w^{-1}\right]  \right]  $, then $f\left(  z,w\right)  $ and
$f\left(  w,z\right)  $ are well-defined (of course, $f\left(  z,w\right)
=f$), but $f\left(  z,z\right)  $ (in general) is not.

\section{\label{chap.deri}Derivatives}

\subsection{\label{sect.deriv}Derivatives}

Let us now define derivative operators on polynomial-like objects. We shall be
brief, as we assume that the reader knows how to (formally -- i.e., without
recourse to analysis) define derivatives of polynomials, and much of the
theory will be analogous for the other objects.\footnote{Our treatment of
derivatives will, however, be complicated by the fact that we are allowing
$\mathbb{F}$ to be an arbitrary commutative ring, not necessarily a field of
characteristic $0$.}

First, let $U$ be a $\mathbb{F}$-module, and $z$ a symbol. Then, we can define
an endomorphism $\partial_{z}$ of the $\mathbb{F}$-module $U\left[  z\right]
$ by setting%
\[
\partial_{z}\left(  \sum_{i\in\mathbb{N}}u_{i}z^{i}\right)  =\sum
_{i\in\mathbb{N}}iu_{i}z^{i-1}%
\]
for every $\sum_{i\in\mathbb{N}}u_{i}z^{i}\in U\left[  z\right]  $ (with
$u_{i}\in U$)\ \ \ \ \footnote{Let us see why this is well-defined. Fix
$\sum_{i\in\mathbb{N}}u_{i}z^{i}\in U\left[  z\right]  $ (with $u_{i}\in U$).
It is clear that $\sum_{i\in\mathbb{N}}iu_{i}z^{i-1}\in U\left[
z,z^{-1}\right]  $ (since all but finitely many $i\in\mathbb{N}$ satisfy
$u_{i}=0$), but we need to prove that $\sum_{i\in\mathbb{N}}iu_{i}z^{i-1}\in
U\left[  z\right]  $. For this it is clearly sufficient to show that
$iu_{i}z^{i-1}\in U\left[  z\right]  $ for every $i\in\mathbb{N}$. So let
$i\in\mathbb{N}$. We need to show that $iu_{i}z^{i-1}\in U\left[  z\right]  $.
But:
\par
\begin{itemize}
\item for $i\geq1$, this is obvious (because for $i\geq1$, we have
$i-1\in\mathbb{N}$).
\par
\item for $i<1$, this follows from $\underbrace{i}_{\substack{=0\\\text{(since
}i<1\text{)}}}u_{i}z^{i-1}=0$.
\end{itemize}
\par
Thus, $iu_{i}z^{i-1}\in U\left[  z\right]  $ is proven, qed.}. This
endomorphism $\partial_{z}$ is often denoted by $\dfrac{\partial}{\partial z}%
$. Analogous endomorphisms $\partial_{z}$ of $U\left[  z,z^{-1}\right]  $, of
$U\left[  \left[  z\right]  \right]  $, of $U\left(  \left(  z\right)
\right)  $ and of $U\left[  \left[  z,z^{-1}\right]  \right]  $ are defined in
the same way (except that $\mathbb{N}$ is replaced by $\mathbb{Z}$ when we are
considering $U\left[  z,z^{-1}\right]  $, $U\left(  \left(  z\right)  \right)
$ and $U\left[  \left[  z,z^{-1}\right]  \right]  $). These endomorphisms are
all continuous; they are furthermore $U$-linear derivations when $U$ is a
commutative $\mathbb{F}$-algebra.\footnote{In the case of $U\left[  \left[
z,z^{-1}\right]  \right]  $, this means that $\partial_{z}$ is $U$-linear and
satisfies $\partial_{z}\left(  fg\right)  =\left(  \partial_{z}f\right)
g+f\left(  \partial_{z}g\right)  $ for any $f\in U\left[  z,z^{-1}\right]  $
and $g\in U\left[  \left[  z,z^{-1}\right]  \right]  $.} Finally, these
endomorphisms $\partial_{z}$ form commutative diagrams with each other --
e.g., the restriction of the endomorphism $\partial_{z}$ of $U\left[
z,z^{-1}\right]  $ to $U\left[  z\right]  $ is precisely the endomorphism
$\partial_{z}$ of $U\left[  z\right]  $.

When there are multiple variables $x_{j}$ (for $j\in J$) instead of a single
variable $z$, an endomorphism $\partial_{x_{j}}$ is defined for each of
them\footnote{The reader probably knows these endomorphisms on polynomial
rings, and will not be surprised that they behave similarly for other
polynomial-like objects.}; it is given by%
\[
\partial_{x_{j}}\left(  \sum_{\mathbf{i}}u_{\mathbf{i}}\mathbf{x}^{\mathbf{i}%
}\right)  =\sum_{\mathbf{i}}\mathbf{i}_{j}u_{\mathbf{i}-\mathbf{e}_{j}%
}\mathbf{x}^{\mathbf{i}},
\]
where we are using the following notations: $\mathbf{i}_{j}$ denotes the
$j$-th entry of the family $\mathbf{i}$, and $\mathbf{e}_{j}$ denotes the
family $\left(  \delta_{k,j}\right)  _{k\in J}$ (and the subtraction in
$\mathbf{i}-\mathbf{e}_{j}$ is componentwise). The endomorphisms
$\partial_{x_{j}}$ for different $j$ commute.

\subsection{Hasse-Schmidt derivatives $\partial_{z}^{\left(  n\right)  }$}

We can also generalize the endomorphisms $\partial_{z}$ of $U\left[  z\right]
$, $U\left[  z,z^{-1}\right]  $, $U\left[  \left[  z\right]  \right]  $ and
$U\left(  \left(  z\right)  \right)  $ in another direction:

\begin{definition}
Let $n\in\mathbb{N}$. Let $U$ be an $\mathbb{F}$-module. We define an
endomorphism $\partial_{z}^{\left(  n\right)  }$ of the $\mathbb{F}$-module
$U\left[  z\right]  $ by setting
\[
\partial_{z}^{\left(  n\right)  }\left(  \sum_{i\in\mathbb{N}}u_{i}%
z^{i}\right)  =\sum_{i\in\mathbb{N}}\dbinom{i}{n}u_{i}z^{i-n}%
\]
for every $\sum_{i\in\mathbb{N}}u_{i}z^{i}\in U\left[  z\right]  $ (with
$u_{i}\in U$)\ \ \ \ \footnotemark. Again, we can define endomorphisms
$\partial_{z}^{\left(  n\right)  }$ of $U\left[  z,z^{-1}\right]  $, of
$U\left[  \left[  z\right]  \right]  $, of $U\left(  \left(  z\right)
\right)  $ and of $U\left[  \left[  z,z^{-1}\right]  \right]  $ according to
the same rule.
\end{definition}

\footnotetext{Let us see why this is well-defined. Fix $\sum_{i\in\mathbb{N}%
}u_{i}z^{i}\in U\left[  z\right]  $ (with $u_{i}\in U$). It is clear that
$\sum_{i\in\mathbb{N}}\dbinom{i}{n}u_{i}z^{i-n}\in U\left[  z,z^{-1}\right]  $
(since all but finitely many $i\in\mathbb{N}$ satisfy $u_{i}=0$), but we need
to prove that $\sum_{i\in\mathbb{N}}\dbinom{i}{n}u_{i}z^{i-n}\in U\left[
z\right]  $. For this it is clearly sufficient to show that $\dbinom{i}%
{n}u_{i}z^{i-n}\in U\left[  z\right]  $ for every $i\in\mathbb{N}$. So let
$i\in\mathbb{N}$. We need to show that $\dbinom{i}{n}u_{i}z^{i-n}\in U\left[
z\right]  $. But:
\par
\begin{itemize}
\item for $i\geq n$, this is obvious (because for $i\geq n$, we have
$i-n\in\mathbb{N}$).
\par
\item for $i<n$, this follows from $\underbrace{\dbinom{i}{n}}%
_{\substack{=0\\\text{(since }i<n\text{ and }i\in\mathbb{N}\text{)}}%
}u_{i}z^{i-n}=0$.
\end{itemize}
\par
Thus, $\dbinom{i}{n}u_{i}z^{i-n}\in U\left[  z\right]  $ is proven, qed.}These
endomorphisms $\partial_{n}^{\left(  z\right)  }$ form commutative diagrams
with each other -- e.g., the restriction of the endomorphism $\partial
_{z}^{\left(  n\right)  }$ of $U\left[  z,z^{-1}\right]  $ to $U\left[
z\right]  $ is precisely the endomorphism $\partial_{z}^{\left(  n\right)  }$
of $U\left[  z\right]  $. Moreover, they satisfy the following properties:

\begin{proposition}
\label{prop.delta.deriv}Let $U$ be an $\mathbb{F}$-module. Let $A$ be any of
the five $\mathbb{F}$-modules $U\left[  z\right]  $, $U\left[  z,z^{-1}%
\right]  $, $U\left[  \left[  z\right]  \right]  $, $U\left(  \left(
z\right)  \right)  $ and $U\left[  \left[  z,z^{-1}\right]  \right]  $. Then,%
\begin{align}
\partial_{z}^{\left(  0\right)  }  &  =\operatorname*{id}%
;\label{eq.prop.delta.deriv.0}\\
\partial_{z}^{\left(  1\right)  }  &  =\partial_{z}%
;\label{eq.prop.delta.deriv.1}\\
n!\partial_{z}^{\left(  n\right)  }  &  =\left(  \partial_{z}\right)
^{n}\ \ \ \ \ \ \ \ \ \ \text{for all }n\in\mathbb{N}%
;\label{eq.prop.delta.deriv.n}\\
\partial_{z}^{\left(  n\right)  }\circ\partial_{z}^{\left(  m\right)  }  &
=\dbinom{n+m}{n}\partial_{z}^{\left(  n+m\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{for all }n\in\mathbb{N}\text{ and }m\in\mathbb{N}.
\label{eq.prop.delta.deriv.nm}%
\end{align}

\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.delta.deriv}.]We WLOG assume that $A=U\left[
\left[  z,z^{-1}\right]  \right]  $ (since the other four $\mathbb{F}$-modules
$U\left[  z\right]  $, $U\left[  z,z^{-1}\right]  $, $U\left[  \left[
z\right]  \right]  $ and $U\left(  \left(  z\right)  \right)  $ are
$\mathbb{F}$-submodules of $U\left[  \left[  z,z^{-1}\right]  \right]  $, and
their respective endomorphisms $\partial_{z}^{\left(  n\right)  }$ are
restrictions of the endomorphism $\partial_{z}^{\left(  n\right)  }$ of
$U\left[  \left[  z,z^{-1}\right]  \right]  $).

For every $\sum_{i\in\mathbb{N}}u_{i}z^{i}\in A$ (with all $u_{i}\in U$), we
have%
\begin{align*}
\partial_{z}^{\left(  0\right)  }\left(  \sum_{i\in\mathbb{N}}u_{i}%
z^{i}\right)   &  =\sum_{i\in\mathbb{N}}\underbrace{\dbinom{i}{0}}_{=1}%
u_{i}\underbrace{z^{i-0}}_{=z^{i}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\partial_{z}^{\left(  0\right)  }\right) \\
&  =\sum_{i\in\mathbb{N}}u_{i}z^{i}=\operatorname*{id}\left(  \sum
_{i\in\mathbb{N}}u_{i}z^{i}\right)  .
\end{align*}
Hence, $\partial_{z}^{\left(  0\right)  }=\operatorname*{id}$. A similar
argument shows that $\partial_{z}^{\left(  1\right)  }=\partial_{z}$ (here we
use the fact that $\dbinom{i}{1}=i$ for each $i\in\mathbb{N}$).

Let us now prove (\ref{eq.prop.delta.deriv.nm}). Indeed, let $n\in\mathbb{N}$
and $m\in\mathbb{N}$. We need to prove that $\partial_{z}^{\left(  n\right)
}\circ\partial_{z}^{\left(  m\right)  }=\dbinom{n+m}{n}\partial_{z}^{\left(
n+m\right)  }$. It is clearly enough to show that
\begin{equation}
\left(  \partial_{z}^{\left(  n\right)  }\circ\partial_{z}^{\left(  m\right)
}\right)  \left(  a\right)  =\dbinom{n+m}{n}\partial_{z}^{\left(  n+m\right)
}\left(  a\right)  \label{pf.prop.delta.deriv.goal4}%
\end{equation}
for every $a\in A$.

\textit{Proof of (\ref{pf.prop.delta.deriv.goal4}):} Let $a\in A$. We need to
prove the equality (\ref{pf.prop.delta.deriv.goal4}). Both sides of this
equality are continuous with respect to $a$ (where \textquotedblleft
continuous\textquotedblright\ is to be understood with respect to the topology
we introduced on $A$). Hence, we can WLOG assume that $a\in U\left[
z,z^{-1}\right]  $ (since $U\left[  z,z^{-1}\right]  $ is dense in $U\left[
\left[  z,z^{-1}\right]  \right]  =A$). Assume this. Now, both sides of the
equality (\ref{pf.prop.delta.deriv.goal4}) are $\mathbb{F}$-linear in $a$.
Hence (and since $a\in U\left[  z,z^{-1}\right]  $), we can WLOG assume that
$a$ belongs to the family $\left(  uz^{q}\right)  _{u\in U;\ q\in\mathbb{Z}}$
(since the family $\left(  uz^{q}\right)  _{u\in U;\ q\in\mathbb{Z}}$ spans
the $\mathbb{F}$-module $U\left[  z,z^{-1}\right]  $). Assume this. Then,
$a=uz^{p}$ for some $u\in U$ and $p\in\mathbb{Z}$. Consider these $u$ and $p$.
Now, the definition of $\partial_{z}^{\left(  n\right)  }$ yields%
\begin{align*}
\partial_{z}^{\left(  m\right)  }\left(  uz^{p}\right)   &  =\dbinom{p}%
{m}uz^{p-m};\ \ \ \ \ \ \ \ \ \ \partial_{z}^{\left(  n\right)  }\left(
z^{p-m}\right)  =\dbinom{p-m}{n}uz^{\left(  p-m\right)  -n};\\
\partial_{z}^{\left(  n+m\right)  }\left(  uz^{p}\right)   &  =\dbinom{p}%
{n+m}uz^{p-\left(  n+m\right)  }.
\end{align*}
But it is easy to see that $\dbinom{p}{m}\dbinom{p-m}{n}=\dbinom{n+m}%
{n}\dbinom{p}{n+m}$\ \ \ \ \footnote{\textit{Proof.} The definition of
$\dbinom{p}{m}$ yields
\begin{equation}
\dbinom{p}{m}=\dfrac{p\left(  p-1\right)  \cdots\left(  p-m+1\right)  }{m!}.
\label{pf.prop.delta.deriv.goal4.pf.3}%
\end{equation}
The definition of $\dbinom{p-m}{n}$ yields
\begin{align*}
\dbinom{p-m}{n}  &  =\dfrac{\left(  p-m\right)  \left(  \left(  p-m\right)
-1\right)  \cdots\left(  \left(  p-m\right)  -n+1\right)  }{n!}\\
&  =\dfrac{\left(  p-m\right)  \left(  p-m-1\right)  \cdots\left(
p-m-n+1\right)  }{n!}.
\end{align*}
Multiplying (\ref{pf.prop.delta.deriv.goal4.pf.3}) with this equality, we
obtain%
\begin{align*}
\dbinom{p}{m}\dbinom{p-m}{n}  &  =\dfrac{p\left(  p-1\right)  \cdots\left(
p-m+1\right)  }{m!}\cdot\dfrac{\left(  p-m\right)  \left(  p-m-1\right)
\cdots\left(  p-m-n+1\right)  }{n!}\\
&  =\dfrac{1}{n!m!}\underbrace{\left(  p\left(  p-1\right)  \cdots\left(
p-m+1\right)  \right)  \cdot\left(  \left(  p-m\right)  \left(  p-m-1\right)
\cdots\left(  p-m-n+1\right)  \right)  }_{=p\left(  p-1\right)  \cdots\left(
p-m-n+1\right)  }\\
&  =\dfrac{1}{n!m!}p\left(  p-1\right)  \cdots\left(  \underbrace{p-m-n+1}%
_{=p-\left(  n+m\right)  +1}\right)  =\dfrac{1}{n!m!}p\left(  p-1\right)
\cdots\left(  p-\left(  n+m\right)  +1\right) \\
&  =\dfrac{\left(  n+m\right)  !}{n!m!}\cdot\dfrac{p\left(  p-1\right)
\cdots\left(  p-\left(  n+m\right)  +1\right)  }{\left(  n+m\right)  !}.
\end{align*}
Comparing this with%
\begin{align*}
&  \underbrace{\dbinom{n+m}{n}}_{=\dfrac{\left(  n+m\right)  !}{n!\left(
\left(  n+m\right)  -n\right)  !}=\dfrac{\left(  n+m\right)  !}{n!m!}%
}\underbrace{\dbinom{p}{n+m}}_{\substack{=\dfrac{p\left(  p-1\right)
\cdots\left(  p-\left(  n+m\right)  +1\right)  }{\left(  n+m\right)
!}\\\text{(by the definition of }\dbinom{p}{n+m}\text{)}}}\\
&  =\dfrac{\left(  n+m\right)  !}{n!m!}\cdot\dfrac{p\left(  p-1\right)
\cdots\left(  p-\left(  n+m\right)  +1\right)  }{\left(  n+m\right)  !},
\end{align*}
we obtain $\dbinom{p}{m}\dbinom{p-m}{n}=\dbinom{n+m}{n}\dbinom{p}{n+m}$, qed.
\par
It might seem that a simpler proof could be obtained using the identity
$\dbinom{p}{m}=\dfrac{p!}{m!\left(  p-m\right)  !}$; but this identity only
holds for $p\geq0$. This type of argument can be salvaged, but we prefer the
proof given above.}.

Now,%
\begin{align*}
&  \left(  \partial_{z}^{\left(  n\right)  }\circ\partial_{z}^{\left(
m\right)  }\right)  \left(  \underbrace{a}_{=uz^{p}}\right) \\
&  =\left(  \partial_{z}^{\left(  n\right)  }\circ\partial_{z}^{\left(
m\right)  }\right)  \left(  uz^{p}\right)  =\partial_{z}^{\left(  n\right)
}\left(  \underbrace{\partial_{z}^{\left(  m\right)  }\left(  uz^{p}\right)
}_{=\dbinom{p}{m}uz^{p-m}}\right)  =\dbinom{p}{m}\underbrace{\partial
_{z}^{\left(  n\right)  }\left(  uz^{p-m}\right)  }_{=\dbinom{p-m}%
{n}uz^{\left(  p-m\right)  -n}}\\
&  =\underbrace{\dbinom{p}{m}\dbinom{p-m}{n}}_{=\dbinom{n+m}{n}\dbinom{p}%
{n+m}}u\underbrace{z^{\left(  p-m\right)  -n}}_{=z^{p-\left(  n+m\right)  }%
}=\dbinom{n+m}{n}\underbrace{\dbinom{p}{n+m}uz^{p-\left(  n+m\right)  }%
}_{=\partial_{z}^{\left(  n+m\right)  }\left(  uz^{p}\right)  }\\
&  =\dbinom{n+m}{n}\partial_{z}^{\left(  n+m\right)  }\left(
\underbrace{uz^{p}}_{=a}\right)  =\dbinom{n+m}{n}\partial_{z}^{\left(
n+m\right)  }\left(  a\right)  .
\end{align*}
This proves (\ref{pf.prop.delta.deriv.goal4}). Thus,
(\ref{eq.prop.delta.deriv.nm}) is proven.

It remains to prove (\ref{eq.prop.delta.deriv.n}). But this can be shown
straightforwardly by induction over $n$: The induction base ($n=0$) is dealt
with by (\ref{eq.prop.delta.deriv.0}), while the induction step follows from%
\begin{align*}
\underbrace{\partial_{z}}_{\substack{=\partial_{z}^{\left(  1\right)
}\\\text{(by (\ref{eq.prop.delta.deriv.1}))}}}\circ\partial_{z}^{\left(
n-1\right)  }  &  =\partial_{z}^{\left(  1\right)  }\circ\partial_{z}^{\left(
n-1\right)  }=\underbrace{\dbinom{1+\left(  n-1\right)  }{1}}_{=n}%
\underbrace{\partial_{z}^{\left(  1+\left(  n-1\right)  \right)  }}%
_{=\partial_{z}^{\left(  n\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.prop.delta.deriv.nm}),
applied to }1\text{ and }n-1\text{ instead of }n\text{ and }m\right) \\
&  =\partial_{z}^{\left(  n\right)  }.
\end{align*}
This completes the proof of Proposition \ref{prop.delta.deriv}.
\end{proof}

The equality (\ref{eq.prop.delta.deriv.n}) yields that $\partial_{z}^{\left(
n\right)  }=\dfrac{\left(  \partial_{z}\right)  ^{n}}{n!}$ in the case when
$\mathbb{F}$ is a $\mathbb{Q}$-algebra; thus, the endomorphisms $\partial
_{z}^{\left(  n\right)  }$ are of little significance in this case. But they
become useful when one wants to prove results over arbitrary fields (or
commutative rings) -- where they, for instance, allow one to construct a
Taylor formula without denominators. Endomorphisms akin to $\partial
_{z}^{\left(  n\right)  }$ can also be defined in the case of multiple
variables; they will be denoted by $\partial_{x_{j}}^{\left(  n\right)  }$.

\begin{proposition}
\label{prop.delta.commutator}Let $U$ be an $\mathbb{F}$-module. Let $A$ be any
of the five $\mathbb{F}$-modules $U\left[  z\right]  $, $U\left[
z,z^{-1}\right]  $, $U\left[  \left[  z\right]  \right]  $, $U\left(  \left(
z\right)  \right)  $ and $U\left[  \left[  z,z^{-1}\right]  \right]  $. Let
$L_{z}$ be the $\mathbb{F}$-linear map $A\rightarrow A$ which sends every
$a\in A$ to $z\cdot a\in A$.\ \ \ \ \footnotemark

Define $\partial_{z}^{\left(  -1\right)  }$ as the zero map $0:A\rightarrow
A$. Then, $\left[  L_{z},\partial_{z}^{\left(  n\right)  }\right]
=-\partial_{z}^{\left(  n-1\right)  }$ for every $n\in\mathbb{N}$. Here,
$\left[  \cdot,\cdot\right]  $ denotes the commutator of two endomorphisms of
$A$; thus, $\left[  L_{z},\partial_{z}^{\left(  n\right)  }\right]
=L_{z}\circ\partial_{z}^{\left(  n\right)  }-\partial_{z}^{\left(  n\right)
}\circ L_{z}$.
\end{proposition}

\footnotetext{Here, $z\cdot a$ is defined because $A$ is an $\mathbb{F}\left[
z\right]  $-module.}

\begin{proof}
[Proof of Proposition \ref{prop.delta.commutator}.]Let $n\in\mathbb{N}$. We
need to prove that $\left[  L_{z},\partial_{z}^{\left(  n\right)  }\right]
=-\partial_{z}^{\left(  n-1\right)  }$. If $n=0$, then this is obvious
(because $\partial_{z}^{\left(  0\right)  }=\operatorname*{id}$ and
$\partial_{z}^{\left(  -1\right)  }=0$). Hence, for the rest of this proof, we
WLOG assume that $n\neq0$.

We also WLOG assume that $A=U\left[  \left[  z,z^{-1}\right]  \right]  $,
because the other four $\mathbb{F}$-modules $U\left[  z\right]  $, $U\left[
z,z^{-1}\right]  $, $U\left[  \left[  z\right]  \right]  $ and $U\left(
\left(  z\right)  \right)  $ are $\mathbb{F}$-submodules of $U\left[  \left[
z,z^{-1}\right]  \right]  $ and their respective $\partial_{z}^{\left(
m\right)  }$ maps are restrictions of the $\partial_{z}^{\left(  m\right)  }$
map of $U\left[  \left[  z,z^{-1}\right]  \right]  $.

We need to prove that $\left[  L_{z},\partial_{z}^{\left(  n\right)  }\right]
=-\partial_{z}^{\left(  n-1\right)  }$. In other words, we need to prove that%
\begin{equation}
\left[  L_{z},\partial_{z}^{\left(  n\right)  }\right]  \left(  a\right)
=-\partial_{z}^{\left(  n-1\right)  }\left(  a\right)
\label{pf.prop.delta.commutator.goal}%
\end{equation}
for every $a\in A$. So let us fix $a\in A$.

We want to prove the equality (\ref{pf.prop.delta.commutator.goal}). Both
sides of this equality are continuous with respect to $a$ (where
\textquotedblleft continuous\textquotedblright\ is to be understood with
respect to the topology we introduced on $A$). Hence, we can WLOG assume that
$a\in U\left[  z,z^{-1}\right]  $ (since $U\left[  z,z^{-1}\right]  $ is dense
in $U\left[  \left[  z,z^{-1}\right]  \right]  =A$). Assume this. Now, both
sides of the equality (\ref{pf.prop.delta.commutator.goal}) are $\mathbb{F}%
$-linear in $a$. Hence (and since $a\in U\left[  z,z^{-1}\right]  $), we can
WLOG assume that $a$ belongs to the family $\left(  uz^{q}\right)  _{u\in
U;\ q\in\mathbb{Z}}$ (since the family $\left(  uz^{q}\right)  _{u\in
U;\ q\in\mathbb{Z}}$ spans the $\mathbb{F}$-module $U\left[  z,z^{-1}\right]
$). Assume this. Then, $a=uz^{q}$ for some $u\in U$ and $q\in\mathbb{Z}$.
Consider these $u$ and $q$. Now, the definition of $\partial_{z}^{\left(
n\right)  }$ yields $\partial_{z}^{\left(  n\right)  }\left(  uz^{q}\right)
=\dbinom{q}{n}uz^{q-n}$. Also, the definition of $\partial_{z}^{\left(
n\right)  }$ yields%
\begin{align*}
\partial_{z}^{\left(  n\right)  }\left(  uz^{q+1}\right)   &
=\underbrace{\dbinom{q+1}{n}}_{\substack{=\dbinom{q}{n}+\dbinom{q}%
{n-1}\\\text{(by the recursion of}\\\text{the binomial coefficients)}%
}}uz^{\left(  q+1\right)  -n}=\left(  \dbinom{q}{n}+\dbinom{q}{n-1}\right)
uz^{\left(  q+1\right)  -n}\\
&  =\left(  \dbinom{q}{n}+\dbinom{q}{n-1}\right)  uz^{q-n+1}=\dbinom{q}%
{n}uz^{q-n+1}+\dbinom{q}{n-1}uz^{q-n+1}.
\end{align*}
Finally, the definition of $\partial_{z}^{\left(  n-1\right)  }$ yields
$\partial_{z}^{\left(  n-1\right)  }\left(  uz^{q}\right)  =\dbinom{q}%
{n-1}uz^{q-\left(  n-1\right)  }=\dbinom{q}{n-1}uz^{q-n+1}$. Now,%
\begin{align*}
&  \underbrace{\left[  L_{z},\partial_{z}^{\left(  n\right)  }\right]
}_{=L_{z}\circ\partial_{z}^{\left(  n\right)  }-\partial_{z}^{\left(
n\right)  }\circ L_{z}}\left(  \underbrace{a}_{=uz^{q}}\right) \\
&  =\left(  L_{z}\circ\partial_{z}^{\left(  n\right)  }-\partial_{z}^{\left(
n\right)  }\circ L_{z}\right)  \left(  uz^{q}\right) \\
&  =\left(  L_{z}\circ\partial_{z}^{\left(  n\right)  }\right)  \left(
uz^{q}\right)  -\left(  \partial_{z}^{\left(  n\right)  }\circ L_{z}\right)
\left(  uz^{q}\right) \\
&  =L_{z}\left(  \underbrace{\partial_{z}^{\left(  n\right)  }\left(
uz^{q}\right)  }_{=\dbinom{q}{n}uz^{q-n}}\right)  -\partial_{z}^{\left(
n\right)  }\left(  \underbrace{L_{z}\left(  uz^{q}\right)  }%
_{\substack{=zuz^{q}\\\text{(by the definition of }L_{z}\text{)}}}\right) \\
&  =\underbrace{L_{z}\left(  \dbinom{q}{n}uz^{q-n}\right)  }%
_{\substack{=z\dbinom{q}{n}uz^{q-n}\\\text{(by the definition of }%
L_{z}\text{)}}}-\partial_{z}^{\left(  n\right)  }\left(  \underbrace{zuz^{q}%
}_{=uz^{q+1}}\right) \\
&  =\underbrace{z\dbinom{q}{n}uz^{q-n}}_{=\dbinom{q}{n}uz^{q-n+1}%
}-\underbrace{\partial_{z}^{\left(  n\right)  }\left(  uz^{q+1}\right)
}_{=\dbinom{q}{n}uz^{q-n+1}+\dbinom{q}{n-1}uz^{q-n+1}}\\
&  =\dbinom{q}{n}uz^{q-n+1}-\left(  \dbinom{q}{n}uz^{q-n+1}+\dbinom{q}%
{n-1}uz^{q-n+1}\right) \\
&  =-\underbrace{\dbinom{q}{n-1}uz^{q-n+1}}_{=\partial_{z}^{\left(
n-1\right)  }\left(  uz^{q}\right)  }=-\partial_{z}^{\left(  n-1\right)
}\left(  \underbrace{uz^{q}}_{=a}\right)  =-\partial_{z}^{\left(  n-1\right)
}\left(  a\right)  .
\end{align*}
This proves (\ref{pf.prop.delta.commutator.goal}). Thus, Proposition
\ref{prop.delta.commutator} is proven.
\end{proof}

\begin{remark}
\label{rmk.Lz.abuse}It is a common abuse of notation to write $z$ for what was
denoted by $L_{z}$ in Proposition \ref{prop.delta.commutator}. This notation
(which we will too use later in these notes) must be used with caution,
however, since it can lead to confusion in certain cases (i.e., when
$U=\mathbb{F}$, then $\partial_{z}z$ might be understood either as the element
$\partial_{z}\left(  z\right)  $ of $\mathbb{F}\left[  z\right]  $ or as the
endomorphism $\partial_{z}\circ z=\partial_{z}\circ L_{z}$ of $\mathbb{F}%
\left[  z\right]  $). We hope that context will clarify all resulting ambiguities.
\end{remark}

When $U$ is a commutative $\mathbb{F}$-algebra, the endomorphisms
$\partial_{z}^{\left(  n\right)  }$ of $U\left[  z\right]  $, $U\left[
z,z^{-1}\right]  $, of $U\left[  \left[  z\right]  \right]  $, of $U\left(
\left(  z\right)  \right)  $ and of $U\left[  \left[  z,z^{-1}\right]
\right]  $ are $U$-linear. Moreover, in this case, the sequence $\left(
\partial_{z}^{\left(  0\right)  },\partial_{z}^{\left(  1\right)  }%
,\partial_{z}^{\left(  2\right)  },\ldots\right)  $ is an example of what is
known as a \textquotedblleft Hasse-Schmidt derivation\textquotedblright\ (or
\textquotedblleft higher derivation\textquotedblright). Let us now define this notion.

\subsection{Hasse-Schmidt derivations in general}

\begin{definition}
Let $A$ and $B$ be two $\mathbb{F}$-algebras. A \textit{Hasse-Schmidt
derivation} from $A$ to $B$ means a sequence $\left(  D_{0},D_{1},D_{2}%
,\ldots\right)  $ of $\mathbb{F}$-linear maps $A\rightarrow B$ such that
$D_{0}\left(  1\right)  =1$ and such that every $n\in\mathbb{N}$, $a\in A$ and
$b\in A$ satisfy%
\begin{equation}
D_{n}\left(  ab\right)  =\sum_{i=0}^{n}D_{i}\left(  a\right)  D_{n-i}\left(
b\right)  . \label{def.HSD.deriv.ab}%
\end{equation}
We abbreviate the notion \textquotedblleft Hasse-Schmidt
derivation\textquotedblright\ by \textquotedblleft\textit{HSD}%
\textquotedblright.
\end{definition}

\begin{definition}
Let $A$ be an $\mathbb{F}$-algebra. An HSD $\left(  D_{0},D_{1},D_{2}%
,\ldots\right)  $ from $A$ to $A$ is said to be \textit{divided-powers} (we
use this word as an adjective) if and only if it satisfies $D_{0}%
=\operatorname*{id}$ and%
\begin{equation}
D_{n}\circ D_{m}=\dbinom{n+m}{n}D_{n+m}\ \ \ \ \ \ \ \ \ \ \text{for all }%
n\in\mathbb{N}\text{ and }m\in\mathbb{N}. \label{def.HSD.deriv.nm}%
\end{equation}

\end{definition}

We notice that the term \textquotedblleft Hasse-Schmidt
derivation\textquotedblright\ is used by different authors in slightly
different meanings, and the notion of a \textquotedblleft divided-powers
Hasse-Schmidt derivation\textquotedblright\ is the scribe's invention.

Now, we claim:

\begin{proposition}
\label{prop.delta.hasse}Let $U$ be a commutative $\mathbb{F}$-algebra.

\textbf{(a)} The sequence $\left(  \partial_{z}^{\left(  0\right)  }%
,\partial_{z}^{\left(  1\right)  },\partial_{z}^{\left(  2\right)  }%
,\ldots\right)  $ is a divided-powers HSD from $A$ to $A$ whenever $A$ is one
of the $\mathbb{F}$-algebras $U\left[  z\right]  $, $U\left[  z,z^{-1}\right]
$, $U\left[  \left[  z\right]  \right]  $ and $U\left(  \left(  z\right)
\right)  $.

\textbf{(b)} Also,%
\[
\partial_{z}^{\left(  n\right)  }\left(  ab\right)  =\sum_{i=0}^{n}%
\partial_{z}^{\left(  i\right)  }\left(  a\right)  \partial_{z}^{\left(
n-i\right)  }\left(  b\right)
\]
for any $n\in\mathbb{N}$, any $a\in U\left[  z,z^{-1}\right]  $ and any $b\in
U\left[  \left[  z,z^{-1}\right]  \right]  $.

\textbf{(c)} We have $\partial_{z}\left(  ab\right)  =\partial_{z}\left(
a\right)  b+a\partial_{z}\left(  b\right)  $ for any $a\in U\left[
z,z^{-1}\right]  $ and any $b\in U\left[  \left[  z,z^{-1}\right]  \right]  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.delta.hasse}.]\textbf{(a)} Let $A$ be one of
the four $\mathbb{F}$-algebras $U\left[  z\right]  $, $U\left[  z,z^{-1}%
\right]  $, $U\left[  \left[  z\right]  \right]  $ and $U\left(  \left(
z\right)  \right)  $. Let us first show that $\left(  \partial_{z}^{\left(
0\right)  },\partial_{z}^{\left(  1\right)  },\partial_{z}^{\left(  2\right)
},\ldots\right)  $ is a HSD from $A$ to $A$. In order to do so, we need to
show that $\partial_{z}^{\left(  0\right)  }\left(  1\right)  =1$ and that
every $n\in\mathbb{N}$, $a\in A$ and $b\in A$ satisfy%
\begin{equation}
\partial_{z}^{\left(  n\right)  }\left(  ab\right)  =\sum_{i=0}^{n}%
\partial_{z}^{\left(  i\right)  }\left(  a\right)  \partial_{z}^{\left(
n-i\right)  }\left(  b\right)  . \label{pf.prop.delta.hasse.a.1}%
\end{equation}
It is clear that $\partial_{z}^{\left(  0\right)  }\left(  1\right)  =1$ (due
to (\ref{eq.prop.delta.deriv.0})), so it remains to prove
(\ref{pf.prop.delta.hasse.a.1}).

\textit{Proof of (\ref{pf.prop.delta.hasse.a.1}):} Let $n\in\mathbb{N}$, $a\in
A$ and $b\in A$. We WLOG assume that $A=U\left(  \left(  z\right)  \right)  $,
since the three other $\mathbb{F}$-algebras $U\left[  z\right]  $, $U\left[
z,z^{-1}\right]  $, $U\left[  \left[  z\right]  \right]  $ are subalgebras of
$U\left(  \left(  z\right)  \right)  $. We need to prove the equality
(\ref{pf.prop.delta.hasse.a.1}). Both sides of this equality are continuous
with respect to each of $a$ and $b$ (where \textquotedblleft
continuous\textquotedblright\ is to be understood with respect to the topology
we introduced on $A$). Hence, we can WLOG assume that each of $a$ and $b$
belongs to $U\left[  z,z^{-1}\right]  $ (since $U\left[  z,z^{-1}\right]  $ is
dense in $U\left(  \left(  z\right)  \right)  $). Assume this. Now, both sides
of the equality (\ref{pf.prop.delta.hasse.a.1}) are $U$-linear in each of $a$
and $b$. Hence (and since $a$ and $b$ belong to $U\left[  z,z^{-1}\right]  $),
we can WLOG assume that each of $a$ and $b$ belongs to the family $\left(
z^{m}\right)  _{m\in\mathbb{Z}}$ (since $\left(  z^{m}\right)  _{m\in
\mathbb{Z}}$ is a basis of the $U$-module $U\left[  z,z^{-1}\right]  $).
Assume this. Then, $a=z^{p}$ and $b=z^{q}$ for some $p\in\mathbb{Z}$ and
$q\in\mathbb{Z}$. Consider these $p$ and $q$.

Now, the left hand side of (\ref{pf.prop.delta.hasse.a.1}) simplifies to%
\begin{equation}
\partial_{z}^{\left(  n\right)  }\left(  \underbrace{a}_{=z^{p}}%
\underbrace{b}_{=z^{q}}\right)  =\partial_{z}^{\left(  n\right)  }\left(
\underbrace{z^{p}z^{q}}_{=z^{p+q}}\right)  =\partial_{z}^{\left(  n\right)
}\left(  z^{p+q}\right)  =\dbinom{p+q}{n}z^{p+q-n}
\label{pf.prop.delta.hasse.a.1.pf.1}%
\end{equation}
(by the definition of $\partial_{z}^{\left(  n\right)  }$), while the right
hand side of (\ref{pf.prop.delta.hasse.a.1}) simplifies to%
\begin{align}
&  \sum_{i=0}^{n}\partial_{z}^{\left(  i\right)  }\left(  \underbrace{a}%
_{=z^{p}}\right)  \partial_{z}^{\left(  n-i\right)  }\left(  \underbrace{b}%
_{=z^{q}}\right) \nonumber\\
&  =\sum_{i=0}^{n}\underbrace{\partial_{z}^{\left(  i\right)  }\left(
z^{p}\right)  }_{\substack{=\dbinom{p}{i}z^{p-i}\\\text{(by the definition of
}\partial_{z}^{\left(  i\right)  }\text{)}}}\underbrace{\partial_{z}^{\left(
n-i\right)  }\left(  z^{q}\right)  }_{\substack{=\dbinom{q}{n-i}z^{q-\left(
n-i\right)  }\\\text{(by the definition of }\partial_{z}^{\left(  n-i\right)
}\text{)}}}\nonumber\\
&  =\sum_{i=0}^{n}\dbinom{p}{i}z^{p-i}\dbinom{q}{n-i}z^{q-\left(  n-i\right)
}=\sum_{i=0}^{n}\dbinom{p}{i}\dbinom{q}{n-i}\underbrace{z^{p-i}z^{q-\left(
n-i\right)  }}_{=z^{\left(  p-i\right)  +\left(  q-\left(  n-i\right)
\right)  }=z^{p+q-n}}\nonumber\\
&  =\sum_{i=0}^{n}\dbinom{p}{i}\dbinom{q}{n-i}z^{p+q-n}.
\label{pf.prop.delta.hasse.a.1.pf.2}%
\end{align}
But the Vandermonde convolution identity says that
\[
\dbinom{p+q}{n}=\sum_{i=0}^{n}\dbinom{p}{i}\dbinom{q}{n-i}.
\]
Hence, (\ref{pf.prop.delta.hasse.a.1.pf.1}) becomes%
\begin{align*}
\partial_{z}^{\left(  n\right)  }\left(  ab\right)   &  =\underbrace{\dbinom
{p+q}{n}}_{=\sum_{i=0}^{n}\dbinom{p}{i}\dbinom{q}{n-i}}z^{p+q-n}\\
&  =\sum_{i=0}^{n}\dbinom{p}{i}\dbinom{q}{n-i}z^{p+q-n}=\sum_{i=0}^{n}%
\partial_{z}^{\left(  i\right)  }\left(  a\right)  \partial_{z}^{\left(
n-i\right)  }\left(  b\right)
\end{align*}
(by (\ref{pf.prop.delta.hasse.a.1.pf.2})). Hence,
(\ref{pf.prop.delta.hasse.a.1}) is proven.

