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\ihead{Regular elements and lcm-coprimality}
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\begin{document}

\title{Regular elements of a ring, monic polynomials and \textquotedblleft
lcm-coprimality\textquotedblright}
\author{Darij Grinberg}
\date{
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\maketitle
\tableofcontents

\section{The statements}

\subsection{The main theorems}

\begin{convention}
In this paper, the word \textquotedblleft ring\textquotedblright\ will always
mean \textquotedblleft ring with unity\textquotedblright.

Furthermore, the letter $\mathbb{N}$ shall always mean the set $\left\{
0,1,2,\ldots\right\}  $.
\end{convention}

Consider the following fact:

\begin{theorem}
\label{thm.div.all}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$. Let
$f\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ be a polynomial in the $n$
indeterminates $X_{1},X_{2},\ldots,X_{n}$ over $A$. Assume that $f$ is
divisible by $X_{i}-X_{j}$ for every $\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$. Then, $f$ is divisible by
$\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  $.

(Here, the symbol \textquotedblleft$\prod_{1\leq i<j\leq n}$\textquotedblright%
\ is an abbreviation for \textquotedblleft$\prod_{\substack{\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2};\\i<j}}$\textquotedblright.
Thus, when $n=0$ or $n=1$, the product $\prod_{1\leq i<j\leq n}\left(
X_{i}-X_{j}\right)  $ is an empty product and therefore equals $1$.)
\end{theorem}

Theorem \ref{thm.div.all} is rather obvious in the case when $A$ is a unique
factorization domain (because then, $A\left[  X_{1},X_{2},\ldots,X_{n}\right]
$ is also a unique factorization domain\footnote{This follows from one of the
many results known as \textquotedblleft Gauss's theorem\textquotedblright%
\ (e.g., \cite[Corollary 8.21]{Knapp2016}, applied $n$ times).}). However, I
would not call Theorem \ref{thm.div.all} obvious in the general case. One of
the goals of this note is to prove Theorem \ref{thm.div.all} in full generality.

Actually, I shall prove the following more general fact:

\begin{theorem}
\label{thm.div.G}Let $A$ be a commutative ring with unity. Let $n\in
\mathbb{N}$. Let $G$ be a subset of the set $\left\{  \left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}  $.

Let $f\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ be a polynomial in the
$n$ indeterminates $X_{1},X_{2},\ldots,X_{n}$ over $A$. Assume that $f$ is
divisible by $X_{i}-X_{j}$ for every $\left(  i,j\right)  \in G$. Then, $f$ is
divisible by $\prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  $.
\end{theorem}

Clearly, Theorem \ref{thm.div.all} is the particular case of Theorem
\ref{thm.div.G} obtained when $G=\left\{  \left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}  $. Theorem \ref{thm.div.G} is
also evident when $A$ is a unique factorization domain, but not in the general case.

\subsection{Applications}

First of all, why is Theorem \ref{thm.div.all} (and thus, by extension,
Theorem \ref{thm.div.G}) useful? Here are four applications:

\begin{itemize}
\item The \textit{Vandermonde determinant formula} (in one of its many forms)
is the statement that if $n\in\mathbb{N}$, and if $x_{1},x_{2},\ldots,x_{n}$
are $n$ elements of a commutative ring $A$, then%
\begin{equation}
\det\left(
\begin{array}
[c]{ccccc}%
1 & 1 & 1 & \cdots & 1\\
x_{1} & x_{2} & x_{3} & \cdots & x_{n}\\
x_{1}^{2} & x_{2}^{2} & x_{3}^{2} & \cdots & x_{n}^{2}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
x_{1}^{n-1} & x_{2}^{n-1} & x_{3}^{n-1} & \cdots & x_{n}^{n-1}%
\end{array}
\right)  =\prod_{1\leq i<j\leq n}\left(  x_{j}-x_{i}\right)  . \label{eq.vdm}%
\end{equation}
One of the shortest proofs of this fact (given, e.g., in \cite[\S 17.1]%
{Garrett09}) proceeds by observing that both the left hand side and the right
hand side of (\ref{eq.vdm}) are polynomials in the variables $x_{1}%
,x_{2},\ldots,x_{n}$, whence we can replace the elements $x_{1},x_{2}%
,\ldots,x_{n}$ by the indeterminates $X_{1},X_{2},\ldots,X_{n}$ in the
polynomial ring $\mathbb{Z}\left[  X_{1},X_{2},\ldots,X_{n}\right]  $; but
once this has been done, we can observe that the left hand side of
(\ref{eq.vdm}) is divisible by $X_{i}-X_{j}$ for every $\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$, and therefore (by
Theorem \ref{thm.div.all}) is also divisible by $\prod_{1\leq i<j\leq
n}\left(  X_{i}-X_{j}\right)  $. This is not the whole proof, but the rest of
the proof (degree considerations as well as a comparison of a single
coefficient) is not difficult (see \cite[\S 17.1]{Garrett09} for details).
What matters for us is that this argument uses Theorem \ref{thm.div.all}
(although only in the case when $A=\mathbb{Z}$; as we have said, this is an
easy case, since $\mathbb{Z}$ is a unique factorization domain).

\item A similar proof that uses Theorem \ref{thm.div.G} (although, again, only
in a simple case in which $A$ is a unique factorization domain) is \cite[proof
of Lemma 5.15.3]{EGHLSVY11}\footnote{This is the proof of Lemma 4.48 in the
arXiv draft of \cite{EGHLSVY11}.}.

\item Theorem \ref{thm.div.all} is used in \cite[proof of Lemma (6.1)]%
{LLPT95}\footnote{Notice that the $a_{p}$ in \cite{LLPT95} are the $X_{p}$ in
Theorem \ref{thm.div.all}.}. In fact, \cite[proof of Lemma (6.1)]{LLPT95}
involves an argument saying that \textquotedblleft Clearly, $f$ is divisible
by all differences $a_{q}-a_{p}$ for $p<q$ if and only if $f$ is divisible by
the product of all the differences\textquotedblright. I believe the word
\textquotedblleft Clearly\textquotedblright\ is inappropriate in this
argument, and should be replaced by an application of Theorem
\ref{thm.div.all}.

\item Theorem \ref{thm.div.all} is used in \cite[proof of Proposition
2.6.4]{HopfComb}. Indeed, \cite[proof of Proposition 2.6.4]{HopfComb} involves
an argument saying that the polynomial $f\left(  \mathbf{x}\right)  $ must
\textquotedblleft be divisible by $x_{i}-x_{j}$, so divisible by the entire
product \newline$\prod_{1\leq i<j\leq n}\left(  x_{i}-x_{j}\right)  =a_{\rho}%
$\textquotedblright. Again, this argument only uses Theorem \ref{thm.div.all}
in the particular case when $A$ is a unique factorization domain (because
\cite[Proposition 2.6.4]{HopfComb} is only stated for $\mathbf{k}$ being
$\mathbb{Z}$ or a field of characteristic $\neq2$; but all such $\mathbf{k}$
are unique factorization domains). However, knowing that Theorem
\ref{thm.div.all} holds for arbitrary $A$, we can extend \cite[Proposition
2.6.4]{HopfComb} to the case of $\mathbf{k}$ being an arbitrary commutative
ring (as long as $\Lambda^{\operatorname*{sgn}}$ is defined appropriately).
\end{itemize}

On our way to the proofs of Theorem \ref{thm.div.all} and Theorem
\ref{thm.div.G}, we shall develop some basics of commutative algebra from
scratch: most importantly, division with remainder by a monic polynomial, and
the fact that monic polynomials are non-zero-divisors (or, as we call them,
regular elements). This is, of course, perfectly well-known in the case of
univariate polynomials over a field; but in the general case, it is rarely
discussed in detail in the literature.

We shall then explore the surroundings of these results: alternative proofs,
analogues for formal power series, applications to other symmetric polynomials.

\subsection{Plan of this paper}

In Section \ref{sect.regular}, we shall define the notion of \textquotedblleft
regular elements\textquotedblright\ (also known as non-zero-divisors) and show
some very basic properties (such as the fact that a product of regular
elements is again regular).

In Section \ref{sect.monic}, we shall review (with proofs) the basic theory of
monic polynomials over a commutative ring, including the fact that division
with remainder always works when we are dividing by a monic polynomial (even
if the base ring is not a field). We will show that monic polynomials are
regular, and prove a first step (Corollary \ref{cor.pol.intersect}) towards
Theorem \ref{thm.div.G}.

In Section \ref{sect.mulpol}, we will start studying polynomial rings
$A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ in multiple indeterminates, and
we will prove some basic properties of the polynomials $X_{i}$ and
$X_{i}-X_{j}$ (with $i\neq j$) in these rings. In particular, we will show
that these polynomials are regular, and have a \textquotedblleft
coprimality\textquotedblright\ property (the particular case of Theorem
\ref{thm.div.G} for $\left\vert G\right\vert =2$).

In Section \ref{sect.lcm-copri}, we shall prove the crucial Proposition
\ref{prop.copri.prod} about when an intersection of principal ideals of a
commutative ring equals the product of these ideals. This will let us derive
the full Theorem \ref{thm.div.G} (and thus Theorem \ref{thm.div.all}) from
this \textquotedblleft coprimality\textquotedblright\ property.

In Section \ref{sect.pow}, we will start extending our results about
polynomials to formal power series. Some things will extend easily, but many
will change, either requiring new proofs or ceasing to be valid in the new
setting. We will show that analogues of Theorem \ref{thm.div.G} and Theorem
\ref{thm.div.all} for power series still hold, but our proofs no longer apply
to this case.

In Section \ref{sect.lcm-coprim-reg}, we will give a second proof of
Proposition \ref{prop.copri.prod}, this time using quotient rings.

In Section \ref{sect.subst}, we shall rewrite Theorem \ref{thm.div.all} in
terms of substitutions of variables.

In Section \ref{sect.symmetric}, we shall apply Theorem \ref{thm.div.all} to
the construction of symmetric polynomials.

In Section \ref{sect.pow.more} (work in progress), we shall continue our study
of univariate power series started in Section \ref{sect.pow}.

\subsection{Prerequisites}

I have written this paper with the express purpose of being as accessible as
it possibly can; only basic notions and facts of abstract algebra (rings,
polynomials, formal power series, nilpotence) are assumed. Even the use of
quotient rings has been kept to a minimum. The occasional counterexample uses
more advanced notions, but counterexamples can be skipped without loss.

\section{\label{sect.regular}Regular elements (a.k.a. non-zero-divisors)}

\subsection{Definition}

We begin with a basic notation:

\begin{definition}
\label{def.reg}Let $A$ be a commutative ring. Let $a\in A$. The element $a$ of
$A$ is said to be \textit{regular} if and only if every $x\in A$ satisfying
$ax=0$ satisfies $x=0$.

Instead of saying that $a$ is regular, one can also say that \textquotedblleft%
$a$ is \textit{cancellable}\textquotedblright, or that \textquotedblleft$a$ is
a \textit{non-zero-divisor}\textquotedblright.
\end{definition}

This notion of \textquotedblleft regular\textquotedblright\ elements has
nothing to do with various other notions of \textquotedblleft
regularity\textquotedblright\ in commutative algebra (for example, it is
completely unrelated to the notion of a \textquotedblleft von Neumann regular
element\textquotedblright\ of a ring). It might sound like a bad idea to
employ a word like \textquotedblleft regular\textquotedblright\ that has
already seen so much different uses; however, we are not really adding a new
conflicting meaning for this word, because the word is already being used in
this meaning by various authors (among them, the authors of \cite{LLPT95}),
and because our use of \textquotedblleft regular\textquotedblright\ is closely
related to the standard notion of a \textquotedblleft regular
sequence\textquotedblright\ in a commutative ring\footnote{Namely: An element
$a$ of a commutative ring $A$ is regular if and only if the one-element
sequence $\left(  a\right)  $ is regular.}.

Many authors (for example, Knapp in \cite{Knapp2016}) define a \textit{zero
divisor} in a commutative ring $A$ to be a nonzero element of $A$ that is not
regular.\footnote{Some authors drop the \textquotedblleft
nonzero\textquotedblright\ requirement in this definition; so they count $0$
as a zero divisor, provided $A$ is not a trivial ring.} Thus, at least in
classical logic, regular elements are the same as elements that are not zero
divisors (with the possible exception of $0$). I find the notion of a
\textquotedblleft zero divisor\textquotedblright\ less natural than that of a
regular element (it is the regular elements, not the zero divisors, that
usually exhibit the nicer behavior), and it is much less suitable for
constructive logic (as it muddies the waters with an unnecessary negation),
but it appears to be more popular for traditional reasons.

\subsection{Properties of regular elements}

Let me state some basic properties of regular elements:

\begin{proposition}
\label{prop.reg.ab}Let $A$ be a commutative ring. Let $a\in A$ and $b\in A$ be
two regular elements of $A$. Then, the element $ab$ of $A$ is regular.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.reg.ab}.]Every $x\in A$ satisfying $\left(
ab\right)  x=0$ satisfies $a\left(  bx\right)  =\left(  ab\right)  x=0$, thus
$bx=0$ (since $a$ is regular), and therefore $x=0$ (since $b$ is regular). In
other words, $ab$ is regular. This proves Proposition \ref{prop.reg.ab}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.reg.ab}.]The element $a$ is regular if and
only if every $x\in A$ satisfying $ax=0$ satisfies $x=0$ (by the definition of
\textquotedblleft regular\textquotedblright). Hence,%
\begin{equation}
\text{every }x\in A\text{ satisfying }ax=0\text{ satisfies }x=0
\label{pf.prop.reg.ab.a-reg}%
\end{equation}
(since the element $a$ is regular).

The element $b$ is regular if and only if every $x\in A$ satisfying $bx=0$
satisfies $x=0$ (by the definition of \textquotedblleft
regular\textquotedblright). Hence,%
\begin{equation}
\text{every }x\in A\text{ satisfying }bx=0\text{ satisfies }x=0
\label{pf.prop.reg.ab.b-reg}%
\end{equation}
(since the element $b$ is regular).

Now, every $x\in A$ satisfying $\left(  ab\right)  x=0$ satisfies
$x=0$\ \ \ \ \footnote{\textit{Proof.} Let $x\in A$ be such that $\left(
ab\right)  x=0$. We must prove that $x=0$.
\par
We have $a\left(  bx\right)  =\left(  ab\right)  x=0$. Hence, $bx=0$ (by
(\ref{pf.prop.reg.ab.a-reg}), applied to $bx$ instead of $x$). Thus, $x=0$ (by
(\ref{pf.prop.reg.ab.b-reg})). Qed.}.

But the element $ab$ is regular if and only if every $x\in A$ satisfying
$\left(  ab\right)  x=0$ satisfies $x=0$ (by the definition of
\textquotedblleft regular\textquotedblright). Hence, the element $ab$ is
regular (since every $x\in A$ satisfying $\left(  ab\right)  x=0$ satisfies
$x=0$). This proves Proposition \ref{prop.reg.ab}.
\end{proof}
\end{verlong}

\begin{proposition}
\label{prop.reg.prod}Let $A$ be a commutative ring. Let $G$ be a finite set.
For every $g\in G$, let $a_{g}$ be a regular element of $A$. Then, the element
$\prod_{g\in G}a_{g}$ of $A$ is regular.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.reg.prod}.]This follows by induction on
$\left\vert G\right\vert $, using Proposition \ref{prop.reg.ab} in the
induction step\footnote{The induction base involves showing that the element
$1$ of $A$ is regular; but this is obvious.}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.reg.prod}.]We shall prove Proposition
\ref{prop.reg.prod} by induction on $\left\vert I\right\vert $:

\textit{Induction base:} Proposition \ref{prop.reg.prod} holds in the case
when $\left\vert G\right\vert =0$\ \ \ \ \footnote{\textit{Proof.} Let $A$,
$G$ and $a_{g}$ be as in Proposition \ref{prop.reg.prod}. Assume that
$\left\vert G\right\vert =0$. Thus, $G=\varnothing$. Hence, $\prod_{g\in
G}a_{g}=\left(  \text{empty product}\right)  =1$.
\par
Now, every element $x\in A$ satisfying $1x=0$ satisfies $x=0$ (because from
$1x=0$, we obtain $x=1x=0$).
\par
But the element $1$ of $A$ is regular if every $x\in A$ satisfying $1x=0$
satisfies $x=0$ (by the definition of \textquotedblleft
regular\textquotedblright). Hence, the element $1$ of $A$ is regular (since
every element $x\in A$ satisfying $1x=0$ satisfies $x=0$). In other words, the
element $\prod_{g\in G}a_{g}$ of $A$ is regular (since $\prod_{g\in G}a_{g}%
=1$).
\par
Now, forget that we fixed $A$, $G$ and $a_{g}$. We thus have shown that if
$A$, $G$ and $a_{g}$ are as in Proposition \ref{prop.reg.prod}, and if
$\left\vert G\right\vert =0$, then the element $\prod_{g\in G}a_{g}$ of $A$ is
regular. In other words, Proposition \ref{prop.reg.prod} holds in the case
when $\left\vert G\right\vert =0$. Qed.}. This completes the induction base.

\textit{Induction step:} Let $N$ be a positive integer. Assume that
Proposition \ref{prop.reg.prod} holds in the case when $\left\vert
G\right\vert =N-1$. We must prove that Proposition \ref{prop.reg.prod} holds
in the case when $\left\vert G\right\vert =N$.

We have assumed that Proposition \ref{prop.reg.prod} holds in the case when
$\left\vert G\right\vert =N-1$. In other words, the following fact holds:

\begin{statement}
\textit{Fact 1:} Let $A$ be a commutative ring. Let $G$ be a finite set such
that $\left\vert G\right\vert =N-1$. For every $g\in G$, let $a_{g}$ be a
regular element of $A$. Then, the element $\prod_{g\in G}a_{g}$ of $A$ is regular.
\end{statement}

Now, let $A$, $G$ and $a_{g}$ be as in Proposition \ref{prop.reg.prod}. Assume
that $\left\vert G\right\vert =N$. We have $\left\vert G\right\vert =N>0$
(since $N$ is a positive integer). Thus, the set $G$ is nonempty. Hence, there
exists some $h\in G$. Consider this $h$.

Recall that for every $g\in G$, the element $a_{g}$ is a regular element of
$A$. Applying this to $g=h$, we conclude that $a_{h}$ is a regular element of
$A$.

From $h\in G$, we obtain $\left\vert G\setminus\left\{  h\right\}  \right\vert
=\underbrace{\left\vert G\right\vert }_{=N}-1=N-1$.

Recall that for every $g\in G$, the element $a_{g}$ is a regular element of
$A$. Hence, for every $g\in G\setminus\left\{  h\right\}  $, the element
$a_{g}$ is a regular element of $A$ (since $g\in G\setminus\left\{  h\right\}
\subseteq G$). Hence, Fact 1 (applied to $G\setminus\left\{  h\right\}  $
instead of $G$) yields that the element $\prod_{g\in G\setminus\left\{
h\right\}  }a_{g}$ of $A$ is regular.

But we have $h\in G$. Hence, $\prod_{g\in G}a_{g}=a_{h}\prod_{g\in
G\setminus\left\{  h\right\}  }a_{g}$ (here, we have split off the factor for
$g=h$ from the product). But both $a_{h}$ and $\prod_{g\in G\setminus\left\{
h\right\}  }a_{g}$ are regular elements of $A$. Thus, Proposition
\ref{prop.reg.ab} (applied to $a_{h}$ and $\prod_{g\in G\setminus\left\{
h\right\}  }a_{g}$ instead of $a$ and $b$) shows that the element $a_{h}%
\prod_{g\in G\setminus\left\{  h\right\}  }a_{g}$ of $A$ is regular. In other
words, the element $\prod_{g\in G}a_{g}$ of $A$ is regular (since $\prod_{g\in
G}a_{g}=a_{h}\prod_{g\in G\setminus\left\{  h\right\}  }a_{g}$).