This shows that $\left(  \partial_{z}^{\left(  0\right)  },\partial
_{z}^{\left(  1\right)  },\partial_{z}^{\left(  2\right)  },\ldots\right)  $
is a HSD from $A$ to $A$. It remains to show that this HSD is divided-powers.
For this, we need to check that $\partial_{z}^{\left(  0\right)
}=\operatorname*{id}$ and%
\[
\partial_{z}^{\left(  n\right)  }\circ\partial_{z}^{\left(  m\right)
}=\dbinom{n+m}{n}\partial_{z}^{\left(  n+m\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{for all }n\in\mathbb{N}\text{ and }m\in\mathbb{N}.
\]
But this follows immediately from (\ref{eq.prop.delta.deriv.0}) and
(\ref{eq.prop.delta.deriv.nm}). This completes the proof of Proposition
\ref{prop.delta.hasse} \textbf{(a)}.

\textbf{(b)} This is proven similarly to our above proof of
(\ref{pf.prop.delta.hasse.a.1}).

\textbf{(c)} Let $a\in U\left[  z,z^{-1}\right]  $ and any $b\in U\left[
\left[  z,z^{-1}\right]  \right]  $. Applying Proposition
\ref{prop.delta.hasse} \textbf{(b)} to $n=1$, we obtain%
\begin{align*}
\partial_{z}^{\left(  1\right)  }\left(  ab\right)   &  =\sum_{i=0}%
^{1}\partial_{z}^{\left(  i\right)  }\left(  a\right)  \partial_{z}^{\left(
1-i\right)  }\left(  b\right)  =\underbrace{\partial_{z}^{\left(  0\right)  }%
}_{=\operatorname*{id}}\left(  a\right)  \underbrace{\partial_{z}^{\left(
1-0\right)  }}_{=\partial_{z}^{\left(  1\right)  }=\partial_{z}}\left(
b\right)  +\underbrace{\partial_{z}^{\left(  1\right)  }}_{=\partial_{z}%
}\left(  a\right)  \underbrace{\partial_{z}^{\left(  1-1\right)  }}%
_{=\partial_{z}^{\left(  0\right)  }=\operatorname*{id}}\left(  b\right) \\
&  =\operatorname*{id}\left(  a\right)  \partial_{z}\left(  b\right)
+\partial_{z}\left(  a\right)  \operatorname*{id}\left(  b\right)
=a\partial_{z}\left(  b\right)  +\partial_{z}\left(  a\right)  b=\partial
_{z}\left(  a\right)  b+a\partial_{z}\left(  b\right)  .
\end{align*}
Since $\partial_{z}^{\left(  1\right)  }=\partial_{z}$, this rewrites as
$\partial_{z}\left(  ab\right)  =\partial_{z}\left(  a\right)  b+a\partial
_{z}\left(  b\right)  $. This proves Proposition \ref{prop.delta.hasse}
\textbf{(c)}.
\end{proof}

We notice in passing that when $\mathbb{F}$ is a $\mathbb{Q}$%
-algebra\footnote{for example, a field of characteristic $0$}, and
$\partial:A\rightarrow A$ is any derivation of $A$, then $\left(
\operatorname*{id},\dfrac{1}{1!}\partial,\dfrac{1}{2!}\partial^{2},\dfrac
{1}{3!}\partial^{3},\ldots\right)  $ is a divided-powers HSD from $A$ to $A$.
(This is very easy to prove.) A kind of converse holds as well (and not just
in characteristic $0$):

\begin{remark}
\label{rmk.divpowHSD.n!}Let $A$ be an $\mathbb{F}$-algebra. Let $\left(
D_{0},D_{1},D_{2},\ldots\right)  $ be a divided-powers HSD from $A$ to $A$.

\textbf{(a)} We have $n!D_{n}=\left(  D_{1}\right)  ^{n}$ for every positive
$n\in\mathbb{N}$.

\textbf{(b)} Assume that $D_{0}=\operatorname*{id}$. Then, $n!D_{n}=\left(
D_{1}\right)  ^{n}$ for every $n\in\mathbb{N}$.

\textbf{(c)} Assume that $D_{0}=\operatorname*{id}$, and that $\mathbb{F}$ is
a $\mathbb{Q}$-algebra. Then, $\left(  D_{0},D_{1},D_{2},\ldots\right)  $ can
be reconstructed from $D_{1}$ alone.
\end{remark}

\begin{proof}
[Proof of Remark \ref{rmk.divpowHSD.n!}.]\textbf{(a)} We prove Remark
\ref{rmk.divpowHSD.n!} \textbf{(a)} by induction over $n$. The induction base
($n=1$) is obvious (because if $n=1$, then $n!D_{n}=1!D_{1}=D_{1}=\left(
D_{1}\right)  ^{1}=\left(  D_{1}\right)  ^{n}$). For the induction step, fix a
positive $N\in\mathbb{N}$. Assume that Remark \ref{rmk.divpowHSD.n!}
\textbf{(a)} holds for $n=N$. We now need to show that Remark
\ref{rmk.divpowHSD.n!} \textbf{(a)} holds for $n=N+1$. We have%
\begin{align*}
D_{N}\circ D_{1}  &  =\underbrace{\dbinom{N+1}{N}}_{=N+1}D_{N+1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{def.HSD.deriv.nm}), applied to
}n=N\text{ and }m=1\right) \\
&  =\left(  N+1\right)  D_{N+1},
\end{align*}
so that $N!D_{N}\circ D_{1}=\underbrace{N!\left(  N+1\right)  }_{=\left(
N+1\right)  !}D_{N+1}=\left(  N+1\right)  !D_{N+1}$. Compared with
\newline$\underbrace{N!D_{N}}_{\substack{=\left(  D_{1}\right)  ^{N}%
\\\text{(since Remark \ref{rmk.divpowHSD.n!} \textbf{(a)}}\\\text{holds for
}n=N\text{)}}}\circ D_{1}=\left(  D_{1}\right)  ^{N}\circ D_{1}=\left(
D_{1}\right)  ^{N+1}$, this yields $\left(  N+1\right)  !D_{N+1}=\left(
D_{1}\right)  ^{N+1}$. In other words, Remark \ref{rmk.divpowHSD.n!}
\textbf{(a)} holds for $n=N+1$. This completes the induction step, and thus
Remark \ref{rmk.divpowHSD.n!} \textbf{(a)} is proven.

\textbf{(b)} Remark \ref{rmk.divpowHSD.n!} \textbf{(a)} yields that
$n!D_{n}=\left(  D_{1}\right)  ^{n}$ for every positive $n\in\mathbb{N}$. It
thus remains to show that $n!D_{n}=\left(  D_{1}\right)  ^{n}$ for $n=0$. This
follows immediately from the assumption that $D_{0}=\operatorname*{id}$.

\textbf{(c)} This follows from Remark \ref{rmk.divpowHSD.n!} \textbf{(b)},
since $n!$ is invertible in $\mathbb{F}$ for every $n\in\mathbb{N}$.
\end{proof}

Proposition \ref{prop.delta.hasse} \textbf{(a)} shows that the sequence
$\left(  \partial_{z}^{\left(  0\right)  },\partial_{z}^{\left(  1\right)
},\partial_{z}^{\left(  2\right)  },\ldots\right)  $ is a divided-powers HSD
from $A$ to $A$ whenever $A$ is one of the $\mathbb{F}$-algebras $U\left[
z\right]  $, $U\left[  z,z^{-1}\right]  $, $U\left[  \left[  z\right]
\right]  $ and $U\left(  \left(  z\right)  \right)  $. Similar statements can
be proven (in the same way) for polynomial-like objects in multiple variables:
For example, if $A$ is one of the $\mathbb{F}$-algebras $U\left[  z,w\right]
$, $U\left[  z,z^{-1},w,w^{-1}\right]  $, $U\left[  \left[  z,w\right]
\right]  $ and $U\left(  \left(  z,w\right)  \right)  $, then both sequences
$\left(  \partial_{z}^{\left(  0\right)  },\partial_{z}^{\left(  1\right)
},\partial_{z}^{\left(  2\right)  },\ldots\right)  $ and $\left(  \partial
_{w}^{\left(  0\right)  },\partial_{w}^{\left(  1\right)  },\partial
_{w}^{\left(  2\right)  },\ldots\right)  $ are divided-powers HSDs from $A$ to
$A$.

To summarize, the notions \textquotedblleft HSD\textquotedblright\ and
\textquotedblleft divided-powers HSD\textquotedblright\ abstract some
properties of the families $\left(  \partial_{z}^{\left(  0\right)  }%
,\partial_{z}^{\left(  1\right)  },\partial_{z}^{\left(  2\right)  }%
,\ldots\right)  $ of operators on the four $\mathbb{F}$-algebras $U\left[
z\right]  $, $U\left[  z,z^{-1}\right]  $, $U\left[  \left[  z\right]
\right]  $ and $U\left(  \left(  z\right)  \right)  $.\ \ \ \ \footnote{We
could also define a notion of \textquotedblleft HSD on
modules\textquotedblright\ to abstract properties of the family $\left(
\partial_{z}^{\left(  0\right)  },\partial_{z}^{\left(  1\right)  }%
,\partial_{z}^{\left(  2\right)  },\ldots\right)  $ of operators on the
$\mathbb{F}$-module $U\left[  \left[  z,z^{-1}\right]  \right]  $, such as
Proposition \ref{prop.delta.hasse} \textbf{(b)}. We won't need this
generality, however (or so I hope).} We will see the use of this generality
when we prove a general fact (Corollary \ref{cor.HSD.extend-to-loc.div}) that
allows us to extend a divided-powers HSD to a localization of its domain. This
will allows us, for example, to extend the operators $\partial_{z}^{\left(
0\right)  }$, $\partial_{z}^{\left(  1\right)  }$, $\partial_{z}^{\left(
2\right)  }$, $\ldots$ to any localization of $U\left[  z,w\right]  $ or
$U\left[  \left[  z,w\right]  \right]  $. This gives us a purely algebraic
reason why derivatives are well-defined for rational functions in two
variables; this is a fact that is commonly used but rarely proven in literature.

Before we can do this, we will introduce another viewpoint on HSDs, which will
make them easier to deal with.

\subsection{Hasse-Schmidt derivations from $A$ to $B$ as algebra maps
$A\rightarrow B\left[  \left[  t\right]  \right]  $}

A sequence $\left(  D_{0},D_{1},D_{2},\ldots\right)  $ of linear maps from one
$\mathbb{F}$-module $P$ to another $\mathbb{F}$-module $Q$ can be described by
a single map $P\rightarrow Q\left[  \left[  t\right]  \right]  $, and it turns
out that the properties of the former sequence to be an HSD or a
divided-powers HSD translate into some rather simple conditions on the latter
map. Let us see this in some detail.

\begin{definition}
\label{def.HSD.powerser}Let $P$ and $Q$ be two $\mathbb{F}$-modules. Let $t$
be a symbol. Let $\mathbf{D}=\left(  D_{0},D_{1},D_{2},\ldots\right)  $ be a
sequence of $\mathbb{F}$-linear maps from $P$ to $Q$. Then, we define an
$\mathbb{F}$-linear map $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle
}:P\rightarrow Q\left[  \left[  t\right]  \right]  $ by%
\[
\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  p\right)
=\sum_{i\in\mathbb{N}}D_{i}\left(  p\right)  t^{i}%
\ \ \ \ \ \ \ \ \ \ \text{for all }p\in P.
\]

\end{definition}

\begin{theorem}
\label{thm.HSD.powerser}Let $A$ and $B$ be two $\mathbb{F}$-algebras. Let $t$
be a symbol. Let $\mathbf{D}=\left(  D_{0},D_{1},D_{2},\ldots\right)  $ be a
sequence of $\mathbb{F}$-linear maps from $A$ to $B$. Then, $\mathbf{D}$ is an
HSD if and only if $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle
}:A\rightarrow B\left[  \left[  t\right]  \right]  $ is an $\mathbb{F}%
$-algebra homomorphism.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.HSD.powerser}.]$\Longrightarrow:$ Assume that
$\mathbf{D}$ is an HSD. We need to show that $\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }:A\rightarrow B\left[  \left[  t\right]
\right]  $ is an $\mathbb{F}$-algebra homomorphism.

Let $a\in A$ and $b\in A$. We are going to show that $\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  ab\right)  =\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  a\right)  \cdot\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  b\right)  $. Indeed, the definition of
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  ab\right)  $
yields%
\begin{align}
\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  ab\right)   &
=\sum_{i\in\mathbb{N}}D_{i}\left(  ab\right)  t^{i}=\sum_{n\in\mathbb{N}%
}\underbrace{D_{n}\left(  ab\right)  }_{\substack{=\sum_{i=0}^{n}D_{i}\left(
a\right)  D_{n-i}\left(  b\right)  \\\text{(by (\ref{def.HSD.deriv.ab}), since
}\mathbf{D}\\\text{is an HSD)}}}t^{n}\label{pf.thm.HSD.powerser.1}\\
&  =\sum_{n\in\mathbb{N}}\left(  \sum_{i=0}^{n}D_{i}\left(  a\right)
D_{n-i}\left(  b\right)  \right)  t^{n}.\nonumber
\end{align}
Compared with%
\begin{align}
\underbrace{\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(
a\right)  }_{\substack{=\sum_{i\in\mathbb{N}}D_{i}\left(  a\right)
t^{i}\\\text{(by the definition of }\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }\left(  a\right)  \text{)}}}\cdot
\underbrace{\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(
b\right)  }_{\substack{=\sum_{i\in\mathbb{N}}D_{i}\left(  b\right)
t^{i}\\\text{(by the definition of }\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }\left(  b\right)  \text{)}}}  &  =\left(  \sum_{i\in
\mathbb{N}}D_{i}\left(  a\right)  t^{i}\right)  \left(  \sum_{i\in\mathbb{N}%
}D_{i}\left(  b\right)  t^{i}\right) \nonumber\\
&  =\sum_{n\in\mathbb{N}}\left(  \sum_{i=0}^{n}D_{i}\left(  a\right)
D_{n-i}\left(  b\right)  \right)  t^{n}, \label{pf.thm.HSD.powerser.2}%
\end{align}
this yields $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(
ab\right)  =\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(
a\right)  \cdot\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(
b\right)  $. Thus, $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$
preserves multiplication.

But we are not done yet -- an $\mathbb{F}$-algebra homomorphism has to also
preserve the unity. So let us show that $\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }\left(  1\right)  =1$. The definition of
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  1\right)  $
yields $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  1\right)
=\sum_{i\in\mathbb{N}}D_{i}\left(  1\right)  t^{i}$. This is a power series in
$t$ whose constant coefficient is $D_{0}\left(  1\right)  =1$ (since
$\mathbf{D}$ is an HSD); every such power series is invertible. Thus,
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  1\right)  $ is
invertible. But since $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$
preserves multiplication, we have $\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }\left(  1\cdot1\right)  =\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }\left(  1\right)  \cdot\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }\left(  1\right)  $. Hence, $\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  1\right)  \cdot\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  1\right)  =\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  1\cdot1\right)  =\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  1\right)  $. We can cancel
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  1\right)  $ out
of this equality (since $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle
}\left(  1\right)  $ is invertible), and obtain $\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  1\right)  =1$. Combined with the fact
that $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$ preserves
multiplication, this yields that $\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }$ is an $\mathbb{F}$-algebra homomorphism. This proves the
$\Longrightarrow$ direction of Theorem \ref{thm.HSD.powerser}.

$\Longleftarrow:$ Assume that $\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }:A\rightarrow B\left[  \left[  t\right]  \right]  $ is an
$\mathbb{F}$-algebra homomorphism. Then, $\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }\left(  1\right)  =1$. Compared with $\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  1\right)  =\sum_{i\in\mathbb{N}}%
D_{i}\left(  1\right)  t^{i}$ (by the definition of $\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  1\right)  $), this yields $\sum
_{i\in\mathbb{N}}D_{i}\left(  1\right)  t^{i}=1$. Comparing the constant
coefficients on both sides of this equality, we obtain $D_{0}\left(  1\right)
=1$.

Also, let $a\in A$ and $b\in A$ satisfy%
\begin{align*}
\sum_{n\in\mathbb{N}}D_{n}\left(  ab\right)  t^{n}  &  =\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  ab\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.thm.HSD.powerser.1})}\right) \\
&  =\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  a\right)
\cdot\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  b\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }\text{ is an }\mathbb{F}\text{-algebra homomorphism}\right) \\
&  =\sum_{n\in\mathbb{N}}\left(  \sum_{i=0}^{n}D_{i}\left(  a\right)
D_{n-i}\left(  b\right)  \right)  t^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.HSD.powerser.2})}\right)  .
\end{align*}
Comparing coefficients on both sides of this equality, we obtain%
\[
D_{n}\left(  ab\right)  =\sum_{i=0}^{n}D_{i}\left(  a\right)  D_{n-i}\left(
b\right)
\]
for every $a\in A$, every $b\in A$ and every $n\in\mathbb{N}$. Combined with
$D_{0}\left(  1\right)  =1$, this yields that $\mathbf{D}$ is an HSD. This
completes the proof of the $\Longleftarrow$ direction of Theorem
\ref{thm.HSD.powerser}.
\end{proof}

\begin{theorem}
\label{thm.HSD.powerser.div}Let $A$ be an $\mathbb{F}$-algebra. Let $t$ and
$u$ be two distinct symbols. Let $\mathbf{D}=\left(  D_{0},D_{1},D_{2}%
,\ldots\right)  $ be an HSD from $A$ to $A$.

Recall that whenever $P$ and $Q$ are two $\mathbb{F}$-modules, any
$\mathbb{F}$-linear map $\phi:P\rightarrow Q$ gives rise to an $\mathbb{F}%
$-linear map $\phi\left[  \left[  t\right]  \right]  :P\left[  \left[
t\right]  \right]  \rightarrow Q\left[  \left[  t\right]  \right]  $ (which
just applies the map $\phi$ to every coefficient). Thus, the map
$\widetilde{\mathbf{D}}_{\left\langle u\right\rangle }:A\rightarrow A\left[
\left[  u\right]  \right]  $ gives rise to a map $\widetilde{\mathbf{D}%
}_{\left\langle u\right\rangle }\left[  \left[  t\right]  \right]  :A\left[
\left[  t\right]  \right]  \rightarrow\left(  A\left[  \left[  u\right]
\right]  \right)  \left[  \left[  t\right]  \right]  $.

We also define a map $\widetilde{\mathbf{D}}_{\left\langle u+t\right\rangle
}:A\rightarrow\left(  A\left[  \left[  u\right]  \right]  \right)  \left[
\left[  t\right]  \right]  $ by%
\[
\widetilde{\mathbf{D}}_{\left\langle u+t\right\rangle }\left(  p\right)
=\sum_{i\in\mathbb{N}}D_{i}\left(  p\right)  \left(  u+t\right)
^{i}\ \ \ \ \ \ \ \ \ \ \text{for all }p\in A.
\]
(This is well-defined since $\left(  A\left[  \left[  u\right]  \right]
\right)  \left[  \left[  t\right]  \right]  \cong A\left[  \left[  u,t\right]
\right]  $.)

Let $\pi_{t}:A\left[  \left[  t\right]  \right]  \rightarrow A$ be the
projection map which sends every formal power series to its constant coefficient.

\textbf{(a)} The maps $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$,
$\widetilde{\mathbf{D}}_{\left\langle u\right\rangle }$,
$\widetilde{\mathbf{D}}_{\left\langle u\right\rangle }\left[  \left[
t\right]  \right]  $, $\widetilde{\mathbf{D}}_{\left\langle u\right\rangle
}\left[  \left[  t\right]  \right]  \circ\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }$ and $\widetilde{\mathbf{D}}_{\left\langle u+t\right\rangle
}$ are $\mathbb{F}$-algebra homomorphisms.

\textbf{(b)} The HSD $\mathbf{D}$ is divided-powers if and only if%
\[
\pi_{t}\circ\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }%
=\operatorname*{id}\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \widetilde{\mathbf{D}}_{\left\langle u\right\rangle
}\left[  \left[  t\right]  \right]  \circ\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }=\widetilde{\mathbf{D}}_{\left\langle u+t\right\rangle }.
\]

\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.HSD.powerser.div}.]\textbf{(a)} We know that
$\mathbf{D}$ is an HSD. Thus, the map $\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }$ is an $\mathbb{F}$-algebra homomorphism (according to the
$\Longrightarrow$ direction of Theorem \ref{thm.HSD.powerser}). Similarly,
$\widetilde{\mathbf{D}}_{\left\langle u\right\rangle }$ and
$\widetilde{\mathbf{D}}_{\left\langle u+t\right\rangle }$ are $\mathbb{F}%
$-algebra homomorphisms\footnote{To be completely fair, the proof for
$\widetilde{\mathbf{D}}_{\left\langle u+t\right\rangle }$ is slightly
different from the proof of $\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }$ because $u+t$ is not a single variable like $t$ but a sum of
two variables. But the proof still goes through with the obvious changes,
because we still have $\left(  \sum_{i\in\mathbb{N}}\alpha_{i}\left(
u+t\right)  ^{i}\right)  \left(  \sum_{i\in\mathbb{N}}\beta_{i}\left(
u+t\right)  ^{i}\right)  =\sum_{n\in\mathbb{N}}\left(  \sum_{i=0}^{n}%
\alpha_{i}\beta_{n-i}\right)  \left(  u+t\right)  ^{n}$ for any two families
$\left(  \alpha_{i}\right)  _{i\in\mathbb{N}}\in A^{\mathbb{N}}$ and $\left(
\beta_{i}\right)  _{i\in\mathbb{N}}\in A^{\mathbb{N}}$.}. Since
$\widetilde{\mathbf{D}}_{\left\langle u\right\rangle }$ is an $\mathbb{F}%
$-algebra homomorphism, we see that the map $\widetilde{\mathbf{D}%
}_{\left\langle u\right\rangle }\left[  \left[  t\right]  \right]  $ is an
$\mathbb{F}$-algebra homomorphism, and thus the composition
$\widetilde{\mathbf{D}}_{\left\langle u\right\rangle }\left[  \left[
t\right]  \right]  \circ\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$
is an $\mathbb{F}$-algebra homomorphism (being the composition of the
$\mathbb{F}$-algebra homomorphisms $\widetilde{\mathbf{D}}_{\left\langle
u\right\rangle }\left[  \left[  t\right]  \right]  $ and
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$). This proves Theorem
\ref{thm.HSD.powerser.div} \textbf{(a)}.

\textbf{(b)} $\Longrightarrow:$ Assume that $\mathbf{D}$ is divided-powers.
Then, $D_{0}=\operatorname*{id}$ and the equality (\ref{def.HSD.deriv.nm})
holds (due to the definition of \textquotedblleft
divided-powers\textquotedblright). Now, let $p\in A$. Then,%
\begin{align}
&  \left(  \widetilde{\mathbf{D}}_{\left\langle u\right\rangle }\left[
\left[  t\right]  \right]  \circ\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }\right)  \left(  p\right) \nonumber\\
&  =\left(  \widetilde{\mathbf{D}}_{\left\langle u\right\rangle }\left[
\left[  t\right]  \right]  \right)  \left(  \underbrace{\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  p\right)  }_{\substack{=\sum
_{i\in\mathbb{N}}D_{i}\left(  p\right)  t^{i}\\\text{(by the definition of
}\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\text{)}}}\right)
=\left(  \widetilde{\mathbf{D}}_{\left\langle u\right\rangle }\left[  \left[
t\right]  \right]  \right)  \left(  \sum_{i\in\mathbb{N}}D_{i}\left(
p\right)  t^{i}\right) \nonumber\\
&  =\sum_{i\in\mathbb{N}}\widetilde{\mathbf{D}}_{\left\langle u\right\rangle
}\left(  D_{i}\left(  p\right)  \right)  t^{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\widetilde{\mathbf{D}}_{\left\langle
u\right\rangle }\left[  \left[  t\right]  \right]  \right) \nonumber\\
&  =\sum_{m\in\mathbb{N}}\underbrace{\widetilde{\mathbf{D}}_{\left\langle
u\right\rangle }\left(  D_{m}\left(  p\right)  \right)  }_{\substack{=\sum
_{i\in\mathbb{N}}D_{i}\left(  D_{m}\left(  p\right)  \right)  u^{i}\\\text{(by
the definition of }\widetilde{\mathbf{D}}_{\left\langle u\right\rangle
}\text{)}}}t^{m}=\sum_{m\in\mathbb{N}}\sum_{i\in\mathbb{N}}D_{i}\left(
D_{m}\left(  p\right)  \right)  u^{i}t^{m}\nonumber\\
&  =\underbrace{\sum_{m\in\mathbb{N}}\sum_{n\in\mathbb{N}}}_{=\sum_{\left(
n,m\right)  \in\mathbb{N}^{2}}}\underbrace{D_{n}\left(  D_{m}\left(  p\right)
\right)  }_{=\left(  D_{n}\circ D_{m}\right)  \left(  p\right)  }u^{n}%
t^{m}=\sum_{\left(  n,m\right)  \in\mathbb{N}^{2}}\underbrace{\left(
D_{n}\circ D_{m}\right)  }_{\substack{=\dbinom{n+m}{n}D_{n+m}\\\text{(by
(\ref{def.HSD.deriv.nm}))}}}\left(  p\right)  u^{n}t^{m}%
\label{pf.thm.HSD.powerser.div.1}\\
&  =\sum_{\left(  n,m\right)  \in\mathbb{N}^{2}}\dbinom{n+m}{n}D_{n+m}\left(
p\right)  u^{n}t^{m}.\nonumber
\end{align}
Compared with%
\begin{align}
&  \widetilde{\mathbf{D}}_{\left\langle u+t\right\rangle }\left(  p\right)
\nonumber\\
&  =\sum_{i\in\mathbb{N}}D_{i}\left(  p\right)  \underbrace{\left(
u+t\right)  ^{i}}_{\substack{=\sum_{n=0}^{i}\dbinom{i}{n}u^{n}t^{i-n}%
\\\text{(by the binomial formula)}}}=\sum_{i\in\mathbb{N}}D_{i}\left(
p\right)  \underbrace{\sum_{n=0}^{i}\dbinom{i}{n}u^{n}t^{i-n}}%
_{\substack{=\sum_{\substack{\left(  n,m\right)  \in\mathbb{N}^{2}%
;\\n+m=i}}\dbinom{i}{n}u^{n}t^{m}\\\text{(here, we substituted }\left(
n,m\right)  \text{ for }\left(  n,i-n\right)  \text{)}}}\nonumber\\
&  =\sum_{i\in\mathbb{N}}D_{i}\left(  p\right)  \sum_{\substack{\left(
n,m\right)  \in\mathbb{N}^{2};\\n+m=i}}\dbinom{i}{n}u^{n}t^{m}=\sum
_{i\in\mathbb{N}}\sum_{\substack{\left(  n,m\right)  \in\mathbb{N}%
^{2};\\n+m=i}}\underbrace{\dbinom{i}{n}}_{\substack{=\dbinom{n+m}%
{n}\\\text{(since }i=n+m\text{)}}}\underbrace{D_{i}\left(  p\right)
}_{\substack{=D_{n+m}\left(  p\right)  \\\text{(since }i=n+m\text{)}}%
}u^{n}t^{m}\nonumber\\
&  =\underbrace{\sum_{i\in\mathbb{N}}\sum_{\substack{\left(  n,m\right)
\in\mathbb{N}^{2};\\n+m=i}}}_{=\sum_{\left(  n,m\right)  \in\mathbb{N}^{2}}%
}\dbinom{n+m}{n}D_{n+m}\left(  p\right)  u^{n}t^{m}\nonumber\\
&  =\sum_{\left(  n,m\right)  \in\mathbb{N}^{2}}\dbinom{n+m}{n}D_{n+m}\left(
p\right)  u^{n}t^{m}, \label{pf.thm.HSD.powerser.div.2}%
\end{align}
this yields $\left(  \widetilde{\mathbf{D}}_{\left\langle u\right\rangle
}\left[  \left[  t\right]  \right]  \circ\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }\right)  \left(  p\right)  =\widetilde{\mathbf{D}%
}_{\left\langle u+t\right\rangle }\left(  p\right)  $. Since we have proven
this for every $p\in A$, we thus conclude that $\widetilde{\mathbf{D}%
}_{\left\langle u\right\rangle }\left[  \left[  t\right]  \right]
\circ\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }%
=\widetilde{\mathbf{D}}_{\left\langle u+t\right\rangle }$.

We shall now show that $\pi_{t}\circ\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }=\operatorname*{id}$. Every $p\in A$ satisfies%
\[
\left(  \pi_{t}\circ\widetilde{\mathbf{D}}_{\left\langle t\right\rangle
}\right)  \left(  p\right)  =\pi_{t}\left(  \underbrace{\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  p\right)  }_{\substack{=\sum
_{i\in\mathbb{N}}D_{i}\left(  p\right)  t^{i}\\\text{(by the definition of
}\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\text{)}}}\right)
=\pi_{t}\left(  \sum_{i\in\mathbb{N}}D_{i}\left(  p\right)  t^{i}\right)
=D_{0}\left(  p\right)
\]
(by the definition of $\pi_{t}$). Thus, $\pi_{t}\circ\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }=D_{0}=\operatorname*{id}$. Hence, we have
shown
\[
\pi_{t}\circ\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }%
=\operatorname*{id}\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \widetilde{\mathbf{D}}_{\left\langle u\right\rangle
}\left[  \left[  t\right]  \right]  \circ\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }=\widetilde{\mathbf{D}}_{\left\langle u+t\right\rangle }.
\]
This proves the $\Longrightarrow$ direction of Theorem
\ref{thm.HSD.powerser.div} \textbf{(b)}.

$\Longleftarrow:$ Assume that $\pi_{t}\circ\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }=\operatorname*{id}$ and
$\widetilde{\mathbf{D}}_{\left\langle u\right\rangle }\left[  \left[
t\right]  \right]  \circ\widetilde{\mathbf{D}}_{\left\langle t\right\rangle
}=\widetilde{\mathbf{D}}_{\left\langle u+t\right\rangle }$. Let $p\in A$.
Then, (\ref{pf.thm.HSD.powerser.div.1}) yields%
\begin{align*}
&  \sum_{\left(  n,m\right)  \in\mathbb{N}^{2}}\left(  D_{n}\circ
D_{m}\right)  \left(  p\right)  u^{n}t^{m}\\
&  =\underbrace{\left(  \widetilde{\mathbf{D}}_{\left\langle u\right\rangle
}\left[  \left[  t\right]  \right]  \circ\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }\right)  }_{=\widetilde{\mathbf{D}}_{\left\langle
u+t\right\rangle }}\left(  p\right)  =\widetilde{\mathbf{D}}_{\left\langle
u+t\right\rangle }\left(  p\right)  =\sum_{\left(  n,m\right)  \in
\mathbb{N}^{2}}\dbinom{n+m}{n}D_{n+m}\left(  p\right)  u^{n}t^{m}%
\end{align*}
(by (\ref{pf.thm.HSD.powerser.div.2})). This can be viewed as an identity
between power series in $\mathbb{F}\left[  \left[  u,t\right]  \right]  $.
Thus, comparing coefficients before $u^{n}t^{m}$ in this identity, we obtain%
\begin{equation}
\left(  D_{n}\circ D_{m}\right)  \left(  p\right)  =\dbinom{n+m}{n}%
D_{n+m}\left(  p\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }n\in
\mathbb{N}\text{ and }m\in\mathbb{N}. \label{pf.thm.HSD.powerser.div.5}%
\end{equation}


Let us now forget that we fixed $p$. We thus have shown
(\ref{pf.thm.HSD.powerser.div.5}) for every $p\in A$. In other words,
$D_{n}\circ D_{m}=\dbinom{n+m}{n}D_{n+m}$ for all $n\in\mathbb{N}$ and
$m\in\mathbb{N}$. Thus, (\ref{def.HSD.deriv.nm}) holds. We shall now show that
$D_{0}=\operatorname*{id}$. Indeed, we have $\pi_{t}\circ\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }=D_{0}$ (this can be proven just as in the
proof of the $\Longrightarrow$ direction). Thus, $D_{0}=\pi_{t}\circ
\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }=\operatorname*{id}$.
Thus, we have shown that $D_{0}=\operatorname*{id}$ and that
(\ref{def.HSD.deriv.nm}) holds. In other words, $\mathbf{D}$ is
divided-powers. This proves the $\Longleftarrow$ direction of Theorem
\ref{thm.HSD.powerser.div} \textbf{(b)}. Theorem \ref{thm.HSD.powerser.div}
\textbf{(b)} is thus proven.
\end{proof}

Theorem \ref{thm.HSD.powerser} shows that HSDs are just algebra homomorphisms
(to a different target) in disguise; this makes dealing with them a lot
easier. Here is one sample application:

\begin{corollary}
\label{cor.HSD.equal-on-gens}Let $A$ and $B$ be two $\mathbb{F}$-algebras. Let
$\mathbf{D}=\left(  D_{0},D_{1},D_{2},\ldots\right)  $ and $\mathbf{E}=\left(
E_{0},E_{1},E_{2},\ldots\right)  $ be two HSDs from $A$ to $B$.

\textbf{(a)} If $a$ is an invertible element of $A$ such that every
$n\in\mathbb{N}$ satisfies $D_{n}\left(  a\right)  =E_{n}\left(  a\right)  $,
then every $n\in\mathbb{N}$ satisfies $D_{n}\left(  a^{-1}\right)
=E_{n}\left(  a^{-1}\right)  $.

\textbf{(b)} Let $G$ be a subset of $A$ such that $G$ generates $A$ as an
$\mathbb{F}$-algebra. Assume that $D_{n}\left(  g\right)  =E_{n}\left(
g\right)  $ for every $g\in G$ and every $n\in\mathbb{N}$. Then,
$\mathbf{D}=\mathbf{E}$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.HSD.equal-on-gens}.]Let $t$ be a new symbol.
Consider the $\mathbb{F}$-linear maps $\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }:A\rightarrow B\left[  \left[  t\right]  \right]  $ and
$\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }:A\rightarrow B\left[
\left[  t\right]  \right]  $ according to Definition \ref{def.HSD.powerser}.
Then, Theorem \ref{thm.HSD.powerser} yields that $\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }$ is an $\mathbb{F}$-algebra homomorphism
(since $\mathbf{D}$ is an HSD). Similarly, $\widetilde{\mathbf{E}%
}_{\left\langle t\right\rangle }$ is an $\mathbb{F}$-algebra homomorphism.

\textbf{(a)} Let $a$ be an invertible element of $A$ such that every
$n\in\mathbb{N}$ satisfies $D_{n}\left(  a\right)  =E_{n}\left(  a\right)  $.
The definition of $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$
yields%
\[
\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  a\right)
=\sum_{i\in\mathbb{N}}\underbrace{D_{i}\left(  a\right)  }_{\substack{=E_{i}%
\left(  a\right)  \\\text{(since every }n\in\mathbb{N}\\\text{satisfies }%
D_{n}\left(  a\right)  =E_{n}\left(  a\right)  \text{)}}}t^{i}=\sum
_{i\in\mathbb{N}}E_{i}\left(  a\right)  t^{i}=\widetilde{\mathbf{E}%
}_{\left\langle t\right\rangle }\left(  a\right)
\]
(by the definition of $\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }%
$). But since $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$ is an
$\mathbb{F}$-algebra homomorphism, we have $\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  a^{-1}\right)  =\left(
\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  a\right)
\right)  ^{-1}$. Similarly, $\widetilde{\mathbf{E}}_{\left\langle
t\right\rangle }\left(  a^{-1}\right)  =\left(  \widetilde{\mathbf{E}%
}_{\left\langle t\right\rangle }\left(  a\right)  \right)  ^{-1}$. Thus,%
\begin{align*}
\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  a^{-1}\right)
&  =\left(  \widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(
a\right)  \right)  ^{-1}=\left(  \widetilde{\mathbf{E}}_{\left\langle
t\right\rangle }\left(  a\right)  \right)  ^{-1}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(
a\right)  =\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }\left(
a\right)  \right) \\
&  =\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }\left(
a^{-1}\right)  .
\end{align*}
Since $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(
a^{-1}\right)  =\sum_{i\in\mathbb{N}}D_{i}\left(  a^{-1}\right)  t^{i}$ (by
the definition of $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$) and
$\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }\left(  a^{-1}\right)
=\sum_{i\in\mathbb{N}}E_{i}\left(  a^{-1}\right)  t^{i}$ (similarly), this
rewrites as $\sum_{i\in\mathbb{N}}D_{i}\left(  a^{-1}\right)  t^{i}=\sum
_{i\in\mathbb{N}}E_{i}\left(  a^{-1}\right)  t^{i}$. Comparing coefficients in
this equality, we conclude that every $n\in\mathbb{N}$ satisfies $D_{n}\left(
a^{-1}\right)  =E_{n}\left(  a^{-1}\right)  $. This proves Corollary
\ref{cor.HSD.equal-on-gens} \textbf{(a)}.

\textbf{(b)} For every $g\in G$, we have%
\[
\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  g\right)
=\sum_{i\in\mathbb{N}}\underbrace{D_{i}\left(  g\right)  }_{\substack{=E_{i}%
\left(  g\right)  \\\text{(since every }n\in\mathbb{N}\\\text{satisfies }%
D_{n}\left(  g\right)  =E_{n}\left(  g\right)  \text{)}}}t^{i}=\sum
_{i\in\mathbb{N}}E_{i}\left(  g\right)  t^{i}=\widetilde{\mathbf{E}%
}_{\left\langle t\right\rangle }\left(  g\right)
\]
(by the definition of $\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }%
$). Hence, the two $\mathbb{F}$-algebra homomorphisms $\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }$ and $\widetilde{\mathbf{E}}_{\left\langle
t\right\rangle }$ are equal to each other on $S$. Since the set $S$ generates
the $\mathbb{F}$-algebra $A$, this yields that these two homomorphisms are
identical. That is, we have $\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }=\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }$. Now,
let $a\in A$. Then, the definition of $\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }$ yields%
\[
\sum_{i\in\mathbb{N}}D_{i}\left(  a\right)  t^{i}%
=\underbrace{\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }%
}_{=\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }}\left(  a\right)
=\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }\left(  a\right)
=\sum_{i\in\mathbb{N}}E_{i}\left(  a\right)  t^{i}%
\]
(by the definition of $\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }%
$). Comparing coefficients, we obtain $D_{n}\left(  a\right)  =E_{n}\left(
a\right)  $ for every $n\in\mathbb{N}$. Since this holds for every $a\in A$,
we can thus conclude that $D_{n}=E_{n}$ for every $n\in\mathbb{N}$. In other
words, $\mathbf{D}=\mathbf{E}$. This proves Corollary
\ref{cor.HSD.equal-on-gens} \textbf{(b)}.
\end{proof}

As another application of Theorem \ref{thm.HSD.powerser}, let us show a
(denominator-free!) formal version of Taylor's expansion formula:

\begin{theorem}
\label{thm.taylor}Let $U$ be a commutative $\mathbb{F}$-algebra. Let $z$ and
$t$ be two distinct symbols. Then,%
\[
a\left(  z+t\right)  =\sum_{n\in\mathbb{N}}t^{n}\partial_{z}^{\left(
n\right)  }a\left(  z\right)  \ \ \ \ \ \ \ \ \ \ \text{in }U\left[  \left[
z,t\right]  \right]
\]
for any $a\left(  z\right)  \in U\left[  \left[  z\right]  \right]  $.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.taylor}.]Set $A=U\left[  \left[  z\right]  \right]
$. We let $\mathbf{D}$ denote the sequence $\left(  \partial_{z}^{\left(
0\right)  },\partial_{z}^{\left(  1\right)  },\partial_{z}^{\left(  2\right)
},\ldots\right)  $ of endomorphisms of $A$. Proposition \ref{prop.delta.hasse}
\textbf{(a)} shows that this sequence $\mathbf{D}$ is an HSD from $A$ to $A$.
Theorem \ref{thm.HSD.powerser} (applied to $B=A$) thus yields that
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }:A\rightarrow A\left[
\left[  t\right]  \right]  $ is an $\mathbb{F}$-algebra homomorphism.

We define a map $\mathbf{e}:A\rightarrow U\left[  \left[  z,t\right]  \right]
$ by%
\[
\left(  \mathbf{e}\left(  a\left(  z\right)  \right)  =a\left(  z+t\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\left(  z\right)  \in A\right)  .
\]
It is easy to see that this map $\mathbf{e}$ is an $\mathbb{F}$-algebra
homomorphism\footnote{Indeed, this is because $\mathbf{e}$ is a substitution
homomorphism. Explicitly, for instance, we see that the map $\mathbf{e}$
respects multiplication because any $a\left(  z\right)  \in A$ and $b\left(
z\right)  \in A$ satisfy%
\begin{align*}
\mathbf{e}\left(  a\left(  z\right)  b\left(  z\right)  \right)   &  =\left(
a\left(  z\right)  b\left(  z\right)  \right)  \left(  z+t\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{e}\right) \\
&  =\underbrace{a\left(  z+t\right)  }_{\substack{=\mathbf{e}\left(  a\left(
z\right)  \right)  \\\text{(by the definition of }\mathbf{e}\text{)}%
}}\underbrace{b\left(  z+t\right)  }_{\substack{=\mathbf{e}\left(  b\left(
z\right)  \right)  \\\text{(by the definition of }\mathbf{e}\text{)}%
}}=\mathbf{e}\left(  a\left(  z\right)  \right)  \mathbf{e}\left(  b\left(
z\right)  \right)  .
\end{align*}
}. Moreover, $\mathbf{e}$ is continuous (with respect to the usual topologies
on $A$ and $U\left[  \left[  z,t\right]  \right]  $).