Now, forget that we fixed $A$, $G$ and $a_{g}$. We thus have shown that if
$A$, $G$ and $a_{g}$ are as in Proposition \ref{prop.reg.prod}, and if
$\left\vert G\right\vert =N$, then the element $\prod_{g\in G}a_{g}$ of $A$ is
regular. In other words, Proposition \ref{prop.reg.prod} holds in the case
when $\left\vert G\right\vert =N$. This completes the induction step. Thus,
the induction proof of Proposition \ref{prop.reg.prod} is complete.
\end{proof}
\end{verlong}

The following trivial proposition just says that the regularity of an element
is unchanged under ring isomorphisms:

\begin{proposition}
\label{prop.reg.iso}Let $A$ and $B$ be two commutative rings. Let
$f:A\rightarrow B$ be a ring isomorphism. Let $a$ be a regular element of $A$.
Then, $f\left(  a\right)  $ is a regular element of $B$.
\end{proposition}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.reg.iso}.]The element $a$ is regular if and
only if every $x\in A$ satisfying $ax=0$ satisfies $x=0$ (by the definition of
\textquotedblleft regular\textquotedblright). Hence,%
\begin{equation}
\text{every }x\in A\text{ satisfying }ax=0\text{ satisfies }x=0
\label{pf.prop.reg.iso.reg-a}%
\end{equation}
(since the element $a$ is regular).

Now, every $x\in B$ satisfying $f\left(  a\right)  x=0$ satisfies
$x=0$\ \ \ \ \footnote{\textit{Proof.} Let $x\in B$ be such that $f\left(
a\right)  x=0$. We must prove that $x=0$.
\par
The map $f$ is an isomorphism. Hence, $f^{-1}\left(  x\right)  $ is a
well-defined element of $A$. Moreover, $f$ is a ring homomorphism; thus,
$f\left(  af^{-1}\left(  x\right)  \right)  =f\left(  a\right)
\underbrace{f\left(  f^{-1}\left(  x\right)  \right)  }_{=x}=f\left(
a\right)  x=0=f\left(  0\right)  $ (since $f$ is a ring homomorphism).
\par
But $f$ is an isomorphism, therefore bijective, therefore injective. Hence,
from $f\left(  af^{-1}\left(  x\right)  \right)  =f\left(  0\right)  $, we
obtain $af^{-1}\left(  x\right)  =0$. Thus, (\ref{pf.prop.reg.iso.reg-a})
(applied to $f^{-1}\left(  x\right)  $ instead of $x$) yields $f^{-1}\left(
x\right)  =0$. Hence, $f\left(  f^{-1}\left(  x\right)  \right)  =f\left(
0\right)  =0$ (since $f$ is a ring homomorphism). Comparing this with
$f\left(  f^{-1}\left(  x\right)  \right)  =x$, we obtain $x=0$. Qed.}.

The element $f\left(  a\right)  $ of $B$ is regular if and only if every $x\in
B$ satisfying $f\left(  a\right)  x=0$ satisfies $x=0$ (by the definition of
\textquotedblleft regular\textquotedblright). Hence, the element $f\left(
a\right)  $ of $B$ is regular (since every $x\in B$ satisfying $f\left(
a\right)  x=0$ satisfies $x=0$). This proves Proposition \ref{prop.reg.iso}.
\end{proof}
\end{verlong}

The next proposition is also fairly trivial:

\begin{proposition}
\label{prop.reg.sub}Let $B$ be a commutative ring. Let $A$ be a subring of
$B$. Let $a$ be an element of $A$ such that $a$ is a regular element of $B$.
Then, $a$ is a regular element of $A$.
\end{proposition}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.reg.sub}.]The element $a$ is a regular element
of $B$ if and only if every $x\in B$ satisfying $ax=0$ satisfies $x=0$ (by the
definition of \textquotedblleft regular\textquotedblright). Hence,%
\begin{equation}
\text{every }x\in B\text{ satisfying }ax=0\text{ satisfies }x=0
\label{pf.prop.reg.sub.1}%
\end{equation}
(since the element $a$ is a regular element of $B$).

Now, let $x\in A$ be such that $ax=0$. Then, $x\in A\subseteq B$ (since $A$ is
a subring of $B$). Hence, $x=0$ (by (\ref{pf.prop.reg.sub.1})).

Now, forget that we fixed $x$. We thus have shown that every $x\in A$
satisfying $ax=0$ satisfies $x=0$.

The element $a$ is a regular element of $A$ if and only if every $x\in A$
satisfying $ax=0$ satisfies $x=0$ (by the definition of \textquotedblleft
regular\textquotedblright). Hence, the element $a$ is a regular element of $A$
(since every $x\in A$ satisfying $ax=0$ satisfies $x=0$). This proves
Proposition \ref{prop.reg.sub}.
\end{proof}
\end{verlong}

\section{\label{sect.monic}Monic polynomials and division with remainder}

\subsection{Monic polynomials, $\left[  X^{n}\right]  p$ and $A\left[
X\right]  _{\leq n}$}

Now, let me discuss monic polynomials over a commutative ring $A$. This is not
an introduction into the notion of polynomials; we will just introduce the
nonstandard notations that we will need, and prove a few basic facts.

\begin{definition}
\label{def.pol.coeff}Let $A$ be a commutative ring.

\textbf{(a)} If $p\in A\left[  X\right]  $ is a polynomial in some
indeterminate $X$ over $A$, and if $n\in\mathbb{N}$, then $\left[
X^{n}\right]  p$ will denote the coefficient of $X^{n}$ in $p$. For example,%
\begin{align*}
\left[  X^{3}\right]  \left(  2X^{4}+5X^{3}+7X+2\right)   &  =5;\\
\left[  X^{4}\right]  \left(  X^{2}+1\right)   &  =0;\\
\left[  X^{0}\right]  \left(  3X+7\right)   &  =7.
\end{align*}


Clearly, every polynomial $p\in A\left[  X\right]  $ satisfies $p=\sum
_{n\in\mathbb{N}}\left(  \left[  X^{n}\right]  p\right)  X^{n}$. (The sum
$\sum_{n\in\mathbb{N}}\left(  \left[  X^{n}\right]  p\right)  X^{n}$ is
well-defined, because all but finitely many among its addends are $0$.)

\textbf{(b)} If $n\in\mathbb{Z}$, then we define an $A$-submodule $A\left[
X\right]  _{\leq n}$ of $A\left[  X\right]  $ as follows:%
\[
A\left[  X\right]  _{\leq n}=\left\{  p\in A\left[  X\right]  \ \mid\ \left[
X^{m}\right]  p=0\text{ for every }m\in\mathbb{N}\text{ satisfying
}m>n\right\}  .
\]
(In other words, $A\left[  X\right]  _{\leq n}$ is the set of all polynomials
$p\in A\left[  X\right]  $ having degree $\leq n$, provided that we understand
the degree of the zero polynomial to be a symbol $-\infty$ that is smaller
than any integer. However, we want to avoid using the concept of
\textquotedblleft degree\textquotedblright, so we are using the above
definition of $A\left[  X\right]  _{\leq n}$ instead.)

\textbf{(c)} Let $n\in\mathbb{N}$. A polynomial $p\in A\left[  X\right]  $ is
said to be \textit{monic of degree }$n$ if and only if it satisfies $\left[
X^{n}\right]  p=1$ and%
\[
\left(  \left[  X^{m}\right]  p=0\text{ for every }m\in\mathbb{N}\text{
satisfying }m>n\right)  .
\]
Instead of saying that $p$ \textquotedblleft is monic of degree $n$%
\textquotedblright, we can also say that $p$ \textquotedblleft is a monic
polynomial of degree $n$\textquotedblright.
\end{definition}

\begin{remark}
I am going to avoid the notion of the \textquotedblleft
degree\textquotedblright\ of a polynomial, since it is (in my opinion)
inferior to working with $A\left[  X\right]  _{\leq n}$ (for various reasons:
it is not defined when $A$ the trivial ring; it is not preserved by ring
homomorphisms; it depends on the vanishing of some coefficients and is
therefore not generally meaningful in constructive mathematics; it requires
some care in handling $\deg0$). Nevertheless, I will use the terminology
\textquotedblleft monic of degree $n$\textquotedblright\ introduced in
Definition \ref{def.pol.coeff}; the way I have defined it above, it is
independent of the notion of degree. Please be aware of the following quirk of
this terminology: If $A$ is the trivial ring, then there exists only one
polynomial $p\in A\left[  X\right]  $, and this polynomial is monic of degree
$n$ for every $n\in\mathbb{N}$. Thus, a polynomial $p$ can be monic of degree
$n$ for many different $n$. (But this only happens when $A$ is trivial.)
\end{remark}

\begin{noncompile}
If $p\in A\left[  X\right]  $ is a nonzero polynomial, then the
\textit{degree} of $p$ is defined to be the largest $n\in\mathbb{N}$
satisfying $\left[  X^{n}\right]  p\neq0$. The degree of the zero polynomial
$0\in A\left[  X\right]  $ is defined to be $-\infty$ (where $-\infty$ is a
symbol that behaves according to the rules $-\infty<a$ and $\left(
-\infty\right)  +a=-\infty$ for all $a\in\mathbb{Z}$).
\end{noncompile}

\subsection{Basic rules for polynomials}

\begin{vershort}
Recall how polynomials are added, scaled and multiplied:
\end{vershort}

\begin{verlong}
The following proposition will be used without mention:
\end{verlong}

\begin{proposition}
\label{prop.pol.+*}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$.

\textbf{(a)} Every $p\in A\left[  X\right]  $ and $q\in A\left[  X\right]  $
satisfy $\left[  X^{n}\right]  \left(  p+q\right)  =\left[  X^{n}\right]
p+\left[  X^{n}\right]  q$.

\textbf{(b)} Every $\lambda\in A$ and $p\in A\left[  X\right]  $ satisfy
$\left[  X^{n}\right]  \left(  \lambda p\right)  =\lambda\left[  X^{n}\right]
p$.