Due to the definition of $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle
}$, we see that every $a\left(  z\right)  \in A$ satisfies%
\begin{equation}
\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  a\left(
z\right)  \right)  =\sum_{i\in\mathbb{N}}\partial_{z}^{\left(  i\right)
}\left(  a\left(  z\right)  \right)  t^{i}=\sum_{n\in\mathbb{N}}\partial
_{z}^{\left(  n\right)  }\left(  a\left(  z\right)  \right)  t^{n}=\sum
_{n\in\mathbb{N}}t^{n}\partial_{z}^{\left(  n\right)  }a\left(  z\right)  ,
\label{pf.thm.taylor.1}%
\end{equation}
Applying this to $a\left(  z\right)  =z$, we obtain%
\begin{align*}
\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  z\right)   &
=\sum_{n\in\mathbb{N}}t^{n}\partial_{z}^{\left(  n\right)  }%
z=\underbrace{t^{0}}_{=1}\underbrace{\partial_{z}^{\left(  0\right)  }%
z}_{\substack{=z\\\text{(by the}\\\text{definition of }\partial_{z}^{\left(
0\right)  }\text{)}}}+\underbrace{t^{1}}_{=t}\underbrace{\partial_{z}^{\left(
1\right)  }\left(  z\right)  }_{\substack{=1\\\text{(by the}\\\text{definition
of }\partial_{z}^{\left(  1\right)  }\text{)}}}+\sum_{\substack{n\in
\mathbb{N};\\n\geq2}}t^{n}\underbrace{\partial_{z}^{\left(  n\right)  }%
z}_{\substack{=0\\\text{(by the}\\\text{definition of }\partial_{z}^{\left(
n\right)  }\text{,}\\\text{since }n\geq2\text{)}}}\\
&  =z+t+\underbrace{\sum_{\substack{n\in\mathbb{N};\\n\geq2}}t^{n}0}%
_{=0}=z+t=\mathbf{e}\left(  z\right)
\end{align*}
(since the definition of $\mathbf{e}$ yields $\mathbf{e}\left(  z\right)
=z+t$).

Recall that $U\left[  z\right]  $ is an $\mathbb{F}$-subalgebra of $U\left[
\left[  z\right]  \right]  =A$. The restrictions of the maps
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$ and $\mathbf{e}$ to
this $\mathbb{F}$-subalgebra $U\left[  z\right]  $ are $\mathbb{F}$-algebra
homomorphisms\footnote{since $\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }$ and $\mathbf{e}$ are $\mathbb{F}$-algebra homomorphisms},
and are equal to each other on the generating set $\left\{  z\right\}  $ of
the $\mathbb{F}$-algebra $U\left[  z\right]  $ (since $\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  z\right)  =\mathbf{e}\left(  z\right)
$). Hence, these restrictions are equal.

We recall that $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$ is a
map $A\rightarrow A\left[  \left[  t\right]  \right]  $. Using the canonical
isomorphism $\underbrace{A}_{=U\left[  \left[  z\right]  \right]  }\left[
\left[  t\right]  \right]  =\left(  U\left[  \left[  z\right]  \right]
\right)  \left[  \left[  t\right]  \right]  \cong U\left[  \left[  z,t\right]
\right]  $, we regard $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$
as a map $A\rightarrow U\left[  \left[  z,t\right]  \right]  $. It is easy to
see that this map $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$ is
continuous\footnote{This is because the $z^{m}t^{n}$-coefficient of
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  a\left(
z\right)  \right)  $ depends only on the $z^{m+n}t^{n}$-coefficient of
$a\left(  z\right)  $. (This follows from (\ref{pf.thm.taylor.1}).)}. Now, the
two maps $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$ and
$\mathbf{e}$ are continuous, and their restrictions to the $\mathbb{F}%
$-subalgebra $U\left[  z\right]  $ are equal. This yields that the two maps
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$ and $\mathbf{e}$ must
be equal (since $U\left[  z\right]  $ is a dense subset of $U\left[  \left[
z\right]  \right]  =A$). In other words, $\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }=\mathbf{e}$. Thus, every $a\left(  z\right)  \in A$ satisfies
$\underbrace{\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }%
}_{=\mathbf{e}}\left(  a\left(  z\right)  \right)  =\mathbf{e}\left(  a\left(
z\right)  \right)  =a\left(  z+t\right)  $. Since $A=U\left[  \left[
z\right]  \right]  $ and $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle
}\left(  a\left(  z\right)  \right)  =\sum_{n\in\mathbb{N}}t^{n}\partial
_{z}^{\left(  n\right)  }a\left(  z\right)  $, this rewrites as follows: Every
$a\left(  z\right)  \in A$ satisfies $\sum_{n\in\mathbb{N}}t^{n}\partial
_{z}^{\left(  n\right)  }a\left(  z\right)  =a\left(  z+t\right)  $. This
proves Theorem \ref{thm.taylor}.
\end{proof}

\subsection{Extending Hasse-Schmidt derivations to localizations}

A more interesting application of Theorem \ref{thm.HSD.powerser} allows us to
extend Hasse-Schmidt derivations to localizations of commutative $\mathbb{F}%
$-algebras. A similar result holds for usual (not Hasse-Schmidt) derivations,
but we shall only state the one for HSDs:

\begin{corollary}
\label{cor.HSD.extend-to-loc}Let $A$ and $B$ be two commutative $\mathbb{F}%
$-algebras, and let $\mathbf{D}=\left(  D_{0},D_{1},D_{2},\ldots\right)  $ be
an HSD from $A$ to $B$. Let $S$ be a multiplicatively closed subset of $A$.
Assume that $D_{0}\left(  s\right)  $ is an invertible element of $B$ for
every $s\in S$. Then, there exists a unique HSD $\mathbf{D}^{\prime}=\left(
D_{0}^{\prime},D_{1}^{\prime},D_{2}^{\prime},\ldots\right)  $ from $S^{-1}A$
to $B$ such that every $a\in A$ and $n\in\mathbb{N}$ satisfy $D_{n}^{\prime
}\left(  \dfrac{a}{1}\right)  =D_{n}\left(  a\right)  $.
\end{corollary}

Before we prove this corollary, let us recall some basic properties of localization:

\begin{proposition}
\label{prop.localization.univ}Let $A$ be a commutative $\mathbb{F}$-algebra.
Let $S$ be a multiplicatively closed subset of $A$. Let $\iota:A\rightarrow
S^{-1}A$ be the canonical $\mathbb{F}$-algebra homomorphism from $A$ to
$S^{-1}A$ sending each $a\in A$ to $\dfrac{a}{1}\in S^{-1}A$. (This $\iota$
might and might not be injective.)

Let $B$ be any commutative $\mathbb{F}$-algebra.

\textbf{(a)} If $\phi:A\rightarrow B$ is an $\mathbb{F}$-algebra homomorphism
with the property that $\left(  \phi\left(  a\right)  \text{ is an invertible
element of }B\text{ for every }s\in S\right)  $, then there exists a unique
$\mathbb{F}$-algebra homomorphism $\phi^{\prime}:S^{-1}A\rightarrow B$ such
that $\phi^{\prime}\circ\iota=\phi$.

\textbf{(b)} If $\phi$ and $\psi$ are two $\mathbb{F}$-algebra homomorphisms
$S^{-1}A\rightarrow B$ satisfying $\phi\circ\iota=\psi\circ\iota$, then
$\phi=\psi$.
\end{proposition}

Notice that Proposition \ref{prop.localization.univ} \textbf{(a)} is the
universal property of the localization $S^{-1}A$ as a commutative $\mathbb{F}%
$-algebra. Proposition \ref{prop.localization.univ} \textbf{(b)} says (in
categorical language) that $\iota:A\rightarrow S^{-1}A$ is an epimorphism in
the category of commutative $\mathbb{F}$-algebras.\footnote{It is the simplest
example of epimorphisms in this category which are not surjective -- or at
least not usually surjective.}

\begin{proof}
[Proof of Proposition \ref{prop.localization.univ}.]\textbf{(a)} Proposition
\ref{prop.localization.univ} \textbf{(a)} is well-known (as we said, it is a
standard universal property), so we omit its proof.

\textbf{(b)} Let $\phi$ and $\psi$ be two $\mathbb{F}$-algebra homomorphisms
$A\rightarrow B$ satisfying $\phi\circ\iota=\psi\circ\iota$. Let $x\in
S^{-1}A$. Then, $x=\dfrac{p}{q}$ for some $p\in A$ and $q\in S$. Consider
these $p$ and $q$. We have $x=\dfrac{p}{q}=\underbrace{\dfrac{p}{1}}%
_{=\iota\left(  p\right)  }\cdot\left(  \underbrace{\dfrac{q}{1}}%
_{=\iota\left(  q\right)  }\right)  ^{-1}=\iota\left(  p\right)  \cdot\left(
\iota\left(  q\right)  \right)  ^{-1}$, thus%
\begin{align*}
\phi\left(  x\right)   &  =\phi\left(  \iota\left(  p\right)  \cdot\left(
\iota\left(  q\right)  \right)  ^{-1}\right)  =\underbrace{\phi\left(
\iota\left(  p\right)  \right)  }_{\substack{=\left(  \phi\circ\iota\right)
\left(  p\right)  =\left(  \psi\circ\iota\right)  \left(  p\right)
\\\text{(since }\phi\circ\iota=\psi\circ\iota\text{)}}}\cdot\left(
\underbrace{\phi\left(  \iota\left(  q\right)  \right)  }_{\substack{=\left(
\phi\circ\iota\right)  \left(  q\right)  =\left(  \psi\circ\iota\right)
\left(  q\right)  \\\text{(since }\phi\circ\iota=\psi\circ\iota\text{)}%
}}\right)  ^{-1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\phi\text{ is an }\mathbb{F}%
\text{-algebra homomorphism}\right) \\
&  =\left(  \psi\circ\iota\right)  \left(  p\right)  \cdot\left(  \left(
\psi\circ\iota\right)  \left(  q\right)  \right)  ^{-1}.
\end{align*}
Compared with%
\begin{align*}
\psi\left(  x\right)   &  =\psi\left(  \iota\left(  p\right)  \cdot\left(
\iota\left(  q\right)  \right)  ^{-1}\right)  =\underbrace{\psi\left(
\iota\left(  p\right)  \right)  }_{=\left(  \psi\circ\iota\right)  \left(
p\right)  }\cdot\left(  \underbrace{\psi\left(  \iota\left(  q\right)
\right)  }_{=\left(  \psi\circ\iota\right)  \left(  q\right)  }\right)
^{-1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\phi\text{ is an }\mathbb{F}%
\text{-algebra homomorphism}\right) \\
&  =\left(  \psi\circ\iota\right)  \left(  p\right)  \cdot\left(  \left(
\psi\circ\iota\right)  \left(  q\right)  \right)  ^{-1},
\end{align*}
this yields $\phi\left(  x\right)  =\psi\left(  x\right)  $. Let us now forget
that we fixed $x$. We thus have proven that $\phi\left(  x\right)
=\psi\left(  x\right)  $ for every $x\in S^{-1}A$. In other words, $\phi=\psi
$. This proves Proposition \ref{prop.localization.univ} \textbf{(b)}.
\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor.HSD.equal-on-gens}.]Let $\iota:A\rightarrow
S^{-1}A$ be the canonical $\mathbb{F}$-algebra homomorphism from $A$ to
$S^{-1}A$ sending each $a\in A$ to $\dfrac{a}{1}\in S^{-1}A$. (This $\iota$
might and might not be injective.)

Let $t$ be a new symbol. We must show that there exists a unique HSD
$\mathbf{D}^{\prime}=\left(  D_{0}^{\prime},D_{1}^{\prime},D_{2}^{\prime
},\ldots\right)  $ from $S^{-1}A$ to $B$ such that every $a\in A$ and
$n\in\mathbb{N}$ satisfy $D_{n}^{\prime}\left(  \dfrac{a}{1}\right)
=D_{n}\left(  a\right)  $. In other words, we must show that there exists a
unique HSD $\mathbf{D}^{\prime}=\left(  D_{0}^{\prime},D_{1}^{\prime}%
,D_{2}^{\prime},\ldots\right)  $ from $S^{-1}A$ to $B$ such that
$\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }\circ
\iota=\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$%
\ \ \ \ \footnote{This is because of the following the logical equivalence:%
\begin{align*}
&  \ \left(  \text{every }a\in A\text{ and }n\in\mathbb{N}\text{ satisfy
}D_{n}^{\prime}\left(  \dfrac{a}{1}\right)  =D_{n}\left(  a\right)  \right) \\
&  \Longleftrightarrow\ \left(  \text{every }a\in A\text{ satisfies
}\underbrace{\left(  \text{every }n\in\mathbb{N}\text{ satisfies }%
D_{n}^{\prime}\left(  \dfrac{a}{1}\right)  =D_{n}\left(  a\right)  \right)
}_{\Longleftrightarrow\left(  \sum_{n\in\mathbb{N}}D_{n}^{\prime}\left(
\dfrac{a}{1}\right)  t^{n}=\sum_{n\in\mathbb{N}}D_{n}\left(  a\right)
t^{n}\text{ in }B\left[  \left[  t\right]  \right]  \right)  }\right) \\
&  \Longleftrightarrow\ \left(  \text{every }a\in A\text{ satisfies
}\underbrace{\sum_{n\in\mathbb{N}}D_{n}^{\prime}\left(  \dfrac{a}{1}\right)
t^{n}}_{\substack{=\widetilde{\mathbf{D}^{\prime}}_{\left\langle
t\right\rangle }\left(  \dfrac{a}{1}\right)  \\\text{(by the definition of
}\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }\text{)}%
}}=\underbrace{\sum_{n\in\mathbb{N}}D_{n}\left(  a\right)  t^{n}%
}_{\substack{=\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(
a\right)  \\\text{(by the definition of }\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }\text{)}}}\text{ in }B\left[  \left[  t\right]  \right]
\right) \\
&  \Longleftrightarrow\ \left(  \text{every }a\in A\text{ satisfies
}\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }\left(
\underbrace{\dfrac{a}{1}}_{\substack{=\iota\left(  a\right)  \\\text{(by the
definition of }\iota\text{)}}}\right)  =\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }\left(  a\right)  \right) \\
&  \Longleftrightarrow\ \left(  \text{every }a\in A\text{ satisfies
}\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }\left(
\iota\left(  a\right)  \right)  =\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }\left(  a\right)  \right)  \ \Longleftrightarrow\ \left(
\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }\circ
\iota=\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\right)  .
\end{align*}
}. We will now prove the uniqueness and the existence parts of this statement separately.

\textit{Uniqueness:} We will show that there exists at most one HSD
$\mathbf{D}^{\prime}=\left(  D_{0}^{\prime},D_{1}^{\prime},D_{2}^{\prime
},\ldots\right)  $ from $S^{-1}A$ to $B$ such that $\widetilde{\mathbf{D}%
^{\prime}}_{\left\langle t\right\rangle }\circ\iota=\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }$. Indeed, let $\mathbf{E}$ and $\mathbf{F}$
be two such HSDs $\mathbf{D}^{\prime}$. We are going to show that
$\mathbf{E}=\mathbf{F}$.

We know that $\mathbf{E}$ and $\mathbf{F}$ are two HSDs $\mathbf{D}^{\prime
}=\left(  D_{0}^{\prime},D_{1}^{\prime},D_{2}^{\prime},\ldots\right)  $ from
$S^{-1}A$ to $B$ such that $\widetilde{\mathbf{D}^{\prime}}_{\left\langle
t\right\rangle }\circ\iota=\widetilde{\mathbf{D}}_{\left\langle t\right\rangle
}$. In other words, $\mathbf{E}$ and $\mathbf{F}$ are two HSDs from $S^{-1}A$
to $B$ such that $\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }%
\circ\iota=\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$ and
$\widetilde{\mathbf{F}}_{\left\langle t\right\rangle }\circ\iota
=\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$.

Theorem \ref{thm.HSD.powerser} shows that $\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }:A\rightarrow B\left[  \left[  t\right]
\right]  $ is an $\mathbb{F}$-algebra homomorphism (since $\mathbf{D}$ is an
HSD). Similarly, $\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }%
:S^{-1}A\rightarrow B\left[  \left[  t\right]  \right]  $ and
$\widetilde{\mathbf{F}}_{\left\langle t\right\rangle }:S^{-1}A\rightarrow
B\left[  \left[  t\right]  \right]  $ are $\mathbb{F}$-algebra homomorphisms.
Thus, Proposition \ref{prop.localization.univ} \textbf{(b)} (applied to
$B\left[  \left[  t\right]  \right]  $, $\widetilde{\mathbf{E}}_{\left\langle
t\right\rangle }$ and $\widetilde{\mathbf{F}}_{\left\langle t\right\rangle }$
instead of $B$, $\phi$ and $\psi$) yields $\widetilde{\mathbf{E}%
}_{\left\langle t\right\rangle }=\widetilde{\mathbf{F}}_{\left\langle
t\right\rangle }$ (since $\widetilde{\mathbf{E}}_{\left\langle t\right\rangle
}\circ\iota=\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }%
=\widetilde{\mathbf{F}}_{\left\langle t\right\rangle }\circ\iota$).

Now, let $x\in S^{-1}A$. Then, the definition of $\widetilde{\mathbf{E}%
}_{\left\langle t\right\rangle }$ yields%
\[
\sum_{n\in\mathbb{N}}E_{n}\left(  x\right)  t^{n}%
=\underbrace{\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }%
}_{=\widetilde{\mathbf{F}}_{\left\langle t\right\rangle }}\left(  x\right)
=\widetilde{\mathbf{F}}_{\left\langle t\right\rangle }\left(  x\right)
=\sum_{n\in\mathbb{N}}F_{n}\left(  x\right)  t^{n}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\widetilde{\mathbf{F}}_{\left\langle
t\right\rangle }\right)  .
\]
Comparing coefficients in this equality, we obtain $E_{n}\left(  x\right)
=F_{n}\left(  x\right)  $ for every $n\in\mathbb{N}$.

Now, let us forget that we fixed $x$. We thus have shown that $E_{n}\left(
x\right)  =F_{n}\left(  x\right)  $ for every $n\in\mathbb{N}$ and every $x\in
S^{-1}A$. In other words, $\mathbf{E}=\mathbf{F}$. This completes the proof of
the uniqueness statement of Corollary \ref{cor.HSD.extend-to-loc}.

\textit{Existence:} We will now show that there exists at least one HSD
$\mathbf{D}^{\prime}=\left(  D_{0}^{\prime},D_{1}^{\prime},D_{2}^{\prime
},\ldots\right)  $ from $S^{-1}A$ to $B$ such that $\widetilde{\mathbf{D}%
^{\prime}}_{\left\langle t\right\rangle }\circ\iota=\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }$. Indeed, we are going to construct such an
HSD $\mathbf{D}^{\prime}$ explicitly.

Theorem \ref{thm.HSD.powerser} shows that $\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }:A\rightarrow B\left[  \left[  t\right]
\right]  $ is an $\mathbb{F}$-algebra homomorphism (since $\mathbf{D}$ is an
HSD). Moreover, $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(
s\right)  $ is an invertible element of $B\left[  \left[  t\right]  \right]  $
for every $s\in S$\ \ \ \ \footnote{\textit{Proof.} Let $s\in S$. Then,
$D_{0}\left(  s\right)  $ is an invertible element of $B$ (by one of the
hypotheses of Corollary \ref{cor.HSD.extend-to-loc}). Now, the definition of
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$ yields
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  s\right)
=\sum_{n\in\mathbb{N}}D_{n}\left(  s\right)  t^{n}$. This is a power series
whose constant coefficient $D_{0}\left(  s\right)  $ is invertible. It is
known that every such power series is invertible. Thus, this power series
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  s\right)  $ is
invertible in $B\left[  \left[  t\right]  \right]  $, qed.}. Thus, we can
apply Proposition \ref{prop.localization.univ} \textbf{(a)} to $B\left[
\left[  t\right]  \right]  $ and $\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }$ instead of $B$ and $\phi$. As a consequence, we conclude
that there exists a unique $\mathbb{F}$-algebra homomorphism $\phi^{\prime
}:S^{-1}A\rightarrow B\left[  \left[  t\right]  \right]  $ such that
$\phi^{\prime}\circ\iota=\widetilde{\mathbf{D}}_{\left\langle t\right\rangle
}$. Consider this $\phi^{\prime}$. Now, for every $n\in\mathbb{N}$, we define
an $\mathbb{F}$-linear map $D_{n}^{\prime}:S^{-1}A\rightarrow B$ as follows:
For every $x\in S^{-1}A$, let $D_{n}^{\prime}\left(  x\right)  $ be the
coefficient of $t^{n}$ in the power series $\phi^{\prime}\left(  x\right)  $.
Then, for every $x\in S^{-1}A$, we have%
\begin{equation}
\phi^{\prime}\left(  x\right)  =\sum_{n\in\mathbb{N}}D_{n}^{\prime}\left(
x\right)  t^{n}. \label{pf.cor.HSD.extend-to-loc.exi.5}%
\end{equation}


Now, let $\mathbf{D}^{\prime}$ denote the sequence $\left(  D_{0}^{\prime
},D_{1}^{\prime},D_{2}^{\prime},\ldots\right)  $ of $\mathbb{F}$-linear maps
from $S^{-1}A$ to $B$. We shall now show that $\mathbf{D}^{\prime}$ is an HSD
from $S^{-1}A$ to $B$ such that $\widetilde{\mathbf{D}^{\prime}}_{\left\langle
t\right\rangle }\circ\iota=\widetilde{\mathbf{D}}_{\left\langle t\right\rangle
}$. Once this is proven, the existence statement of Corollary
\ref{cor.HSD.extend-to-loc} will be proven, and thus the proof of Corollary
\ref{cor.HSD.extend-to-loc} will be complete.

For every $x\in S^{-1}A$, we have%
\begin{align*}
\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }\left(
x\right)   &  =\sum_{n\in\mathbb{N}}D_{n}^{\prime}\left(  x\right)
t^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }\right) \\
&  =\phi^{\prime}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.cor.HSD.extend-to-loc.exi.5})}\right)  .
\end{align*}
Hence, $\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }%
=\phi^{\prime}$. Thus, $\widetilde{\mathbf{D}^{\prime}}_{\left\langle
t\right\rangle }:S^{-1}A\rightarrow B\left[  \left[  t\right]  \right]  $ is
an $\mathbb{F}$-algebra homomorphism. Thus, Theorem \ref{thm.HSD.powerser}
(applied to $S^{-1}A$, $D_{i}^{\prime}$ and $\mathbf{D}^{\prime}$ instead of
$A$, $D_{i}$ and $\mathbf{D}$) yields that $\mathbf{D}^{\prime}$ is an HSD.
Also, that $\underbrace{\widetilde{\mathbf{D}^{\prime}}_{\left\langle
t\right\rangle }}_{=\phi^{\prime}}\circ\iota=\phi^{\prime}\circ\iota
=\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$. This completes the
proof of Corollary \ref{cor.HSD.extend-to-loc}.
\end{proof}

\begin{definition}
\label{def.HSD.compose}Let $A$, $B$ and $C$ be $\mathbb{F}$-algebras.

\textbf{(a)} If $\mathbf{D}=\left(  D_{0},D_{1},D_{2},\ldots\right)  $ is an
HSD from $A$ to $B$, and if $\phi:B\rightarrow C$ is an $\mathbb{F}$-algebra
homomorphism, then we denote by $\phi\circ\mathbf{D}$ the sequence $\left(
\phi\circ D_{0},\phi\circ D_{1},\phi\circ D_{2},\ldots\right)  $ of
$\mathbb{F}$-linear maps $A\rightarrow C$. It is easy to see that this
sequence $\phi\circ\mathbf{D}$ is an HSD from $A$ to $C$.

\textbf{(b)} If $\mathbf{D}=\left(  D_{0},D_{1},D_{2},\ldots\right)  $ is an
HSD from $B$ to $C$, and if $\psi:A\rightarrow B$ is an $\mathbb{F}$-algebra
homomorphism, then we denote by $\mathbf{D}\circ\psi$ the sequence $\left(
D_{0}\circ\psi,D_{1}\circ\psi,D_{2}\circ\psi,\ldots\right)  $ of $\mathbb{F}%
$-linear maps $A\rightarrow C$. It is easy to see that this sequence
$\mathbf{D}\circ\psi$ is an HSD from $A$ to $C$.
\end{definition}

\begin{corollary}
\label{cor.HSD.extend-to-loc.div}Let $A$ be a commutative $\mathbb{F}%
$-algebra, and let $\mathbf{D}=\left(  D_{0},D_{1},D_{2},\ldots\right)  $ be a
divided-powers HSD from $A$ to $A$. Let $S$ be a multiplicatively closed
subset of $A$.

\textbf{(a)} There exists a unique HSD $\mathbf{D}^{\prime}=\left(
D_{0}^{\prime},D_{1}^{\prime},D_{2}^{\prime},\ldots\right)  $ from $S^{-1}A$
to $S^{-1}A$ such that every $a\in A$ and $n\in\mathbb{N}$ satisfy
$D_{n}^{\prime}\left(  \dfrac{a}{1}\right)  =\dfrac{D_{n}\left(  a\right)
}{1}$.

\textbf{(b)} This HSD $\mathbf{D}^{\prime}$ is divided-powers.

\textbf{(c)} Let $\iota:A\rightarrow S^{-1}A$ be the canonical $\mathbb{F}%
$-algebra homomorphism from $A$ to $S^{-1}A$ sending each $a\in A$ to
$\dfrac{a}{1}\in S^{-1}A$. Then, $\mathbf{D}^{\prime}\circ\iota=\iota
\circ\mathbf{D}$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.HSD.extend-to-loc.div}.]Let $\iota:A\rightarrow
S^{-1}A$ be the canonical $\mathbb{F}$-algebra homomorphism from $A$ to
$S^{-1}A$ sending each $a\in A$ to $\dfrac{a}{1}\in S^{-1}A$. (This $\iota$
might and might not be injective.)

Recall that $\iota\circ\mathbf{D}$ denotes the sequence $\left(  \iota\circ
D_{0},\iota\circ D_{1},\iota\circ D_{2},\ldots\right)  $ of maps from $A$ to
$S^{-1}A$. It is easy to see that $\iota\circ\mathbf{D}$ is an HSD from $A$ to
$S^{-1}A$ (since $\mathbf{D}$ is an HSD from $A$ to $A$, and since $\iota$ is
an $\mathbb{F}$-algebra homomorphism $A\rightarrow S^{-1}A$).

We have $D_{0}=\operatorname*{id}$ (since the HSD $\mathbf{D}$ is divided-powers).

\textbf{(a)} We know that $\left(  \iota\circ D_{0}\right)  \left(  s\right)
$ is an invertible element of $S^{-1}A$ for every $s\in S$%
\ \ \ \ \footnote{\textit{Proof.} Let $s\in S$. Then, $D_{0}%
=\operatorname*{id}$, so that $D_{0}\left(  s\right)  =\operatorname*{id}%
\left(  s\right)  =s$. Thus, $\left(  \iota\circ D_{0}\right)  \left(
s\right)  =\iota\left(  \underbrace{D_{0}\left(  s\right)  }_{=s}\right)
=\iota\left(  s\right)  =\dfrac{s}{1}$. This is clearly invertible in
$S^{-1}A$ (since $s\in S$). Thus, $\left(  \iota\circ D_{0}\right)  \left(
s\right)  $ is invertible in $S^{-1}A$, qed.}. Hence, Corollary
\ref{cor.HSD.extend-to-loc} (applied to $S^{-1}A$, $\iota\circ\mathbf{D}$ and
$\iota\circ D_{i}$ instead of $B$, $\mathbf{D}$ and $D_{i}$) yields that there
exists a unique HSD $\mathbf{D}^{\prime}=\left(  D_{0}^{\prime},D_{1}^{\prime
},D_{2}^{\prime},\ldots\right)  $ from $S^{-1}A$ to $S^{-1}A$ such that every
$a\in A$ and $n\in\mathbb{N}$ satisfy $D_{n}^{\prime}\left(  \dfrac{a}%
{1}\right)  =\left(  \iota\circ D_{n}\right)  \left(  a\right)  $. In other
words, there exists a unique HSD $\mathbf{D}^{\prime}=\left(  D_{0}^{\prime
},D_{1}^{\prime},D_{2}^{\prime},\ldots\right)  $ from $S^{-1}A$ to $S^{-1}A$
such that every $a\in A$ and $n\in\mathbb{N}$ satisfy $D_{n}^{\prime}\left(
\dfrac{a}{1}\right)  =\dfrac{D_{n}\left(  a\right)  }{1}$ (because every $a\in
A$ and $n\in\mathbb{N}$ satisfies $\left(  \iota\circ D_{n}\right)  \left(
a\right)  =\iota\left(  D_{n}\left(  a\right)  \right)  =\dfrac{D_{n}\left(
a\right)  }{1}$). This proves Corollary \ref{cor.HSD.extend-to-loc.div}
\textbf{(a)}.

\textbf{(b)} Consider this HSD $\mathbf{D}^{\prime}$. Let us write it in the
form $\left(  D_{0}^{\prime},D_{1}^{\prime},D_{2}^{\prime},\ldots\right)  $.
Then, every $a\in A$ and $n\in\mathbb{N}$ satisfy%
\begin{equation}
D_{n}^{\prime}\left(  \dfrac{a}{1}\right)  =\dfrac{D_{n}\left(  a\right)  }{1}
\label{pf.cor.HSD.extend-to-loc.div.b.defD'}%
\end{equation}
(according to the definition of $\mathbf{D}^{\prime}$). In other words,
\begin{equation}
D_{n}^{\prime}\circ\iota=\iota\circ D_{n}\ \ \ \ \ \ \ \ \ \ \text{for every
}n\in\mathbb{N} \label{pf.cor.HSD.extend-to-loc.div.b.D'1}%
\end{equation}
\ \footnote{\textit{Proof.} Let $n\in\mathbb{N}$. Every $a\in A$ satisfies
$\iota\left(  a\right)  =\dfrac{a}{1}$ (by the definition of $\iota$) and
$\iota\left(  D_{n}\left(  a\right)  \right)  =\dfrac{D_{n}\left(  a\right)
}{1}$ (by the definition of $\iota$). Hence, every $a\in A$ satisfies%
\begin{align*}
\left(  D_{n}^{\prime}\circ\iota\right)  \left(  a\right)   &  =D_{n}^{\prime
}\left(  \underbrace{\iota\left(  a\right)  }_{=\dfrac{a}{1}}\right)
=D_{n}^{\prime}\left(  \dfrac{a}{1}\right)  =\dfrac{D_{n}\left(  a\right)
}{1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.cor.HSD.extend-to-loc.div.b.defD'})}\right) \\
&  =\iota\left(  D_{n}\left(  a\right)  \right)  =\left(  \iota\circ
D_{n}\right)  \left(  a\right)  .
\end{align*}
Thus, $D_{n}^{\prime}\circ\iota=\iota\circ D_{n}$. This proves
(\ref{pf.cor.HSD.extend-to-loc.div.b.D'1}).}.

Whenever $P$ and $Q$ are two $\mathbb{F}$-modules, $t$ is a symbol, and
$\phi:P\rightarrow Q$ is an $\mathbb{F}$-linear map, we shall denote by
$\phi\left[  \left[  t\right]  \right]  $ the $\mathbb{F}$-linear map
$P\left[  \left[  t\right]  \right]  \rightarrow Q\left[  \left[  t\right]
\right]  $ which sends every $\sum_{n\in\mathbb{N}}p_{n}t^{n}\in P\left[
\left[  t\right]  \right]  $ (with $p_{n}\in P$) to $\sum_{n\in\mathbb{N}%
}q_{n}t^{n}\in Q\left[  \left[  t\right]  \right]  $. Notice that if $P$ and
$Q$ are $\mathbb{F}$-algebras and $\phi$ is an $\mathbb{F}$-algebra
homomorphism, then $\phi\left[  \left[  t\right]  \right]  $ is an
$\mathbb{F}$-algebra homomorphism as well.

Let $t$ and $u$ be two distinct symbols. Define the maps
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$,
$\widetilde{\mathbf{D}}_{\left\langle u\right\rangle }$ and
$\widetilde{\mathbf{D}}_{\left\langle u+t\right\rangle }$ as in Theorem
\ref{thm.HSD.powerser.div}. Furthermore, define the maps
$\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }$,
$\widetilde{\mathbf{D}^{\prime}}_{\left\langle u\right\rangle }$ and
$\widetilde{\mathbf{D}^{\prime}}_{\left\langle u+t\right\rangle }$ in the same
way (but for $\mathbf{D}^{\prime}$, $D_{i}^{\prime}$ and $S^{-1}A$ instead of
$\mathbf{D}$, $D_{i}$ and $A$).

Theorem \ref{thm.HSD.powerser.div} \textbf{(a)} yields that the maps
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$,
$\widetilde{\mathbf{D}}_{\left\langle u\right\rangle }$,
$\widetilde{\mathbf{D}}_{\left\langle u\right\rangle }\left[  \left[
t\right]  \right]  $, $\widetilde{\mathbf{D}}_{\left\langle u\right\rangle
}\left[  \left[  t\right]  \right]  \circ\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }$ and $\widetilde{\mathbf{D}}_{\left\langle u+t\right\rangle
}$ are $\mathbb{F}$-algebra homomorphisms. The same Theorem
\ref{thm.HSD.powerser.div} \textbf{(a)} (but applied to $\mathbf{D}^{\prime}$,
$D_{i}^{\prime}$ and $S^{-1}A$ instead of $\mathbf{D}$, $D_{i}$ and $A$)
yields that the maps $\widetilde{\mathbf{D}^{\prime}}_{\left\langle
t\right\rangle }$, $\widetilde{\mathbf{D}^{\prime}}_{\left\langle
u\right\rangle }$, $\widetilde{\mathbf{D}^{\prime}}_{\left\langle
u\right\rangle }\left[  \left[  t\right]  \right]  $, $\widetilde{\mathbf{D}%
^{\prime}}_{\left\langle u\right\rangle }\left[  \left[  t\right]  \right]
\circ\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }$ and
$\widetilde{\mathbf{D}^{\prime}}_{\left\langle u+t\right\rangle }$ are
$\mathbb{F}$-algebra homomorphisms.

We will use the notation $\pi_{t}$ to denote both the projection map $A\left[
\left[  t\right]  \right]  \rightarrow A$ which sends every formal power
series to its constant coefficient, and the projection map $\left(
S^{-1}A\right)  \left[  \left[  t\right]  \right]  \rightarrow S^{-1}A$ which
sends every formal power series to its constant coefficient. This will not
cause any confusion, because the two maps have distinct domains. Both of these
two maps $\pi_{t}$ are $\mathbb{F}$-algebra homomorphisms.

Recall that $\mathbf{D}$ is divided-powers. Hence, Theorem
\ref{thm.HSD.powerser.div} \textbf{(b)} yields that%
\begin{equation}
\pi_{t}\circ\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }%
=\operatorname*{id}\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \widetilde{\mathbf{D}}_{\left\langle u\right\rangle
}\left[  \left[  t\right]  \right]  \circ\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }=\widetilde{\mathbf{D}}_{\left\langle u+t\right\rangle }.
\label{pf.cor.HSD.extend-to-loc.div.b.givens}%
\end{equation}


Our goal is to prove that $\mathbf{D}^{\prime}$ is divided-powers. According
to Theorem \ref{thm.HSD.powerser.div} \textbf{(b)} (applied to $\mathbf{D}%
^{\prime}$, $D_{i}^{\prime}$ and $S^{-1}A$ instead of $\mathbf{D}$, $D_{i}$
and $A$), this boils down to proving that%
\begin{equation}
\pi_{t}\circ\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle
}=\operatorname*{id}\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \widetilde{\mathbf{D}^{\prime}}_{\left\langle
u\right\rangle }\left[  \left[  t\right]  \right]  \circ\widetilde{\mathbf{D}%
^{\prime}}_{\left\langle t\right\rangle }=\widetilde{\mathbf{D}^{\prime}%
}_{\left\langle u+t\right\rangle }.
\label{pf.cor.HSD.extend-to-loc.div.b.goals}%
\end{equation}


So let us now prove (\ref{pf.cor.HSD.extend-to-loc.div.b.goals}).

We have $\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }%
\circ\iota=\iota\left[  \left[  t\right]  \right]  \circ\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }$\ \ \ \ \footnote{\textit{Proof.} For every
$a\in A$, we have%
\begin{align*}
\left(  \widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }%
\circ\iota\right)  \left(  a\right)   &  =\widetilde{\mathbf{D}^{\prime}%
}_{\left\langle t\right\rangle }\left(  \iota\left(  a\right)  \right)
=\sum_{n\in\mathbb{N}}\underbrace{D_{n}^{\prime}\left(  \iota\left(  a\right)
\right)  }_{=\left(  D_{n}^{\prime}\circ\iota\right)  \left(  a\right)  }%
t^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }\right) \\
&  =\sum_{n\in\mathbb{N}}\underbrace{\left(  D_{n}^{\prime}\circ\iota\right)
}_{\substack{=\iota\circ D_{n}\\\text{(by
(\ref{pf.cor.HSD.extend-to-loc.div.b.D'1}))}}}\left(  a\right)  t^{n}%
=\sum_{n\in\mathbb{N}}\underbrace{\left(  \iota\circ D_{n}\right)  \left(
a\right)  }_{=\iota\left(  D_{n}\left(  a\right)  \right)  }t^{n}\\
&  =\sum_{n\in\mathbb{N}}\iota\left(  D_{n}\left(  a\right)  \right)
t^{n}=\left(  \iota\left[  \left[  t\right]  \right]  \right)  \left(
\underbrace{\sum_{n\in\mathbb{N}}D_{n}\left(  a\right)  t^{n}}%
_{\substack{=\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(
a\right)  \\\text{(by the definition of }\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }\text{)}}}\right)  =\left(  \iota\left[  \left[  t\right]
\right]  \right)  \left(  \widetilde{\mathbf{D}}_{\left\langle t\right\rangle
}\left(  a\right)  \right) \\
&  =\left(  \iota\left[  \left[  t\right]  \right]  \circ\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\right)  \left(  a\right)  .
\end{align*}
Thus, $\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }%
\circ\iota=\iota\left[  \left[  t\right]  \right]  \circ\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }$, qed.}. Similarly, $\widetilde{\mathbf{D}%
^{\prime}}_{\left\langle u\right\rangle }\circ\iota=\left(  \iota\left[
\left[  u\right]  \right]  \right)  \circ\widetilde{\mathbf{D}}_{\left\langle
u\right\rangle }$ and $\widetilde{\mathbf{D}^{\prime}}_{\left\langle
u+t\right\rangle }\circ\iota=\left(  \iota\left[  \left[  u\right]  \right]
\right)  \left[  \left[  t\right]  \right]  \circ\widetilde{\mathbf{D}%
}_{\left\langle u+t\right\rangle }$\ \ \ \ \footnote{When proving the latter
equality, we need to use%
\begin{equation}
\sum_{n\in\mathbb{N}}\iota\left(  D_{n}\left(  a\right)  \right)  \left(
u+t\right)  ^{n}=\left(  \left(  \iota\left[  \left[  u\right]  \right]
\right)  \left[  \left[  t\right]  \right]  \right)  \left(  \sum
_{n\in\mathbb{N}}D_{n}\left(  a\right)  \left(  u+t\right)  ^{n}\right)
\label{pf.cor.HSD.extend-to-loc.div.b.ulem1}%
\end{equation}
instead of $\sum_{n\in\mathbb{N}}\iota\left(  D_{n}\left(  a\right)  \right)
t^{n}=\left(  \iota\left[  \left[  t\right]  \right]  \right)  \left(
\sum_{n\in\mathbb{N}}D_{n}\left(  a\right)  t^{n}\right)  $. If you do not
find (\ref{pf.cor.HSD.extend-to-loc.div.b.ulem1}) obvious, you can check it
easily as follows: Regard $S^{-1}A$ as an $A$-algebra via the map
$\iota:A\rightarrow S^{-1}A$; then $\left(  S^{-1}A\right)  \left[  \left[
u\right]  \right]  $ and $\left(  \left(  S^{-1}A\right)  \left[  \left[
u\right]  \right]  \right)  \left[  \left[  t\right]  \right]  $ become
$A$-algebras as well. The map $\iota$ is an $A$-algebra homomorphism, and
therefore so are the maps $\iota\left[  \left[  u\right]  \right]  $ and
$\left(  \iota\left[  \left[  u\right]  \right]  \right)  \left[  \left[
t\right]  \right]  $. Moreover, the map $\left(  \iota\left[  \left[
u\right]  \right]  \right)  \left[  \left[  t\right]  \right]  $ is continuous
when regarded as a map $A\left[  \left[  u,t\right]  \right]  \rightarrow
\left(  S^{-1}A\right)  \left[  \left[  u,t\right]  \right]  $ (since it just
acts on each coefficient separately). Thus, $\left(  \iota\left[  \left[
u\right]  \right]  \right)  \left[  \left[  t\right]  \right]  $ is a
continuous $A$-algebra homomorphism, and therefore we have%
\begin{align*}
\left(  \left(  \iota\left[  \left[  u\right]  \right]  \right)  \left[
\left[  t\right]  \right]  \right)  \left(  \sum_{n\in\mathbb{N}}D_{n}\left(
a\right)  \left(  u+t\right)  ^{n}\right)   &  =\sum_{n\in\mathbb{N}}%
D_{n}\left(  a\right)  \left(  \underbrace{\left(  \left(  \iota\left[
\left[  u\right]  \right]  \right)  \left[  \left[  t\right]  \right]
\right)  \left(  u+t\right)  }_{=u+t}\right)  ^{n}\\
&  =\sum_{n\in\mathbb{N}}\underbrace{D_{n}\left(  a\right)  \left(
u+t\right)  ^{n}}_{\substack{=\iota\left(  D_{n}\left(  a\right)  \right)
\left(  u+t\right)  ^{n}\\\text{(since }S^{-1}A\text{ becomes an
}A\text{-module via }\iota\text{)}}}=\sum_{n\in\mathbb{N}}\iota\left(
D_{n}\left(  a\right)  \right)  \left(  u+t\right)  ^{n}.
\end{align*}
This proves (\ref{pf.cor.HSD.extend-to-loc.div.b.ulem1}).}. Thus,%
\begin{align*}
\widetilde{\mathbf{D}^{\prime}}_{\left\langle u\right\rangle }\left[  \left[
t\right]  \right]  \circ\underbrace{\widetilde{\mathbf{D}^{\prime}%
}_{\left\langle t\right\rangle }\circ\iota}_{=\iota\left[  \left[  t\right]
\right]  \circ\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }}  &
=\underbrace{\widetilde{\mathbf{D}^{\prime}}_{\left\langle u\right\rangle
}\left[  \left[  t\right]  \right]  \circ\iota\left[  \left[  t\right]
\right]  }_{=\left(  \widetilde{\mathbf{D}^{\prime}}_{\left\langle
u\right\rangle }\circ\iota\right)  \left[  \left[  t\right]  \right]  }%
\circ\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }=\underbrace{\left(
\widetilde{\mathbf{D}^{\prime}}_{\left\langle u\right\rangle }\circ
\iota\right)  }_{=\left(  \iota\left[  \left[  u\right]  \right]  \right)
\circ\widetilde{\mathbf{D}}_{\left\langle u\right\rangle }}\left[  \left[
t\right]  \right]  \circ\widetilde{\mathbf{D}}_{\left\langle t\right\rangle
}\\
&  =\underbrace{\left(  \left(  \iota\left[  \left[  u\right]  \right]
\right)  \circ\widetilde{\mathbf{D}}_{\left\langle u\right\rangle }\right)
\left[  \left[  t\right]  \right]  }_{=\left(  \iota\left[  \left[  u\right]
\right]  \right)  \left[  \left[  t\right]  \right]  \circ
\widetilde{\mathbf{D}}_{\left\langle u\right\rangle }\left[  \left[  t\right]
\right]  }\circ\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }=\left(
\iota\left[  \left[  u\right]  \right]  \right)  \left[  \left[  t\right]
\right]  \circ\underbrace{\widetilde{\mathbf{D}}_{\left\langle u\right\rangle
}\left[  \left[  t\right]  \right]  \circ\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }}_{\substack{=\widetilde{\mathbf{D}}_{\left\langle
u+t\right\rangle }\\\text{(by (\ref{pf.cor.HSD.extend-to-loc.div.b.givens}))}%
}}\\
&  =\left(  \iota\left[  \left[  u\right]  \right]  \right)  \left[  \left[
t\right]  \right]  \circ\widetilde{\mathbf{D}}_{\left\langle u+t\right\rangle
}=\widetilde{\mathbf{D}^{\prime}}_{\left\langle u+t\right\rangle }\circ\iota.
\end{align*}
Therefore, Proposition \ref{prop.localization.univ} \textbf{(b)} (applied to
$B=\left(  \left(  S^{-1}A\right)  \left[  \left[  u\right]  \right]  \right)
\left[  \left[  t\right]  \right]  $, $\phi=\widetilde{\mathbf{D}^{\prime}%
}_{\left\langle u\right\rangle }\left[  \left[  t\right]  \right]
\circ\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }$ and
$\psi=\widetilde{\mathbf{D}^{\prime}}_{\left\langle u+t\right\rangle }$)
yields that $\widetilde{\mathbf{D}^{\prime}}_{\left\langle u\right\rangle
}\left[  \left[  t\right]  \right]  \circ\widetilde{\mathbf{D}^{\prime}%
}_{\left\langle t\right\rangle }=\widetilde{\mathbf{D}^{\prime}}_{\left\langle
u+t\right\rangle }$ (since we know that $\widetilde{\mathbf{D}^{\prime}%
}_{\left\langle u\right\rangle }\left[  \left[  t\right]  \right]
\circ\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }$ and
$\widetilde{\mathbf{D}^{\prime}}_{\left\langle u+t\right\rangle }$ are
$\mathbb{F}$-algebra homomorphisms). This proves the second claim of
(\ref{pf.cor.HSD.extend-to-loc.div.b.goals}).

Also, $\pi_{t}\circ\widetilde{\mathbf{D}^{\prime}}_{\left\langle
t\right\rangle }$ is an $\mathbb{F}$-algebra homomorphism (since $\pi_{t}$ and
$\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle }$ are
$\mathbb{F}$-algebra homomorphisms), and we have%
\begin{align*}
\pi_{t}\circ\underbrace{\widetilde{\mathbf{D}^{\prime}}_{\left\langle
t\right\rangle }\circ\iota}_{=\iota\left[  \left[  t\right]  \right]
\circ\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }}  &
=\underbrace{\pi_{t}\circ\iota\left[  \left[  t\right]  \right]
}_{\substack{=\iota\circ\pi_{t}\\\text{(this follows from the}%
\\\text{definitions of }\pi_{t}\text{ and }\iota\left[  \left[  t\right]
\right]  \text{)}}}\circ\widetilde{\mathbf{D}}_{\left\langle t\right\rangle
}=\iota\circ\underbrace{\pi_{t}\circ\widetilde{\mathbf{D}}_{\left\langle
t\right\rangle }}_{\substack{=\operatorname*{id}\\\text{(by
(\ref{pf.cor.HSD.extend-to-loc.div.b.givens}))}}}\\
&  =\iota=\operatorname*{id}\circ\iota.
\end{align*}
Therefore, Proposition \ref{prop.localization.univ} \textbf{(b)} (applied to
$B=S^{-1}A$, $\phi=\pi_{t}\circ\widetilde{\mathbf{D}^{\prime}}_{\left\langle
t\right\rangle }$ and $\psi=\operatorname*{id}$) yields that $\pi_{t}%
\circ\widetilde{\mathbf{D}^{\prime}}_{\left\langle t\right\rangle
}=\operatorname*{id}$ (since we know that $\pi_{t}\circ\widetilde{\mathbf{D}%
^{\prime}}_{\left\langle t\right\rangle }$ and $\operatorname*{id}$ are
$\mathbb{F}$-algebra homomorphisms). This proves the first claim of
(\ref{pf.cor.HSD.extend-to-loc.div.b.goals}). Hence,
(\ref{pf.cor.HSD.extend-to-loc.div.b.goals}) is proven. As explained above,
this shows that $\mathbf{D}^{\prime}$ is divided-powers. This proves Corollary
\ref{cor.HSD.extend-to-loc.div} \textbf{(b)}.

\textbf{(c)} The definition of $\mathbf{D}^{\prime}\circ\iota$ yields%
\begin{align*}
\mathbf{D}^{\prime}\circ\iota &  =\left(  D_{0}^{\prime}\circ\iota
,D_{1}^{\prime}\circ\iota,D_{2}^{\prime}\circ\iota,\ldots\right)  =\left(
\iota\circ D_{0},\iota\circ D_{1},\iota\circ D_{2},\ldots\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{due to
(\ref{pf.cor.HSD.extend-to-loc.div.b.D'1})}\right) \\
&  =\iota\circ\mathbf{D}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\iota\circ\mathbf{D}\right)  .
\end{align*}
This proves Corollary \ref{cor.HSD.extend-to-loc.div} \textbf{(c)}.
\end{proof}

\begin{definition}
\label{def.HSD.lift}In the situation of Corollary
\ref{cor.HSD.extend-to-loc.div}, we shall refer to the unique HSD
$\mathbf{D}^{\prime}$ constructed in Corollary \ref{cor.HSD.extend-to-loc.div}
\textbf{(a)} as the \textit{lift of the HSD }$\mathbf{D}$ \textit{to the
localization }$S^{-1}A$.
\end{definition}

\begin{corollary}
\label{cor.HSD.extend-to-loc.epi}Let $A$ and $B$ be two commutative
$\mathbb{F}$-algebras. Let $S$ be a multiplicatively closed subset of $A$. Let
$\iota:A\rightarrow S^{-1}A$ be the canonical $\mathbb{F}$-algebra
homomorphism from $A$ to $S^{-1}A$ sending each $a\in A$ to $\dfrac{a}{1}\in
S^{-1}A$.

Let $\mathbf{E}$ and $\mathbf{F}$ be two HSDs from $S^{-1}A$ to $B$ such that
$\mathbf{E}\circ\iota=\mathbf{F}\circ\iota$. (Recall that $\mathbf{E}%
\circ\iota$ and $\mathbf{F}\circ\iota$ are defined according to Definition
\ref{def.HSD.compose}.) Then, $\mathbf{E}=\mathbf{F}$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.HSD.extend-to-loc.epi}.]Write $\mathbf{E}$ and
$\mathbf{F}$ in the forms $\mathbf{E}=\left(  E_{0},E_{1},E_{2},\ldots\right)
$ and $\mathbf{F}=\left(  F_{0},F_{1},F_{2},\ldots\right)  $, respectively. We
have $\mathbf{E}\circ\iota=\mathbf{F}\circ\iota$. Thus, $E_{n}\circ\iota
=F_{n}\circ\iota$ for every $n\in\mathbb{N}$\ \ \ \ \footnote{\textit{Proof.}
The definition of $\mathbf{E}\circ\iota$ yields $\mathbf{E}\circ\iota=\left(
E_{0}\circ\iota,E_{1}\circ\iota,E_{2}\circ\iota,\ldots\right)  $, so that%
\[
\left(  E_{0}\circ\iota,E_{1}\circ\iota,E_{2}\circ\iota,\ldots\right)
=\mathbf{E}\circ\iota=\mathbf{F}\circ\iota=\left(  F_{0}\circ\iota,F_{1}%
\circ\iota,F_{2}\circ\iota,\ldots\right)
\]
(by the definition of $\mathbf{F}\circ\iota$). In other words, $E_{n}%
\circ\iota=F_{n}\circ\iota$ for every $n\in\mathbb{N}$.}. Now, let $t$ be a
new symbol. Define $\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }$ and
$\widetilde{\mathbf{F}}_{\left\langle t\right\rangle }$ as in Theorem
\ref{thm.HSD.powerser}. Then, Theorem \ref{thm.HSD.powerser} (applied to
$S^{-1}A$, $\mathbf{E}$ and $E_{i}$ instead of $A$, $\mathbf{D}$ and $D_{i}$)
yields that $\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }%
:S^{-1}A\rightarrow B\left[  \left[  t\right]  \right]  $ is an $\mathbb{F}%
$-algebra homomorphism (since $\mathbf{E}$ is an HSD). Similarly,
$\widetilde{\mathbf{F}}_{\left\langle t\right\rangle }:S^{-1}A\rightarrow
B\left[  \left[  t\right]  \right]  $ is an $\mathbb{F}$-algebra homomorphism.

Every $x\in A$ satisfies
\begin{align*}
\left(  \widetilde{\mathbf{F}}_{\left\langle t\right\rangle }\circ
\iota\right)  \left(  x\right)   &  =\widetilde{\mathbf{F}}_{\left\langle
t\right\rangle }\left(  \iota\left(  x\right)  \right)  =\sum_{n\in\mathbb{N}%
}\underbrace{F_{n}\left(  \iota\left(  x\right)  \right)  }_{=\left(
F_{n}\circ\iota\right)  \left(  x\right)  }t^{n}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\widetilde{\mathbf{F}}_{\left\langle
t\right\rangle }\right) \\
&  =\sum_{n\in\mathbb{N}}\left(  F_{n}\circ\iota\right)  \left(  x\right)
t^{n}.
\end{align*}
Similarly, every $x\in A$ satisfies $\left(  \widetilde{\mathbf{E}%
}_{\left\langle t\right\rangle }\circ\iota\right)  \left(  x\right)
=\sum_{n\in\mathbb{N}}\left(  E_{n}\circ\iota\right)  \left(  x\right)  t^{n}%
$. Thus, every $x\in A$ satisfies
\[
\left(  \widetilde{\mathbf{E}}_{\left\langle t\right\rangle }\circ
\iota\right)  \left(  x\right)  =\sum_{n\in\mathbb{N}}\underbrace{\left(
E_{n}\circ\iota\right)  }_{=F_{n}\circ\iota}\left(  x\right)  t^{n}=\sum
_{n\in\mathbb{N}}\left(  F_{n}\circ\iota\right)  \left(  x\right)
t^{n}=\left(  \widetilde{\mathbf{F}}_{\left\langle t\right\rangle }\circ
\iota\right)  \left(  x\right)  .
\]
In other words, $\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }%
\circ\iota=\widetilde{\mathbf{F}}_{\left\langle t\right\rangle }\circ\iota$.
Thus, Proposition \ref{prop.localization.univ} \textbf{(b)} (applied to
$B\left[  \left[  t\right]  \right]  $, $\widetilde{\mathbf{E}}_{\left\langle
t\right\rangle }$ and $\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }$
instead of $B$, $\phi$ and $\psi$) yields $\widetilde{\mathbf{E}%
}_{\left\langle t\right\rangle }=\widetilde{\mathbf{F}}_{\left\langle
t\right\rangle }$. Now, every $x\in S^{-1}A$ satisfies%
\begin{align*}
\sum_{n\in\mathbb{N}}E_{n}\left(  x\right)  t^{n}  &
=\underbrace{\widetilde{\mathbf{E}}_{\left\langle t\right\rangle }%
}_{=\widetilde{\mathbf{F}}_{\left\langle t\right\rangle }}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\widetilde{\mathbf{E}%
}_{\left\langle t\right\rangle }\right) \\
&  =\widetilde{\mathbf{F}}_{\left\langle t\right\rangle }\left(  x\right)
=\sum_{n\in\mathbb{N}}F_{n}\left(  x\right)  t^{n}.
\end{align*}
Comparing coefficients in this equality, we conclude that $E_{n}\left(
x\right)  =F_{n}\left(  x\right)  $ for every $x\in S^{-1}A$ and
$n\in\mathbb{N}$. In other words, $E_{n}=F_{n}$ for every $n\in\mathbb{N}$. In
other words, $\mathbf{E}=\mathbf{F}$. This proves Corollary
\ref{cor.HSD.extend-to-loc.epi}.
\end{proof}

\subsection{Residues}

If $U$ is a $\mathbb{F}$-module, and if $f\in U\left[  \left[  z,z^{-1}%
\right]  \right]  $, then we let $\operatorname*{Res}fdz$ denote the $\left(
-1\right)  $-st coefficient of $f$. That is,%
\[
\operatorname*{Res}\left(  \sum_{i\in\mathbb{Z}}u_{i}z^{i}\right)  dz=u_{-1}.
\]
Notice that neither \textquotedblleft$d$\textquotedblright\ not
\textquotedblleft$dz$\textquotedblright\ should be regarded as standalone
objects in the expression \textquotedblleft$\operatorname*{Res}fdz$%
\textquotedblright; they are part of the notation.\footnote{This notation, of
course, imitates the \textquotedblleft residue at $0$\textquotedblright%
\ notation from complex analysis. But it is purely formal and elementary and
works over any commutative ring $\mathbb{F}$.}

A similar notation can be defined for $U$-valued formal distributions in
multiple variables. Indeed, the case of multiple variables can be reduced to
the case of one variable as follows: Let $U$ be a $\mathbb{F}$-module, and let
$\mathbf{x}=\left(  x_{j}\right)  _{j\in J}$ be a family of indeterminates.
Let $k\in J$, and let $\widetilde{\mathbf{x}}_{k}$ denote the family $\left(
x_{j}\right)  _{j\in J\setminus\left\{  k\right\}  }$ of indeterminates (that
is, the family $\mathbf{x}$ with $x_{k}$ removed). Then, we can identify
$U\left[  \left[  \mathbf{x},\mathbf{x}^{-1}\right]  \right]  $ with $\left(
U\left[  \left[  \widetilde{\mathbf{x}}_{k},\left(  \widetilde{\mathbf{x}}%
_{k}\right)  ^{-1}\right]  \right]  \right)  \left[  \left[  x_{k},x_{k}%
^{-1}\right]  \right]  $\ \ \ \ \footnote{This is similar to how we can
identify $U\left[  \mathbf{x}\right]  $ with $\left(  U\left[
\widetilde{\mathbf{x}}_{k}\right]  \right)  \left[  x_{k}\right]  $ because a
polynomial in the variables $\mathbf{x}$ can be regarded as a polynomial in
the variable $x_{k}$ whose coefficients are polynomials in the remaining
variables $\widetilde{\mathbf{x}}_{k}$. Similarly it is possible to identify
$U\left[  \mathbf{x},\mathbf{x}^{-1}\right]  $ with $\left(  U\left[
\widetilde{\mathbf{x}}_{k},\left(  \widetilde{\mathbf{x}}_{k}\right)
^{-1}\right]  \right)  \left[  x_{k},x_{k}^{-1}\right]  $, and to identify
$U\left[  \left[  \mathbf{x}\right]  \right]  $ with $\left(  U\left[  \left[
\widetilde{\mathbf{x}}_{k}\right]  \right]  \right)  \left[  \left[
x_{k}\right]  \right]  $. (But we cannot identify $U\left(  \left(
\mathbf{x}\right)  \right)  $ with $\left(  U\left(  \left(
\widetilde{\mathbf{x}}_{k}\right)  \right)  \right)  \left(  \left(
x_{k}\right)  \right)  $.)}. Hence, $f\in U\left[  \left[  \mathbf{x}%
,\mathbf{x}^{-1}\right]  \right]  $, an element $\operatorname*{Res}fdx_{k}$
of $U\left[  \left[  \widetilde{\mathbf{x}}_{k},\left(  \widetilde{\mathbf{x}%
}_{k}\right)  ^{-1}\right]  \right]  $ is defined (because $f\in U\left[
\left[  \mathbf{x},\mathbf{x}^{-1}\right]  \right]  =\left(  U\left[  \left[
\widetilde{\mathbf{x}}_{k},\left(  \widetilde{\mathbf{x}}_{k}\right)
^{-1}\right]  \right]  \right)  \left[  \left[  x_{k},x_{k}^{-1}\right]
\right]  $).

We notice a simple fact:

\begin{proposition}
\label{prop.residues.del}Let $U$ be an $\mathbb{F}$-module. Let $f\in U\left[
\left[  z,z^{-1}\right]  \right]  $.

\textbf{(a)} We have $\operatorname*{Res}\left(  \partial_{z}^{\left(
n\right)  }f\right)  dz=0$ for every positive integer $n$.

\textbf{(b)} We have $\operatorname*{Res}\left(  \partial_{z}f\right)  dz=0$.

\textbf{(c)} For every $g\in U\left[  z,z^{-1}\right]  $, we have
\begin{equation}
\operatorname*{Res}f\partial_{z}\left(  g\right)  dz=-\operatorname*{Res}%
\partial_{z}\left(  f\right)  gdz \label{eq.prop.residues.del.1}%
\end{equation}
and%
\begin{equation}
\operatorname*{Res}g\partial_{z}\left(  f\right)  dz=-\operatorname*{Res}%
\partial_{z}\left(  g\right)  fdz \label{eq.prop.residues.del.2}%
\end{equation}

\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.residues.del}.]\textbf{(a)} Let $n$ be a
positive integer. Then, $n-1\in\mathbb{N}$ and $n>n-1$. Hence, $\dbinom
{n-1}{n}=0$.

Let us write $f$ in the form $f=\sum_{i\in\mathbb{Z}}u_{i}z^{i}$ for some
$\left(  u_{i}\right)  _{i\in\mathbb{Z}}\in U^{\mathbb{Z}}$. Then, the
definition of $\partial_{z}^{\left(  n\right)  }$ yields $\partial
_{z}^{\left(  n\right)  }f=\sum_{i\in\mathbb{Z}}\dbinom{i}{n}u_{i}z^{i-n}%
=\sum_{i\in\mathbb{Z}}\dbinom{i+n}{n}u_{i+n}z^{i}$ (here, we have substituted
$i+n$ for $i$ in the sum). Hence, the definition of $\operatorname*{Res}%
\left(  \partial_{z}^{\left(  n\right)  }f\right)  dz$ yields%
\[
\operatorname*{Res}\left(  \partial_{z}^{\left(  n\right)  }f\right)
dz=\underbrace{\dbinom{-1+n}{n}}_{=\dbinom{n-1}{n}=0}u_{-1+n}=0.
\]
This proves Proposition \ref{prop.residues.del} \textbf{(a)}.

\textbf{(b)} Proposition \ref{prop.residues.del} \textbf{(a)} (applied to
$n=1$) yields $\operatorname*{Res}\left(  \partial_{z}^{\left(  1\right)
}f\right)  dz=0$. Since $\partial_{z}^{\left(  1\right)  }=\partial_{z}$, this
yields $\operatorname*{Res}\left(  \partial_{z}f\right)  dz=0$. This proves
Proposition \ref{prop.residues.del} \textbf{(b)}.

\textbf{(c)} Let $g\in U\left[  z,z^{-1}\right]  $. Proposition
\ref{prop.delta.hasse} \textbf{(c)} (applied to $a=g$ and $b=f$) yields
$\partial_{z}\left(  gf\right)  =\partial_{z}\left(  g\right)  f+g\partial
_{z}\left(  f\right)  $. Hence, $\partial_{z}\left(  g\right)  f=\partial
_{z}\left(  gf\right)  -g\partial_{z}\left(  f\right)  $. Now,%
\begin{align*}
\operatorname*{Res}\underbrace{f\partial_{z}\left(  g\right)  }_{=\partial
_{z}\left(  g\right)  f=\partial_{z}\left(  gf\right)  -g\partial_{z}\left(
f\right)  }dz  &  =\operatorname*{Res}\left(  \partial_{z}\left(  gf\right)
-g\partial_{z}\left(  f\right)  \right)  dz\\
&  =\underbrace{\operatorname*{Res}\partial_{z}\left(  gf\right)
dz}_{\substack{=0\\\text{(by Proposition \ref{prop.residues.del}
\textbf{(b)},}\\\text{applied to }gf\text{ instead of }f\text{)}%
}}-\operatorname*{Res}\underbrace{g\partial_{z}\left(  f\right)  }%
_{=\partial_{z}\left(  f\right)  g}dz\\
&  =-\operatorname*{Res}\partial_{z}\left(  f\right)  gdz.
\end{align*}
This proves (\ref{eq.prop.residues.del.1}). Now,%
\begin{align*}
\operatorname*{Res}\underbrace{g\partial_{z}\left(  f\right)  }_{=\partial
_{z}\left(  f\right)  g}dz  &  =\operatorname*{Res}\partial_{z}\left(
f\right)  gdz=-\operatorname*{Res}\underbrace{f\partial_{z}\left(  g\right)
}_{=\partial_{z}\left(  g\right)  f}dz\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{eq.prop.residues.del.1})}\right) \\
&  =-\operatorname*{Res}\partial_{z}\left(  g\right)  fdz.
\end{align*}
This proves (\ref{eq.prop.residues.del.2}), and thus completes the proof of
Proposition \ref{prop.residues.del} \textbf{(c)}.
\end{proof}

\subsection{\label{sect.diffop}Differential operators}

\begin{definition}
\label{def.Lh}If $U$ is an $\mathbb{F}$-module, and if $h\in\mathbb{F}\left[
\left[  w,w^{-1}\right]  \right]  $, then we shall write $L_{h}$ for the map
$U\left[  \left[  w,w^{-1}\right]  \right]  \rightarrow U\left[  \left[
w,w^{-1}\right]  \right]  ,\ a\mapsto ha$. (Notice that $U$ is implicit in
this notation; therefore it sometimes leads to several different maps being
denoted by $L_{h}$. But this rarely makes any troubles, since these maps can
be distinguished by their domains.)

Also, if $U$ is an $\mathbb{F}$-module, and if $h\in U\left[  \left[
w,w^{-1}\right]  \right]  $, then we shall write $L_{h}$ for the map
$\mathbb{F}\left[  \left[  w,w^{-1}\right]  \right]  \rightarrow U\left[
\left[  w,w^{-1}\right]  \right]  ,\ a\mapsto ha$. (Again, $U$ is implicit,
and we rely on context to clarify what we mean by $L_{h}$.)

When $U$ is an $\mathbb{F}$-module, and when $h$ belongs to either
$\mathbb{F}\left[  \left[  w,w^{-1}\right]  \right]  $ or $U\left[  \left[
w,w^{-1}\right]  \right]  $, we are often going to abbreviate the map $L_{h}$
by $h$ (by abuse of notation). Thus, $h\left(  a\right)  =ha$ for every $h$.
This is a generalization of the abuse of notation we mentioned in Remark
\ref{rmk.Lz.abuse}.
\end{definition}

We next define a (purely algebraic) notion of differential operators -- one of
many reasonable such notions.

If $U$ is an $\mathbb{F}$-module, and if $\left(  c_{j}\left(  w\right)
\right)  _{j\in\mathbb{N}}$ is a family of formal distributions in $U\left[
\left[  w,w^{-1}\right]  \right]  $ with the property that all but finitely
many $j\in\mathbb{N}$ satisfy $c_{j}\left(  w\right)  =0$, then $\sum
_{j\in\mathbb{N}}c_{j}\left(  w\right)  \partial_{w}^{\left(  j\right)  }$ is
a well-defined $\mathbb{F}$-linear map $\mathbb{F}\left[  w,w^{-1}\right]
\rightarrow U\left[  \left[  w,w^{-1}\right]  \right]  $ (where $c_{j}\left(
w\right)  \partial_{w}^{\left(  j\right)  }$ means $c_{j}\left(  w\right)
\circ\partial_{w}^{\left(  j\right)  }$, and where $c_{j}\left(  w\right)  $
abbreviates $L_{c_{j}\left(  w\right)  }$, according to Definition
\ref{def.Lh}).\footnote{This follows from the fact that $U\left[  \left[
w,w^{-1}\right]  \right]  $ is an $\mathbb{F}\left[  w,w^{-1}\right]
$-module.} These maps will be called \textit{differential operators of finite
order into }$U$. In other words:

\begin{definition}
\label{def.diffop}Let $U$ be an $\mathbb{F}$-module. A \textit{differential
operator of finite order into }$U$ means a map $\mathbb{F}\left[
w,w^{-1}\right]  \rightarrow U\left[  \left[  w,w^{-1}\right]  \right]  $
which has the form $\sum_{j\in\mathbb{N}}c_{j}\left(  w\right)  \partial
_{w}^{\left(  j\right)  }$ for a family $\left(  c_{j}\left(  w\right)
\right)  _{j\in\mathbb{N}}$ of formal distributions in $U\left[  \left[
w,w^{-1}\right]  \right]  $ with the property that all but finitely many
$j\in\mathbb{N}$ satisfy $c_{j}\left(  w\right)  =0$.
\end{definition}

\begin{remark}
Maps of the form $\sum_{j\in\mathbb{N}}c_{j}\left(  w\right)  \partial
_{w}^{\left(  j\right)  }$ are well-defined even if we do not require that all
but finitely many $j\in\mathbb{N}$ satisfy $c_{j}\left(  w\right)  =0$. We
shall not make use of them in this generality, however.
\end{remark}

It is clear that (for given $U$) the differential operators of finite order
into $U$ form an $\mathbb{F}$-module. This $\mathbb{F}$-module contains the
operator $c\left(  w\right)  $ for each $w\in U\left[  \left[  w,w^{-1}%
\right]  \right]  $ (recall that \textquotedblleft the operator $c\left(
w\right)  $\textquotedblright\ really means the map $L_{c\left(  w\right)
}:\mathbb{F}\left[  w,w^{-1}\right]  \rightarrow U\left[  \left[
w,w^{-1}\right]  \right]  $), and when $U=\mathbb{F}$, it also contains the
operator $\partial_{w}^{\left(  j\right)  }$ for every $j\in\mathbb{N}$.

We first notice that the \textquotedblleft coefficients\textquotedblright%
\ $c_{j}\left(  w\right)  $ of a differential operator $\sum_{j\in\mathbb{N}%
}c_{j}\left(  w\right)  \partial_{w}^{\left(  j\right)  }$ are uniquely
determined by the operator. More precisely:

\begin{proposition}
\label{prop.diffop.unique}Let $U$ be an $\mathbb{F}$-module. Let $D$ be a
differential operator of finite order into $U$. Then, there exists a unique
family $\left(  c_{j}\left(  w\right)  \right)  _{j\in\mathbb{N}}$ of formal
distributions in $U\left[  \left[  w,w^{-1}\right]  \right]  $ with the
properties that

\begin{itemize}
\item all but finitely many $j\in\mathbb{N}$ satisfy $c_{j}\left(  w\right)
=0$;

\item we have $D=\sum_{j\in\mathbb{N}}c_{j}\left(  w\right)  \partial
_{w}^{\left(  j\right)  }$.
\end{itemize}
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.diffop.unique}.]The existence of such a family
follows from the fact that $D$ is a differential operator of finite order into
$U$. It thus remains to prove its uniqueness. In other words, it remains to
prove that if $\left(  c_{j}^{\left\langle 1\right\rangle }\left(  w\right)
\right)  _{j\in\mathbb{N}}$ and $\left(  c_{j}^{\left\langle 2\right\rangle
}\left(  w\right)  \right)  _{j\in\mathbb{N}}$ are two families $\left(
c_{j}\left(  w\right)  \right)  _{j\in\mathbb{N}}$ with the two properties
stated in Proposition \ref{prop.diffop.unique}, then $\left(  c_{j}%
^{\left\langle 1\right\rangle }\left(  w\right)  \right)  _{j\in\mathbb{N}%
}=\left(  c_{j}^{\left\langle 2\right\rangle }\left(  w\right)  \right)
_{j\in\mathbb{N}}$.

Thus, let $\left(  c_{j}^{\left\langle 1\right\rangle }\left(  w\right)
\right)  _{j\in\mathbb{N}}$ and $\left(  c_{j}^{\left\langle 2\right\rangle
}\left(  w\right)  \right)  _{j\in\mathbb{N}}$ be two families $\left(
c_{j}\left(  w\right)  \right)  _{j\in\mathbb{N}}$ with the two properties
stated in Proposition \ref{prop.diffop.unique}.

For every $j\in\mathbb{N}$, define a formal distribution $s_{j}\left(
w\right)  \in U\left[  \left[  w,w^{-1}\right]  \right]  $ by%
\[
s_{j}\left(  w\right)  =c_{j}^{\left\langle 1\right\rangle }\left(  w\right)
-c_{j}^{\left\langle 2\right\rangle }\left(  w\right)  .
\]
It is easy to show that
\begin{equation}
0=\sum_{j\in\mathbb{N}}s_{j}\left(  w\right)  \partial_{w}^{\left(  j\right)
}. \label{pf.prop.diffop.unique.3}%
\end{equation}
\footnote{\textit{Proof.} The family $\left(  c_{j}^{\left\langle
1\right\rangle }\left(  w\right)  \right)  _{j\in\mathbb{N}}$ satisfies the
second of the two properties stated in Proposition \ref{prop.diffop.unique}.
In other words, we have%
\begin{equation}
D=\sum_{j\in\mathbb{N}}c_{j}^{\left\langle 1\right\rangle }\left(  w\right)
\partial_{w}^{\left(  j\right)  } \label{pf.prop.diffop.unique.1}%
\end{equation}
The same argument, applied to the family $\left(  c_{j}^{\left\langle
2\right\rangle }\left(  w\right)  \right)  _{j\in\mathbb{N}}$ instead of
$\left(  c_{j}^{\left\langle 1\right\rangle }\left(  w\right)  \right)
_{j\in\mathbb{N}}$, yields%
\[
D=\sum_{j\in\mathbb{N}}c_{j}^{\left\langle 2\right\rangle }\left(  w\right)
\partial_{w}^{\left(  j\right)  }.
\]
Subtracting this equality from (\ref{pf.prop.diffop.unique.1}), we obtain%
\begin{align*}
0  &  =\sum_{j\in\mathbb{N}}c_{j}^{\left\langle 1\right\rangle }\left(
w\right)  \partial_{w}^{\left(  j\right)  }-\sum_{j\in\mathbb{N}}%
c_{j}^{\left\langle 2\right\rangle }\left(  w\right)  \partial_{w}^{\left(
j\right)  }=D=\sum_{j\in\mathbb{N}}\underbrace{\left(  c_{j}^{\left\langle
1\right\rangle }\left(  w\right)  -c_{j}^{\left\langle 2\right\rangle }\left(
w\right)  \right)  }_{\substack{=s_{j}\left(  w\right)  \\\text{(since }%
s_{j}\left(  w\right)  \text{ was defined as}\\c_{j}^{\left\langle
1\right\rangle }\left(  w\right)  -c_{j}^{\left\langle 2\right\rangle }\left(
w\right)  \text{)}}}\partial_{w}^{\left(  j\right)  }\\
&  =\sum_{j\in\mathbb{N}}s_{j}\left(  w\right)  \partial_{w}^{\left(
j\right)  }.
\end{align*}
This proves (\ref{pf.prop.diffop.unique.3}).}

We will now show that every $n\in\mathbb{N}$ satisfies
\begin{equation}
s_{n}\left(  w\right)  =0. \label{pf.prop.diffop.unique.hook}%
\end{equation}


\textit{Proof of (\ref{pf.prop.diffop.unique.hook}):} We shall prove
(\ref{pf.prop.diffop.unique.hook}) by strong induction over $n$.

So let $N\in\mathbb{N}$. Assume that (\ref{pf.prop.diffop.unique.hook}) holds
for each $n\in\mathbb{N}$ satisfying $n<N$. We need to show that
(\ref{pf.prop.diffop.unique.hook}) holds for $n=N$ as well.

We know that (\ref{pf.prop.diffop.unique.hook}) holds for each $n\in
\mathbb{N}$ satisfying $n<N$. In other words,
\begin{equation}
s_{n}\left(  w\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for each }n\in
\mathbb{N}\text{ satisfying }n<N. \label{pf.prop.diffop.unique.hook.pf.hyp}%
\end{equation}


Applying both sides of the equality (\ref{pf.prop.diffop.unique.3}) to the
Laurent polynomial $w^{N}\in\mathbb{F}\left[  w,w^{-1}\right]  $, we obtain%
\begin{align*}
0 &  =\sum_{j\in\mathbb{N}}s_{j}\left(  w\right)  \partial_{w}^{\left(
j\right)  }\left(  w^{N}\right)  \\
&  =\sum_{j=0}^{N-1}\underbrace{s_{j}\left(  w\right)  }%
_{\substack{=0\\\text{(by (\ref{pf.prop.diffop.unique.hook.pf.hyp}%
),}\\\text{applied to }n=j\text{)}}}\partial_{w}^{\left(  j\right)  }\left(
w^{N}\right)  +\sum_{j\geq N}s_{j}\left(  w\right)  \underbrace{\partial
_{w}^{\left(  j\right)  }\left(  w^{N}\right)  }_{\substack{=\dbinom{N}%
{j}w^{N-j}\\\text{(by the definition of }\partial_{w}^{\left(  j\right)
}\text{)}}}\\
&  =\underbrace{\sum_{j=0}^{N-1}0\partial_{w}^{\left(  j\right)  }\left(
w^{N}\right)  }_{=0}+\sum_{j\geq N}s_{j}\left(  w\right)  \dbinom{N}{j}%
w^{N-j}=\sum_{j\geq N}s_{j}\left(  w\right)  \dbinom{N}{j}w^{N-j}\\
&  =s_{N}\left(  w\right)  \underbrace{\dbinom{N}{N}}_{=1}\underbrace{w^{N-N}%
}_{=1}+\sum_{j>N}s_{j}\left(  w\right)  \underbrace{\dbinom{N}{j}%
}_{\substack{=0\\\text{(since }N<j\text{)}}}w^{N-j}\\
&  =s_{N}\left(  w\right)  +\underbrace{\sum_{j=0}^{N-1}s_{j}\left(  w\right)
0w^{N-j}}_{=0}=s_{N}\left(  w\right)  .
\end{align*}
Thus, $s_{N}\left(  w\right)  =0$. In other words,
(\ref{pf.prop.diffop.unique.hook}) holds for $n=N$. This completes the
induction step.

Now, we have proven (\ref{pf.prop.diffop.unique.hook}) by induction. Thus,
every $n\in\mathbb{N}$ satisfies $s_{n}\left(  w\right)  =0$. In other words,
every $n\in\mathbb{N}$ satisfies $c_{n}^{\left\langle 1\right\rangle }\left(
w\right)  -c_{n}^{\left\langle 2\right\rangle }\left(  w\right)  =0$ (since
$s_{n}\left(  w\right)  $ is defined to be $c_{n}^{\left\langle 1\right\rangle
}\left(  w\right)  -c_{n}^{\left\langle 2\right\rangle }\left(  w\right)  $).
In other words, every $n\in\mathbb{N}$ satisfies $c_{n}^{\left\langle
1\right\rangle }\left(  w\right)  =c_{n}^{\left\langle 2\right\rangle }\left(
w\right)  $. Hence, $\left(  c_{j}^{\left\langle 1\right\rangle }\left(
w\right)  \right)  _{j\in\mathbb{N}}=\left(  c_{j}^{\left\langle
2\right\rangle }\left(  w\right)  \right)  _{j\in\mathbb{N}}$. This completes
the proof of Proposition \ref{prop.diffop.unique}.
\end{proof}

Next, we shall discuss composition of differential operators. Two differential
operators of finite order into $U$ do not necessarily have a well-defined
composition, even if $U=\mathbb{F}$ (because they send Laurent polynomials to
formal distributions, but they cannot be applied to formal distributions).
However, a composition can be defined in a particular setting:

\begin{definition}
\label{def.diffop.laur}Let $U$ be an $\mathbb{F}$-module.

\textbf{(a)} Let $D$ be a differential operator of finite order into $U$. We
can then write $D$ in the form $\sum_{j\in\mathbb{N}}c_{j}\left(  w\right)
\partial_{w}^{\left(  j\right)  }$ for a family $\left(  c_{j}\left(
w\right)  \right)  _{j\in\mathbb{N}}$ of formal distributions in $U\left[
\left[  w,w^{-1}\right]  \right]  $ with the property that all but finitely
many $j\in\mathbb{N}$ satisfy $c_{j}\left(  w\right)  =0$. Consider this
family. (Proposition \ref{prop.diffop.unique} says that this family is
unique.) We say that $D$ is \textit{Laurent} if all $c_{j}\left(  w\right)  $
are Laurent polynomials (i.e., belong to $\mathbb{F}\left[  w,w^{-1}\right]  $).

It is clear that if $D$ is Laurent, then $D$ has the following two properties:

\begin{itemize}
\item The image of $D$ is contained in $U\left[  w,w^{-1}\right]  $.

\item The operator $D$ can be uniquely extended to a continuous $\mathbb{F}%
$-linear map $\mathbb{F}\left[  \left[  w,w^{-1}\right]  \right]  \rightarrow
U\left[  \left[  w,w^{-1}\right]  \right]  $ (which is again defined as
$\sum_{j\in\mathbb{N}}c_{j}\left(  w\right)  \partial_{w}^{\left(  j\right)
}$), since $U\left[  \left[  w,w^{-1}\right]  \right]  $ is an $\mathbb{F}%
\left[  w,w^{-1}\right]  $-module. We denote this extension by $D$ again.
\end{itemize}

\textbf{(b)} Let $D$ be any differential operator of finite order into $U$,
and let $E$ be any differential operator of finite order into $\mathbb{F}$. We
cannot define a map $D\circ E$ in general.