\textbf{(c)} Every $p\in A\left[  X\right]  $ and $q\in A\left[  X\right]  $
satisfy $\left[  X^{n}\right]  \left(  pq\right)  =\sum_{k=0}^{n}\left(
\left[  X^{k}\right]  p\right)  \cdot\left(  \left[  X^{n-k}\right]  q\right)
$.
\end{proposition}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.pol.+*}.]\textbf{(a)} Proposition
\ref{prop.pol.+*} \textbf{(a)} follows from the definition of $p+q$.

\textbf{(b)} Proposition \ref{prop.pol.+*} \textbf{(b)} follows from the
definition of $\lambda p$.

\textbf{(c)} Proposition \ref{prop.pol.+*} \textbf{(c)} follows from the
definition of $pq$.
\end{proof}
\end{verlong}

Let us next state a few trivial facts:

\begin{lemma}
\label{lem.pol.XkXn}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$.

\textbf{(a)} We have $\left[  X^{n}\right]  \left(  X^{n}\right)  =1$.

\textbf{(b)} For every $k\in\mathbb{N}$ satisfying $k\neq n$, we have $\left[
X^{k}\right]  \left(  X^{n}\right)  =0$.
\end{lemma}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.pol.XkXn}.]\textbf{(a)} Clearly, $\left[
X^{n}\right]  \left(  X^{n}\right)  $ is the coefficient of $X^{n}$ in $X^{n}$
(by the definition of $\left[  X^{n}\right]  \left(  X^{n}\right)  $). Since
this coefficient is $1$, we thus conclude that $\left[  X^{n}\right]  \left(
X^{n}\right)  $ is $1$. This proves Lemma \ref{lem.pol.XkXn} \textbf{(a)}.

\textbf{(b)} Let $k\in\mathbb{N}$ be such that $k\neq n$. Clearly, $\left[
X^{k}\right]  \left(  X^{n}\right)  $ is the coefficient of $X^{k}$ in $X^{n}$
(by the definition of $\left[  X^{k}\right]  \left(  X^{n}\right)  $). Since
this coefficient is $0$ (because $k\neq n$), we thus conclude that $\left[
X^{k}\right]  \left(  X^{n}\right)  $ is $0$. In other words, $\left[
X^{k}\right]  \left(  X^{n}\right)  =0$. This proves Lemma \ref{lem.pol.XkXn}
\textbf{(b)}.
\end{proof}
\end{verlong}

\begin{lemma}
\label{lem.pol.AXn}Let $A$ be a commutative ring. Let $n\in\mathbb{Z}$.

\textbf{(a)} For every $q\in A\left[  X\right]  _{\leq n}$ and every
$m\in\mathbb{N}$ satisfying $m>n$, we have $\left[  X^{m}\right]  q=0$.

\textbf{(b)} Let $q\in A\left[  X\right]  $. Assume that $\left[
X^{m}\right]  q=0$ for every $m\in\mathbb{N}$ satisfying $m>n$. Then, $q\in
A\left[  X\right]  _{\leq n}$.
\end{lemma}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.pol.AXn}.]The definition of $A\left[  X\right]
_{\leq n}$ yields%
\begin{equation}
A\left[  X\right]  _{\leq n}=\left\{  p\in A\left[  X\right]  \ \mid\ \left[
X^{m}\right]  p=0\text{ for every }m\in\mathbb{N}\text{ satisfying
}m>n\right\}  . \label{pf.lem.pol.AXn.1}%
\end{equation}


\textbf{(a)} Let $q\in A\left[  X\right]  _{\leq n}$. Thus,%
\begin{align*}
q  &  \in A\left[  X\right]  _{\leq n}\\
&  =\left\{  p\in A\left[  X\right]  \ \mid\ \left[  X^{m}\right]  p=0\text{
for every }m\in\mathbb{N}\text{ satisfying }m>n\right\}
\end{align*}
(by (\ref{pf.lem.pol.AXn.1})). In other words, $q$ is an element $p$ of
$A\left[  X\right]  $ satisfying \newline$\left(  \left[  X^{m}\right]
p=0\text{ for every }m\in\mathbb{N}\text{ satisfying }m>n\right)  $. In other
words, $q$ is an element of $A\left[  X\right]  $ and satisfies $\left(
\left[  X^{m}\right]  q=0\text{ for every }m\in\mathbb{N}\text{ satisfying
}m>n\right)  $. In particular, we have $\left(  \left[  X^{m}\right]
q=0\text{ for every }m\in\mathbb{N}\text{ satisfying }m>n\right)  $. In other
words, for every $m\in\mathbb{N}$ satisfying $m>n$, we have $\left[
X^{m}\right]  q=0$. This proves Lemma \ref{lem.pol.AXn} \textbf{(a)}.

\textbf{(b)} We know that $q$ is an element of $A\left[  X\right]  $ and
satisfies \newline$\left(  \left[  X^{m}\right]  q=0\text{ for every }%
m\in\mathbb{N}\text{ satisfying }m>n\right)  $. In other words, $q$ is an
element $p$ of $A\left[  X\right]  $ satisfying $\left(  \left[  X^{m}\right]
p=0\text{ for every }m\in\mathbb{N}\text{ satisfying }m>n\right)  $. In other
words,%
\[
q\in\left\{  p\in A\left[  X\right]  \ \mid\ \left[  X^{m}\right]  p=0\text{
for every }m\in\mathbb{N}\text{ satisfying }m>n\right\}  .
\]
In view of (\ref{pf.lem.pol.AXn.1}), this rewrites as $q\in A\left[  X\right]
_{\leq n}$. This proves Lemma \ref{lem.pol.AXn} \textbf{(b)}.
\end{proof}
\end{verlong}

\begin{proposition}
\label{prop.pol.union}Let $A$ be a commutative ring. Then, $A\left[  X\right]
=\bigcup_{n\in\mathbb{N}}A\left[  X\right]  _{\leq n}$.
\end{proposition}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.pol.union}.]Let $q\in A\left[  X\right]  $.
Then, $q$ is a polynomial. In other words, $q=\sum_{k\in\mathbb{N}}q_{k}X^{k}$
for some family $\left(  q_{k}\right)  _{k\in\mathbb{N}}\in A^{\mathbb{N}}$
such that \newline$\left(  \text{all but finitely many }k\in\mathbb{N}\text{
satisfy }q_{k}=0\right)  $. Consider this family $\left(  q_{k}\right)
_{k\in\mathbb{N}}$.

We have $q=\sum_{k\in\mathbb{N}}q_{k}X^{k}$. Therefore, for every
$k\in\mathbb{N}$, the coefficient of $X^{k}$ in $q$ is $q_{k}$ (by the
definition of a coefficient). In other words, for every $k\in\mathbb{N}$, we
have $\left(  \text{the coefficient of }X^{k}\text{ in }q\right)  =q_{k}$.
Hence, for every $k\in\mathbb{N}$, we have%
\begin{align}
\left[  X^{k}\right]  q  &  =\left(  \text{the coefficient of }X^{k}\text{ in
}q\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left[  X^{k}\right]  q\text{ is
defined as the coefficient of }X^{k}\text{ in }q\right) \nonumber\\
&  =q_{k}. \label{pf.prop.pol.union.1}%
\end{align}


We know that all but finitely many $k\in\mathbb{N}$ satisfy $q_{k}=0$. In
other words, there exists a finite subset $K$ of $\mathbb{N}$ such that every
$k\in\mathbb{N}\setminus K$ satisfies $q_{k}=0$. Consider this set $K$.

The set $K$ is finite. Hence, the set $\left\{  0\right\}  \cup K$ is finite.
Moreover, the set $\left\{  0\right\}  \cup K$ is nonempty (since it contains
$0$ (because $0\in\left\{  0\right\}  \subseteq\left\{  0\right\}  \cup K$)).
Recall that every nonempty finite set of integers has a maximum. We can apply
this to the set $\left\{  0\right\}  \cup K$ (since $\left\{  0\right\}  \cup
K$ is a nonempty finite set of integers); thus, we conclude that the set
$\left\{  0\right\}  \cup K$ has a maximum. Let $d$ be this maximum. Clearly,
$d\in\mathbb{N}$\ \ \ \ \footnote{\textit{Proof.} We know that $d$ is the
maximum of the set $\left\{  0\right\}  \cup K$. Hence, every element of
$\left\{  0\right\}  \cup K$ is $\leq d$. In other words, for every
$u\in\left\{  0\right\}  \cup K$, we have $u\leq d$. Applying this to $u=0$,
we obtain $0\leq d$ (since $0\in\left\{  0\right\}  \subseteq\left\{
0\right\}  \cup K$). Hence, $d\in\mathbb{N}$ (since $d$ is an integer), qed.}.

Now, let $m\in\mathbb{N}$ be such that $m>d$. Then, $m\notin K$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $m\in K$. Hence,
$m\in K\subseteq\left\{  0\right\}  \cup K$.
\par
But $d$ is the maximum of the set $\left\{  0\right\}  \cup K$. Hence, every
element of $\left\{  0\right\}  \cup K$ is $\leq d$. In other words, for every
$u\in\left\{  0\right\}  \cup K$, we have $u\leq d$. Applying this to $u=m$,
we obtain $m\leq d$ (since $m\in\left\{  0\right\}  \cup K$). But this
contradicts $m>d$. This contradiction proves that our assumption was wrong,
qed.}. Thus, $m\in\mathbb{N}\setminus K$ (since $m\in\mathbb{N}$). Now, recall
that every $k\in\mathbb{N}\setminus K$ satisfies $q_{k}=0$. Applying this to
$k=m$, we obtain $q_{m}=0$. Now, (\ref{pf.prop.pol.union.1}) (applied to
$k=m$) yields $\left[  X^{m}\right]  q=q_{m}=0$.

Now, forget that we fixed $m$. We thus have shown that $\left[  X^{m}\right]
q=0$ for every $q\in\mathbb{N}$ satisfying $q>d$. Hence, Lemma
\ref{lem.pol.AXn} \textbf{(b)} (applied to $n=d$) yields $q\in A\left[
X\right]  _{\leq d}$.

But $d\in\mathbb{N}$. Hence, $A\left[  X\right]  _{\leq d}$ is a term of the
union $\bigcup_{n\in\mathbb{N}}A\left[  X\right]  _{\leq n}$. Thus, $A\left[
X\right]  _{\leq d}\subseteq\bigcup_{n\in\mathbb{N}}A\left[  X\right]  _{\leq
n}$. Hence, $q\in A\left[  X\right]  _{\leq d}\subseteq\bigcup_{n\in
\mathbb{N}}A\left[  X\right]  _{\leq n}$.

Now, forget that we fixed $q$. We thus have shown that $q\in\bigcup
_{n\in\mathbb{N}}A\left[  X\right]  _{\leq n}$ for each $q\in A\left[
X\right]  $. In other words, $A\left[  X\right]  \subseteq\bigcup
_{n\in\mathbb{N}}A\left[  X\right]  _{\leq n}$. Combining this with the
inclusion $\bigcup_{n\in\mathbb{N}}A\left[  X\right]  _{\leq n}\subseteq
A\left[  X\right]  $ (which is obvious), we obtain $A\left[  X\right]
=\bigcup_{n\in\mathbb{N}}A\left[  X\right]  _{\leq n}$. This proves
Proposition \ref{prop.pol.union}.
\end{proof}
\end{verlong}

\begin{lemma}
\label{lem.pol.monic}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$. Let
$p\in A\left[  X\right]  $ be a monic polynomial of degree $n$. Then:

\textbf{(a)} We have $\left[  X^{n}\right]  p=1$.

\textbf{(b)} We have $\left[  X^{m}\right]  p=0$ for every $m\in\mathbb{N}$
satisfying $m>n$.

\textbf{(c)} We have $p\in A\left[  X\right]  _{\leq n}$.
\end{lemma}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.pol.monic}.]The polynomial $p\in A\left[  X\right]  $
is monic of degree $n$ if and only if it satisfies $\left[  X^{n}\right]  p=1$
and%
\[
\left(  \left[  X^{m}\right]  p=0\text{ for every }m\in\mathbb{N}\text{
satisfying }m>n\right)
\]
(by the definition of \textquotedblleft monic of degree $n$\textquotedblright%
). Since the polynomial $p$ is monic of degree $n$, we can therefore conclude
that it satisfies $\left[  X^{n}\right]  p=1$ and%
\[
\left(  \left[  X^{m}\right]  p=0\text{ for every }m\in\mathbb{N}\text{
satisfying }m>n\right)  .
\]


In particular, we have $\left[  X^{n}\right]  p=1$. This proves Lemma
\ref{lem.pol.monic} \textbf{(a)}.

Also, we have seen that $\left(  \left[  X^{m}\right]  p=0\text{ for every
}m\in\mathbb{N}\text{ satisfying }m>n\right)  $. This proves Lemma
\ref{lem.pol.monic} \textbf{(b)}.

\textbf{(c)} We have shown that $\left(  \left[  X^{m}\right]  p=0\text{ for
every }m\in\mathbb{N}\text{ satisfying }m>n\right)  $. Hence, Lemma
\ref{lem.pol.AXn} \textbf{(b)} (applied to $q=p$) yields $p\in A\left[
X\right]  _{\leq n}$. This proves Lemma \ref{lem.pol.monic} \textbf{(c)}.
\end{proof}
\end{verlong}

\begin{lemma}
\label{lem.pol.monic2}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$. Let
$p\in A\left[  X\right]  _{\leq n}$ be such that $\left[  X^{n}\right]  p=1$.
Then, $p$ is a monic polynomial of degree $n$.
\end{lemma}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.pol.monic2}.]We have $\left[  X^{m}\right]  p=0$ for
every $m\in\mathbb{N}$ satisfying $m>n$ (by Lemma \ref{lem.pol.AXn}
\textbf{(a)}, applied to $q=p$).

The polynomial $p\in A\left[  X\right]  $ is monic of degree $n$ if and only
if it satisfies $\left[  X^{n}\right]  p=1$ and%
\[
\left(  \left[  X^{m}\right]  p=0\text{ for every }m\in\mathbb{N}\text{
satisfying }m>n\right)
\]
(by the definition of \textquotedblleft monic of degree $n$\textquotedblright%
). Hence, the polynomial $p\in A\left[  X\right]  $ is monic of degree $n$
(since we know that it satisfies $\left[  X^{n}\right]  p=1$ and%
\[
\left(  \left[  X^{m}\right]  p=0\text{ for every }m\in\mathbb{N}\text{
satisfying }m>n\right)
\]
). This proves Lemma \ref{lem.pol.monic2}.
\end{proof}
\end{verlong}

\begin{lemma}
\label{lem.pol.nn-1}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$. Let
$q\in A\left[  X\right]  _{\leq n}$. Assume that $\left[  X^{n}\right]  q=0$.
Then, $q\in A\left[  X\right]  _{\leq n-1}$.
\end{lemma}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.pol.nn-1}.]We have $q\in A\left[  X\right]  _{\leq
n}$. Hence,%
\begin{equation}
\left[  X^{m}\right]  q=0\ \ \ \ \ \ \ \ \ \ \text{for every }m\in
\mathbb{N}\text{ satisfying }m>n \label{pf.lem.pol.nn-1.3}%
\end{equation}
(by Lemma \ref{lem.pol.AXn} \textbf{(a)}). Thus,%
\begin{equation}
\left[  X^{m}\right]  q=0\ \ \ \ \ \ \ \ \ \ \text{for every }m\in
\mathbb{N}\text{ satisfying }m>n-1 \label{pf.lem.pol.nn-1.4}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.pol.nn-1.4}):} Let $m\in\mathbb{N}$ be
such that $m>n-1$. We must prove that $\left[  X^{m}\right]  q=0$. We are in
one of the following two cases:
\par
\textit{Case 1:} We have $m=n$.
\par
\textit{Case 2:} We have $m\neq n$.
\par
Let us first consider Case 1. In this case, we have $m=n$. Hence, $\left[
X^{m}\right]  q=\left[  X^{n}\right]  q=0$. Thus, $\left[  X^{m}\right]  q=0$
is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $m\neq n$. But $m>n-1$, and
thus $m\geq\left(  n-1\right)  +1$ (since $m$ and $n-1$ are integers). Hence,
$m\geq\left(  n-1\right)  +1=n$. Combining this with $m\neq n$, we obtain
$m>n$. Hence, (\ref{pf.lem.pol.nn-1.3}) shows that $\left[  X^{m}\right]
q=0$. Thus, $\left[  X^{m}\right]  q=0$ is proven in Case 2.
\par
We have now proven $\left[  X^{m}\right]  q=0$ in each of the two Cases 1 and
2. Since these two Cases cover all possibilities, this shows that $\left[
X^{m}\right]  q=0$ always holds. Thus, (\ref{pf.lem.pol.nn-1.4}) is proven.}.
Hence, $q\in A\left[  X\right]  _{\leq n-1}$ (by Lemma \ref{lem.pol.AXn}
\textbf{(b)}, applied to $n-1$ instead of $n$). This proves Lemma
\ref{lem.pol.nn-1}.
\end{proof}
\end{verlong}

\begin{lemma}
\label{lem.pol.0}Let $A$ be a commutative ring. Then, $A\left[  X\right]
_{\leq-1}=\left\{  0\right\}  $.
\end{lemma}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.pol.0}.]Let $q\in A\left[  X\right]  _{\leq-1}$.
Thus,%
\begin{equation}
\left[  X^{m}\right]  q=0\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\mathbb{N}
\label{pf.lem.pol.0.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.pol.0.1}):} Let $m\in\mathbb{N}$.
Then, $m\geq0>-1$. Hence, Lemma \ref{lem.pol.AXn} \textbf{(a)} (applied to
$n=-1$) yields $\left[  X^{m}\right]  q=0$. This proves (\ref{pf.lem.pol.0.1}%
).}.

Recall that every polynomial $p\in A\left[  X\right]  $ satisfies
$p=\sum_{n\in\mathbb{N}}\left(  \left[  X^{n}\right]  p\right)  X^{n}$.
Applying this to $p=q$, we obtain%
\[
q=\sum_{n\in\mathbb{N}}\underbrace{\left(  \left[  X^{n}\right]  q\right)
}_{\substack{=0\\\text{(by (\ref{pf.lem.pol.0.1}), applied}\\\text{to
}m=n\text{)}}}X^{n}=\sum_{n\in\mathbb{N}}0X^{n}=0.
\]
Hence, $q\in\left\{  0\right\}  $.

Now, forget that we fixed $q$. We thus have shown that $q\in\left\{
0\right\}  $ for every $q\in A\left[  X\right]  _{\leq-1}$. In other words,
$A\left[  X\right]  _{\leq-1}\subseteq\left\{  0\right\}  $. Combining this
with $\left\{  0\right\}  \subseteq A\left[  X\right]  _{\leq-1}$ (which is
obvious, since $A\left[  X\right]  _{\leq-1}$ is an $A$-submodule of $A\left[
X\right]  $), we obtain $A\left[  X\right]  _{\leq-1}=\left\{  0\right\}  $.
This proves Lemma \ref{lem.pol.0}.
\end{proof}
\end{verlong}

\begin{lemma}
\label{lem.pol.sub}Let $A$ be a commutative ring. Let $g$ and $h$ be two
elements of $\mathbb{Z}$ such that $g\leq h$. Then, $A\left[  X\right]  _{\leq
g}\subseteq A\left[  X\right]  _{\leq h}$.
\end{lemma}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.pol.sub}.]Let $q\in A\left[  X\right]  _{\leq g}$.
Hence,%
\begin{equation}
\left[  X^{m}\right]  q=0\ \ \ \ \ \ \ \ \ \ \text{for every }m\in
\mathbb{N}\text{ satisfying }m>g \label{pf.lem.pol.sub.1}%
\end{equation}
(by Lemma \ref{lem.pol.AXn} \textbf{(a)}, applied to $n=g$). Thus,%
\begin{equation}
\left[  X^{m}\right]  q=0\ \ \ \ \ \ \ \ \ \ \text{for every }m\in
\mathbb{N}\text{ satisfying }m>h \label{pf.lem.pol.sub.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.pol.sub.2}):} Let $m\in\mathbb{N}$ be
such that $m>h$. Then, $m>h\geq g$ (since $g\leq h$). Hence,
(\ref{pf.lem.pol.sub.1}) shows that $\left[  X^{m}\right]  q=0$.Thus,
(\ref{pf.lem.pol.sub.2}) is proven.}. Hence, $q\in A\left[  X\right]  _{\leq
h}$ (by Lemma \ref{lem.pol.AXn} \textbf{(b)}, applied to $n=h$). This proves
Lemma \ref{lem.pol.sub}.
\end{proof}
\end{verlong}

Next, we show an easy fact about the product of a polynomial with a monic polynomial:

\begin{lemma}
\label{lem.pol.timesmon}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$.
Let $p\in A\left[  X\right]  $ be a monic polynomial of degree $n$. Let
$g\in\mathbb{N}$. Let $q\in A\left[  X\right]  _{\leq g}$. Then:

\textbf{(a)} We have $pq\in A\left[  X\right]  _{\leq g+n}$.