However, if $E$ is Laurent, then $D\circ E$ is a well-defined map from
$\mathbb{F}\left[  w,w^{-1}\right]  $ to $U\left[  \left[  w,w^{-1}\right]
\right]  $ (since the image of $E$ is contained in $\mathbb{F}\left[
w,w^{-1}\right]  $).

On the other hand, if $D$ is Laurent, then $D\circ E$ is a well-defined map
from $\mathbb{F}\left[  w,w^{-1}\right]  $ to $U\left[  \left[  w,w^{-1}%
\right]  \right]  $ (but here, the $D$ in $D\circ E$ does not mean the
original map $D:\mathbb{F}\left[  w,w^{-1}\right]  \rightarrow U\left[
\left[  w,w^{-1}\right]  \right]  $, but rather the extension of $D$ to a
continuous $\mathbb{F}$-linear map $\mathbb{F}\left[  \left[  w,w^{-1}\right]
\right]  \rightarrow U\left[  \left[  w,w^{-1}\right]  \right]  $).

Altogether, we thus know that there is a well-defined operator $D\circ E$
whenever at least one of $D$ and $E$ is Laurent.
\end{definition}

We shall now prove that the composition of two differential operators is again
a differential operator, provided that it is well-defined:

\begin{proposition}
\label{prop.diffop.compos}Let $U$ be an $\mathbb{F}$-module. Let $D$ be any
differential operator of finite order into $U$. Let $E$ be any differential
operator of finite order into $\mathbb{F}$.

\textbf{(a)} Assume that at least one of $D$ and $E$ is Laurent. Then, the map
$D\circ E$ is a differential operator of finite order into $U$.

\textbf{(b)} If both $D$ and $E$ are Laurent, then $D\circ E$ is also Laurent.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.diffop.compos}.]For every $e\left(  w\right)
\in\mathbb{F}\left[  \left[  w,w^{-1}\right]  \right]  $ and every
$n\in\mathbb{N}$, we have%
\begin{equation}
\partial_{w}^{\left(  n\right)  }\circ e\left(  w\right)  =\sum_{i=0}%
^{n}\left(  \partial_{w}^{\left(  n-i\right)  }\left(  e\left(  w\right)
\right)  \right)  \circ\partial_{w}^{\left(  i\right)  }
\label{pf.prop.diffop.compos.1}%
\end{equation}
(as maps from $\mathbb{F}\left[  w,w^{-1}\right]  $ to $\mathbb{F}\left[
w,w^{-1}\right]  $).\ \ \ \ \footnote{\textit{Proof of
(\ref{pf.prop.diffop.compos.1}):} Let $e\left(  w\right)  \in\mathbb{F}\left[
\left[  w,w^{-1}\right]  \right]  $ and $n\in\mathbb{N}$. Let $g\in
\mathbb{F}\left[  w,w^{-1}\right]  $ be arbitrary. Applying Proposition
\ref{prop.delta.hasse} \textbf{(b)} to $\mathbb{F}$ instead of $U$ (and
renaming the indeterminate $z$ as $w$), we conclude that
\[
\partial_{w}^{\left(  n\right)  }\left(  ab\right)  =\sum_{i=0}^{n}%
\partial_{w}^{\left(  i\right)  }\left(  a\right)  \partial_{w}^{\left(
n-i\right)  }\left(  b\right)
\]
for any $a\in\mathbb{F}\left[  w,w^{-1}\right]  $ and $b\in\mathbb{F}\left[
w,w^{-1}\right]  $. Applying this to $a=g$ and $b=e\left(  w\right)  $, we
obtain%
\[
\partial_{w}^{\left(  n\right)  }\left(  e\left(  w\right)  \cdot g\right)
=\sum_{i=0}^{n}\partial_{w}^{\left(  i\right)  }\left(  g\right)  \partial
_{w}^{\left(  n-i\right)  }\left(  e\left(  w\right)  \right)  =\left(
\sum_{i=0}^{n}\left(  \partial_{w}^{\left(  n-i\right)  }\left(  e\left(
w\right)  \right)  \right)  \circ\partial_{w}^{\left(  i\right)  }\right)
\left(  g\right)  .
\]
Hence,
\begin{equation}
\left(  \partial_{w}^{\left(  n\right)  }\circ e\left(  w\right)  \right)
\left(  g\right)  =\partial_{w}^{\left(  n\right)  }\left(  e\left(  w\right)
\cdot g\right)  =\left(  \sum_{i=0}^{n}\left(  \partial_{w}^{\left(
n-i\right)  }\left(  e\left(  w\right)  \right)  \right)  \circ\partial
_{w}^{\left(  i\right)  }\right)  \left(  g\right)  .
\label{pf.prop.diffop.compos.1.pf.1}%
\end{equation}
\par
Let us now forget that we fixed $g$. We thus have proven
(\ref{pf.prop.diffop.compos.1.pf.1}) for every $g\in\mathbb{F}\left[
w,w^{-1}\right]  $. In other words, we have $\partial_{w}^{\left(  n\right)
}\circ e\left(  w\right)  =\sum_{i=0}^{n}\left(  \partial_{w}^{\left(
n-i\right)  }\left(  e\left(  w\right)  \right)  \right)  \circ\partial
_{w}^{\left(  i\right)  }$. This proves (\ref{pf.prop.diffop.compos.1}).}

Now, $D$ is a differential operator of finite order into $\mathbb{F}$. Hence,
we can write $D$ in the form $D=\sum_{j\in\mathbb{N}}d_{j}\left(  w\right)
\partial_{w}^{\left(  j\right)  }$ for a family of $\left(  d_{j}\left(
w\right)  \right)  _{j\in\mathbb{N}}$ of formal distributions in
$\mathbb{F}\left[  \left[  w,w^{-1}\right]  \right]  $ with the property that
all but finitely many $j\in\mathbb{N}$ satisfy $d_{j}\left(  w\right)  =0$.
Consider this family.

Also, $E$ is a differential operator of finite order into $U$. Hence, we can
write $E$ in the form $E=\sum_{j\in\mathbb{N}}e_{j}\left(  w\right)
\partial_{w}^{\left(  j\right)  }$ for a family of $\left(  e_{j}\left(
w\right)  \right)  _{j\in\mathbb{N}}$ of formal distributions in $U\left[
\left[  w,w^{-1}\right]  \right]  $ with the property that all but finitely
many $j\in\mathbb{N}$ satisfy $e_{j}\left(  w\right)  =0$. Consider this family.

For all $i\in\mathbb{N}$ and $j\in\mathbb{N}$, the product $d_{i}\left(
w\right)  \cdot e_{j}\left(  w\right)  $ in $U\left[  \left[  w,w^{-1}\right]
\right]  $ is well-defined\footnote{\textit{Proof.} We assumed that at least
one of $D$ and $E$ is Laurent. We WLOG assume that $D$ is Laurent (since the
proof in the case when $E$ is Laurent is similar). Then, all $d_{j}\left(
w\right)  $ are Laurent polynomials. That is, $d_{i}\left(  w\right)  \in
U\left[  w,w^{-1}\right]  $ for every $i\in\mathbb{N}$. Hence, $d_{i}\left(
w\right)  \cdot e_{j}\left(  w\right)  $ is well-defined for all
$i\in\mathbb{N}$ and $j\in\mathbb{N}$ (since the product of an element of
$U\left[  w,w^{-1}\right]  $ with an element of $\mathbb{F}\left[  \left[
w,w^{-1}\right]  \right]  $ is always well-defined). Qed.}.

The operator $D\circ E$ is well-defined (since at least one of $D$ and $E$ is
Laurent). We have $D=\sum_{j\in\mathbb{N}}d_{j}\left(  w\right)  \partial
_{w}^{\left(  j\right)  }=\sum_{n\in\mathbb{N}}d_{n}\left(  w\right)
\partial_{w}^{\left(  n\right)  }$ and thus%
\begin{align}
&  \underbrace{D}_{=\sum_{n\in\mathbb{N}}d_{n}\left(  w\right)  \partial
_{w}^{\left(  n\right)  }}\circ\underbrace{E}_{=\sum_{j\in\mathbb{N}}%
e_{j}\left(  w\right)  \partial_{w}^{\left(  j\right)  }}\nonumber\\
&  =\left(  \sum_{n\in\mathbb{N}}d_{n}\left(  w\right)  \partial_{w}^{\left(
n\right)  }\right)  \circ\left(  \sum_{j\in\mathbb{N}}e_{j}\left(  w\right)
\partial_{w}^{\left(  j\right)  }\right) \nonumber\\
&  =\sum_{n\in\mathbb{N}}\sum_{j\in\mathbb{N}}d_{n}\left(  w\right)
\circ\underbrace{\partial_{w}^{\left(  n\right)  }\circ e_{j}\left(  w\right)
}_{\substack{=\sum_{i=0}^{n}\left(  \partial_{w}^{\left(  n-i\right)  }\left(
e_{j}\left(  w\right)  \right)  \right)  \circ\partial_{w}^{\left(  i\right)
}\\\text{(by (\ref{pf.prop.diffop.compos.1}), applied to }e\left(  w\right)
=e_{j}\left(  w\right)  \text{)}}}\circ\partial_{w}^{\left(  j\right)
}\nonumber\\
&  =\sum_{n\in\mathbb{N}}\sum_{j\in\mathbb{N}}d_{n}\left(  w\right)
\circ\left(  \sum_{i=0}^{n}\left(  \partial_{w}^{\left(  n-i\right)  }\left(
e_{j}\left(  w\right)  \right)  \right)  \circ\partial_{w}^{\left(  i\right)
}\right)  \circ\partial_{w}^{\left(  j\right)  }\nonumber\\
&  =\sum_{n\in\mathbb{N}}\sum_{i=0}^{n}\sum_{j\in\mathbb{N}}\underbrace{d_{n}%
\left(  w\right)  \circ\left(  \partial_{w}^{\left(  n-i\right)  }\left(
e_{j}\left(  w\right)  \right)  \right)  }_{=d_{n}\left(  w\right)
\cdot\left(  \partial_{w}^{\left(  n-i\right)  }\left(  e_{j}\left(  w\right)
\right)  \right)  }\circ\underbrace{\partial_{w}^{\left(  i\right)  }%
\circ\partial_{w}^{\left(  j\right)  }}_{\substack{=\dbinom{i+j}{i}%
\partial_{w}^{\left(  i+j\right)  }\\\text{(by (\ref{eq.prop.delta.deriv.nm}),
applied to }i\text{ and }j\text{ instead of}\\n\text{ and }m\text{, and with
}z\text{ renamed as }w\text{)}}}\nonumber\\
&  =\sum_{n\in\mathbb{N}}\sum_{i=0}^{n}\sum_{j\in\mathbb{N}}d_{n}\left(
w\right)  \cdot\left(  \partial_{w}^{\left(  n-i\right)  }\left(  e_{j}\left(
w\right)  \right)  \right)  \circ\left(  \dbinom{i+j}{i}\partial_{w}^{\left(
i+j\right)  }\right) \nonumber\\
&  =\sum_{n\in\mathbb{N}}\sum_{i=0}^{n}\sum_{j\in\mathbb{N}}\dbinom{i+j}%
{i}d_{n}\left(  w\right)  \cdot\left(  \partial_{w}^{\left(  n-i\right)
}\left(  e_{j}\left(  w\right)  \right)  \right)  \circ\partial_{w}^{\left(
i+j\right)  }\nonumber\\
&  =\underbrace{\sum_{n\in\mathbb{N}}\sum_{i=0}^{n}\sum_{\substack{m\in
\mathbb{N};\\m\geq i}}}_{=\sum_{m\in\mathbb{N}}\sum_{i=0}^{m}\sum
_{\substack{n\in\mathbb{N};\\n\geq i}}}\dbinom{m}{i}d_{n}\left(  w\right)
\cdot\left(  \partial_{w}^{\left(  n-i\right)  }\left(  e_{m-i}\left(
w\right)  \right)  \right)  \circ\partial_{w}^{\left(  m\right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }m\text{ for
}i+j\text{ in the third sum}\right) \nonumber\\
&  =\sum_{m\in\mathbb{N}}\sum_{i=0}^{m}\sum_{\substack{n\in\mathbb{N};\\n\geq
i}}\dbinom{m}{i}d_{n}\left(  w\right)  \cdot\left(  \partial_{w}^{\left(
n-i\right)  }\left(  e_{m-i}\left(  w\right)  \right)  \right)  \circ
\partial_{w}^{\left(  m\right)  }\nonumber\\
&  =\sum_{m\in\mathbb{N}}\underbrace{\sum_{i=0}^{m}\dbinom{m}{i}\left(
\sum_{\substack{n\in\mathbb{N};\\n\geq i}}d_{n}\left(  w\right)  \cdot\left(
\partial_{w}^{\left(  n-i\right)  }\left(  e_{m-i}\left(  w\right)  \right)
\right)  \right)  }_{\substack{\in\mathbb{F}\left[  \left[  w,w^{-1}\right]
\right]  \\\text{(because the inner sum has only finitely many nonzero
addends}\\\text{(since all but finitely many }j\in\mathbb{N}\text{ satisfy
}d_{j}\left(  w\right)  =0\text{))}}}\circ\partial_{w}^{\left(  m\right)  }.
\label{pf.prop.diffop.compos.final}%
\end{align}
Since all but finitely many $m\in\mathbb{N}$ satisfy $\sum_{i=0}^{m}\dbinom
{m}{i}\left(  \sum_{\substack{n\in\mathbb{N};\\n\geq i}}d_{n}\left(  w\right)
\cdot\left(  \partial_{w}^{\left(  n-i\right)  }\left(  e_{m-i}\left(
w\right)  \right)  \right)  \right)  =0$\ \ \ \ \footnote{This is easily
derived from the facts that all but finitely many $j\in\mathbb{N}$ satisfy
$d_{j}\left(  w\right)  =0$, and all but finitely many $j\in\mathbb{N}$
satisfy $e_{j}\left(  w\right)  =0$.}, this yields that $D\circ E$ is a
differential operator of finite order into $U$. Proposition
\ref{prop.diffop.compos} \textbf{(a)} is thus proven.

\textbf{(b)} Assume that $D$ and $E$ Laurent. Then, all $d_{j}\left(
w\right)  $ are Laurent polynomials, and all $e_{j}\left(  w\right)  $ are
Laurent polynomials. Now, it is easy to see that%
\[
\sum_{i=0}^{m}\dbinom{m}{i}\left(  \sum_{\substack{n\in\mathbb{N};\\n\geq
i}}d_{n}\left(  w\right)  \cdot\left(  \partial_{w}^{\left(  n-i\right)
}\left(  e_{m-i}\left(  w\right)  \right)  \right)  \right)
\]
is a Laurent polynomial for each $m\in\mathbb{N}$ (since the inner sum
$\sum_{\substack{n\in\mathbb{N};\\n\geq i}}$ in this expression has only
finitely many nonzero terms). Thus, (\ref{pf.prop.diffop.compos.final}) shows
that $D\circ E$ is Laurent. This proves Proposition \ref{prop.diffop.compos}
\textbf{(b)}.
\end{proof}

We now define the notion of an adjoint differential operator:

\begin{definition}
\label{def.diffop.adj}Let $U$ be an $\mathbb{F}$-module. Let $D$ be a
differential operator of finite order into $U$. We can then write $D$ in the
form $\sum_{j\in\mathbb{N}}c_{j}\left(  w\right)  \partial_{w}^{\left(
j\right)  }$ for a family $\left(  c_{j}\left(  w\right)  \right)
_{j\in\mathbb{N}}$ of formal distributions in $U\left[  \left[  w,w^{-1}%
\right]  \right]  $ with the property that all but finitely many
$j\in\mathbb{N}$ satisfy $c_{j}\left(  w\right)  =0$. Consider this family.
(Proposition \ref{prop.diffop.unique} says that this family is unique.)

We denote by $D^{\ast}$ the map
\[
\sum_{j\in\mathbb{N}}\left(  -1\right)  ^{j}\partial_{w}^{\left(  j\right)
}\circ c_{j}\left(  w\right)  :\mathbb{F}\left[  w,w^{-1}\right]  \rightarrow
U\left[  \left[  w,w^{-1}\right]  \right]  ,
\]
which is again a differential operator of finite order into $U$ (due to
Proposition \ref{prop.diffop.compos}). This operator $D^{\ast}$ is called the
\textit{adjoint differential operator} to $D$. If $D$ is Laurent, then
$D^{\ast}$ is Laurent (this follows easily from Proposition
\ref{prop.diffop.compos} \textbf{(b)}).
\end{definition}

We end this section with some results whose proofs we leave to the reader:

\begin{proposition}
\label{prop.diffop.adj}Let $U$ be an $\mathbb{F}$-module. Let $D$ be a Laurent
differential operator of finite order into $U$. Then, any $f\in\mathbb{F}%
\left[  w,w^{-1}\right]  $ and $g\in\mathbb{F}\left[  w,w^{-1}\right]  $
satisfy%
\[
\operatorname*{Res}\left(  D^{\ast}f\right)  gdw=\operatorname*{Res}f\left(
Dg\right)  dw.
\]

\end{proposition}

In view of the fact that we regard $\operatorname*{Res}fgdw$ as a sort of
bilinear pairing between $f$ and $g$ (see Section \ref{sect.pairing} below),
this Proposition \ref{prop.diffop.adj} likely explains where the name
\textquotedblleft adjoint differential operator\textquotedblright\ for
$D^{\ast}$ comes from.

\begin{definition}
\label{def.diffop.Ucoeff}Let $U$ be an $\mathbb{F}$-module. Let $D$ be any
differential operator of finite order into $\mathbb{F}$. We can then write $D$
in the form $\sum_{j\in\mathbb{N}}c_{j}\left(  w\right)  \partial_{w}^{\left(
j\right)  }$ for a family $\left(  c_{j}\left(  w\right)  \right)
_{j\in\mathbb{N}}$ of formal distributions in $\mathbb{F}\left[  \left[
w,w^{-1}\right]  \right]  $ with the property that all but finitely many
$j\in\mathbb{N}$ satisfy $c_{j}\left(  w\right)  =0$. Consider this family.
(Proposition \ref{prop.diffop.unique} says that this family is unique.) Then,
we define a map $D_{\left(  U\right)  }:U\left[  w,w^{-1}\right]  \rightarrow
U\left[  \left[  w,w^{-1}\right]  \right]  $ by $D_{\left(  U\right)  }%
=\sum_{j\in\mathbb{N}}c_{j}\left(  w\right)  \partial_{w}^{\left(  j\right)
}$ (this is the same expression as for $D$, but it is now acting on $U\left[
w,w^{-1}\right]  $).
\end{definition}

\begin{proposition}
\label{prop.diffop.adj.op}Let $U$ be an $\mathbb{F}$-module. Let $D$ be any
differential operator of finite order into $U$. Let $E$ be any differential
operator of finite order into $\mathbb{F}$. Assume that at least one of $D$
and $E$ is Laurent. We can canonically lift the differential operator
$D:\mathbb{F}\left[  w,w^{-1}\right]  \rightarrow\mathbb{F}\left[  \left[
w,w^{-1}\right]  \right]  $ to a map $\widetilde{D}:U\left[  w,w^{-1}\right]
\rightarrow U\left[  \left[  w,w^{-1}\right]  \right]  $ by writing $D$ in the
form $\sum_{j\in\mathbb{N}}c_{j}\left(  w\right)  \partial_{w}^{\left(
j\right)  }$ (with $\left(  c_{j}\left(  w\right)  \right)  _{j\in\mathbb{N}}$
being as usual) and defining $\widetilde{D}$ to be the map $\sum
_{j\in\mathbb{N}}c_{j}\left(  w\right)  \partial_{w}^{\left(  j\right)  }$
(but now acting on $U\left[  w,w^{-1}\right]  $). (This is well-defined
because of Proposition \ref{prop.diffop.unique}.) Notice that if $D$ is
Laurent, then so is $\widetilde{D}$.

Now, we have%
\[
\left(  D\circ E\right)  ^{\ast}=\left(  E^{\ast}\right)  _{U}\circ D^{\ast}.
\]

\end{proposition}

The proof of Proposition \ref{prop.diffop.adj.op} can be obtained easily using
Proposition \ref{prop.diffop.adj}, once one reduces to the case of $D$ and $E$
both being Laurent (by $\mathbb{F}$-linearity and continuity) and of
$U=\mathbb{F}$ (by $U$-linearity).

\section{\label{chap.delta}Locality and the formal $\delta$-function}

\subsection{\label{sect.pairing}Pairing between distributions and polynomials}

In analysis, distributions are often defined as linear functionals on the
space of test functions. Similarly, formal distributions can be regarded as
linear maps acting on polynomials. Specifically, let $U$ be a $\mathbb{F}%
$-module. For every $f\in\mathbb{F}\left[  z,z^{-1}\right]  $ and $a\in
U\left[  \left[  z,z^{-1}\right]  \right]  $, we can set
\begin{equation}
\left\langle f,a\right\rangle =\operatorname*{Res}fadz\in U. \label{pairing.1}%
\end{equation}
When $U=\mathbb{F}$, this value belongs to $\mathbb{F}$, so we have defined a
pairing $\left\langle \cdot,\cdot\right\rangle :\mathbb{F}\left[
z,z^{-1}\right]  \times\mathbb{F}\left[  \left[  z,z^{-1}\right]  \right]
\rightarrow\mathbb{F}$. This pairing leads to a linear map $\mathbb{F}\left[
\left[  z,z^{-1}\right]  \right]  \rightarrow\left(  \mathbb{F}\left[
z,z^{-1}\right]  \right)  ^{\ast}$, which is easily seen to be an isomorphism.

The equality (\ref{pairing.1}) motivates a seemingly weird notation:

\begin{definition}
\label{def.weirdcoeff}If $a\in U\left[  \left[  z,z^{-1}\right]  \right]  $
and $n\in\mathbb{Z}$, then we denote by $a_{\left[  n\right]  }$ the $\left(
-n-1\right)  $-st coefficient of $a$. (Thus, $a=\sum_{n\in\mathbb{Z}%
}a_{\left[  n\right]  }z^{-n-1}$ for every $a\in U\left[  \left[
z,z^{-1}\right]  \right]  $.)
\end{definition}

Why did we choose to have $a_{\left[  n\right]  }$ mean the $\left(
-n-1\right)  $-st coefficient rather than the $n$-th coefficient? The purpose
was to ensure that%
\[
\left\langle z^{n},f\right\rangle =f_{\left[  n\right]  }%
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{Z}.
\]


Similarly, if $a\in U\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $ and
$\left(  m,n\right)  \in\mathbb{Z}^{2}$, then $a_{\left[  m,n\right]  }$ shall
denote the coefficient of $a$ before $z^{-m-1}w^{-n-1}$.

[Notice that I am using the notations $a_{\left[  n\right]  }$ and $a_{\left[
m,n\right]  }$ instead of the more common notations $a_{\left(  n\right)  }$
and $a_{\left(  m,n\right)  }$ to avoid conflict with the \textquotedblleft%
$_{\left(  n\right)  }$\textquotedblright\ notation for vertex-algebra products.]

\subsection{Local formal distributions}

Let $U$ be an $\mathbb{F}$-module.

\begin{definition}
\label{def.local}A formal distribution $a\in U\left[  \left[  z,z^{-1}%
,w,w^{-1}\right]  \right]  $ is said to be \textit{local} if $\left(
z-w\right)  ^{N}a\left(  z,w\right)  =0$ for some $N\in\mathbb{N}$.
\end{definition}

(Recall once again that $a\left(  z,w\right)  $ is just another way to say
$a$, and that the $\mathbb{F}\left[  z,w\right]  $-module $U\left[  \left[
z,z^{-1},w,w^{-1}\right]  \right]  $ has torsion, so that $\left(  z-w\right)
^{N}a\left(  z,w\right)  =0$ does not force $a\left(  z,w\right)  =0$.)

\begin{proposition}
\label{exe.3.1}Let $a=a\left(  z,w\right)  \in U\left[  \left[  z,z^{-1}%
,w,w^{-1}\right]  \right]  $ be local. Then:

\textbf{(a)} The formal distribution $a\left(  w,z\right)  $ is also local.

\textbf{(b)} The formal distributions $\partial_{z}a\left(  z,w\right)  $ and
$\partial_{w}a\left(  z,w\right)  $ are local as well.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{exe.3.1}.]We know that $a$ is local. Thus, $\left(
z-w\right)  ^{N}a\left(  z,w\right)  =0$ for some $N\in\mathbb{N}$. Consider
this $N$.

\textbf{(a)} Substituting $w$ and $z$ for $z$ and $w$ in $\left(  z-w\right)
^{N}a\left(  z,w\right)  =0$, we obtain $\left(  w-z\right)  ^{N}a\left(
w,z\right)  =0$. Thus,%
\[
\underbrace{\left(  z-w\right)  ^{N}}_{=\left(  -\left(  w-z\right)  \right)
^{N}=\left(  -1\right)  ^{N}\left(  w-z\right)  ^{N}}a\left(  w,z\right)
=\left(  -1\right)  ^{N}\underbrace{\left(  w-z\right)  ^{N}a\left(
w,z\right)  }_{=0}=0.
\]
This shows that $a\left(  w,z\right)  $ is local. Proposition \ref{exe.3.1}
\textbf{(a)} is proven.

\textbf{(b)} We have%
\[
\partial_{z}\left(  \underbrace{\left(  z-w\right)  ^{N+1}}_{=\left(
z-w\right)  \left(  z-w\right)  ^{N}}a\left(  z,w\right)  \right)
=\partial_{z}\left(  \left(  z-w\right)  \underbrace{\left(  z-w\right)
^{N}a\left(  z,w\right)  }_{=0}\right)  =0,
\]
thus%
\begin{align*}
0  &  =\partial_{z}\left(  \left(  z-w\right)  ^{N+1}a\left(  z,w\right)
\right) \\
&  =\underbrace{\left(  \partial_{z}\left(  z-w\right)  ^{N+1}\right)
}_{=\left(  N+1\right)  \left(  z-w\right)  ^{N}}a\left(  z,w\right)  +\left(
z-w\right)  ^{N+1}\partial_{z}a\left(  z,w\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\partial_{z}\text{ is a derivation}\right) \\
&  =\left(  N+1\right)  \underbrace{\left(  z-w\right)  ^{N}a\left(
z,w\right)  }_{=0}+\left(  z-w\right)  ^{N+1}\partial_{z}a\left(  z,w\right)
=\left(  z-w\right)  ^{N+1}\partial_{z}a\left(  z,w\right)  .
\end{align*}
Hence, $\left(  z-w\right)  ^{N+1}\partial_{z}a\left(  z,w\right)  =0$. This
shows that $\partial_{z}a\left(  z,w\right)  $ is local. Similarly,
$\partial_{w}a\left(  z,w\right)  $ is local. Proposition \ref{exe.3.1}
\textbf{(b)} is thus proven.
\end{proof}

\subsection{The formal $\delta$-function}

\begin{example}
We define an element $\delta\left(  z-w\right)  $ of $\mathbb{F}\left[
\left[  z,z^{-1},w,w^{-1}\right]  \right]  $ as follows:%
\[
\delta\left(  z-w\right)  =\sum_{n\in\mathbb{Z}}z^{-n-1}w^{n}.
\]
(The notation \textquotedblleft$\delta\left(  z-w\right)  $\textquotedblright%
\ is slightly confusing; the minus should be interpreted as a symbolic piece
of the notation rather than as a subtraction sign. Of course, it is supposed
to resemble the subtraction sign, as the whole formal distribution
$\delta\left(  z-w\right)  $ is supposed to resemble the delta function of
$z-w$ as a distribution on the $z$-$w$-plane. Two other notations for
$\delta\left(  z-w\right)  $ are $\delta\left(  z,w\right)  $ and
$\delta\left(  z/w\right)  $.)

The $\mathbb{F}$-valued formal distribution $\delta\left(  z-w\right)  $ is
called the \textit{formal }$\delta$\textit{-function}. Notice that
$\delta\left(  z-w\right)  $ is symmetric in $z$ and $w$: We have%
\begin{align*}
\delta\left(  z-w\right)   &  =\sum_{n\in\mathbb{Z}}z^{-n-1}w^{n}=\sum
_{n\in\mathbb{Z}}z^{n}w^{-n-1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }-n-1\text{ for
}n\text{ in the sum}\right) \\
&  =\sum_{n\in\mathbb{Z}}w^{-n-1}z^{n}=\delta\left(  w-z\right)  .
\end{align*}

\end{example}

The next proposition (whose proof is immediate) shows some motivation for the
study of $\delta\left(  z-w\right)  $ and for its name:

\begin{proposition}
\label{prop.delta.props}\textbf{(a)} For every $m\in\mathbb{Z}$, we have%
\begin{align}
\delta\left(  z-w\right)  z^{m}  &  =\left(  \sum_{n\in\mathbb{Z}}%
z^{-n-1}w^{n}\right)  z^{m}=\sum_{n\in\mathbb{Z}}z^{-n+m-1}w^{n}=\sum
_{n\in\mathbb{Z}}z^{-n+1}w^{n+m}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }n+m\text{ for
}n\text{ in the sum}\right) \nonumber\\
&  =\left(  \sum_{n\in\mathbb{Z}}z^{-n-1}w^{n}\right)  w^{m}=\delta\left(
z-w\right)  w^{m}. \label{eq.prop.delta.props.1}%
\end{align}


\textbf{(b)} We have%
\begin{equation}
\operatorname*{Res}\delta\left(  z-w\right)  dz=1.
\label{eq.prop.delta.props.2}%
\end{equation}


\textbf{(c)} For every $\varphi\in\mathbb{F}\left[  z,z^{-1}\right]  $, we
have%
\begin{equation}
\delta\left(  z-w\right)  \varphi\left(  z\right)  =\delta\left(  z-w\right)
\varphi\left(  w\right)  . \label{eq.prop.delta.props.3}%
\end{equation}
(This follows from (\ref{eq.prop.delta.props.1}), because linearity allows us
to WLOG assume that $\varphi$ is a monomial.)

\textbf{(d)} For every $\varphi\in\mathbb{F}\left[  z,z^{-1}\right]  $, we
have%
\begin{align}
\operatorname*{Res}\underbrace{\delta\left(  z-w\right)  \varphi\left(
z\right)  }_{\substack{=\delta\left(  z-w\right)  \varphi\left(  w\right)
\\\text{(by (\ref{eq.prop.delta.props.3}))}}}dz  &  =\operatorname*{Res}%
\delta\left(  z-w\right)  \varphi\left(  w\right)  dz\nonumber\\
&  =\varphi\left(  w\right)  \underbrace{\operatorname*{Res}\delta\left(
z-w\right)  dz}_{=1}=\varphi\left(  w\right)  . \label{eq.prop.delta.props.4}%
\end{align}
One can view this equality as a defining property for the formal $\delta$-function.

\textbf{(e)} Applying (\ref{eq.prop.delta.props.1}) to $m=1$, we obtain
$\delta\left(  z-w\right)  z=\delta\left(  z-w\right)  w$, so that $\left(
z-w\right)  \delta\left(  z-w\right)  =0$. Thus, the formal distribution
$\delta\left(  z-w\right)  $ is local.
\end{proposition}

The formal distribution $\delta\left(  z-w\right)  $ is interesting in that it
can be multiplied not only with Laurent polynomials, but also with univariate
formal distributions:

\begin{proposition}
\label{exe.3.2}Let $U$ be a $\mathbb{F}$-module. Let $a\left(  z\right)  \in
U\left[  \left[  z,z^{-1}\right]  \right]  $.

\textbf{(a)} The formal distributions $\delta\left(  z-w\right)  a\left(
z\right)  \in U\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $ and
$\delta\left(  z-w\right)  a\left(  w\right)  \in U\left[  \left[
z,z^{-1},w,w^{-1}\right]  \right]  $ are well-defined (although there is no
general notion of a product of a formal distribution in $U\left[  \left[
z,z^{-1},w,w^{-1}\right]  \right]  $ with a formal distribution in $U\left[
\left[  z,z^{-1}\right]  \right]  $).

\textbf{(b)} We have%
\begin{equation}
\delta\left(  z-w\right)  a\left(  z\right)  =\delta\left(  z-w\right)
a\left(  w\right)  . \label{eq.prop.delta.props.3'}%
\end{equation}
(In other words, the equality (\ref{eq.prop.delta.props.3}) holds not only for
$\varphi\in\mathbb{F}\left[  z,z^{-1}\right]  $, but also more generally for
$\varphi\in U\left[  \left[  z,z^{-1}\right]  \right]  $.)

\textbf{(c)} We have%
\begin{equation}
\operatorname*{Res}\delta\left(  z-w\right)  a\left(  z\right)  dz=a\left(
w\right)  . \label{eq.prop.delta.props.4'}%
\end{equation}

\end{proposition}

\begin{proof}
[Proof of Proposition \ref{exe.3.2}.]\textbf{(a)} There are essentially two
ways to prove that $\delta\left(  z-w\right)  a\left(  z\right)  $ is
well-defined. One is to directly compute it:%
\[
\underbrace{\delta\left(  z-w\right)  }_{=\sum_{n\in\mathbb{Z}}z^{-n-1}w^{n}%
}a\left(  z\right)  =\sum_{n\in\mathbb{Z}}z^{-n-1}w^{n}a\left(  z\right)
=\sum_{n\in\mathbb{Z}}\underbrace{z^{-n-1}a\left(  z\right)  }%
_{\substack{\text{a well-defined element}\\\text{of }U\left[  \left[
z,z^{-1}\right]  \right]  }}w^{n}.
\]
The other way is to argue that $\delta\left(  z-w\right)  =\sum_{n\in
\mathbb{Z}}z^{-n-1}w^{n}$ belongs not only to $U\left[  \left[  z,z^{-1}%
,w,w^{-1}\right]  \right]  $, but also to the smaller $\mathbb{F}$-module
$\left(  U\left[  z,z^{-1}\right]  \right)  \left[  \left[  w,w^{-1}\right]
\right]  $ (which is embedded in $U\left[  \left[  z,z^{-1},w,w^{-1}\right]
\right]  $ via the identification of $U\left[  \left[  z,z^{-1},w,w^{-1}%
\right]  \right]  $ with $\left(  U\left[  \left[  z,z^{-1}\right]  \right]
\right)  \left[  \left[  w,w^{-1}\right]  \right]  $)\ \ \ \ \footnote{This
$\mathbb{F}$-module $\left(  U\left[  z,z^{-1}\right]  \right)  \left[
\left[  w,w^{-1}\right]  \right]  $ can be characterized as the set of all $u
= \sum_{\left(  i,j\right)  \in\mathbb{Z}^{2}} u_{i,j} z^{i} w^{j}$ (with all
$u_{i,j} \in U$) having the property that, for every $j \in\mathbb{Z}$, all
but finitely many $i \in\mathbb{Z}$ satisfy $u_{i,j} = 0$. Keep in mind that
how many these ``finitely many'' are can depend on $j$. Thus, $\left(
U\left[  z,z^{-1}\right]  \right)  \left[  \left[  w,w^{-1}\right]  \right]  $
is not the same as $\left(  U \left[  \left[  w,w^{-1}\right]  \right]
\right)  \left[  z,z^{-1}\right]  $.}, and that any element of the latter
$\mathbb{F}$-module can be multiplied with any element of $U\left[  \left[
z,z^{-1}\right]  \right]  $.

Either way, we have shown that $\delta\left(  z-w\right)  a\left(  z\right)  $
is well-defined. A similar argument shows that $\delta\left(  z-w\right)
a\left(  w\right)  $ is well-defined. Proposition \ref{exe.3.2} \textbf{(a)}
is proven.

\textbf{(b)} Essentially, we can obtain (\ref{eq.prop.delta.props.3'}) by
breaking up $a\left(  z\right)  $ into an infinite sum of monomials and
applying (\ref{eq.prop.delta.props.1}). Let us give some details on this
argument to make sure that everything is well-defined: Write $a\left(
z\right)  =\sum_{m\in\mathbb{Z}}a_{m}z^{m}$. Then, $a\left(  w\right)
=\sum_{m\in\mathbb{Z}}a_{m}w^{m}$ and%
\begin{align*}
\delta\left(  z-w\right)  \underbrace{a\left(  z\right)  }_{=\sum
_{m\in\mathbb{Z}}a_{m}z^{m}}  &  =\delta\left(  z-w\right)  \cdot\left(
\sum_{m\in\mathbb{Z}}a_{m}z^{m}\right)  =\sum_{m\in\mathbb{Z}}a_{m}%
\underbrace{\delta\left(  z-w\right)  z^{m}}_{\substack{=\delta\left(
z-w\right)  w^{m}\\\text{(by (\ref{eq.prop.delta.props.1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{check that distributivity does hold
here!}\right) \\
&  =\sum_{m\in\mathbb{Z}}a_{m}\delta\left(  z-w\right)  w^{m}=\delta\left(
z-w\right)  \underbrace{\left(  \sum_{m\in\mathbb{Z}}a_{m}w^{m}\right)
}_{=a\left(  w\right)  }=\delta\left(  z-w\right)  a\left(  w\right)  .
\end{align*}
This proves Proposition \ref{exe.3.2} \textbf{(b)}.

\textbf{(c)} Proposition \ref{exe.3.2} \textbf{(c)} follows from Proposition
\ref{exe.3.2} \textbf{(b)} in the same way as (\ref{eq.prop.delta.props.4})
follows from (\ref{eq.prop.delta.props.3}).
\end{proof}

We shall soon (in Section \ref{sect.iziw}) see another way to construct
$\delta\left(  z-w\right)  $. But before we get there, we need to introduce a
few more $\mathbb{F}$-algebras.

\subsection{The rings $U\left(  \left(  z,w/z\right)  \right)  $ and $U\left(
\left(  w,z/w\right)  \right)  $}

Recall that, in Section \ref{sect.polynomials.multivar}, we defined an
$\mathbb{F}$-algebra $U\left(  \left(  z,w\right)  \right)  $ of Laurent
series in $z$ and $w$ for every $\mathbb{F}$-algebra $U$; it is explicitly
given by (\ref{eq.polynomials.multivar.laurent}). Now, for every $\mathbb{F}%
$-algebra $U$, let us define an $\mathbb{F}$-algebra $U\left(  \left(
z,w/z\right)  \right)  $, which will satisfy
\[
U\left(  \left(  z,w\right)  \right)  \subseteq U\left(  \left(  z,w/z\right)
\right)  \subseteq U\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]
\]
(and these containments are strict, unless $U=0$). In analytical terms,
$U\left(  \left(  z,w/z\right)  \right)  $ is the $\mathbb{F}$-algebra of
\textquotedblleft$U$-valued formal Laurent series expanded in the domain
$\left\vert z\right\vert >\left\vert w\right\vert $\textquotedblright\ (or,
maybe better said, in the domain $1>\left\vert z\right\vert >\left\vert
w\right\vert $). Rather than dwelling on what this means, we are going to
define this $\mathbb{F}$-algebra purely algebraically.

\begin{definition}
\label{def.F((z,w/z))}Let $U$ be an $\mathbb{F}$-module. Let $a$ and $b$ be
two new (unrelated and distinct) symbols. Then, $U\left(  \left(
z,w/z\right)  \right)  $ denotes the image of the continuous $\mathbb{F}%
$-linear map%
\begin{align*}
U\left(  \left(  a,b\right)  \right)   &  \rightarrow U\left[  \left[
z,z^{-1},w,w^{-1}\right]  \right]  ,\\
a^{i}b^{j}  &  \mapsto z^{i}\left(  w/z\right)  ^{j}=z^{i}w^{j}z^{-j}%
=z^{i-j}w^{j}.
\end{align*}

\end{definition}

Informally, this means that $U\left(  \left(  z,w/z\right)  \right)  $ can be
obtained as follows: Imagine for a moment that $w/z$ is a formal variable. and
consider all Laurent series in $z$ and $w/z$. Then, transform these series
into elements of $U\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $ by
replacing each $\left(  w/z\right)  ^{j}$ by $w^{j}z^{-j}$. The set of all
possible results is $U\left(  \left(  z,w/z\right)  \right)  $.

It is easy to see that
\begin{align}
U\left(  \left(  z,w/z\right)  \right)   &  =\left\{  \sum_{\left(
i,j\right)  \in\mathbb{Z}^{2}}u_{\left(  i,j\right)  }z^{i}w^{j}\in U\left[
\left[  z,z^{-1},w,w^{-1}\right]  \right]  \ \mid\text{\ all }u_{\left(
i,j\right)  }\in U\text{, and}\right. \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \phantom{iii}\text{there exists an }%
N\in\mathbb{Z}\text{ such that all }\left(  i,j\right)  \in\mathbb{Z}%
^{2}\right. \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \phantom{\sum_i^i}\text{satisfying }%
\min\left\{  i+j,j\right\}  <N\text{ satisfy }u_{\left(  i,j\right)
}=0\right\}  \label{eq.F((z,w/z)).explicit}%
\end{align}
as subsets of $U\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $. From
this, it follows easily that $U\left(  \left(  z,w\right)  \right)  \subseteq
U\left(  \left(  z,w/z\right)  \right)  $.