\textbf{(b)} We have $\left[  X^{g+n}\right]  \left(  pq\right)  =\left[
X^{g}\right]  q$.
\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.pol.timesmon}.]Let $m\in\mathbb{N}$ be such that
$m\geq g+n$. For every $k\in\left\{  0,1,\ldots,n-1\right\}  $, we have%
\begin{equation}
\left[  X^{m-k}\right]  q=0 \label{pf.lem.pol.timesmon.short.1}%
\end{equation}
\footnote{\textit{Proof of (\textit{\ref{pf.lem.pol.timesmon.short.1}}):} Let
$k\in\left\{  0,1,\ldots,n-1\right\}  $. Then, $\underbrace{m}_{\geq
g+n}-\underbrace{k}_{\leq n-1<n}>\left(  g+n\right)  -n=g\geq0$, so that
$m-k\in\mathbb{N}$. Also, $m-k>g$; hence, Lemma \ref{lem.pol.AXn} \textbf{(a)}
(applied to $g$ and $m-k$ instead of $n$ and $m$) yields $\left[
X^{m-k}\right]  q=0$. This proves (\ref{pf.lem.pol.timesmon.short.1}).}. For
every $k\in\left\{  n+1,n+2,\ldots,m\right\}  $, we have%
\begin{equation}
\left[  X^{k}\right]  p=0 \label{pf.lem.pol.timesmon.short.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.pol.timesmon.short.1}):} Let
$k\in\left\{  n+1,n+2,\ldots,m\right\}  $. Thus, $k>n$. Also, $k\in\mathbb{N}%
$. Hence, Lemma \ref{lem.pol.monic} \textbf{(b)} (applied to $k$ instead of
$m$) shows that $\left[  X^{k}\right]  p=0$. This proves
(\ref{pf.lem.pol.timesmon.short.2}).}. Now, $0\leq n$ (since $n\in\mathbb{N}%
$); also, from $g\in\mathbb{N}$, we obtain $n\leq g+n\leq m$. Proposition
\ref{prop.pol.+*} \textbf{(c)} (applied to $m$ instead of $n$) yields%
\begin{align}
\left[  X^{m}\right]  \left(  pq\right)   &  =\sum_{k=0}^{m}\left(  \left[
X^{k}\right]  p\right)  \cdot\left(  \left[  X^{m-k}\right]  q\right)
\nonumber\\
&  =\sum_{k=0}^{n-1}\left(  \left[  X^{k}\right]  p\right)  \cdot
\underbrace{\left(  \left[  X^{m-k}\right]  q\right)  }%
_{\substack{=0\\\text{(by (\ref{pf.lem.pol.timesmon.short.1}))}}}+\sum
_{k=n}^{m}\left(  \left[  X^{k}\right]  p\right)  \cdot\left(  \left[
X^{m-k}\right]  q\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }0\leq n\leq m\right) \nonumber\\
&  =\underbrace{\sum_{k=0}^{n-1}\left(  \left[  X^{k}\right]  p\right)
\cdot0}_{=0}+\sum_{k=n}^{m}\left(  \left[  X^{k}\right]  p\right)
\cdot\left(  \left[  X^{m-k}\right]  q\right) \nonumber\\
&  =\sum_{k=n}^{m}\left(  \left[  X^{k}\right]  p\right)  \cdot\left(  \left[
X^{m-k}\right]  q\right) \nonumber\\
&  =\underbrace{\left(  \left[  X^{n}\right]  p\right)  }%
_{\substack{=1\\\text{(by Lemma \ref{lem.pol.monic} \textbf{(a)})}}%
}\cdot\left(  \left[  X^{m-n}\right]  q\right)  +\sum_{k=n+1}^{m}%
\underbrace{\left(  \left[  X^{k}\right]  p\right)  }_{\substack{=0\\\text{(by
(\ref{pf.lem.pol.timesmon.short.2}))}}}\cdot\left(  \left[  X^{m-k}\right]
q\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\leq m\right) \nonumber\\
&  =\left[  X^{m-n}\right]  q+\underbrace{\sum_{k=n+1}^{m}0\cdot\left(
\left[  X^{m-k}\right]  q\right)  }_{=0}=\left[  X^{m-n}\right]  q.
\label{pf.lem.pol.timesmon.short.5}%
\end{align}


Now, forget that we fixed $m$. We thus have proven
(\ref{pf.lem.pol.timesmon.short.5}) for every $m\in\mathbb{N}$ satisfying
$m\geq g+n$.

\textbf{(a)} For every $m\in\mathbb{N}$ satisfying $m>g+n$, we have%
\begin{align*}
\left[  X^{m}\right]  \left(  pq\right)   &  =\left[  X^{m-n}\right]
q\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.pol.timesmon.short.5}%
)}\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Lemma \ref{lem.pol.AXn} \textbf{(a)}, applied to }m-n\text{ and }g\\
\text{instead of }m\text{ and }n\\
\text{(since }q\in A\left[  X\right]  _{\leq g}\text{ and }m-n>g\text{ (since
}m>g+n\text{))}%
\end{array}
\right)  .
\end{align*}
Hence, Lemma \ref{lem.pol.AXn} \textbf{(b)} (applied to $g+n$ and $pq$ instead
of $n$ and $q$) shows that $pq\in A\left[  X\right]  _{\leq g+n}$. This proves
Lemma \ref{lem.pol.timesmon} \textbf{(a)}.

\textbf{(b)} Applying (\ref{pf.lem.pol.timesmon.short.5}) to $m=g+n$, we find
$\left[  X^{g+n}\right]  \left(  pq\right)  =\left[  X^{g+n-n}\right]
q=\left[  X^{g}\right]  q$. This proves Lemma \ref{lem.pol.timesmon}
\textbf{(b)}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.pol.timesmon}.]For every $k\in\mathbb{N}$ satisfying
$k>n$, we have%
\begin{equation}
\left[  X^{k}\right]  p=0 \label{pf.lem.pol.timesmon.1}%
\end{equation}
(by Lemma \ref{lem.pol.monic} \textbf{(b)}, applied to $m=k$).

\textbf{(a)} Let $m\in\mathbb{N}$ be such that $m>g+n$. Then, $m-n>g$ (since
$m>g+n$), so that $m-n>g\geq0$ (since $g\in\mathbb{N}$) and thus $m>n$. Hence,
$n<m$.

For every $k\in\left\{  0,1,\ldots,n\right\}  $, we have%
\begin{equation}
\left[  X^{m-k}\right]  q=0 \label{pf.lem.pol.timesmon.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.pol.timesmon.2}):} Let $k\in\left\{
0,1,\ldots,n\right\}  $. Thus, $0\leq k\leq n$. Now, $\underbrace{m}%
_{>n}-\underbrace{k}_{\leq n}>n-n=0$, so that $m-k\in\mathbb{N}$. Furthermore,
$\underbrace{m}_{>g+n}-\underbrace{k}_{\leq n}>\left(  g+n\right)  -n=g$.
Hence, Lemma \ref{lem.pol.AXn} \textbf{(a)} (applied to $g$ and $m-k$ instead
of $n$ and $m$) yields $\left[  X^{m-k}\right]  q=0$. This proves
(\ref{pf.lem.pol.timesmon.2}).}. For every $k\in\left\{  n+1,n+2,\ldots
,m\right\}  $, we have%
\begin{equation}
\left[  X^{k}\right]  p=0 \label{pf.lem.pol.timesmon.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.pol.timesmon.3}):} Let $k\in\left\{
n+1,n+2,\ldots,m\right\}  $. Thus, $k>n$. Also, $k\in\mathbb{N}$. Hence,
(\ref{pf.lem.pol.timesmon.1}) shows that $\left[  X^{k}\right]  p=0$. This
proves (\ref{pf.lem.pol.timesmon.3}).}.

Now, Proposition \ref{prop.pol.+*} \textbf{(c)} (applied to $m$ instead of
$n$) yields%
\begin{align*}
\left[  X^{m}\right]  \left(  pq\right)   &  =\sum_{k=0}^{m}\left(  \left[
X^{k}\right]  p\right)  \cdot\left(  \left[  X^{m-k}\right]  q\right) \\
&  =\sum_{k=0}^{n}\left(  \left[  X^{k}\right]  p\right)  \cdot
\underbrace{\left(  \left[  X^{m-k}\right]  q\right)  }%
_{\substack{=0\\\text{(by (\ref{pf.lem.pol.timesmon.2}))}}}+\sum_{k=n+1}%
^{m}\underbrace{\left(  \left[  X^{k}\right]  p\right)  }%
_{\substack{=0\\\text{(by (\ref{pf.lem.pol.timesmon.3}))}}}\cdot\left(
\left[  X^{m-k}\right]  q\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }0\leq n<m\right) \\
&  =\underbrace{\sum_{k=0}^{n}\left(  \left[  X^{k}\right]  p\right)  \cdot
0}_{=0}+\underbrace{\sum_{k=n+1}^{m}0\cdot\left(  \left[  X^{m-k}\right]
q\right)  }_{=0}=0+0=0.
\end{align*}


Now, forget that we fixed $m$. We thus have shown that $\left[  X^{m}\right]
\left(  pq\right)  =0$ for every $m\in\mathbb{N}$ satisfying $m>g+n$. Hence,
Lemma \ref{lem.pol.AXn} \textbf{(b)} (applied to $g+n$ and $pq$ instead of $n$
and $q$) shows that $pq\in A\left[  X\right]  _{\leq g+n}$. This proves Lemma
\ref{lem.pol.timesmon} \textbf{(a)}.