It is easy to see that
\[
\partial_{z}\left(  U\left(  \left(  z,w/z\right)  \right)  \right)  \subseteq
U\left(  \left(  z,w/z\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \partial_{w}\left(  U\left(  \left(  z,w/z\right)
\right)  \right)  \subseteq U\left(  \left(  z,w/z\right)  \right)
\]
\footnote{\textit{Proof.} Let us introduce a new notation: If $N\in\mathbb{Z}%
$, and if $u=\sum_{\left(  i,j\right)  \in\mathbb{Z}^{2}}u_{\left(
i,j\right)  }z^{i}w^{j}$ is an element of $U\left[  \left[  z,z^{-1}%
,w,w^{-1}\right]  \right]  $ (with all $u_{\left(  i,j\right)  }\in U$), then
we say that $u$ is $N$\textit{-downbounded} if all $\left(  i,j\right)
\in\mathbb{Z}^{2}$ satisfying $\min\left\{  i+j,j\right\}  <N$ satisfy
$u_{\left(  i,j\right)  }=0$. Then, (\ref{eq.F((z,w/z)).explicit}) says that%
\[
U\left(  \left(  z,w/z\right)  \right)  =\left\{  u\in U\left[  \left[
z,z^{-1},w,w^{-1}\right]  \right]  \ \mid\ u\text{ is }N\text{-downbounded for
some }N\in\mathbb{Z}\right\}  .
\]
Hence, $\partial_{z}\left(  U\left(  \left(  z,w/z\right)  \right)  \right)
\subseteq U\left(  \left(  z,w/z\right)  \right)  $ follows from the trivial
observation that for every $N\in\mathbb{Z}$ and every $N$-downbounded element
$u$ of $U\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $, the element
$\partial_{z}u$ is $\left(  N-1\right)  $-downbounded. Similarly,
$\partial_{w}\left(  U\left(  \left(  z,w/z\right)  \right)  \right)
\subseteq U\left(  \left(  z,w/z\right)  \right)  $ can be shown.}. Hence, the
endomorphisms $\partial_{z}$ and $\partial_{w}$ of $U\left[  \left[
z,z^{-1},w,w^{-1}\right]  \right]  $ restrict to endomorphisms $\partial_{z}$
and $\partial_{w}$ of $U\left(  \left(  z,w/z\right)  \right)  $.

More generally, we can see that every $n\in\mathbb{N}$ satisfies%
\[
\partial_{z}^{\left(  n\right)  }\left(  U\left(  \left(  z,w/z\right)
\right)  \right)  \subseteq U\left(  \left(  z,w/z\right)  \right)
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \partial_{w}^{\left(
n\right)  }\left(  U\left(  \left(  z,w/z\right)  \right)  \right)  \subseteq
U\left(  \left(  z,w/z\right)  \right)  .
\]
Hence, for every $n\in\mathbb{N}$, the endomorphisms $\partial_{z}^{\left(
n\right)  }$ and $\partial_{w}^{\left(  n\right)  }$ of $U\left[  \left[
z,z^{-1},w,w^{-1}\right]  \right]  $ restrict to endomorphisms $\partial
_{z}^{\left(  n\right)  }$ and $\partial_{w}^{\left(  n\right)  }$ of
$U\left(  \left(  z,w/z\right)  \right)  $.

When $U$ is a $\mathbb{F}$-algebra, then any two elements of $U\left(  \left(
z,w/z\right)  \right)  $ can be multiplied\footnote{The multiplication, of
course, is defined as usual:%
\[
\left(  \sum_{\left(  i,j\right)  \in\mathbb{Z}^{2}}u_{\left(  i,j\right)
}z^{i}w^{j}\right)  \left(  \sum_{\left(  i,j\right)  \in\mathbb{Z}^{2}%
}v_{\left(  i,j\right)  }z^{i}w^{j}\right)  =\sum_{\left(  i,j\right)
\in\mathbb{Z}^{2}}\left(  \sum_{\left(  i^{\prime},j^{\prime}\right)
\in\mathbb{Z}^{2}}u_{\left(  i^{\prime},j^{\prime}\right)  }v_{\left(
i-i^{\prime},j-j^{\prime}\right)  }\right)  z^{i}w^{j}.
\]
What needs to be checked is that this definition makes sense, i.e., that the
inner sum $\sum_{\left(  i^{\prime},j^{\prime}\right)  \in\mathbb{Z}^{2}%
}u_{\left(  i^{\prime},j^{\prime}\right)  }v_{\left(  i-i^{\prime}%
,j-j^{\prime}\right)  }$ is well-defined and the outer sum $\sum_{\left(
i,j\right)  \in\mathbb{Z}^{2}}\left(  \sum_{\left(  i^{\prime},j^{\prime
}\right)  \in\mathbb{Z}^{2}}u_{\left(  i^{\prime},j^{\prime}\right)
}v_{\left(  i-i^{\prime},j-j^{\prime}\right)  }\right)  z^{i}w^{j}$ is an
element of $U\left(  \left(  z,w/z\right)  \right)  $. This can be done either
directly (using (\ref{eq.F((z,w/z)).explicit})), or using the fact that the
continuous $\mathbb{F}$-linear map%
\begin{align*}
U\left(  \left(  a,b\right)  \right)   &  \rightarrow U\left(  \left(
z,w/z\right)  \right)  ,\\
a^{i}b^{j}  &  \mapsto z^{i-j}w^{j}%
\end{align*}
is an $\mathbb{F}$-module isomorphism and preserves multiplication. We leave
this to the reader.}. Thus, in this case, $U\left(  \left(  z,w/z\right)
\right)  $ becomes an $\mathbb{F}$-algebra. Clearly, $U\left(  \left(
z,w\right)  \right)  $ is an $\mathbb{F}$-subalgebra of $U\left(  \left(
z,w/z\right)  \right)  $. The endomorphisms $\partial_{z}^{\left(  n\right)
}$ and $\partial_{w}^{\left(  n\right)  }$ of $U\left(  \left(  z,w/z\right)
\right)  $ satisfy the following analogue of Proposition
\ref{prop.delta.hasse} \textbf{(a)}:

\begin{proposition}
\label{prop.delta.hasse.U((z,w/z))}Let $U$ be a commutative $\mathbb{F}%
$-algebra. The sequence $\left(  \partial_{z}^{\left(  0\right)  }%
,\partial_{z}^{\left(  1\right)  },\partial_{z}^{\left(  2\right)  }%
,\ldots\right)  $ is a divided-powers HSD from $U\left(  \left(  z,w/z\right)
\right)  $ to $U\left(  \left(  z,w/z\right)  \right)  $. Also, the sequence
$\left(  \partial_{w}^{\left(  0\right)  },\partial_{w}^{\left(  1\right)
},\partial_{w}^{\left(  2\right)  },\ldots\right)  $ is a divided-powers HSD
from $U\left(  \left(  z,w/z\right)  \right)  $ to $U\left(  \left(
z,w/z\right)  \right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.delta.hasse.U((z,w/z))}.]The proof of
Proposition \ref{prop.delta.hasse.U((z,w/z))} is analogous to that of
Proposition \ref{prop.delta.hasse} \textbf{(a)}; the fact that we now have two
variables instead of one does not require any significant changes (since the
$\partial_{z}^{\left(  n\right)  }$ commute with multiplication by $w$, and
the $\partial_{w}^{\left(  n\right)  }$ commute with multiplication by $z$).
We leave the details to the reader.
\end{proof}

When $U$ is a commutative $\mathbb{F}$-algebra, the $\mathbb{F}$-algebra
$U\left(  \left(  z,w/z\right)  \right)  $ is commutative and actually a $U$-algebra.

We finally notice that we can define an $\mathbb{F}$-module $U\left(  \left(
w,z/w\right)  \right)  $ analogously to how we defined $U\left(  \left(
z,w/z\right)  \right)  $ above. Again, this is a commutative $\mathbb{F}%
$-algebra if $U$ is a commutative $\mathbb{F}$-algebra, and satisfies similar
properties as $U\left(  \left(  z,w/z\right)  \right)  $. It is easy to see
that $U\left(  \left(  w,z/w\right)  \right)  \cap U\left(  \left(
z,w/z\right)  \right)  =U\left(  \left(  z,w\right)  \right)  $.

\subsection{\label{sect.iziw}Another point of view on $\delta\left(
z-w\right)  $: the $i_{z}$ and $i_{w}$ operators}

\begin{definition}
\label{def.R}Let $R$ be the localization of the ring $\mathbb{F}\left[
z,z^{-1},w,w^{-1}\right]  $ at the multiplicatively closed subset $\left\{
\left(  z-w\right)  ^{k}\ \mid\ k\in\mathbb{N}\right\}  $. In other words, $R$
is the localization of the ring $\mathbb{F}\left[  z,w\right]  $ at the
multiplicatively closed subset $\left\{  z^{a}w^{b}\left(  z-w\right)
^{c}\ \mid\ a,b,c\in\mathbb{N}\right\}  $. In other words, $R$ is the ring of
rational functions in $z$ and $w$ whose denominators are of the form
$z^{a}w^{b}\left(  z-w\right)  ^{c}$ with $a,b,c\in\mathbb{N}$. (In
geometrical language, $R$ is the ring of all rational functions in $z$ and $w$
with poles only at $z=0$, at $w=0$ and at $z=w$.)
\end{definition}

It is easy to see that the derivative operators $\partial_{z}$ and
$\partial_{w}$, and also the operators $\partial_{z}^{\left(  n\right)  }$ and
$\partial_{w}^{\left(  n\right)  }$ for each $n\in\mathbb{N}$, can be lifted
to $R$ from $\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  $. In more precise
terms, this means the following:

Recall that $R$ is a localization of the ring $\mathbb{F}\left[
z,z^{-1},w,w^{-1}\right]  $: namely, $R=\left\{  \left(  z-w\right)
^{k}\ \mid\ k\in\mathbb{N}\right\}  ^{-1}\mathbb{F}\left[  z,z^{-1}%
,w,w^{-1}\right]  $. Hence, the lift of the divided-powers HSD $\left(
\partial_{z}^{\left(  0\right)  },\partial_{z}^{\left(  1\right)  }%
,\partial_{z}^{\left(  2\right)  },\ldots\right)  $ to the localization
\newline$R=\left\{  \left(  z-w\right)  ^{k}\ \mid\ k\in\mathbb{N}\right\}
^{-1}\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  $ is well-defined (according
to Definition \ref{def.HSD.lift}). We shall denote this lift again by $\left(
\partial_{z}^{\left(  0\right)  },\partial_{z}^{\left(  1\right)  }%
,\partial_{z}^{\left(  2\right)  },\ldots\right)  $, since the restrictions of
these maps $\partial_{z}^{\left(  n\right)  }$ to $\mathbb{F}\left[
z,z^{-1},w,w^{-1}\right]  $ are the $\partial_{z}^{\left(  n\right)  }$ we
already know. Corollary \ref{cor.HSD.extend-to-loc.div} \textbf{(b)} shows
that this lift $\left(  \partial_{z}^{\left(  0\right)  },\partial
_{z}^{\left(  1\right)  },\partial_{z}^{\left(  2\right)  },\ldots\right)  $
is a divided-powers HSD. Similarly, there is a well-defined lift of the
divided-powers HSD $\left(  \partial_{w}^{\left(  0\right)  },\partial
_{w}^{\left(  1\right)  },\partial_{w}^{\left(  2\right)  },\ldots\right)  $
to the localization $R=\left\{  \left(  z-w\right)  ^{k}\ \mid\ k\in
\mathbb{N}\right\}  ^{-1}\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  $; this
lift will be denoted by $\left(  \partial_{w}^{\left(  0\right)  }%
,\partial_{w}^{\left(  1\right)  },\partial_{w}^{\left(  2\right)  }%
,\ldots\right)  $ and is again a divided-powers HSD.

We thus have defined operators $\partial_{z}^{\left(  0\right)  }$,
$\partial_{z}^{\left(  1\right)  }$, $\partial_{z}^{\left(  2\right)  }$,
$\ldots$ and operators $\partial_{w}^{\left(  0\right)  }$, $\partial
_{w}^{\left(  1\right)  }$, $\partial_{w}^{\left(  2\right)  }$, $\ldots$ on
$R$. We denote the operator $\partial_{z}^{\left(  1\right)  }$ by
$\partial_{z}$, and we denote the operator $\partial_{w}^{\left(  1\right)  }$
by $\partial_{w}$. (Of course, this is analogous to the fact that
$\partial_{z}^{\left(  1\right)  }=\partial_{z}$ and $\partial_{w}^{\left(
1\right)  }=\partial_{w}$ for all the other domains we have introduced these
operators on.)

The operators $\partial_{z}^{\left(  0\right)  }$, $\partial_{z}^{\left(
1\right)  }$, $\partial_{z}^{\left(  2\right)  }$, $\ldots$ on $R$ satisfy the
equalities (\ref{eq.prop.delta.deriv.0}), (\ref{eq.prop.delta.deriv.1}),
(\ref{eq.prop.delta.deriv.n}) and (\ref{eq.prop.delta.deriv.nm}%
)\footnote{\textit{Proof.} The operators $\partial_{z}^{\left(  0\right)  }$,
$\partial_{z}^{\left(  1\right)  }$, $\partial_{z}^{\left(  2\right)  }$,
$\ldots$ on $R$ satisfy the equalities (\ref{eq.prop.delta.deriv.0}) and
(\ref{eq.prop.delta.deriv.nm}) because the HSD $\left(  \partial_{z}^{\left(
0\right)  },\partial_{z}^{\left(  1\right)  },\partial_{z}^{\left(  2\right)
},\ldots\right)  $ is divided-powers. They satisfy the equality
(\ref{eq.prop.delta.deriv.1}) by definition of $\partial_{z}$. Finally, Remark
\ref{rmk.divpowHSD.n!} \textbf{(b)} (applied to $R$ and $\partial_{z}^{\left(
i\right)  }$ instead of $A$ and $D_{i}$) yields that $n!\partial_{z}^{\left(
n\right)  }=\left(  \partial_{z}^{\left(  1\right)  }\right)  ^{n}$ for every
$n\in\mathbb{N}$. Since $\partial_{z}^{\left(  1\right)  }=\partial_{z}$, this
rewrites as $n!\partial_{z}^{\left(  n\right)  }=\left(  \partial_{z}\right)
^{n}$ for every $n\in\mathbb{N}$. Thus, the operators $\partial_{z}^{\left(
0\right)  }$, $\partial_{z}^{\left(  1\right)  }$, $\partial_{z}^{\left(
2\right)  }$, $\ldots$ on $R$ satisfy the equality
(\ref{eq.prop.delta.deriv.n}) as well.}. An analogous statement holds for the
operators $\partial_{w}^{\left(  0\right)  }$, $\partial_{w}^{\left(
1\right)  }$, $\partial_{w}^{\left(  2\right)  }$, $\ldots$ on $R$. When
$\mathbb{F}$ is a $\mathbb{Q}$-algebra, we can use
(\ref{eq.prop.delta.deriv.n}) to rewrite all $\partial_{z}^{\left(  n\right)
}$ in terms of $\partial_{z}$, but in the general case the $\partial
_{z}^{\left(  n\right)  }$ carry some information that is not in $\partial
_{z}$.

We shall now define an $\mathbb{F}$-algebra homomorphism $i_{z}:R\rightarrow
\mathbb{F}\left(  \left(  z,w/z\right)  \right)  $ which will serve as an
algebraic analogue of what would be called \textquotedblleft expanding a
rational function in the domain $\left\vert z\right\vert >\left\vert
w\right\vert $\textquotedblright\ in complex analysis. Namely, we define an
$\mathbb{F}$-algebra homomorphism $i_{z}:R\rightarrow\mathbb{F}\left(  \left(
z,w/z\right)  \right)  $ by sending%
\[
z\mapsto z,\ \ \ \ \ \ \ \ \ \ w\mapsto w.
\]
In order to see that this is well-defined (according to the universal property
of localization), we need to show that the elements $z$, $w$ and $z-w$ of the
ring $\mathbb{F}\left(  \left(  z,w/z\right)  \right)  $ are invertible. This
can be directly checked: The element $z$ is invertible since $z^{-1}%
\in\mathbb{F}\left(  \left(  z,w/z\right)  \right)  $. The element $w$ is
invertible since $w^{-1}=z^{-1}\left(  w/z\right)  ^{-1}\in\mathbb{F}\left(
\left(  z,w/z\right)  \right)  $. The element $z-w$ is invertible since
$\dfrac{1}{z}\dfrac{1}{1-\dfrac{w}{z}}\in\mathbb{F}\left(  \left(
z,w/z\right)  \right)  $ is its inverse (because $z-w=z\left(  1-\dfrac{w}%
{z}\right)  $). So the well-definedness of $i_{z}$ is proven. Since
$\mathbb{F}\left(  \left(  z,w/z\right)  \right)  \subseteq\mathbb{F}\left[
\left[  z,z^{-1},w,w^{-1}\right]  \right]  $, we can regard $i_{z}$ as an
$\mathbb{F}$-linear map $i_{z}:R\rightarrow\mathbb{F}\left[  \left[
z,z^{-1},w,w^{-1}\right]  \right]  $, but this point of view prevents us from
speaking of $i_{z}$ as an $\mathbb{F}$-algebra homomorphism (since
$\mathbb{F}\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $ is not an
$\mathbb{F}$-algebra).

Explicitly, the map $i_{z}$ sends Laurent polynomials in $\mathbb{F}\left[
z,z^{-1},w,w^{-1}\right]  $ to themselves (regarded as elements of
$\mathbb{F}\left(  \left(  z,w/z\right)  \right)  \subseteq\mathbb{F}\left[
\left[  z,z^{-1},w,w^{-1}\right]  \right]  $), while sending $\dfrac{1}{z-w}$
to%
\begin{align}
i_{z}\dfrac{1}{z-w}  &  =\dfrac{1}{z}\dfrac{1}{1-\dfrac{w}{z}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\dfrac{1}{z}\dfrac{1}{1-\dfrac{w}{z}%
}\text{ is the inverse of }z-w\text{ in }\mathbb{F}\left(  \left(
z,w/z\right)  \right)  \right) \nonumber\\
&  =\dfrac{1}{z}\left(  1+\dfrac{w}{z}+\dfrac{w^{2}}{z^{2}}+\cdots\right)
=\sum_{n\geq0}z^{-n-1}w^{n}\label{iz.frac}\\
&  \in\mathbb{F}\left(  \left(  z,w/z\right)  \right)  \subseteq
\mathbb{F}\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  .\nonumber
\end{align}
The map $i_{z}$ is an $\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  $-algebra
homomorphism from $R$ to $\mathbb{F}\left(  \left(  z,w/z\right)  \right)  $,
and thus an $\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  $-module homomorphism
from $R$ to $\mathbb{F}\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $.

We can similarly define an $\mathbb{F}$-algebra homomorphism $i_{w}%
:R\rightarrow\mathbb{F}\left(  \left(  w,z/w\right)  \right)  $ which is
analogous to \textquotedblleft expansion in the domain $\left\vert
w\right\vert >\left\vert z\right\vert $\textquotedblright. It sends Laurent
polynomials in $\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  $ to themselves,
while sending $\dfrac{1}{z-w}$ to%
\begin{align}
i_{w}\dfrac{1}{z-w}  &  =\dfrac{1}{w}\dfrac{1}{\dfrac{z}{w}-1}=-\dfrac{1}%
{w}\dfrac{1}{1-\dfrac{z}{w}}\nonumber\\
&  =\dfrac{1}{w}\left(  1+\dfrac{z}{w}+\dfrac{z^{2}}{w^{2}}+\cdots\right)
=-\sum_{n\geq0}z^{n}w^{-n-1}=-\sum_{n<0}z^{-n-1}w^{n}\label{iw.frac}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }-n-1\text{ for
}n\text{ in the sum}\right) \nonumber\\
&  \in\mathbb{F}\left(  \left(  w,z/w\right)  \right)  \subseteq
\mathbb{F}\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  .\nonumber
\end{align}
The map $i_{w}$ is an $\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  $-algebra
homomorphism from $R$ to $\mathbb{F}\left(  \left(  w,z/w\right)  \right)  $,
and thus an $\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  $-module homomorphism
from $R$ to $\mathbb{F}\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $.

Subtracting (\ref{iw.frac}) from (\ref{iz.frac}), we obtain%
\begin{align*}
i_{z}\dfrac{1}{z-w}-i_{w}\dfrac{1}{z-w}  &  =\sum_{n\geq0}z^{-n-1}%
w^{n}-\left(  -\sum_{n<0}z^{-n-1}w^{n}\right) \\
&  =\sum_{n\geq0}z^{-n-1}w^{n}+\sum_{n<0}z^{-n-1}w^{n}=\sum_{n\in\mathbb{Z}%
}z^{-n-1}w^{n}=\delta\left(  z-w\right)  .
\end{align*}
Thus,%
\begin{equation}
\delta\left(  z-w\right)  =i_{z}\dfrac{1}{z-w}-i_{w}\dfrac{1}{z-w}.
\label{eq.prop.delta.props.5}%
\end{equation}


\begin{proposition}
\label{exe.3.3}\textbf{(a)} The maps $i_{z}$ and $i_{w}$ commute with the
operators $\partial_{z}$ and $\partial_{w}$. In other words,%
\begin{align*}
i_{z}\left(  \partial_{z}q\right)   &  =\partial_{z}\left(  i_{z}q\right)
;\ \ \ \ \ \ \ \ \ \ i_{z}\left(  \partial_{w}q\right)  =\partial_{w}\left(
i_{z}q\right)  ;\\
i_{w}\left(  \partial_{z}q\right)   &  =\partial_{z}\left(  i_{w}q\right)
;\ \ \ \ \ \ \ \ \ \ i_{w}\left(  \partial_{w}q\right)  =\partial_{w}\left(
i_{w}q\right)
\end{align*}
for every $q\in R$.

\textbf{(b)} The maps $i_{z}$ and $i_{w}$ commute with the operators
$\partial_{z}^{\left(  n\right)  }$ and $\partial_{w}^{\left(  n\right)  }$
for every $n\in\mathbb{N}$. (These operators were defined in Section
\ref{sect.deriv}.) In other words,%
\begin{align*}
i_{z}\left(  \partial_{z}^{\left(  n\right)  }q\right)   &  =\partial
_{z}^{\left(  n\right)  }\left(  i_{z}q\right)  ;\ \ \ \ \ \ \ \ \ \ i_{z}%
\left(  \partial_{w}^{\left(  n\right)  }q\right)  =\partial_{w}^{\left(
n\right)  }\left(  i_{z}q\right)  ;\\
i_{w}\left(  \partial_{z}^{\left(  n\right)  }q\right)   &  =\partial
_{z}^{\left(  n\right)  }\left(  i_{w}q\right)  ;\ \ \ \ \ \ \ \ \ \ i_{w}%
\left(  \partial_{w}^{\left(  n\right)  }q\right)  =\partial_{w}^{\left(
n\right)  }\left(  i_{w}q\right)
\end{align*}
for every $q\in R$ and $n\in\mathbb{N}$.

\textbf{(c)} The maps $i_{z}$ and $i_{w}$ are $\mathbb{F}\left[
z,z^{-1},w,w^{-1}\right]  $-linear. In other words, they commute with
multiplication by elements of $\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  $.
In other words,%
\[
i_{z}\left(  pq\right)  =pi_{z}\left(  q\right)  ;\ \ \ \ \ \ \ \ \ \ i_{w}%
\left(  pq\right)  =pi_{w}\left(  q\right)
\]
for every $q\in R$ and $p\in\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{exe.3.3}.]\textbf{(c)} The map $i_{z}$ is an
$\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  $-algebra homomorphism (since it
is an $\mathbb{F}$-algebra homomorphism which sends $z$ to $z$ and $w$ to
$w$), thus $\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  $-linear. This proves
Proposition \ref{exe.3.3} \textbf{(c)}.

\textbf{(b)} Let $q\in R$ and $n\in\mathbb{N}$. We shall only prove the
identity $i_{z}\left(  \partial_{z}^{\left(  n\right)  }q\right)
=\partial_{z}^{\left(  n\right)  }\left(  i_{z}q\right)  $. The other three
identities that are claimed in Proposition \ref{exe.3.3} can be shown in an
analogous manner.

Let $A=\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  $, and let $S$ be the
multiplicatively closed subset $\left\{  \left(  z-w\right)  ^{k}\ \mid
\ k\in\mathbb{N}\right\}  $ of $A$. Then, $R=S^{-1}A$ (by the definition of
$R$). Let $\iota:A\rightarrow S^{-1}A$ be the canonical $\mathbb{F}$-algebra
homomorphism from $A$ to $S^{-1}A$ sending each $a\in A$ to $\dfrac{a}{1}\in
S^{-1}A$. Thus, $\iota$ is an $\mathbb{F}$-algebra homomorphism from $A$ to
$R$ (since $R=S^{-1}A$).

We define three HSDs:

\begin{itemize}
\item Let $\mathbf{D}_{P}$ be the HSD $\left(  \partial_{z}^{\left(  0\right)
},\partial_{z}^{\left(  1\right)  },\partial_{z}^{\left(  2\right)  }%
,\ldots\right)  $ from $A$ to $A$.

\item Let $\mathbf{D}_{R}$ be the HSD $\left(  \partial_{z}^{\left(  0\right)
},\partial_{z}^{\left(  1\right)  },\partial_{z}^{\left(  2\right)  }%
,\ldots\right)  $ from $R$ to $R$.

\item Let $\mathbf{D}_{L}$ be the HSD $\left(  \partial_{z}^{\left(  0\right)
},\partial_{z}^{\left(  1\right)  },\partial_{z}^{\left(  2\right)  }%
,\ldots\right)  $ from $\mathbb{F}\left(  \left(  z,w/z\right)  \right)  $ to
$\mathbb{F}\left(  \left(  z,w/z\right)  \right)  $.
\end{itemize}

We recall that the HSD $\mathbf{D}_{R}$ was defined as the lift of the HSD
$\mathbf{D}_{P}$ to the localization $S^{-1}A=R$. Hence, we can apply
Corollary \ref{cor.HSD.extend-to-loc.div} \textbf{(c)} to $\mathbf{D}_{P}$ and
$\mathbf{D}_{R}$ instead of $\mathbf{D}$ and $\mathbf{D}^{\prime}$. As the
result, we obtain $\mathbf{D}_{R}\circ\iota=\iota\circ\mathbf{D}_{P}$ (where
$\mathbf{D}_{R}\circ\iota$ and $\iota\circ\mathbf{D}_{P}$ are defined
according to Definition \ref{def.HSD.compose}).

Recall that $i_{z}:R\rightarrow\mathbb{F}\left(  \left(  z,w/z\right)
\right)  $ is an $\mathbb{F}$-algebra homomorphism. Hence, two HSDs
$i_{z}\circ\mathbf{D}_{R}$ and $\mathbf{D}_{L}\circ i_{z}$ from $R$ to
$\mathbb{F}\left(  \left(  z,w/z\right)  \right)  $ are defined (according to
Definition \ref{def.HSD.compose}). We shall now show that these two HSDs are equal.

Let $\operatorname*{inc}$ denote the canonical inclusion map $A\rightarrow
\mathbb{F}\left(  \left(  z,w/z\right)  \right)  $. Then, $i_{z}\circ
\iota=\operatorname*{inc}$ (since the map $i_{z}$ sends Laurent polynomials in
$\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  $ to themselves). Moreover,
$\operatorname*{inc}$ is an $\mathbb{F}$-algebra homomorphism, so that two
HSDs $\operatorname*{inc}\circ\mathbf{D}_{P}$ and $\mathbf{D}_{L}%
\circ\operatorname*{inc}$ are well-defined. It is easy to see that
$\operatorname*{inc}\circ\mathbf{D}_{P}=\mathbf{D}_{L}\circ\operatorname*{inc}%
$\ \ \ \ \footnote{\textit{Proof.} Proving that $\operatorname*{inc}%
\circ\mathbf{D}_{P}=\mathbf{D}_{L}\circ\operatorname*{inc}$ is tantamount to
showing that $\operatorname*{inc}\circ\partial_{z}^{\left(  n\right)
}=\partial_{z}^{\left(  n\right)  }\circ\operatorname*{inc}$ for every
$n\in\mathbb{N}$ (because of how $\operatorname*{inc}\circ\mathbf{D}_{P}$ and
$\mathbf{D}_{L}\circ\operatorname*{inc}$ are defined). This is equivalent to
showing that the restriction of the endomorphism $\partial_{z}^{\left(
n\right)  }$ of $\mathbb{F}\left(  \left(  z,w/z\right)  \right)  $ to $A$ is
precisely the endomorphism $\partial_{z}^{\left(  n\right)  }$ of $A$. But the
latter claim is obvious (just remember that both endomorphisms $\partial
_{z}^{\left(  n\right)  }$ were defined by the same formula).}. Now,%
\[
i_{z}\circ\underbrace{\mathbf{D}_{R}\circ\iota}_{=\iota\circ\mathbf{D}_{P}%
}=\underbrace{i_{z}\circ\iota}_{=\operatorname*{inc}}\circ\mathbf{D}%
_{P}=\operatorname*{inc}\circ\mathbf{D}_{P}=\mathbf{D}_{L}\circ
\underbrace{\operatorname*{inc}}_{=i_{z}\circ\iota}=\mathbf{D}_{L}\circ
i_{z}\circ\iota.
\]
Thus, Corollary \ref{cor.HSD.extend-to-loc.epi} (applied to $\mathbb{F}\left(
\left(  z,w/z\right)  \right)  $, $i_{z}\circ\mathbf{D}_{R}$ and
$\mathbf{D}_{L}\circ i_{z}$ instead of $B$, $\mathbf{E}$ and $\mathbf{F}$)
yields $i_{z}\circ\mathbf{D}_{R}=\mathbf{D}_{L}\circ i_{z}$. But the
definition of $i_{z}\circ\mathbf{D}_{R}$ yields%
\[
\left(  i_{z}\circ\partial_{z}^{\left(  0\right)  },i_{z}\circ\partial
_{z}^{\left(  1\right)  },i_{z}\circ\partial_{z}^{\left(  2\right)  }%
,\ldots\right)  =i_{z}\circ\mathbf{D}_{R}=\mathbf{D}_{L}\circ i_{z}=\left(
\partial_{z}^{\left(  0\right)  }\circ i_{z},\partial_{z}^{\left(  1\right)
}\circ i_{z},\partial_{z}^{\left(  2\right)  }\circ i_{z},\ldots\right)
\]
(by the definition of $\mathbf{D}_{L}\circ i_{z}$). Hence, $i_{z}\circ
\partial_{z}^{\left(  n\right)  }=\partial_{z}^{\left(  n\right)  }\circ
i_{z}$. Thus,$i_{z}\left(  \partial_{z}^{\left(  n\right)  }q\right)
=\underbrace{\left(  i_{z}\circ\partial_{z}^{\left(  n\right)  }\right)
}_{=\partial_{z}^{\left(  n\right)  }\circ i_{z}}\left(  q\right)  =\left(
\partial_{z}^{\left(  n\right)  }\circ i_{z}\right)  \left(  q\right)
=\partial_{z}^{\left(  n\right)  }\left(  i_{z}q\right)  $. This completes our
proof of Proposition \ref{exe.3.3} \textbf{(b)}.

\begin{noncompile}
Here are some pieces of a computational proof I wanted to write.

Let us first show that%
\begin{equation}
i_{z}\left(  z^{a}w^{b}\left(  z-w\right)  ^{c}\right)  =\sum_{m\geq0}%
\dbinom{c}{m}\left(  -1\right)  ^{m}z^{a+c-m-1}w^{b+m} \label{pf.exe.3.3.a.0}%
\end{equation}
for all $a,b,c\in\mathbb{Z}$.

\textit{Proof of (\ref{pf.exe.3.3.a.0}):} Let $a,b,c\in\mathbb{Z}$. Recall
that $i_{z}$ is an $\mathbb{F}$-algebra homomorphism which sends $z$ to $z$
and sends $w$ to $w$. Thus,%
\begin{equation}
i_{z}\left(  z^{a}w^{b}\left(  z-w\right)  ^{c}\right)  =z^{a}w^{b}\left(
z-w\right)  ^{c}, \label{pf.exe.3.3.a.0.pf.1}%
\end{equation}
where the $\left(  z-w\right)  ^{c}$ on the right hand side is to be computed
in the ring $\mathbb{F}\left(  \left(  z,w/z\right)  \right)  $. Let us
compute $\left(  z-w\right)  ^{c}$ in the ring $\mathbb{F}\left(  \left(
z,w/z\right)  \right)  $: We have%
\begin{align*}
\left(  \underbrace{z-w}_{=z\left(  1-\dfrac{w}{z}\right)  }\right)  ^{c}  &
=\left(  z\left(  1-\dfrac{w}{z}\right)  \right)  ^{c}=z^{c}%
\underbrace{\left(  1-\dfrac{w}{z}\right)  ^{c}}_{\substack{=\sum_{m\geq
0}\dbinom{c}{m}\left(  -1\right)  ^{m}\left(  \dfrac{w}{z}\right)
^{m}\\\text{(by Newton's binomial formula)}}}\\
&  =z^{c}\sum_{m\geq0}\dbinom{c}{m}\left(  -1\right)  ^{m}\left(  \dfrac{w}%
{z}\right)  ^{m}=\sum_{m\geq0}\dbinom{c}{m}\left(  -1\right)  ^{m}%
\underbrace{\left(  \dfrac{w}{z}\right)  ^{m}z^{c}}_{=z^{c-m}w^{m}}\\
&  =\sum_{m\geq0}\dbinom{c}{m}\left(  -1\right)  ^{m}z^{c-m}w^{m}.
\end{align*}
Thus, (\ref{pf.exe.3.3.a.0.pf.1}) becomes%
\begin{align*}
i_{z}\left(  z^{a}w^{b}\left(  z-w\right)  ^{c}\right)   &  =z^{a}%
w^{b}\underbrace{\left(  z-w\right)  ^{c}}_{=\sum_{m\geq0}\dbinom{c}{m}\left(
-1\right)  ^{m}z^{c-m}w^{m}}=z^{a}w^{b}\sum_{m\geq0}\dbinom{c}{m}\left(
-1\right)  ^{m}z^{c-m}w^{m}\\
&  =\sum_{m\geq0}\dbinom{c}{m}\left(  -1\right)  ^{m}z^{a+c-m-1}w^{b+m}.
\end{align*}
This proves (\ref{pf.exe.3.3.a.0}).

Now, let $n\in\mathbb{N}$. Furthermore, let $q\in R$. We shall prove the
equality $i_{z}\left(  \partial_{z}^{\left(  n\right)  }q\right)
=\partial_{z}^{\left(  n\right)  }\left(  i_{z}q\right)  $.

This equality is $\mathbb{F}$-linear in $q$. Hence, we can WLOG assume that
$q=z^{a}w^{b}\left(  z-w\right)  ^{c}$ for some $a,b,c\in\mathbb{Z}$ (because
the $\mathbb{F}$-module $R$ is spanned by elements of the form $z^{a}%
w^{b}\left(  z-w\right)  ^{c}$ with $a,b,c\in\mathbb{Z}$). Assume this, and
consider these $a,b,c$. Now, it is easy to see that%
\[
\partial_{z}^{\left(  n\right)  }q=\sum_{k=0}^{n}\dbinom{a}{k}\dbinom{c}%
{n-k}z^{a-k}w^{b}\left(  z-w\right)  ^{c-\left(  n-k\right)  }.
\]
(Not sure about this actually.)
\end{noncompile}

\textbf{(a)} Proposition \ref{exe.3.3} \textbf{(a)} follows from Proposition
\ref{exe.3.3} \textbf{(b)} (applied to $n=1$), since $\partial_{z}^{\left(
1\right)  }=\partial_{z}$ and $\partial_{w}^{\left(  1\right)  }=\partial_{w}$.
\end{proof}

\subsection{Further properties of $\delta\left(  z-w\right)  $}

Next, we observe:

\begin{proposition}
\label{prop.dwdelta}\textbf{(a)} For every $k\in\mathbb{N}$, we have
$\partial_{w}^{\left(  k\right)  }\dfrac{1}{z-w}=\dfrac{1}{\left(  z-w\right)
^{k+1}}$ in $R$.

\textbf{(b)} For every $k\in\mathbb{N}$, we have%
\begin{align}
\partial_{w}^{\left(  k\right)  }\delta\left(  z-w\right)   &  =i_{z}\dfrac
{1}{\left(  z-w\right)  ^{k+1}}-i_{w}\dfrac{1}{\left(  z-w\right)  ^{k+1}%
}\label{eq.6a}\\
&  =\sum_{n\in\mathbb{Z}}\dbinom{n}{k}w^{n-k}z^{-n-1} \label{eq.6b}%
\end{align}
in $\mathbb{F}\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.dwdelta}.]\textbf{(a)} When $\mathbb{F}$ is a
$\mathbb{Q}$-algebra, one can easily prove Proposition \ref{prop.dwdelta}
\textbf{(a)} by induction over $k$ using (\ref{eq.prop.delta.deriv.n}).
However, we want to prove it in general. So let us take a different approach.

Let $\mathbf{D}$ denote the HSD $\left(  \partial_{w}^{\left(  0\right)
},\partial_{w}^{\left(  1\right)  },\partial_{w}^{\left(  2\right)  }%
,\ldots\right)  $ of $R$. Let $t$ be a new symbol. Define an $\mathbb{F}%
$-linear map $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle
}:R\rightarrow R\left[  \left[  t\right]  \right]  $ as in Definition
\ref{def.HSD.powerser}. Theorem \ref{thm.HSD.powerser} then yields that
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$ is an $\mathbb{F}%
$-algebra homomorphism (since $\mathbf{D}$ is an HSD). But the definition of
$\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }$ yields%
\begin{align*}
\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  z-w\right)   &
=\sum_{n\in\mathbb{N}}\partial_{w}^{\left(  n\right)  }\left(  z-w\right)
t^{n}=\left(  z-w\right)  +1t\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\partial_{w}^{\left(  n\right)
}\left(  z-w\right)  =\left\{
\begin{array}
[c]{c}%
z-w,\text{ if }n=0;\\
1,\text{ if }n=1;\\
0,\text{ otherwise}%
\end{array}
\right.  \text{ for all }n\in\mathbb{N}\right) \\
&  =\left(  z-w\right)  +t.
\end{align*}
Since $\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(
z-w\right)  $ is an $\mathbb{F}$-algebra homomorphism, we have
\begin{align*}
\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  \left(
z-w\right)  ^{-1}\right)   &  =\left(  \underbrace{\widetilde{\mathbf{D}%
}_{\left\langle t\right\rangle }\left(  z-w\right)  }_{=\left(  z-w\right)
+t}\right)  ^{-1}=\left(  \left(  z-w\right)  +t\right)  ^{-1}\\
&  =\left(  z-w\right)  ^{-1}\underbrace{\left(  1+\dfrac{t}{z-w}\right)
^{-1}}_{=\sum_{n\in\mathbb{N}}\left(  \dfrac{t}{z-w}\right)  ^{n}}=\left(
z-w\right)  ^{-1}\sum_{n\in\mathbb{N}}\left(  \dfrac{t}{z-w}\right)  ^{n}\\
&  =\sum_{n\in\mathbb{N}}\underbrace{\left(  z-w\right)  ^{-1}\left(
\dfrac{t}{z-w}\right)  ^{n}}_{=\dfrac{1}{\left(  z-w\right)  ^{n+1}}t^{n}%
}=\sum_{n\in\mathbb{N}}\dfrac{1}{\left(  z-w\right)  ^{n+1}}t^{n}.
\end{align*}
Compared with%
\[
\widetilde{\mathbf{D}}_{\left\langle t\right\rangle }\left(  \left(
z-w\right)  ^{-1}\right)  =\sum_{n\in\mathbb{N}}\partial_{w}^{\left(
n\right)  }\left(  \left(  z-w\right)  ^{-1}\right)  t^{n},
\]
this yields $\sum_{n\in\mathbb{N}}\partial_{w}^{\left(  n\right)  }\left(
\left(  z-w\right)  ^{-1}\right)  t^{n}=\sum_{n\in\mathbb{N}}\dfrac{1}{\left(
z-w\right)  ^{n+1}}t^{n}$. Comparing coefficients in this equality, we obtain
$\partial_{w}^{\left(  n\right)  }\left(  \left(  z-w\right)  ^{-1}\right)
=\dfrac{1}{\left(  z-w\right)  ^{n+1}}$ for all $n\in\mathbb{N}$. In other
words, $\partial_{w}^{\left(  k\right)  }\left(  \left(  z-w\right)
^{-1}\right)  =\dfrac{1}{\left(  z-w\right)  ^{k+1}}$ for all $k\in\mathbb{N}%
$. This proves Proposition \ref{prop.dwdelta} \textbf{(a)}.