\textbf{(b)} We have $g+n\in\mathbb{N}$ (since $g\in\mathbb{N}$ and
$n\in\mathbb{N}$) and $g+n\geq n$ (since $g\geq0$ (since $g\in\mathbb{N}$))
and $n\geq0$ (since $n\in\mathbb{N}$). Thus, $g+n\geq n\geq0$, so that $0\leq
n\leq g+n$. For every $k\in\left\{  0,1,\ldots,n-1\right\}  $, we have%
\begin{equation}
\left[  X^{g+n-k}\right]  q=0 \label{pf.lem.pol.timesmon.4}%
\end{equation}
\footnote{\textit{Proof of (\textit{\ref{pf.lem.pol.timesmon.4}}):} Let
$k\in\left\{  0,1,\ldots,n-1\right\}  $. Thus, $0\leq k\leq n-1$. Now,
$g+n-\underbrace{k}_{\leq n-1<n}>g+n-n=g\geq0$, so that $g+n-k\in\mathbb{N}$.
Furthermore, $g+n-k>g$. Hence, Lemma \ref{lem.pol.AXn} \textbf{(a)} (applied
to $g$ and $g+n-k$ instead of $n$ and $m$) yields $\left[  X^{g+n-k}\right]
q=0$. This proves (\ref{pf.lem.pol.timesmon.4}).}. For every $k\in\left\{
n+1,n+2,\ldots,g+n\right\}  $, we have%
\begin{equation}
\left[  X^{k}\right]  p=0 \label{pf.lem.pol.timesmon.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.pol.timesmon.5}):} Let $k\in\left\{
n+1,n+2,\ldots,g+n\right\}  $. Thus, $k>n$. Also, $k\in\mathbb{N}$. Hence,
(\ref{pf.lem.pol.timesmon.1}) shows that $\left[  X^{k}\right]  p=0$. This
proves (\ref{pf.lem.pol.timesmon.5}).}.

Now, Proposition \ref{prop.pol.+*} \textbf{(c)} (applied to $g+n$ instead of
$n$) yields%
\begin{align*}
\left[  X^{g+n}\right]  \left(  pq\right)   &  =\sum_{k=0}^{g+n}\left(
\left[  X^{k}\right]  p\right)  \cdot\left(  \left[  X^{g+n-k}\right]
q\right) \\
&  =\sum_{k=0}^{n-1}\left(  \left[  X^{k}\right]  p\right)  \cdot
\underbrace{\left(  \left[  X^{g+n-k}\right]  q\right)  }%
_{\substack{=0\\\text{(by (\ref{pf.lem.pol.timesmon.4}))}}}+\sum_{k=n}%
^{g+n}\left(  \left[  X^{k}\right]  p\right)  \cdot\left(  \left[
X^{g+n-k}\right]  q\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }0\leq n\leq g+n\right) \\
&  =\underbrace{\sum_{k=0}^{n-1}\left(  \left[  X^{k}\right]  p\right)
\cdot0}_{=0}+\sum_{k=n}^{g+n}\left(  \left[  X^{k}\right]  p\right)
\cdot\left(  \left[  X^{g+n-k}\right]  q\right) \\
&  =\sum_{k=n}^{g+n}\left(  \left[  X^{k}\right]  p\right)  \cdot\left(
\left[  X^{g+n-k}\right]  q\right) \\
&  =\underbrace{\left(  \left[  X^{n}\right]  p\right)  }%
_{\substack{=1\\\text{(by Lemma \ref{lem.pol.monic} \textbf{(a)})}}%
}\cdot\underbrace{\left(  \left[  X^{g+n-n}\right]  q\right)  }%
_{\substack{=\left[  X^{g}\right]  q\\\text{(since }g+n-n=g\text{)}}%
}+\sum_{k=n+1}^{g+n}\underbrace{\left(  \left[  X^{k}\right]  p\right)
}_{\substack{=0\\\text{(by (\ref{pf.lem.pol.timesmon.5}))}}}\cdot\left(
\left[  X^{g+n-k}\right]  q\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }k=n\text{ from the sum}\\
\text{(since }n\leq g+n\text{)}%
\end{array}
\right) \\
&  =\left[  X^{g}\right]  q+\underbrace{\sum_{k=n+1}^{g+n}0\cdot\left(
\left[  X^{g+n-k}\right]  q\right)  }_{=0}=\left[  X^{g}\right]  q.
\end{align*}
This proves Lemma \ref{lem.pol.timesmon} \textbf{(b)}.
\end{proof}
\end{verlong}

\subsection{Monic polynomials are regular}

The next proposition says that if we multiply a monic polynomial $p$ of some
degree $n$ with a polynomial $q$ and the result turns out to have degree $<n$
(that is, formally speaking: $pq\in A\left[  X\right]  _{\leq n-1}$), then $q$
must have been $0$ to begin with. This will help us prove that monic
polynomials are regular.

\begin{proposition}
\label{prop.pol.q=0}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$. Let
$p\in A\left[  X\right]  $ be a monic polynomial of degree $n$. Let $q\in
A\left[  X\right]  $ be such that $pq\in A\left[  X\right]  _{\leq n-1}$.
Then, $q=0$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.pol.q=0}.]Renaming the index $n$ as $d$ in
Proposition \ref{prop.pol.union}, we obtain $A\left[  X\right]  =\bigcup
_{d\in\mathbb{N}}A\left[  X\right]  _{\leq d}$.

Now, $q\in A\left[  X\right]  =\bigcup_{d\in\mathbb{N}}A\left[  X\right]
_{\leq d}$. Hence, there exists some $d\in\mathbb{N}$ such that $q\in A\left[
X\right]  _{\leq d}$. Consider this $d$.

We shall now show that%
\begin{equation}
q\in A\left[  X\right]  _{\leq d-i}\ \ \ \ \ \ \ \ \ \ \text{for every }%
i\in\left\{  0,1,\ldots,d+1\right\}  . \label{pf.prop.pol.q=0.main}%
\end{equation}


\textit{Proof of (\ref{pf.prop.pol.q=0.main}):} We shall prove
(\ref{pf.prop.pol.q=0.main}) by induction over $i$:

\begin{vershort}
\textit{Induction base:} We have $q\in A\left[  X\right]  _{\leq d}$. In other
words, (\ref{pf.prop.pol.q=0.main}) holds for $i=0$.
\end{vershort}

\begin{verlong}
\textit{Induction base:} We have $q\in A\left[  X\right]  _{\leq d}=A\left[
X\right]  _{\leq d-0}$ (since $d=d-0$). In other words,
(\ref{pf.prop.pol.q=0.main}) holds for $i=0$. This completes the induction base.
\end{verlong}

\textit{Induction step:} Let $j\in\left\{  0,1,\ldots,d+1\right\}  $ be
positive. Assume that (\ref{pf.prop.pol.q=0.main}) holds for $i=j-1$. We now
must prove that (\ref{pf.prop.pol.q=0.main}) holds for $i=j$.

We have assumed that (\ref{pf.prop.pol.q=0.main}) holds for $i=j-1$. In other
words, we have $q\in A\left[  X\right]  _{\leq d-\left(  j-1\right)  }$.

\begin{vershort}
Set $g=d-\left(  j-1\right)  $. Then, $g=d-\left(  j-1\right)  =\left(
d+1\right)  -j\geq0$ (since $j\leq d+1$), so that $g\in\mathbb{N}$.
\end{vershort}

\begin{verlong}
We have $j\in\left\{  0,1,\ldots,d+1\right\}  $, thus $0\leq j\leq d+1$. From
$j\leq d+1$, we obtain $j-1\leq d$, so that $d-\left(  j-1\right)  \geq0$. Set
$g=d-\left(  j-1\right)  $. Thus, $g=d-\left(  j-1\right)  \geq0$, so that
$g\in\mathbb{N}$.
\end{verlong}

Lemma \ref{lem.pol.timesmon} \textbf{(b)} yields $\left[  X^{g+n}\right]
\left(  pq\right)  =\left[  X^{g}\right]  q$. But $\underbrace{g}_{\geq
0}+n\geq n>n-1$. Hence, Lemma \ref{lem.pol.AXn} \textbf{(a)} (applied to $pq$,
$n-1$ and $g+n$ instead of $q$, $n$ and $m$) yields $\left[  X^{g+n}\right]
\left(  pq\right)  =0$ (since $pq\in A\left[  X\right]  _{\leq n-1}$).
Comparing this with $\left[  X^{g+n}\right]  \left(  pq\right)  =\left[
X^{g}\right]  q$, we obtain $\left[  X^{g}\right]  q=0$.

But $q\in A\left[  X\right]  _{\leq d-\left(  j-1\right)  }=A\left[  X\right]
_{\leq g}$ (since $d-\left(  j-1\right)  =g$). Lemma \ref{lem.pol.nn-1}
(applied to $g$ instead of $n$) thus shows that $q\in A\left[  X\right]
_{\leq g-1}$ (since $\left[  X^{g}\right]  q=0$). Since $\underbrace{g}%
_{=d-\left(  j-1\right)  }-1=d-\left(  j-1\right)  -1=d-j$, this rewrites as
$q\in A\left[  X\right]  _{\leq d-j}$. In other words,
(\ref{pf.prop.pol.q=0.main}) holds for $i=j$. This completes the induction
step. Thus, the induction proof of (\ref{pf.prop.pol.q=0.main}) is complete.