\textbf{(b)} Let $k\in\mathbb{N}$. Applying the map $\partial_{w}^{\left(
k\right)  }$ to both sides of (\ref{eq.prop.delta.props.5}), we obtain%
\begin{align*}
\partial_{w}^{\left(  k\right)  }\delta\left(  z-w\right)   &
=\underbrace{\partial_{w}^{\left(  k\right)  }\left(  i_{z}\dfrac{1}%
{z-w}\right)  }_{\substack{=i_{z}\left(  \partial_{w}^{\left(  k\right)
}\dfrac{1}{z-w}\right)  \\\text{(by Proposition \ref{exe.3.3} \textbf{(b)})}%
}}-\underbrace{\partial_{w}^{\left(  k\right)  }\left(  i_{w}\dfrac{1}%
{z-w}\right)  }_{\substack{=i_{w}\left(  \partial_{w}^{\left(  k\right)
}\dfrac{1}{z-w}\right)  \\\text{(by Proposition \ref{exe.3.3} \textbf{(b)})}%
}}\\
&  =i_{z}\underbrace{\left(  \partial_{w}^{\left(  k\right)  }\dfrac{1}%
{z-w}\right)  }_{\substack{=\dfrac{1}{\left(  z-w\right)  ^{k+1}}\\\text{(by
Proposition \ref{prop.dwdelta} \textbf{(a)})}}}-i_{w}\underbrace{\left(
\partial_{w}^{\left(  k\right)  }\dfrac{1}{z-w}\right)  }_{\substack{=\dfrac
{1}{\left(  z-w\right)  ^{k+1}}\\\text{(by Proposition \ref{prop.dwdelta}
\textbf{(a)})}}}\\
&  =i_{z}\dfrac{1}{\left(  z-w\right)  ^{k+1}}-i_{w}\dfrac{1}{\left(
z-w\right)  ^{k+1}}.
\end{align*}
On the other hand, applying the map $\partial_{w}^{\left(  k\right)  }$ to
both sides of the equality $\delta\left(  z-w\right)  =\sum_{n\in\mathbb{Z}%
}z^{-n-1}w^{n}$, we obtain%
\[
\partial_{w}^{\left(  k\right)  }\delta\left(  z-w\right)  =\partial
_{w}^{\left(  k\right)  }\sum_{n\in\mathbb{Z}}z^{-n-1}w^{n}=\sum
_{n\in\mathbb{Z}}\dbinom{n}{k}z^{-n-1}w^{n-k}=\sum_{n\in\mathbb{Z}}\dbinom
{n}{k}w^{n-k}z^{-n-1}.
\]
This completes the proof of Proposition \ref{prop.dwdelta} \textbf{(b)}.
\end{proof}

\begin{proposition}
\label{prop.delta.res}\textbf{(a)} We have $\operatorname*{Res}\partial
_{w}^{\left(  n\right)  }\delta\left(  z-w\right)  dz=\delta_{n,0}$ for every
$n\in\mathbb{N}$.

\textbf{(b)} Any $m\in\mathbb{N}$ and $n\in\mathbb{N}$ satisfy%
\[
\left(  z-w\right)  ^{m}\partial_{w}^{\left(  n\right)  }\delta\left(
z-w\right)  =\left\{
\begin{array}
[c]{c}%
\partial_{w}^{\left(  n-m\right)  }\delta\left(  z-w\right)  ,\text{ if }m\leq
n;\\
0,\text{ if }m>n
\end{array}
\right.  .
\]


\textbf{(c)} We have $\delta\left(  z-w\right)  =\delta\left(  w-z\right)  $.
(Here, $\delta\left(  w-z\right)  $ means $\left(  \delta\left(  z-w\right)
\right)  \left(  w,z\right)  $, that is, the result of switching $z$ with $w$
in $\delta\left(  z-w\right)  $.)

\textbf{(d)} We have $\partial_{z}\delta\left(  z-w\right)  =-\partial
_{w}\delta\left(  z-w\right)  $.

\textbf{(e)} We have $\partial_{z}^{\left(  j\right)  }\delta\left(
z-w\right)  =\left(  -1\right)  ^{j}\partial_{w}^{\left(  j\right)  }%
\delta\left(  z-w\right)  $ for every $j\in\mathbb{N}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.delta.res}.]\textbf{(a)} For every
$k\in\mathbb{N}$, we have%
\begin{align*}
\operatorname*{Res}\underbrace{\partial_{w}^{\left(  k\right)  }\delta\left(
z-w\right)  }_{\substack{=\sum_{n\in\mathbb{Z}}\dbinom{n}{k}w^{n-k}%
z^{-n-1}\\\text{(by (\ref{eq.6b}))}}}dz  &  =\operatorname*{Res}\left(
\sum_{n\in\mathbb{Z}}\dbinom{n}{k}w^{n-k}z^{-n-1}\right)
dz=\underbrace{\dbinom{0}{k}}_{=\delta_{k,0}}w^{0-k}\\
&  =\delta_{k,0}w^{0-k}=\delta_{k,0}.
\end{align*}
In other words, for every $n\in\mathbb{N}$, we have $\operatorname*{Res}%
\partial_{w}^{\left(  n\right)  }\delta\left(  z-w\right)  dz=\delta_{n,0}$.
This proves Proposition \ref{prop.delta.res} \textbf{(a)}.

\textbf{(b)} Let $m\in\mathbb{N}$ and $n\in\mathbb{N}$. Applying (\ref{eq.6a})
to $k=n$, we obtain%
\[
\partial_{w}^{\left(  n\right)  }\delta\left(  z-w\right)  =i_{z}\dfrac
{1}{\left(  z-w\right)  ^{n+1}}-i_{w}\dfrac{1}{\left(  z-w\right)  ^{n+1}}.
\]
Multiplying this equality with $\left(  z-w\right)  ^{m}$ from the left, we
obtain%
\begin{align}
\left(  z-w\right)  ^{m}\partial_{w}^{\left(  n\right)  }\delta\left(
z-w\right)   &  =\underbrace{\left(  z-w\right)  ^{m}i_{z}\dfrac{1}{\left(
z-w\right)  ^{n+1}}}_{\substack{=i_{z}\left(  \left(  z-w\right)  ^{m}%
\cdot\dfrac{1}{\left(  z-w\right)  ^{n+1}}\right)  \\\text{(since }i_{z}\text{
is an }\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  \text{-module}%
\\\text{homomorphism)}}}-\underbrace{\left(  z-w\right)  ^{m}i_{w}\dfrac
{1}{\left(  z-w\right)  ^{n+1}}}_{\substack{=i_{w}\left(  \left(  z-w\right)
^{m}\cdot\dfrac{1}{\left(  z-w\right)  ^{n+1}}\right)  \\\text{(since }%
i_{w}\text{ is an }\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  \text{-module}%
\\\text{homomorphism)}}}\nonumber\\
&  =i_{z}\left(  \underbrace{\left(  z-w\right)  ^{m}\cdot\dfrac{1}{\left(
z-w\right)  ^{n+1}}}_{=\dfrac{1}{\left(  z-w\right)  ^{\left(  n-m\right)
+1}}}\right)  -i_{w}\left(  \underbrace{\left(  z-w\right)  ^{m}\cdot\dfrac
{1}{\left(  z-w\right)  ^{n+1}}}_{=\dfrac{1}{\left(  z-w\right)  ^{\left(
n-m\right)  +1}}}\right) \nonumber\\
&  =i_{z}\dfrac{1}{\left(  z-w\right)  ^{\left(  n-m\right)  +1}}-i_{w}%
\dfrac{1}{\left(  z-w\right)  ^{\left(  n-m\right)  +1}}.
\label{pf.prop.delta.res.b.3}%
\end{align}
Now, we must be in one of the following two cases:

\textit{Case 1:} We have $m\leq n$.

\textit{Case 2:} We have $m>n$.

Let us first consider Case 1. In this case, we have $m\leq n$. Thus, applying
(\ref{eq.6a}) to $k=n-m$, we obtain%
\[
\partial_{w}^{\left(  n-m\right)  }\delta\left(  z-w\right)  =i_{z}\dfrac
{1}{\left(  z-w\right)  ^{\left(  n-m\right)  +1}}-i_{w}\dfrac{1}{\left(
z-w\right)  ^{\left(  n-m\right)  +1}}.
\]
Compared with (\ref{pf.prop.delta.res.b.3}), this yields%
\[
\left(  z-w\right)  ^{m}\partial_{w}^{\left(  n\right)  }\delta\left(
z-w\right)  =\partial_{w}^{\left(  n-m\right)  }\delta\left(  z-w\right)
=\left\{
\begin{array}
[c]{c}%
\partial_{w}^{\left(  n-m\right)  }\delta\left(  z-w\right)  ,\text{ if }m\leq
n;\\
0,\text{ if }m>n
\end{array}
\right.
\]
(since $m\leq n$). Thus, Proposition \ref{prop.delta.res} \textbf{(b)} is
proven in Case 1.

Let us next consider Case 2. In this case, we have $m>n$, so that $m-n-1\geq
0$. Now, $\dfrac{1}{\left(  z-w\right)  ^{\left(  n-m\right)  +1}}=\left(
z-w\right)  ^{m-n-1}\in\mathbb{F}\left[  z,z^{-1},w,w^{-1}\right]  $ (since
$m-n-1\geq0$), so that $i_{z}\dfrac{1}{\left(  z-w\right)  ^{\left(
n-m\right)  +1}}=\dfrac{1}{\left(  z-w\right)  ^{\left(  n-m\right)  +1}}$
(since $i_{z}$ sends Laurent polynomials in $\mathbb{F}\left[  z,z^{-1}%
,w,w^{-1}\right]  $ to themselves). Similarly, $i_{w}\dfrac{1}{\left(
z-w\right)  ^{\left(  n-m\right)  +1}}=\dfrac{1}{\left(  z-w\right)  ^{\left(
n-m\right)  +1}}$. Hence, (\ref{pf.prop.delta.res.b.3}) becomes%
\begin{align*}
\left(  z-w\right)  ^{m}\partial_{w}^{\left(  n\right)  }\delta\left(
z-w\right)   &  =\underbrace{i_{z}\dfrac{1}{\left(  z-w\right)  ^{\left(
n-m\right)  +1}}}_{=\dfrac{1}{\left(  z-w\right)  ^{\left(  n-m\right)  +1}}%
}-\underbrace{i_{w}\dfrac{1}{\left(  z-w\right)  ^{\left(  n-m\right)  +1}}%
}_{=\dfrac{1}{\left(  z-w\right)  ^{\left(  n-m\right)  +1}}}\\
&  =0=\left\{
\begin{array}
[c]{c}%
\partial_{w}^{\left(  n-m\right)  }\delta\left(  z-w\right)  ,\text{ if }m\leq
n;\\
0,\text{ if }m>n
\end{array}
\right.
\end{align*}
(since $m>n$). Thus, Proposition \ref{prop.delta.res} \textbf{(b)} is proven
in Case 2. The proof of Proposition \ref{prop.delta.res} \textbf{(b)} is now complete.

\textbf{(c)} We have $\delta\left(  z-w\right)  =\sum_{n\in\mathbb{Z}}%
z^{-n-1}w^{n}$, so that%
\begin{align*}
\delta\left(  w-z\right)   &  =\sum_{n\in\mathbb{Z}}w^{-n-1}z^{n}=\sum
_{n\in\mathbb{Z}}w^{n}z^{-n-1}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted}\\
-n-1\text{ for }n\text{ in the sum}%
\end{array}
\right) \\
&  =\sum_{n\in\mathbb{Z}}z^{-n-1}w^{n}=\delta\left(  z-w\right)  ,
\end{align*}
so that Proposition \ref{prop.delta.res} \textbf{(c)} is proven.

\textbf{(d)} Applying $\partial_{z}$ to both sides of $\delta\left(
z-w\right)  =\sum_{n\in\mathbb{Z}}z^{-n-1}w^{n}$, we obtain%
\[
\partial_{z}\delta\left(  z-w\right)  =\partial_{z}\sum_{n\in\mathbb{Z}%
}z^{-n-1}w^{n}=\sum_{n\in\mathbb{Z}}\left(  -n-1\right)  z^{-n-2}w^{n}%
=\sum_{n\in\mathbb{Z}}\left(  -n\right)  z^{-n-1}w^{n-1}%
\]
(here, we have substituted $n-1$ for $n$ in the sum). Applying $\partial_{w}$
to both sides of $\delta\left(  z-w\right)  =\sum_{n\in\mathbb{Z}}%
z^{-n-1}w^{n}$, we obtain%
\[
\partial_{w}\delta\left(  z-w\right)  =\partial_{w}\sum_{n\in\mathbb{Z}%
}z^{-n-1}w^{n}=\sum_{n\in\mathbb{Z}}nz^{-n-1}w^{n-1}.
\]
Adding these two inequalities together, we obtain%
\begin{align*}
\partial_{z}\delta\left(  z-w\right)  +\partial_{w}\delta\left(  z-w\right)
&  =\sum_{n\in\mathbb{Z}}\left(  -n\right)  z^{-n-1}w^{n-1}+\sum
_{n\in\mathbb{Z}}nz^{-n-1}w^{n-1}\\
&  =\sum_{n\in\mathbb{Z}}\underbrace{\left(  -n+n\right)  }_{=0}%
z^{-n-1}w^{n-1}=0.
\end{align*}
This yields Proposition \ref{prop.delta.res} \textbf{(d)}.

\textbf{(e)} Let $j\in\mathbb{N}$. Every $n\in\mathbb{Z}$ satisfies%
\begin{equation}
\dbinom{j-1-n}{j}=\left(  -1\right)  ^{j}\dbinom{n}{j}.
\label{pf.prop.delta.res.e.upperneg}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.delta.res.e.upperneg}):} Let
$n\in\mathbb{Z}$. Then,%
\begin{align}
&  \left(  j-1-n\right)  \left(  \left(  j-1-n\right)  -1\right)
\cdots\left(  \left(  j-1-n\right)  -j+1\right) \nonumber\\
&  =\left(  j-1-n\right)  \left(  j-2-n\right)  \cdots\left(  j-j-n\right)
\nonumber\\
&  =\left(  -\left(  n-j+1\right)  \right)  \left(  -\left(  n-j+2\right)
\right)  \cdots\left(  -\left(  n-j+j\right)  \right) \nonumber\\
&  =\left(  -1\right)  ^{j}\cdot\underbrace{\left(  n-j+1\right)  \left(
n-j+2\right)  \cdots\left(  n-j+j\right)  }_{\substack{=\left(  n-j+j\right)
\left(  n-j+\left(  j-1\right)  \right)  \cdots\left(  n-j+1\right)
\\=n\left(  n-1\right)  \cdots\left(  n-j+1\right)  }}\nonumber\\
&  =\left(  -1\right)  ^{j}\cdot n\left(  n-1\right)  \cdots\left(
n-j+1\right)  . \label{pf.prop.delta.res.e.upperneg.pf.1}%
\end{align}
Now, the definition of binomial coefficients shows that%
\begin{align*}
\dbinom{j-1-n}{j}  &  =\dfrac{\left(  j-1-n\right)  \left(  \left(
j-1-n\right)  -1\right)  \cdots\left(  \left(  j-1-n\right)  -j+1\right)
}{j!}\\
&  =\dfrac{\left(  -1\right)  ^{j}\cdot n\left(  n-1\right)  \cdots\left(
n-j+1\right)  }{j!}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.delta.res.e.upperneg.pf.1})}\right) \\
&  =\left(  -1\right)  ^{j}\cdot\underbrace{\dfrac{n\left(  n-1\right)
\cdots\left(  n-j+1\right)  }{j!}}_{\substack{=\dbinom{n}{j}\\\text{(by the
definition of }\dbinom{n}{j}\text{)}}}=\left(  -1\right)  ^{j}\dbinom{n}{j}.
\end{align*}
This proves (\ref{pf.prop.delta.res.e.upperneg}).} But applying (\ref{eq.6b})
to $k=j$, we obtain%
\[
\partial_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  =\sum
_{n\in\mathbb{Z}}\dbinom{n}{j}w^{n-j}z^{-n-1}.
\]
Renaming the indeterminates $z$ and $w$ as $w$ and $z$ in this equality, we
obtain%
\begin{align*}
\partial_{z}^{\left(  j\right)  }\delta\left(  w-z\right)   &  =\sum
_{n\in\mathbb{Z}}\dbinom{n}{j}z^{n-j}w^{-n-1}=\sum_{n\in\mathbb{Z}%
}\underbrace{\dbinom{j-1-n}{j}}_{\substack{=\left(  -1\right)  ^{j}\dbinom
{n}{j}\\\text{(by (\ref{pf.prop.delta.res.e.upperneg}))}}}\underbrace{z^{-n-1}%
w^{n-j}}_{=w^{n-j}z^{-n-1}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }j-1-n\text{ for
}n\text{ in the sum}\right) \\
&  =\sum_{n\in\mathbb{Z}}\left(  -1\right)  ^{j}\dbinom{n}{j}w^{n-j}%
z^{-n-1}=\left(  -1\right)  ^{j}\underbrace{\sum_{n\in\mathbb{Z}}\dbinom{n}%
{j}w^{n-j}z^{-n-1}}_{=\partial_{w}^{\left(  j\right)  }\delta\left(
z-w\right)  }\\
&  =\left(  -1\right)  ^{j}\partial_{w}^{\left(  j\right)  }\delta\left(
z-w\right)  .
\end{align*}
This proves Proposition \ref{prop.delta.res} \textbf{(e)}.
\end{proof}

\subsection{The decomposition theorem}

The next fact is known as the \textit{decomposition theorem}:

\begin{theorem}
\label{thm.decomposition}Let $U$ be an $\mathbb{F}$-module. Let $a\left(
z,w\right)  \in U\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $ be a
local $U$-valued formal distribution. Then, $a\left(  z,w\right)  $ uniquely
decomposes as a sum
\[
\sum_{j\in\mathbb{N}}c^{j}\left(  w\right)  \partial_{w}^{\left(  j\right)
}\delta\left(  z-w\right)
\]
with $\left(  c^{j}\left(  w\right)  \right)  _{j\in\mathbb{N}}$ being a
family of formal distributions $c^{j}\left(  w\right)  \in U\left[  \left[
w,w^{-1}\right]  \right]  $ having the property that all but finitely many
$j\in\mathbb{N}$ satisfy $c^{j}\left(  w\right)  =0$.

The formal distributions $c^{j}\left(  w\right)  $ in this decomposition are
given by%
\begin{equation}
c^{j}\left(  w\right)  =\operatorname*{Res}\left(  z-w\right)  ^{j}a\left(
z,w\right)  dz\ \ \ \ \ \ \ \ \ \ \text{for all }j\in\mathbb{N}. \label{eq.8}%
\end{equation}

\end{theorem}

Before we prove this theorem, let us show a lemma:

\begin{lemma}
\label{lem.decomposition.cut}Let $U$ be an $\mathbb{F}$-module. Let $b\left(
z,w\right)  \in U\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $. Assume
that%
\begin{equation}
\operatorname*{Res}\left(  z-w\right)  ^{n}b\left(  z,w\right)
dz=0\ \ \ \ \ \ \ \ \ \ \text{for all }n\in\mathbb{N}.
\label{eq.lem.decomposition.cut.condi}%
\end{equation}


\textbf{(a)} Then, $b\left(  z,w\right)  \in\left(  U\left[  \left[
w,w^{-1}\right]  \right]  \right)  \left[  \left[  z\right]  \right]  $.

\textbf{(b)} Assume that $b\left(  z,w\right)  $ is local. Then, $b\left(
z,w\right)  =0$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.decomposition.cut}.]We have $b\left(  z,w\right)  \in
U\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  =\left(  U\left[  \left[
w,w^{-1}\right]  \right]  \right)  \left[  \left[  z,z^{-1}\right]  \right]
$. Hence, we can write $b\left(  z,w\right)  $ in the form $b\left(
z,w\right)  =\sum_{j\in\mathbb{Z}}z^{j}b_{j}\left(  w\right)  $ for some
$b_{j}\left(  w\right)  \in U\left[  \left[  w,w^{-1}\right]  \right]  $.
Consider these $b_{j}\left(  w\right)  $. Every $n\in\mathbb{N}$ satisfies%
\[
\operatorname*{Res}\underbrace{z^{n}}_{\substack{=\left(  \left(  z-w\right)
+w\right)  ^{n}\\=\sum_{k=0}^{n}\dbinom{n}{k}\left(  z-w\right)  ^{k}w^{n-k}%
}}b\left(  z,w\right)  dz=\sum_{k=0}^{n}\dbinom{n}{k}%
\underbrace{\operatorname*{Res}\left(  z-w\right)  ^{k}b\left(  z,w\right)
dz}_{\substack{=0\\\text{(by (\ref{eq.lem.decomposition.cut.condi}), applied
to }k\text{ instead of }n\text{)}}}w^{n-k}=0.
\]
Compared with%
\[
\operatorname*{Res}z^{n}\underbrace{b\left(  z,w\right)  }_{=\sum
_{j\in\mathbb{Z}}z^{j}b_{j}\left(  w\right)  }dz=\operatorname*{Res}z^{n}%
\sum_{j\in\mathbb{Z}}z^{j}b_{j}\left(  w\right)  dz=\operatorname*{Res}%
\sum_{j\in\mathbb{Z}}z^{n+j}b_{j}\left(  w\right)  dz=b_{-n-1}\left(
w\right)  ,
\]
this yields that $b_{-n-1}\left(  w\right)  =0$ for every $n\in\mathbb{N}$. In
other words, $b_{j}\left(  w\right)  =0$ for every negative $j\in\mathbb{Z}$.
Hence,%
\begin{align*}
b\left(  z,w\right)   &  =\sum_{j\in\mathbb{Z}}z^{j}b_{j}\left(  w\right)
=\sum_{j\in\mathbb{N}}z^{j}b_{j}\left(  w\right)  +\sum_{\substack{j\in
\mathbb{Z};\\j\text{ is negative}}}z^{j}\underbrace{b_{j}\left(  w\right)
}_{=0}=\sum_{j\in\mathbb{N}}z^{j}b_{j}\left(  w\right) \\
&  \in\left(  U\left[  \left[  w,w^{-1}\right]  \right]  \right)  \left[
\left[  z\right]  \right]  .
\end{align*}
This proves Lemma \ref{lem.decomposition.cut} \textbf{(a)}.

\textbf{(b)} Lemma \ref{lem.decomposition.cut} \textbf{(a)} yields $b\left(
z,w\right)  \in\left(  U\left[  \left[  w,w^{-1}\right]  \right]  \right)
\left[  \left[  z\right]  \right]  $. But $\left(  U\left[  \left[
w,w^{-1}\right]  \right]  \right)  \left[  \left[  z\right]  \right]  $ is an
$\mathbb{F}\left[  z,w\right]  $-module, and the element $z-w$ of
$\mathbb{F}\left[  z,w\right]  $ acts as a non-zero-divisor on this module
(i.e., every element $\alpha\in\left(  U\left[  \left[  w,w^{-1}\right]
\right]  \right)  \left[  \left[  z\right]  \right]  $ satisfies $\left(
z-w\right)  \alpha=0$ must itself satisfy $\alpha=0$%
)\ \ \ \ \footnote{\textit{Proof.} Actually, $\left(  U\left[  \left[
w,w^{-1}\right]  \right]  \right)  \left[  \left[  z\right]  \right]  $ is not
just an $\mathbb{F}\left[  z,w\right]  $-module, but also an $\left(
\mathbb{F}\left[  w,w^{-1}\right]  \right)  \left[  \left[  z\right]  \right]
$-module (since $U\left[  \left[  w,w^{-1}\right]  \right]  $ is an
$\mathbb{F}\left[  w,w^{-1}\right]  $-module). The element $z-w$ of $\left(
\mathbb{F}\left[  w,w^{-1}\right]  \right)  \left[  \left[  z\right]  \right]
$ is invertible (since $z-w=w\left(  zw^{-1}-1\right)  $), and thus acts as a
non-zero-divisor on the $\left(  \mathbb{F}\left[  w,w^{-1}\right]  \right)
\left[  \left[  z\right]  \right]  $-module $\left(  U\left[  \left[
w,w^{-1}\right]  \right]  \right)  \left[  \left[  z\right]  \right]  $,
qed.}. But $b\left(  z,w\right)  $ is local. In other words, $\left(
z-w\right)  ^{N}b\left(  z,w\right)  =0$ for some $N\in\mathbb{N}$. Consider
this $N$. We can cancel the $\left(  z-w\right)  ^{N}$ from the equality
$\left(  z-w\right)  ^{N}b\left(  z,w\right)  =0$ (since $z-w$ acts as a
non-zero-divisor on the module $\left(  U\left[  \left[  w,w^{-1}\right]
\right]  \right)  \left[  \left[  z\right]  \right]  $, and $b\left(
z,w\right)  $ belongs to said module). As a result, we obtain $b\left(
z,w\right)  =0$. This proves Lemma \ref{lem.decomposition.cut} \textbf{(b)}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.decomposition}.]\textit{Uniqueness:} Let us first
show the uniqueness of the family $\left(  c^{j}\left(  w\right)  \right)
_{j\in\mathbb{N}}$ satisfying $a\left(  z,w\right)  =\sum_{j\in\mathbb{N}%
}c^{j}\left(  w\right)  \partial_{w}^{\left(  j\right)  }\delta\left(
z-w\right)  $. For this, we let $\left(  c^{j}\left(  w\right)  \right)
_{j\in\mathbb{N}}$ be any family of formal distributions $c^{j}\left(
w\right)  \in U\left[  \left[  w,w^{-1}\right]  \right]  $ having the property
that all but finitely many $j\in\mathbb{N}$ satisfy $c^{j}\left(  w\right)
=0$. Assume that $a\left(  z,w\right)  =\sum_{j\in\mathbb{N}}c^{j}\left(
w\right)  \partial_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  $. We
need to prove that (\ref{eq.8}) holds.

For every $n\in\mathbb{N}$, we have%
\begin{align}
&  \sum_{j\in\mathbb{N}}c^{j}\left(  w\right)  \operatorname*{Res}%
\underbrace{\left(  z-w\right)  ^{n}\partial_{w}^{\left(  j\right)  }%
\delta\left(  z-w\right)  }_{\substack{=\left\{
\begin{array}
[c]{c}%
\partial_{w}^{\left(  j-n\right)  }\delta\left(  z-w\right)  ,\text{ if }n\leq
j;\\
0,\text{ if }n>j
\end{array}
\right.  \\\text{(by Proposition \ref{prop.delta.res} \textbf{(b)}, applied to
}n\text{ and }j\\\text{instead of }m\text{ and }n\text{)}}}dz\nonumber\\
&  =\sum_{j\in\mathbb{N}}c^{j}\left(  w\right)  \operatorname*{Res}\left\{
\begin{array}
[c]{c}%
\partial_{w}^{\left(  j-n\right)  }\delta\left(  z-w\right)  ,\text{ if }n\leq
j;\\
0,\text{ if }n>j
\end{array}
\right.  dz\nonumber\\
&  =\sum_{\substack{j\in\mathbb{N};\\j\geq n}}c^{j}\left(  w\right)
\underbrace{\operatorname*{Res}\partial_{w}^{\left(  j-n\right)  }%
\delta\left(  z-w\right)  dz}_{\substack{=\delta_{j-n,0}\\\text{(by
Proposition \ref{prop.delta.res} \textbf{(a)}, applied to }j-n\text{ instead
of }n\text{)}}}\nonumber\\
&  =\sum_{\substack{j\in\mathbb{N};\\j\geq n}}c^{j}\left(  w\right)
\delta_{j-n,0}=c^{n}\left(  w\right)  . \label{pf.thm.decomposition.1}%
\end{align}
Hence, for every $j\in\mathbb{N}$, we have%
\begin{align*}
&  \operatorname*{Res}\left(  z-w\right)  ^{n}\underbrace{a\left(  z,w\right)
}_{=\sum_{j\in\mathbb{N}}c^{j}\left(  w\right)  \partial_{w}^{\left(
j\right)  }\delta\left(  z-w\right)  }dz=\sum_{j\in\mathbb{N}}c^{j}\left(
w\right)  \operatorname*{Res}\left(  z-w\right)  ^{n}\partial_{w}^{\left(
j\right)  }\delta\left(  z-w\right)  dz\\
&  =c^{n}\left(  w\right)  .
\end{align*}
Renaming $n$ as $j$ and interchanging the two sides of this equality, we
obtain precisely (\ref{eq.8}). Thus, (\ref{eq.8}) holds, and so the uniqueness
of the family $\left(  c^{j}\left(  w\right)  \right)  _{j\in\mathbb{N}}$ is proven.

\textit{Existence:} Now, we shall show that the family $\left(  c^{j}\left(
w\right)  \right)  _{j\in\mathbb{N}}$ defined by (\ref{eq.8}) actually
satisfies $a\left(  z,w\right)  =\sum_{j\in\mathbb{N}}c^{j}\left(  w\right)
\partial_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  $ and has the
property that all but finitely many $j\in\mathbb{N}$ satisfy $c^{j}\left(
w\right)  =0$.

Indeed, the latter property obviously holds (since $a\left(  z,w\right)  $ is
local). So it remains to prove that the family $\left(  c^{j}\left(  w\right)
\right)  _{j\in\mathbb{N}}$ defined by (\ref{eq.8}) satisfies
\begin{equation}
a\left(  z,w\right)  =\sum_{j\in\mathbb{N}}c^{j}\left(  w\right)  \partial
_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  .
\label{pf.thm.decomposition.6}%
\end{equation}
Consider this family $\left(  c^{j}\left(  w\right)  \right)  _{j\in
\mathbb{N}}$. Since $a\left(  z,w\right)  $ is local, we know that every
sufficiently high $j\in\mathbb{N}$ satisfies $\left(  z-w\right)  ^{j}a\left(
z,w\right)  =0$, and therefore%
\begin{equation}
\text{every sufficiently high }j\in\mathbb{N}\text{ satisfies }c^{j}\left(
w\right)  =0. \label{pf.thm.decomposition.7}%
\end{equation}
Thus, the sum $\sum_{j\in\mathbb{N}}c^{j}\left(  w\right)  \partial
_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  $ is well-defined. We set%
\[
b\left(  z,w\right)  =a\left(  z,w\right)  -\sum_{j\in\mathbb{N}}c^{j}\left(
w\right)  \partial_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  .
\]
Then, our goal is to show $b\left(  z,w\right)  =0$ (because this will
immediately yield (\ref{pf.thm.decomposition.6})).

We know that $a\left(  z,w\right)  $ is local. Also, for every $j\in
\mathbb{N}$, the formal distribution $\partial_{w}^{\left(  j\right)  }%
\delta\left(  z-w\right)  $ is local (since Proposition \ref{prop.delta.res}
\textbf{(b)} yields $\left(  z-w\right)  ^{j+1}\partial_{w}^{\left(  j\right)
}\delta\left(  z-w\right)  =0$), and therefore the formal distribution
$c^{j}\left(  w\right)  \partial_{w}^{\left(  j\right)  }\delta\left(
z-w\right)  $ is local as well. Hence, the sum $\sum_{j\in\mathbb{N}}%
c^{j}\left(  w\right)  \partial_{w}^{\left(  j\right)  }\delta\left(
z-w\right)  $ is local as well (since it has only finitely many nonzero
addends -- due to (\ref{pf.thm.decomposition.7}) -- and these addends are
local). Thus, $b\left(  z,w\right)  $ is local (since we have defined
$b\left(  z,w\right)  $ as a difference between the local formal distributions
$a\left(  z,w\right)  $ and $\sum_{j\in\mathbb{N}}c^{j}\left(  w\right)
\partial_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  $).

For every $n\in\mathbb{N}$, we have%
\begin{align*}
&  \operatorname*{Res}\left(  z-w\right)  ^{n}\underbrace{b\left(  z,w\right)
}_{=a\left(  z,w\right)  -\sum_{j\in\mathbb{N}}c^{j}\left(  w\right)
\partial_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  }dz\\
&  =\underbrace{\operatorname*{Res}\left(  z-w\right)  ^{n}a\left(
z,w\right)  dz}_{\substack{=c^{n}\left(  w\right)  \\\text{(by (\ref{eq.8}),
applied to }j=n\text{)}}}-\underbrace{\sum_{j\in\mathbb{N}}c^{j}\left(
w\right)  \operatorname*{Res}\left(  z-w\right)  ^{n}\partial_{w}^{\left(
j\right)  }\delta\left(  z-w\right)  dz}_{\substack{=c^{n}\left(  w\right)
\\\text{(by (\ref{pf.thm.decomposition.1}))}}}\\
&  =c^{n}\left(  w\right)  -c^{n}\left(  w\right)  =0.
\end{align*}
Hence, Lemma \ref{lem.decomposition.cut} \textbf{(b)} yields $b\left(
z,w\right)  =0$ (since $b\left(  z,w\right)  $ is local). This completes the
proof of (\ref{pf.thm.decomposition.6}). Thus, the existence part of Theorem
\ref{thm.decomposition} is proven. The proof of Theorem
\ref{thm.decomposition} is thus complete.
\end{proof}

For the next corollary, we shall use the notations of Section
\ref{sect.diffop}.

\begin{corollary}
\label{exe.3.4}Let $U$ be an $\mathbb{F}$-module. To every $a=a\left(
z,w\right)  \in U\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $, let us
associate the $\mathbb{F}$-linear operator%
\begin{align*}
D_{a}:\mathbb{F}\left[  w,w^{-1}\right]   &  \rightarrow U\left[  \left[
w,w^{-1}\right]  \right]  ,\\
\varphi\left(  w\right)   &  \mapsto\operatorname*{Res}\varphi\left(
z\right)  a\left(  z,w\right)  dz.
\end{align*}


\textbf{(a)} For every $c\left(  w\right)  \in U\left[  \left[  w,w^{-1}%
\right]  \right]  $ and $j\in\mathbb{N}$, we have%
\begin{equation}
D_{c\left(  w\right)  \partial_{w}^{\left(  j\right)  }\delta\left(
z-w\right)  }=c\left(  w\right)  \partial_{w}^{\left(  j\right)  }
\label{eq.exe.3.4.a.1}%
\end{equation}
as maps $\mathbb{F}\left[  w,w^{-1}\right]  \rightarrow U\left[  \left[
w,w^{-1}\right]  \right]  $. (Here, $c\left(  w\right)  \partial_{w}^{\left(
j\right)  }$ means the operator $\partial_{w}^{\left(  j\right)  }$, followed
by multiplication with $c\left(  w\right)  $.) In particular, if
$U=\mathbb{F}$, then $D_{\delta\left(  z-w\right)  }=1$ (the identity map, or,
rather, the canonical inclusion $\mathbb{F}\left[  w,w^{-1}\right]
\rightarrow\mathbb{F}\left[  \left[  w,w^{-1}\right]  \right]  $).

\textbf{(b)} Let $a=a\left(  z,w\right)  \in U\left[  \left[  z,z^{-1}%
,w,w^{-1}\right]  \right]  $. Prove that $a\left(  z,w\right)  $ is local if
and only if $D_{a}$ is a differential operator of finite order. (See
Definition \ref{def.diffop} for the definition of a differential operator of
finite order.)

\textbf{(c)} Let $a=a\left(  z,w\right)  \in U\left[  \left[  z,z^{-1}%
,w,w^{-1}\right]  \right]  $ be local. Then, show that $a\left(  w,z\right)  $
is local and satisfies%
\begin{equation}
D_{a\left(  w,z\right)  }=\left(  D_{a\left(  z,w\right)  }\right)  ^{\ast}.
\label{eq.exe.3.4.c.*}%
\end{equation}
(See Definition \ref{def.diffop.adj} for the meaning of $\left(  D_{a\left(
z,w\right)  }\right)  ^{\ast}$.)

\textbf{(d)} Let $a=a\left(  z,w\right)  \in U\left[  \left[  z,z^{-1}%
,w,w^{-1}\right]  \right]  $. Show that%
\begin{align}
D_{\psi\left(  w\right)  a\left(  z,w\right)  }  &  =\psi\left(  w\right)
\circ D_{a\left(  z,w\right)  }\ \ \ \ \ \ \ \ \ \ \text{for all }\psi\left(
w\right)  \in\mathbb{F}\left[  w,w^{-1}\right]  ;\label{eq.exe.3.4.d.1}\\
D_{\partial_{w}a\left(  z,w\right)  }  &  =\partial_{w}\circ D_{a\left(
z,w\right)  };\label{eq.exe.3.4.d.2}\\
D_{\psi\left(  z\right)  a\left(  z,w\right)  }  &  =D_{a\left(  z,w\right)
}\circ\psi\left(  w\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }\psi\left(
z\right)  \in\mathbb{F}\left[  z,z^{-1}\right]  ;\label{eq.exe.3.4.d.3}\\
D_{\partial_{z}a\left(  z,w\right)  }  &  =-D_{a\left(  z,w\right)  }%
\circ\partial_{w}. \label{eq.exe.3.4.d.4}%
\end{align}

\end{corollary}

\begin{proof}
[Proof of Corollary \ref{exe.3.4}.]\textbf{(a)} Let $c\left(  w\right)  \in
U\left[  \left[  w,w^{-1}\right]  \right]  $ and $j\in\mathbb{N}$. Let
$\varphi\left(  w\right)  \in\mathbb{F}\left[  w,w^{-1}\right]  $. We write
$\varphi\left(  w\right)  $ in the form $\varphi\left(  w\right)  =\sum
_{m\in\mathbb{Z}}\varphi_{m}w^{m}$, where all $\varphi_{m}$ belong to
$\mathbb{F}$ and all but finitely many of these $\varphi_{m}$ are zero. Then,
$\partial_{w}^{\left(  j\right)  }\varphi\left(  w\right)  =\sum
_{m\in\mathbb{Z}}\dbinom{m}{j}\varphi_{m}w^{m-j}$ (by the definition of
$\partial_{w}^{\left(  j\right)  }$).

Now,%
\begin{align*}
D_{c\left(  w\right)  \partial_{w}^{\left(  j\right)  }\delta\left(
z-w\right)  }\left(  \varphi\left(  w\right)  \right)   &
=\operatorname*{Res}\varphi\left(  z\right)  c\left(  w\right)  \partial
_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  dz\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }D_{c\left(
w\right)  \partial_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  }\right)
\\
&  =c\left(  w\right)  \operatorname*{Res}\varphi\left(  z\right)
\partial_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  dz.
\end{align*}
Since%
\begin{align*}
&  \operatorname*{Res}\underbrace{\varphi\left(  z\right)  }_{\substack{=\sum
_{m\in\mathbb{Z}}\varphi_{m}z^{m}\\\text{(since }\varphi\left(  w\right)
=\sum_{m\in\mathbb{Z}}\varphi_{m}w^{m}\text{)}}}\partial_{w}^{\left(
j\right)  }\delta\left(  z-w\right)  dz\\
&  =\operatorname*{Res}\left(  \sum_{m\in\mathbb{Z}}\varphi_{m}z^{m}\right)
\partial_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  dz=\sum
_{m\in\mathbb{Z}}\varphi_{m}\operatorname*{Res}z^{m}\underbrace{\partial
_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  }_{\substack{=\sum
_{n\in\mathbb{Z}}\dbinom{n}{j}w^{n-j}z^{-n-1}\\\text{(by (\ref{eq.6b}),
applied to }k=j\text{)}}}dz\\
&  =\sum_{m\in\mathbb{Z}}\varphi_{m}\operatorname*{Res}\underbrace{z^{m}%
\left(  \sum_{n\in\mathbb{Z}}\dbinom{n}{j}w^{n-j}z^{-n-1}\right)  }%
_{=\sum_{n\in\mathbb{Z}}\dbinom{n}{j}w^{n-j}z^{m-n-1}}dz=\sum_{m\in\mathbb{Z}%
}\varphi_{m}\underbrace{\operatorname*{Res}\sum_{n\in\mathbb{Z}}\dbinom{n}%
{j}w^{n-j}z^{m-n-1}dz}_{=\dbinom{m}{j}w^{m-j}}\\
&  =\sum_{m\in\mathbb{Z}}\varphi_{m}\dbinom{m}{j}w^{m-j}=\sum_{m\in\mathbb{Z}%
}\dbinom{m}{j}\varphi_{m}w^{m-j}=\partial_{w}^{\left(  j\right)  }%
\varphi\left(  w\right)  ,
\end{align*}
this rewrites as%
\[
D_{c\left(  w\right)  \partial_{w}^{\left(  j\right)  }\delta\left(
z-w\right)  }\left(  \varphi\left(  w\right)  \right)  =c\left(  w\right)
\underbrace{\operatorname*{Res}\varphi\left(  z\right)  \partial_{w}^{\left(
j\right)  }\delta\left(  z-w\right)  dz}_{=\partial_{w}^{\left(  j\right)
}\varphi\left(  w\right)  }=c\left(  w\right)  \partial_{w}^{\left(  j\right)
}\varphi\left(  w\right)  .
\]
Now, we forget that we fixed $\varphi\left(  w\right)  $. We thus have shown
that $D_{c\left(  w\right)  \partial_{w}^{\left(  j\right)  }\delta\left(
z-w\right)  }\left(  \varphi\left(  w\right)  \right)  =c\left(  w\right)
\partial_{w}^{\left(  j\right)  }\varphi\left(  w\right)  $ for every
$\varphi\left(  w\right)  \in\mathbb{F}\left[  w,w^{-1}\right]  $. Hence,
$D_{c\left(  w\right)  \partial_{w}^{\left(  j\right)  }\delta\left(
z-w\right)  }=c\left(  w\right)  \partial_{w}^{\left(  j\right)  }$ is proven.
It remains to show that $D_{\delta\left(  z-w\right)  }=1$ if $U=\mathbb{F}$.
But this follows from (\ref{eq.exe.3.4.a.1}), applied to $c\left(  w\right)
=1$ and $j=0$. Proposition \ref{exe.3.4} \textbf{(a)} is proven.

Before we start proving Proposition \ref{exe.3.4} \textbf{(b)}, let us show
two auxiliary results:

\begin{itemize}
\item Every $a=a\left(  z,w\right)  \in U\left[  \left[  z,z^{-1}%
,w,w^{-1}\right]  \right]  $ satisfies%
\begin{equation}
D_{\left(  z-w\right)  a\left(  z,w\right)  }=\left[  D_{a},w\right]  .
\label{pf.exe.3.4.b.lem}%
\end{equation}
Here, $w$ denotes the continuous $\mathbb{F}$-linear map \newline$U\left[
\left[  z,z^{-1},w,w^{-1}\right]  \right]  \rightarrow U\left[  \left[
z,z^{-1},w,w^{-1}\right]  \right]  $ which sends every \newline$b\in U\left[
\left[  z,z^{-1},w,w^{-1}\right]  \right]  $ to $w\cdot b$. (In analogy to
Proposition \ref{prop.delta.commutator}, we should have called it $L_{w}$
instead of $w$, but the notation $w$ is shorter.)]

\textit{Proof of (\ref{pf.exe.3.4.b.lem}):} Let $a=a\left(  z,w\right)  \in
U\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $. Then, every
$\varphi\left(  w\right)  \in\mathbb{F}\left[  w,w^{-1}\right]  $ satisfies%
\begin{align}
&  D_{\left(  z-w\right)  a\left(  z,w\right)  }\left(  \varphi\left(
w\right)  \right) \nonumber\\
&  =\operatorname*{Res}\varphi\left(  z\right)  \left(  z-w\right)  a\left(
z,w\right)  dz\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}D_{\left(  z-w\right)  a\left(  z,w\right)  }\right) \nonumber\\
&  =\operatorname*{Res}\varphi\left(  z\right)  za\left(  z,w\right)
dz-w\operatorname*{Res}\varphi\left(  z\right)  a\left(  z,w\right)
dz\nonumber\\
&  =\operatorname*{Res}z\varphi\left(  z\right)  a\left(  z,w\right)
dz-w\operatorname*{Res}\varphi\left(  z\right)  a\left(  z,w\right)  dz.
\label{pf.exe.3.4.b.lem.pf.1}%
\end{align}
Hence, every $\varphi\left(  w\right)  \in\mathbb{F}\left[  w,w^{-1}\right]  $
satisfies%
\begin{align*}
&  \underbrace{\left[  D_{a},w\right]  }_{=D_{a}\circ w-w\circ D_{a}}\left(
\varphi\left(  w\right)  \right) \\
&  =\left(  D_{a}\circ w-w\circ D_{a}\right)  \left(  \varphi\left(  w\right)
\right)  =\underbrace{D_{a}\left(  w\varphi\left(  w\right)  \right)
}_{\substack{=\operatorname*{Res}z\varphi\left(  z\right)  a\left(
z,w\right)  dz\\\text{(by the definition of }D_{a}\text{)}}%
}-w\underbrace{D_{a}\left(  \varphi\left(  w\right)  \right)  }%
_{\substack{=\operatorname*{Res}\varphi\left(  z\right)  a\left(  z,w\right)
dz\\\text{(by the definition of }D_{a}\text{)}}}\\
&  =\operatorname*{Res}z\varphi\left(  z\right)  a\left(  z,w\right)
dz-w\operatorname*{Res}\varphi\left(  z\right)  a\left(  z,w\right)  dz\\
&  =D_{\left(  z-w\right)  a\left(  z,w\right)  }\left(  \varphi\left(
w\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.exe.3.4.b.lem.pf.1})}\right)  .
\end{align*}
In other words, $\left[  D_{a},w\right]  =D_{\left(  z-w\right)  a\left(
z,w\right)  }$. This proves (\ref{pf.exe.3.4.b.lem}).

\item If an $a\in U\left[  \left[  z,z^{-1},w,w^{-1}\right]  \right]  $
satisfies $D_{a}=0$, then%
\begin{equation}
a=0. \label{pf.exe.3.4.b.lem0}%
\end{equation}


\textit{Proof of (\ref{pf.exe.3.4.b.lem0}):} Let $a\in U\left[  \left[
z,z^{-1},w,w^{-1}\right]  \right]  $ be such that $D_{a}=0$. Write $a$ in the
form $a=\sum_{\left(  n,m\right)  \in\mathbb{Z}^{2}}a_{n,m}z^{n}w^{m}$ with
$a_{n,m}\in U$. Then, every $i\in\mathbb{Z}$ satisfies
\begin{align*}
D_{a}\left(  w^{i}\right)   &  =\operatorname*{Res}z^{i}\underbrace{a\left(
z,w\right)  }_{=a=\sum_{\left(  n,m\right)  \in\mathbb{Z}^{2}}a_{n,m}%
z^{n}w^{m}}dz\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
D_{a}\right) \\
&  =\operatorname*{Res}z^{i}\sum_{\left(  n,m\right)  \in\mathbb{Z}^{2}%
}a_{n,m}z^{n}w^{m}dz=\sum_{m\in\mathbb{Z}}a_{-i-1,m}w^{m}.
\end{align*}
Hence, every $i\in\mathbb{Z}$ satisfies $\sum_{m\in\mathbb{Z}}a_{-i-1,m}%
w^{m}=\underbrace{D_{a}}_{=0}\left(  w^{i}\right)  =0$. Comparing coefficients
in this equality, we conclude that every $i\in\mathbb{Z}$ and $m\in\mathbb{Z}$
satisfy $a_{-i-1,m}=0$. In other words, every $\left(  n,m\right)
\in\mathbb{Z}^{2}$ satisfy $a_{n,m}=0$. Hence, $a=0$. This completes our proof
of (\ref{pf.exe.3.4.b.lem0}).
\end{itemize}

\textbf{(b)} $\Longrightarrow:$ Assume that $a\left(  z,w\right)  $ is local.
Theorem \ref{thm.decomposition} thus yields that $a\left(  z,w\right)  $ can
be written in the form%
\[
\sum_{j\in\mathbb{N}}c^{j}\left(  w\right)  \partial_{w}^{\left(  j\right)
}\delta\left(  z-w\right)
\]
with $\left(  c^{j}\left(  w\right)  \right)  _{j\in\mathbb{N}}$ being a
family of formal distributions $c^{j}\left(  w\right)  \in U\left[  \left[
w,w^{-1}\right]  \right]  $ having the property that all but finitely many
$j\in\mathbb{N}$ satisfy $c^{j}\left(  w\right)  =0$. Consider this $\left(
c^{j}\left(  w\right)  \right)  _{j\in\mathbb{N}}$. Since $a\left(
z,w\right)  =\sum_{j\in\mathbb{N}}c^{j}\left(  w\right)  \partial_{w}^{\left(
j\right)  }\delta\left(  z-w\right)  $, we have%
\begin{align}
D_{a}  &  =\sum_{j\in\mathbb{N}}\underbrace{D_{c^{j}\left(  w\right)
\partial_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  }}%
_{\substack{=c^{j}\left(  w\right)  \partial_{w}^{\left(  j\right)
}\\\text{(by (\ref{eq.exe.3.4.a.1}), applied to }c\left(  w\right)
=c^{j}\left(  w\right)  \text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}D_{a}\text{ depends }\mathbb{F}\text{-linearly on }a\right) \nonumber\\
&  =\sum_{j\in\mathbb{N}}c^{j}\left(  w\right)  \partial_{w}^{\left(
j\right)  }. \label{pf.exe.3.4.b.=>.Da}%
\end{align}
Thus, $D_{a}$ is a differential operator of finite order (since all but
finitely many $j\in\mathbb{N}$ satisfy $c^{j}\left(  w\right)  =0$). This
proves the $\Longrightarrow$ direction of Proposition \ref{exe.3.4}
\textbf{(b)}.

$\Longleftarrow:$ Assume that $D_{a}$ is a differential operator of finite
order. In other words, $D_{a}=\sum_{j\in\mathbb{N}}c_{j}\left(  w\right)
\partial_{w}^{\left(  j\right)  }$ for a family $\left(  c_{j}\left(
w\right)  \right)  _{j\in\mathbb{N}}$ of formal distributions in $U\left[
\left[  w,w^{-1}\right]  \right]  $ with the property that all but finitely
many $j\in\mathbb{N}$ satisfy $c_{j}\left(  w\right)  =0$. Consider this
family $\left(  c_{j}\left(  w\right)  \right)  _{j\in\mathbb{N}}$.

Let us define the $\mathbb{F}$-linear map $w:U\left[  \left[  z,z^{-1}%
,w,w^{-1}\right]  \right]  \rightarrow U\left[  \left[  z,z^{-1}%
,w,w^{-1}\right]  \right]  $ as in (\ref{pf.exe.3.4.b.lem}). We have%
\begin{equation}
\left[  w,\partial_{w}^{\left(  n\right)  }\right]  =-\partial_{w}^{\left(
n-1\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}.
\label{pf.exe.3.4.b.<=.commut}%
\end{equation}
\footnote{Indeed, this is similar to Proposition \ref{prop.delta.commutator};
there are only two differences:
\par
\begin{itemize}
\item We now have two variables $z$ and $w$ instead of a single variable $z$.
But this does not change much because only the variable $w$ is
\textquotedblleft active\textquotedblright. (If we identify $U\left[  \left[
z,z^{-1},w,w^{-1}\right]  \right]  $ with $\left(  U\left[  \left[
z,z^{-1}\right]  \right]  \right)  \left[  \left[  w,w^{-1}\right]  \right]
$, then we are back in the single-variable case.)
\par
\item We are now denoting by $w$ the map that would be denoted by $L_{w}$ in
the terminology of Proposition \ref{prop.delta.commutator}.
\end{itemize}
\par
These two differences are insubstantial, and the proof of Proposition
\ref{prop.delta.commutator} can be easily adapted to prove
(\ref{pf.exe.3.4.b.<=.commut}).}

All but finitely many $j\in\mathbb{N}$ satisfy $c_{j}\left(  w\right)  =0$.
Thus, there exists a $J\in\mathbb{N}$ such that every $j\in\mathbb{N}$
satisfying $j\geq J$ satisfies $c_{j}\left(  w\right)  =0$. Consider this $J$.
Then,%
\begin{equation}
D_{a}=\sum_{j\in\mathbb{N}}c_{j}\left(  w\right)  \partial_{w}^{\left(
j\right)  }=\sum_{j=0}^{J-1}c_{j}\left(  w\right)  \partial_{w}^{\left(
j\right)  } \label{pf.exe.3.4.b.<=.D0}%
\end{equation}
(by the definition of $J$). We are going to prove that $\left(  z-w\right)
^{J}a\left(  z,w\right)  =0$.

Indeed, let us show that%
\begin{equation}
D_{\left(  z-w\right)  ^{k}a\left(  z,w\right)  }=\sum_{j=k}^{J-1}c_{j}\left(
w\right)  \partial_{w}^{\left(  j-k\right)  }\ \ \ \ \ \ \ \ \ \ \text{for
every }k\in\mathbb{N} \label{pf.exe.3.4.b.<=.D1}%
\end{equation}
(where any sum whose lower limit is larger than its upper limit is understood
to be empty).

\textit{Proof of (\ref{pf.exe.3.4.b.<=.D1}):} We shall prove
(\ref{pf.exe.3.4.b.<=.D1}) by induction over $k$.

\textit{Induction base:} For $k=0$, the equality (\ref{pf.exe.3.4.b.<=.D1})
follows immediately from (\ref{pf.exe.3.4.b.<=.D0}). Thus, the induction base
is complete.

\textit{Induction step:} Let $K\in\mathbb{N}$. Assume that
(\ref{pf.exe.3.4.b.<=.D1}) holds for $k=K$. We need to show that
(\ref{pf.exe.3.4.b.<=.D1}) holds for $k=K+1$.

We have%
\begin{align*}
&  D_{\left(  z-w\right)  ^{K+1}a\left(  z,w\right)  }\\
&  =D_{\left(  z-w\right)  \left(  z-w\right)  ^{K}a\left(  z,w\right)
}=\left[  \underbrace{D_{\left(  z-w\right)  ^{K}a\left(  z,w\right)  }%
}_{\substack{=\sum_{j=K}^{J-1}c_{j}\left(  w\right)  \partial_{w}^{\left(
j-K\right)  }\\\text{(since (\ref{pf.exe.3.4.b.<=.D1}) holds for }k=K\text{)}%
}},w\right] \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.exe.3.4.b.lem}), applied to
}\left(  z-w\right)  ^{K}a\left(  z,w\right)  \text{ instead of }a\right) \\
&  =\left[  \sum_{j=K}^{J-1}c_{j}\left(  w\right)  \partial_{w}^{\left(
j-K\right)  },w\right]  =\sum_{j=K}^{J-1}\underbrace{\left[  c_{j}\left(
w\right)  \partial_{w}^{\left(  j-K\right)  },w\right]  }_{\substack{=c_{j}%
\left(  w\right)  \left[  \partial_{w}^{\left(  j-K\right)  },w\right]
\\\text{(since }c_{j}\left(  w\right)  \text{ commutes with }w\text{)}}%
}=\sum_{j=K}^{J-1}c_{j}\left(  w\right)  \underbrace{\left[  \partial
_{w}^{\left(  j-K\right)  },w\right]  }_{=-\left[  w,\partial_{w}^{\left(
j-K\right)  }\right]  }\\
&  =-\sum_{j=K}^{J-1}c_{j}\left(  w\right)  \underbrace{\left[  w,\partial
_{w}^{\left(  j-K\right)  }\right]  }_{\substack{=-\partial_{w}^{\left(
j-K-1\right)  }\\\text{(by (\ref{pf.exe.3.4.b.<=.commut}), applied to
}n=j-K\text{)}}}=\sum_{j=K}^{J-1}c_{j}\left(  w\right)  \partial_{w}^{\left(
j-K-1\right)  }\\
&  =c_{K}\left(  w\right)  \underbrace{\partial_{w}^{\left(  K-K-1\right)  }%
}_{=\partial_{w}^{\left(  -1\right)  }=0}+\sum_{j=K+1}^{J-1}c_{j}\left(
w\right)  \partial_{w}^{\left(  j-K-1\right)  }=\sum_{j=K+1}^{J-1}c_{j}\left(
w\right)  \partial_{w}^{\left(  j-K-1\right)  }.
\end{align*}
In other words, (\ref{pf.exe.3.4.b.<=.D1}) holds for $k=K+1$. This completes
the induction step. Thus, (\ref{pf.exe.3.4.b.<=.D1}) is proven.

Now, applying (\ref{pf.exe.3.4.b.<=.D1}) to $k=J$, we obtain%
\[
D_{\left(  z-w\right)  ^{J}a\left(  z,w\right)  }=\sum_{j=J}^{J-1}c_{j}\left(
w\right)  \partial_{w}^{\left(  j-J\right)  }=\left(  \text{empty sum}\right)
=0.
\]
Thus, (\ref{pf.exe.3.4.b.lem0}) (applied to $\left(  z-w\right)  ^{J}a\left(
z,w\right)  $ instead of $a$) yields that $\left(  z-w\right)  ^{J}a\left(
z,w\right)  =0$. Hence, $a\left(  z,w\right)  $ is local. This proves the
$\Longleftarrow$ direction of Proposition \ref{exe.3.4} \textbf{(b)}.

\textbf{(c)} Clearly, $a\left(  w,z\right)  $ is
local.\footnote{\textit{Proof.} We know that $a\left(  z,w\right)  $ is local.
Thus, there exists an $N\in\mathbb{N}$ such that $\left(  z-w\right)
^{N}a\left(  z,w\right)  =0$. Consider this $N$. Switching the indeterminates
$z$ and $w$ in $\left(  z-w\right)  ^{N}a\left(  z,w\right)  =0$, we obtain
$\left(  w-z\right)  ^{N}a\left(  w,z\right)  =0$. Hence, $\underbrace{\left(
z-w\right)  ^{N}}_{=\left(  -1\right)  ^{N}\left(  w-z\right)  ^{N}}a\left(
w,z\right)  =\left(  -1\right)  ^{N}\underbrace{\left(  w-z\right)
^{N}a\left(  w,z\right)  }_{=0}=0$. Thus, $a\left(  w,z\right)  $ is local,
qed.} It remains to prove that $D_{a\left(  w,z\right)  }=\left(  D_{a\left(
z,w\right)  }\right)  ^{\ast}$. In other words, it remains to prove that
$D_{a\left(  w,z\right)  }=\left(  D_{a}\right)  ^{\ast}$ (since $a\left(
z,w\right)  =a$).

Theorem \ref{thm.decomposition} yields that $a\left(  z,w\right)  $ can be
written in the form%
\[
\sum_{j\in\mathbb{N}}c^{j}\left(  w\right)  \partial_{w}^{\left(  j\right)
}\delta\left(  z-w\right)
\]
with $\left(  c^{j}\left(  w\right)  \right)  _{j\in\mathbb{N}}$ being a
family of formal distributions $c^{j}\left(  w\right)  \in U\left[  \left[
w,w^{-1}\right]  \right]  $ having the property that all but finitely many
$j\in\mathbb{N}$ satisfy $c^{j}\left(  w\right)  =0$. Consider this $\left(
c^{j}\left(  w\right)  \right)  _{j\in\mathbb{N}}$. Theorem
\ref{thm.decomposition} furthermore shows that these $c^{j}\left(  w\right)  $
are given by (\ref{eq.8}).

We have $a\left(  z,w\right)  =\sum_{j\in\mathbb{N}}c^{j}\left(  w\right)
\partial_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  $. Renaming the
indeterminates $z$ and $w$ as $w$ and $z$ in this equality, we obtain%
\begin{align*}
a\left(  w,z\right)   &  =\sum_{j\in\mathbb{N}}c^{j}\left(  z\right)
\partial_{z}^{\left(  j\right)  }\underbrace{\delta\left(  w-z\right)
}_{\substack{=\delta\left(  z-w\right)  \\\text{(by Proposition
\ref{prop.delta.res} \textbf{(c)})}}}=\sum_{j\in\mathbb{N}}c^{j}\left(
z\right)  \underbrace{\partial_{z}^{\left(  j\right)  }\delta\left(
z-w\right)  }_{\substack{=\left(  -1\right)  ^{j}\partial_{w}^{\left(
j\right)  }\delta\left(  z-w\right)  \\\text{(by Proposition
\ref{prop.delta.res} \textbf{(e)})}}}\\
&  =\sum_{j\in\mathbb{N}}c^{j}\left(  z\right)  \left(  -1\right)
^{j}\partial_{w}^{\left(  j\right)  }\delta\left(  z-w\right)  =\sum
_{j\in\mathbb{N}}\left(  -1\right)  ^{j}\underbrace{c^{j}\left(  z\right)
\partial_{w}^{\left(  j\right)  }}_{\substack{=\partial_{w}^{\left(  j\right)
}c^{j}\left(  z\right)  \\\text{(since }c^{j}\left(  z\right)  \text{ is a
polynomial in }z\\\text{and thus commutes with }\partial_{w}^{\left(
j\right)  }\text{)}}}\delta\left(  z-w\right) \\
&  =\sum_{j\in\mathbb{N}}\left(  -1\right)  ^{j}\partial_{w}^{\left(
j\right)  }\left(  c^{j}\left(  z\right)  \delta\left(  z-w\right)  \right)  .
\end{align*}


From (\ref{pf.exe.3.4.b.=>.Da}), we have $D_{a}=\sum_{j\in\mathbb{N}}%
c^{j}\left(  w\right)  \partial_{w}^{\left(  j\right)  }$, so that $\left(
D_{a}\right)  ^{\ast}=\sum_{j\in\mathbb{N}}\left(  -1\right)  ^{j}\partial
_{w}^{\left(  j\right)  }c^{j}\left(  w\right)  $. Now, for every
$\varphi\left(  w\right)  \in\mathbb{F}\left[  w,w^{-1}\right]  $, we have%
\begin{align*}
&  D_{a\left(  w,z\right)  }\left(  \varphi\left(  w\right)  \right) \\
&  =\operatorname*{Res}\varphi\left(  z\right)  \underbrace{a\left(
w,z\right)  }_{=\sum_{j\in\mathbb{N}}\left(  -1\right)  ^{j}\partial
_{w}^{\left(  j\right)  }\left(  c^{j}\left(  z\right)  \delta\left(
z-w\right)  \right)  }dz\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}D_{a\left(  w,z\right)  }\right) \\
&  =\operatorname*{Res}\varphi\left(  z\right)  \left(  \sum_{j\in\mathbb{N}%
}\left(  -1\right)  ^{j}\partial_{w}^{\left(  j\right)  }\left(  c^{j}\left(
z\right)  \delta\left(  z-w\right)  \right)  \right)  dz\\
&  =\sum_{j\in\mathbb{N}}\left(  -1\right)  ^{j}%
\underbrace{\operatorname*{Res}\varphi\left(  z\right)  \partial_{w}^{\left(
j\right)  }\left(  c^{j}\left(  z\right)  \delta\left(  z-w\right)  \right)
dz}_{\substack{=\partial_{w}^{\left(  j\right)  }\operatorname*{Res}%
\varphi\left(  z\right)  c^{j}\left(  z\right)  \delta\left(  z-w\right)
dz\\\text{(since both multiplication with }\varphi\left(  z\right)  \text{
and}\\\text{the map }q\mapsto\operatorname*{Res}qdz\text{ commute with
}\partial_{w}^{\left(  j\right)  }\text{)}}}\\
&  =\sum_{j\in\mathbb{N}}\left(  -1\right)  ^{j}\partial_{w}^{\left(
j\right)  }\underbrace{\operatorname*{Res}\varphi\left(  z\right)
c^{j}\left(  z\right)  \delta\left(  z-w\right)  dz}%
_{\substack{=\operatorname*{Res}\delta\left(  z-w\right)  \varphi\left(
z\right)  c^{j}\left(  z\right)  dz\\=\varphi\left(  w\right)  c^{j}\left(
w\right)  \\\text{(by Proposition \ref{exe.3.2} \textbf{(c)}, applied}%
\\\text{to }\varphi\left(  z\right)  c^{j}\left(  z\right)  \text{ instead of
}a\left(  z\right)  \text{)}}}\\
&  =\sum_{j\in\mathbb{N}}\left(  -1\right)  ^{j}\partial_{w}^{\left(
j\right)  }\left(  \varphi\left(  w\right)  c^{j}\left(  w\right)  \right)
=\underbrace{\left(  \sum_{j\in\mathbb{N}}\left(  -1\right)  ^{j}\partial
_{w}^{\left(  j\right)  }c^{j}\left(  w\right)  \right)  }_{=\left(
D_{a}\right)  ^{\ast}}\left(  \varphi\left(  w\right)  \right)  =\left(
D_{a}\right)  ^{\ast}\left(  \varphi\left(  w\right)  \right)  .
\end{align*}
Hence, $D_{a\left(  w,z\right)  }=\left(  D_{a}\right)  ^{\ast}$. This
completes the proof of Proposition \ref{exe.3.4} \textbf{(c)}.

\textbf{(d)} \textit{Proof of (\ref{eq.exe.3.4.d.1}):} For every $\psi\left(
w\right)  \in\mathbb{F}\left[  w,w^{-1}\right]  $ and $\varphi\left(
w\right)  \in\mathbb{F}\left[  w,w^{-1}\right]  $, we have%
\begin{align*}
D_{\psi\left(  w\right)  a\left(  z,w\right)  }\left(  \varphi\left(
w\right)  \right)   &  =\operatorname*{Res}\varphi\left(  z\right)
\psi\left(  w\right)  a\left(  z,w\right)  dz\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }D_{\psi\left(  w\right)  a\left(  z,w\right)
}\right) \\
&  =\psi\left(  w\right)  \underbrace{\operatorname*{Res}\varphi\left(
z\right)  a\left(  z,w\right)  dz}_{\substack{=D_{a\left(  z,w\right)
}\left(  \varphi\left(  w\right)  \right)  \\\text{(by the definition of
}D_{a\left(  z,w\right)  }\text{)}}}=\psi\left(  w\right)  D_{a\left(
z,w\right)  }\left(  \varphi\left(  w\right)  \right) \\
&  =\left(  \psi\left(  w\right)  \circ D_{a\left(  z,w\right)  }\right)
\left(  \varphi\left(  w\right)  \right)  .
\end{align*}
This yields (\ref{eq.exe.3.4.d.1}).

\textit{Proof of (\ref{eq.exe.3.4.d.2}):} For every $\varphi\left(  w\right)
\in\mathbb{F}\left[  w,w^{-1}\right]  $, we have%
\begin{align*}
D_{\partial_{w}a\left(  z,w\right)  }\left(  \varphi\left(  w\right)  \right)
&  =\operatorname*{Res}\varphi\left(  z\right)  \partial_{w}a\left(
z,w\right)  dz\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}D_{\partial_{w}a\left(  z,w\right)  }\right) \\
&  =\partial_{w}\underbrace{\operatorname*{Res}\varphi\left(  z\right)
a\left(  z,w\right)  dz}_{\substack{=D_{a\left(  z,w\right)  }\left(
\varphi\left(  w\right)  \right)  \\\text{(by the definition of }D_{a\left(
z,w\right)  }\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }q\mapsto
\operatorname*{Res}qdz\text{ commutes with }\partial_{w}\right) \\
&  =\partial_{w}D_{a\left(  z,w\right)  }\left(  \varphi\left(  w\right)
\right)  =\left(  \partial_{w}\circ D_{a\left(  z,w\right)  }\right)  \left(
\varphi\left(  w\right)  \right)  .
\end{align*}
This yields (\ref{eq.exe.3.4.d.2}).

\textit{Proof of (\ref{eq.exe.3.4.d.3}):} For every $\varphi\left(  w\right)
\in\mathbb{F}\left[  w,w^{-1}\right]  $, we have%
\begin{align*}
D_{\psi\left(  z\right)  a\left(  z,w\right)  }\left(  \varphi\left(
w\right)  \right)   &  =\operatorname*{Res}\varphi\left(  z\right)
\psi\left(  z\right)  a\left(  z,w\right)  dz\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }D_{\psi\left(  w\right)  a\left(  z,w\right)
}\right) \\
&  =D_{a\left(  z,w\right)  }\left(  \varphi\left(  w\right)  \psi\left(
w\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}D_{a\left(  z,w\right)  }\right) \\
&  =D_{a\left(  z,w\right)  }\left(  \psi\left(  w\right)  \varphi\left(
w\right)  \right)  =\left(  D_{a\left(  z,w\right)  }\circ\psi\left(
w\right)  \right)  \left(  \varphi\left(  w\right)  \right)  .
\end{align*}
This yields (\ref{eq.exe.3.4.d.3}).

\textit{Proof of (\ref{eq.exe.3.4.d.4}):} For every $\varphi\left(  w\right)
\in\mathbb{F}\left[  w,w^{-1}\right]  $, we have%
\begin{align*}
D_{\partial_{z}a\left(  z,w\right)  }\left(  \varphi\left(  w\right)  \right)
&  =\operatorname*{Res}\varphi\left(  z\right)  \partial_{z}a\left(
z,w\right)  dz\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}D_{\partial_{z}a\left(  z,w\right)  }\right) \\
&  =-\underbrace{\operatorname*{Res}\partial_{z}\left(  \varphi\left(
z\right)  \right)  a\left(  z,w\right)  dz}_{\substack{=D_{a\left(
z,w\right)  }\left(  \partial_{w}\left(  \varphi\left(  w\right)  \right)
\right)  \\\text{(by the definition of }D_{a\left(  z,w\right)  }\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{eq.prop.residues.del.2}), applied to }U\left[  \left[
w,w^{-1}\right]  \right]  \text{, }\varphi\left(  z\right)  \text{ and
}a\left(  z,w\right) \\
\text{instead of }U\text{, }g\text{ and }f
\end{array}
\right) \\
&  =-D_{a\left(  z,w\right)  }\left(  \partial_{w}\left(  \varphi\left(
w\right)  \right)  \right)  =\left(  -D_{a\left(  z,w\right)  }\circ
\partial_{w}\right)  \left(  \varphi\left(  w\right)  \right)  .
\end{align*}
This yields (\ref{eq.exe.3.4.d.4}).
\end{proof}

\section{\label{chap.lie}$j$-th products over Lie (super)algebras}

\subsection{Local pairs over Lie (super)algebras}

Now, assume that $2$ and $3$ are invertible in the ground ring $\mathbb{F}$
(for instance, this holds when $\mathbb{F}$ is a field of characteristic
$\notin\left\{  2,3\right\}  $). Let $\mathfrak{g}$ be a Lie (super)algebra.
We remind the reader that a \textit{Lie superalgebra} is defined as an
$\mathbb{F}$-supermodule $\mathfrak{g}=\mathfrak{g}_{\overline{0}}%
\oplus\mathfrak{g}_{\overline{1}}$ endowed with a Lie bracket $\left[
\cdot,\cdot\right]  :\mathfrak{g}\times\mathfrak{g}\rightarrow\mathfrak{g}$
which is even (i.e., satisfies $\left[  \mathfrak{g}_{\alpha},\mathfrak{g}%
_{\beta}\right]  \subseteq\mathfrak{g}_{\alpha+\beta}$ for all $\alpha
,\beta\in\mathbb{Z}/2\mathbb{Z}$) and satisfies the following axioms (which
are obtained from the axioms in the definition of a Lie algebra by the
Koszul-Quillen rule):\footnote{We remark that Lie superalgebras could also be
defined more generally, without assuming that $2$ and $3$ are invertible in
$\mathbb{F}$. But in this generality, the two axioms below would not be
sufficient to obtain a good theory. For instance, if $\mathbb{F}$ is a field
of characteristic $2$, it is important to require $\left[  a,a\right]  = 0$
for all even elements $a$ of $\mathfrak{g}$ as an extra axiom; when $2$ is
invertible, this follows easily from the antisymmetry.}

\begin{itemize}
\item We have $\left[  a,b\right]  =-\left(  -1\right)  ^{p\left(  a\right)
p\left(  b\right)  }\left[  b,a\right]  $ for any two homogeneous elements $a$
and $b$ of $\mathfrak{g}$. Here, as usual, $p\left(  c\right)  $ denotes the
parity of the homogeneous element $c$.

\item We have $\left[  a,\left[  b,c\right]  \right]  =\left[  \left[
a,b\right]  ,c\right]  +\left(  -1\right)  ^{p\left(  a\right)  p\left(
b\right)  }\left[  b,\left[  a,c\right]  \right]  $ for any three homogeneous
elements $a$, $b$ and $c$ of $\mathfrak{g}$.
\end{itemize}

(Keep in mind that $\left[  a,a\right]  =0$ holds for any $a\in\mathfrak{g}%
_{0}$, as can be easily derived from the first of these two axioms; but in
general, $a\in\mathfrak{g}_{1}$ do not satisfy $\left[  a,a\right]  =0$.)

\begin{example}
\textbf{(a)} If $A$ is any associative superalgebra (i.e., any $\mathbb{Z} / 2
\mathbb{Z}$-graded associative algebra), then $A$ becomes a Lie superalgebra
when equipped with the supercommutator bracket%
\[
\left[  a,b\right]  :=ab-\left(  -1\right)  ^{p\left(  a\right)  p\left(
b\right)  }ba\ \ \ \ \ \ \ \ \ \ \text{for all }a,b\in A.
\]


\textbf{(b)} If $V=V_{\overline{0}}\oplus V_{\overline{1}}$ is an $\mathbb{F}%
$-supermodule, then $\operatorname*{End}V$ (the $\mathbb{F}$-superalgebra of
\textbf{all} $\mathbb{F}$-linear endomorphisms of $V$, not just the even ones)
becomes a Lie superalgebra when equipped with the supercommutator bracket%
\[
\left[  a,b\right]  :=ab-\left(  -1\right)  ^{p\left(  a\right)  p\left(
b\right)  }ba\ \ \ \ \ \ \ \ \ \ \text{for all }a,b\in\operatorname*{End}V.
\]
This Lie superalgebra structure is, of course, a particular case of part
\textbf{(a)} of this example.
\end{example}

\begin{definition}
Let $\mathfrak{g}$ be a Lie (super)algebra.

\textbf{(a)} Two $\mathfrak{g}$-valued formal distributions $a\left(
z\right)  \in\mathfrak{g}\left[  \left[  z,z^{-1}\right]  \right]  $ and
$b\left(  z\right)  \in\mathfrak{g}\left[  \left[  z,z^{-1}\right]  \right]  $
are said to \textit{form a local pair} if the formal distribution $\left[
a\left(  z\right)  ,b\left(  w\right)  \right]  \in\mathfrak{g}\left[  \left[
z,z^{-1},w,w^{-1}\right]  \right]  $ is local.

\textbf{(b)} Let $\left(  a\left(  z\right)  ,b\left(  z\right)  \right)  $ be
a local pair. According to Theorem \ref{thm.decomposition}, we have%
\begin{equation}
\left[  a\left(  z\right)  ,b\left(  w\right)  \right]  =\sum_{j\in\mathbb{N}%
}c^{j}\left(  w\right)  \partial_{w}^{\left(  j\right)  }\delta\left(
z-w\right)  , \label{eq.j-thprod.def.0}%
\end{equation}
where the formal distributions $c^{j}\left(  w\right)  \in\mathfrak{g}\left[
\left[  w,w^{-1}\right]  \right]  $ are given by%
\begin{equation}
c^{j}\left(  w\right)  =\operatorname*{Res}\left(  z-w\right)  ^{j}\left[
a\left(  z\right)  ,b\left(  w\right)  \right]
dz\ \ \ \ \ \ \ \ \ \ \text{for all }j\in\mathbb{N}. \label{eq.j-thprod.def.1}%
\end{equation}
For each $j\in\mathbb{N}$, we shall denote by $a\left(  w\right)  _{\left(
j\right)  }b\left(  w\right)  $ the formal distribution $c^{j}\left(
w\right)  \in\mathfrak{g}\left[  \left[  w,w^{-1}\right]  \right]  $ defined
by (\ref{eq.j-thprod.def.1}). Notice that the notation \textquotedblleft%
$a\left(  w\right)  _{\left(  j\right)  }b\left(  w\right)  $%
\textquotedblright\ is a ternary operator with three arguments: $a\left(
w\right)  $, $b\left(  w\right)  $ and $j$. We shall often use a curried
version of this operator: Namely, given an $a\left(  z\right)  \in
\mathfrak{g}\left[  \left[  z,z^{-1}\right]  \right]  $ and a $j\in\mathbb{N}%
$, we write $a\left(  w\right)  _{\left(  j\right)  }$ for the operator which
takes a $b\left(  w\right)  $ for which $\left(  a\left(  z\right)  ,b\left(
z\right)  \right)  $ is a local pair, and returns the formal distribution
$a\left(  w\right)  _{\left(  j\right)  }b\left(  w\right)  $.

As one would expect, $a\left(  z \right)  _{\left(  j\right)  }b\left(  z
\right)  $ means the result of substituting $z$ for $w$ in the formal
distribution $a\left(  w\right)  _{\left(  j\right)  }b\left(  w\right)  $
(and similarly for every other indeterminate instead of $z$).

We refer to $a\left(  z \right)  _{\left(  j\right)  }b\left(  z \right)  $ as
the $j$\textit{-th product of the formal distributions} $a\left(  z\right)  $
and $b\left(  z\right)  $. For given $j\in\mathbb{N}$, we refer to the
operator which sends every local pair $\left(  a\left(  z\right)  ,b\left(
z\right)  \right)  $ to $a\left(  z \right)  _{\left(  j\right)  }b\left(  z
\right)  $ as the $j$\textit{-th multiplication}.
\end{definition}

The formula (\ref{eq.j-thprod.def.0}) is called the \textit{singular part of
the OPE (operator-product expansion)}.

Let us make an important remark.

\begin{remark}
Let $\mathfrak{g}$ be a Lie (super)algebra, and let $\left(  a\left(
z\right)  ,b\left(  z\right)  \right)  $ be a local pair. We have denoted by
$a\left(  w\right)  _{\left(  j\right)  }b\left(  w\right)  $ the formal
distribution $c^{j}\left(  w\right)  \in\mathfrak{g}\left[  \left[
w,w^{-1}\right]  \right]  $ defined by (\ref{eq.j-thprod.def.1}). Thus,
(\ref{eq.j-thprod.def.0}) becomes%
\begin{align*}
\left[  a\left(  z\right)  ,b\left(  w\right)  \right]   &  =\sum
_{j\in\mathbb{N}}\underbrace{\left(  a\left(  w\right)  _{\left(  j\right)
}b\left(  w\right)  \right)  }_{\substack{=\sum_{n\in\mathbb{Z}}\left(
a\left(  w\right)  _{\left(  j\right)  }b\left(  w\right)  \right)  _{\left[
n\right]  }w^{-n-1}\\\text{(here, we are using the notation of}%
\\\text{Definition \ref{def.weirdcoeff}, for }w\text{ instead of }z\text{)}%
}}\partial_{w}^{\left(  j\right)  }\delta\left(  z-w\right) \\
&  =\sum_{j\in\mathbb{N}}\sum_{n\in\mathbb{Z}}\left(  a\left(  w\right)
_{\left(  j\right)  }b\left(  w\right)  \right)  _{\left[  n\right]  }%
w^{-n-1}\partial_{w}^{\left(  j\right)  }\delta\left(  z-w\right) \\
&  =\sum_{j\in\mathbb{N}}\sum_{r\in\mathbb{Z}}\left(  a\left(  w\right)
_{\left(  j\right)  }b\left(  w\right)  \right)  _{\left[  r\right]  }%
w^{-r-1}\underbrace{\partial_{w}^{\left(  j\right)  }\delta\left(  z-w\right)
}_{\substack{=\sum_{n\in\mathbb{Z}}\dbinom{n}{j}w^{n-j}z^{-n-1}\\\text{(by
(\ref{eq.6b}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}n\text{ as }r\right) \\
&  =\sum_{j\in\mathbb{N}}\sum_{r\in\mathbb{Z}}\left(  a\left(  w\right)
_{\left(  j\right)  }b\left(  w\right)  \right)  _{\left[  r\right]  }%
w^{-r-1}\sum_{n\in\mathbb{Z}}\dbinom{n}{j}w^{n-j}z^{-n-1}\\
&  =\sum_{j\in\mathbb{N}}\sum_{r\in\mathbb{Z}}\sum_{n\in\mathbb{Z}}\dbinom
{n}{j}\left(  a\left(  w\right)  _{\left(  j\right)  }b\left(  w\right)
\right)  _{\left[  r\right]  }\underbrace{w^{-r-1}w^{n-j}z^{-n-1}%
}_{=w^{n-j-r-1}z^{-n-1}}\\
&  =\sum_{j\in\mathbb{N}}\sum_{r\in\mathbb{Z}}\sum_{n\in\mathbb{Z}}\dbinom
{n}{j}\left(  a\left(  w\right)  _{\left(  j\right)  }b\left(  w\right)
\right)  _{\left[  r\right]  }w^{n-j-r-1}z^{-n-1}\\
&  =\sum_{j\in\mathbb{N}}\sum_{r\in\mathbb{Z}}\sum_{m\in\mathbb{Z}}\dbinom
{m}{j}\left(  a\left(  w\right)  _{\left(  j\right)  }b\left(  w\right)
\right)  _{\left[  r\right]  }w^{m-j-r-1}z^{-m-1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}m\text{ as }n\right)  .
\end{align*}
Comparing coefficients in front of $z^{-m-1}w^{-n-1}$ on both sides of this
equality, we conclude that%
\begin{equation}
\left[  a_{\left[  m\right]  },b_{\left[  n\right]  }\right]  =\sum
_{j\in\mathbb{N}}\dbinom{m}{j}\left(  a\left(  w\right)  _{\left(  j\right)
}b\left(  w\right)  \right)  _{\left[  m+n-j\right]  }. \label{eq.12}%
\end{equation}
This equality is an equivalent rewriting of (\ref{eq.j-thprod.def.0}).
\end{remark}


\end{document}