Now, applying (\ref{pf.prop.pol.q=0.main}) to $i=d+1$, we obtain $q\in
A\left[  X\right]  _{\leq d-\left(  d+1\right)  }=A\left[  X\right]  _{\leq
-1}=\left\{  0\right\}  $ (by Lemma \ref{lem.pol.0}). In other words, $q=0$.
This proves Proposition \ref{prop.pol.q=0}.
\end{proof}

\begin{corollary}
\label{cor.pol.u=0}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$. Let
$p\in A\left[  X\right]  $ be a monic polynomial of degree $n$. Let $u\in
A\left[  X\right]  _{\leq n-1}$ be such that $p\mid u$ (in the ring $A\left[
X\right]  $). Then, $u=0$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.pol.u=0}.]We have $p\mid u$. In other words,
there exists some $q\in A\left[  X\right]  $ such that $u=pq$. Consider this
$q$. We have $pq=u\in A\left[  X\right]  _{\leq n-1}$. Thus, Proposition
\ref{prop.pol.q=0} shows that $q=0$. Hence, $u=p\underbrace{q}_{=0}=0$. This
proves Corollary \ref{cor.pol.u=0}.
\end{proof}

\begin{corollary}
\label{cor.pol.monic-reg}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$.
Let $p\in A\left[  X\right]  $ be a monic polynomial of degree $n$. Then, the
element $p$ of $A\left[  X\right]  $ is regular.
\end{corollary}

\begin{vershort}
\begin{proof}
[Proof of Corollary \ref{cor.pol.monic-reg}.]We must prove that $p$ is
regular. In other words, we must prove that every $x\in A\left[  X\right]  $
satisfying $px=0$ satisfies $x=0$ (because this is what it means for $p$ to be regular).

So let $x\in A\left[  X\right]  $ satisfy $px=0$. Then, $px=0\in A\left[
X\right]  _{\leq n-1}$. Hence, Proposition \ref{prop.pol.q=0} (applied to
$q=x$) yields $x=0$. This completes our proof of Corollary
\ref{cor.pol.monic-reg}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Corollary \ref{cor.pol.monic-reg}.]Let $x\in A\left[  X\right]  $ be
such that $px=0$. Recall that $A\left[  X\right]  _{\leq n-1}$ is an
$A$-submodule of $A\left[  X\right]  $; hence, $0\in A\left[  X\right]  _{\leq
n-1}$. Now, $px=0\in A\left[  X\right]  _{\leq n-1}$. Hence, Proposition
\ref{prop.pol.q=0} (applied to $q=x$) yields $x=0$.

Now, forget that we fixed $x$. We thus have shown that every $x\in A\left[
X\right]  $ satisfying $px=0$ satisfies $x=0$.

The element $p$ of $A\left[  X\right]  $ is regular if and only if every $x\in
A\left[  X\right]  $ satisfying $px=0$ satisfies $x=0$ (by the definition of
\textquotedblleft regular\textquotedblright). Hence, the element $p$ of
$A\left[  X\right]  $ is regular (since every $x\in A\left[  X\right]  $
satisfying $px=0$ satisfies $x=0$). This proves Corollary
\ref{cor.pol.monic-reg}.
\end{proof}
\end{verlong}

\subsection{Division with remainder}

Now, we shall state the most important result in this section: division with
remainder by a monic polynomial:

\begin{theorem}
\label{thm.pol.quorem}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$. Let
$p\in A\left[  X\right]  $ be a monic polynomial of degree $n$. Let $f\in
A\left[  X\right]  $. Then, there exists a unique pair $\left(  q,r\right)
\in A\left[  X\right]  \times A\left[  X\right]  _{\leq n-1}$ such that
$f=qp+r$.
\end{theorem}

\begin{remark}
Let $A$, $n$, $p$ and $f$ be as in Theorem \ref{thm.pol.quorem}. Theorem
\ref{thm.pol.quorem} claims that there exists a unique pair $\left(
q,r\right)  \in A\left[  X\right]  \times A\left[  X\right]  _{\leq n-1}$ such
that $f=qp+r$. The two entries $q$ and $r$ of this pair $\left(  q,r\right)  $
are called the \textit{quotient} and the \textit{remainder} (respectively)
\textit{obtained when dividing }$f$ \textit{by }$p$. Note that the remainder
$r$ belongs to $A\left[  X\right]  _{\leq n-1}$ (since $\left(  q,r\right)
\in A\left[  X\right]  \times A\left[  X\right]  _{\leq n-1}$).
\end{remark}

We shall prove the existence and the uniqueness parts of Theorem
\ref{thm.pol.quorem} separately, beginning with the uniqueness part:

\begin{lemma}
\label{lem.pol.quorem.uni}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$.
Let $p\in A\left[  X\right]  $ be a monic polynomial of degree $n$. Let $f\in
A\left[  X\right]  $. Then, there exists \textbf{at most one} pair $\left(
q,r\right)  \in A\left[  X\right]  \times A\left[  X\right]  _{\leq n-1}$ such
that $f=qp+r$.
\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.pol.quorem.uni}.]Let $\left(  q_{1},r_{1}\right)  $
and $\left(  q_{2},r_{2}\right)  $ be two pairs $\left(  q,r\right)  \in
A\left[  X\right]  \times A\left[  X\right]  _{\leq n-1}$ such that $f=qp+r$.
We shall prove that $\left(  q_{1},r_{1}\right)  =\left(  q_{2},r_{2}\right)
$.

We know that $\left(  q_{1},r_{1}\right)  $ is a pair $\left(  q,r\right)  \in
A\left[  X\right]  \times A\left[  X\right]  _{\leq n-1}$ such that $f=qp+r$.
In other words, $\left(  q_{1},r_{1}\right)  $ is a pair in $A\left[
X\right]  \times A\left[  X\right]  _{\leq n-1}$ such that $f=q_{1}p+r_{1}$.
Similarly, $\left(  q_{2},r_{2}\right)  $ is a pair in $A\left[  X\right]
\times A\left[  X\right]  _{\leq n-1}$ such that $f=q_{2}p+r_{2}$.

We have $r_{1}\in A\left[  X\right]  _{\leq n-1}$ (since $\left(  q_{1}%
,r_{1}\right)  \in A\left[  X\right]  \times A\left[  X\right]  _{\leq n-1}$)
and $r_{2}\in A\left[  X\right]  _{\leq n-1}$ (similarly). Thus, $r_{2}%
-r_{1}\in A\left[  X\right]  _{\leq n-1}$. Now, $q_{1}p+r_{1}=f=q_{2}p+r_{2}$.
Hence, $q_{1}p-q_{2}p=r_{2}-r_{1}\in A\left[  X\right]  _{\leq n-1}$. Thus,
$p\left(  q_{1}-q_{2}\right)  =\left(  q_{1}-q_{2}\right)  p=q_{1}p-q_{2}p\in
A\left[  X\right]  _{\leq n-1}$. Proposition \ref{prop.pol.q=0} (applied to
$q=q_{1}-q_{2}$) thus shows that $q_{1}-q_{2}=0$. In other words, $q_{1}%
=q_{2}$. Hence, $r_{2}-r_{1}=\underbrace{q_{1}}_{=q_{2}}p-q_{2}p=q_{2}%
p-q_{2}p=0$, so that $r_{1}=r_{2}$. Now, $\left(  \underbrace{q_{1}}_{=q_{2}%
},\underbrace{r_{1}}_{=r_{2}}\right)  =\left(  q_{2},r_{2}\right)  $.

We thus have shown that if $\left(  q_{1},r_{1}\right)  $ and $\left(
q_{2},r_{2}\right)  $ are two pairs $\left(  q,r\right)  \in A\left[
X\right]  \times A\left[  X\right]  _{\leq n-1}$ such that $f=qp+r$, then
$\left(  q_{1},r_{1}\right)  =\left(  q_{2},r_{2}\right)  $. In other words,
there exists \textbf{at most one} pair $\left(  q,r\right)  \in A\left[
X\right]  \times A\left[  X\right]  _{\leq n-1}$ such that $f=qp+r$. This
proves Lemma \ref{lem.pol.quorem.uni}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.pol.quorem.uni}.]Let $\left(  q_{1},r_{1}\right)  $
and $\left(  q_{2},r_{2}\right)  $ be two pairs $\left(  q,r\right)  \in
A\left[  X\right]  \times A\left[  X\right]  _{\leq n-1}$ such that $f=qp+r$.
We shall prove that $\left(  q_{1},r_{1}\right)  =\left(  q_{2},r_{2}\right)
$.

We know that $\left(  q_{1},r_{1}\right)  $ is a pair $\left(  q,r\right)  \in
A\left[  X\right]  \times A\left[  X\right]  _{\leq n-1}$ such that $f=qp+r$.
In other words, $\left(  q_{1},r_{1}\right)  $ is a pair in $A\left[
X\right]  \times A\left[  X\right]  _{\leq n-1}$ such that $f=q_{1}p+r_{1}$.
We have $\left(  q_{1},r_{1}\right)  \in A\left[  X\right]  \times A\left[
X\right]  _{\leq n-1}$. In other words, $q_{1}\in A\left[  X\right]  $ and
$r_{1}\in A\left[  X\right]  _{\leq n-1}$.

We know that $\left(  q_{2},r_{2}\right)  $ is a pair $\left(  q,r\right)  \in
A\left[  X\right]  \times A\left[  X\right]  _{\leq n-1}$ such that $f=qp+r$.
In other words, $\left(  q_{2},r_{2}\right)  $ is a pair in $A\left[
X\right]  \times A\left[  X\right]  _{\leq n-1}$ such that $f=q_{2}p+r_{2}$.
We have $\left(  q_{2},r_{2}\right)  \in A\left[  X\right]  \times A\left[
X\right]  _{\leq n-1}$. In other words, $q_{2}\in A\left[  X\right]  $ and
$r_{2}\in A\left[  X\right]  _{\leq n-1}$.

From $r_{2}\in A\left[  X\right]  _{\leq n-1}$ and $r_{1}\in A\left[
X\right]  _{\leq n-1}$, we obtain $r_{2}-r_{1}\in A\left[  X\right]  _{\leq
n-1}$ (since $A\left[  X\right]  _{\leq n-1}$ is an $A$-submodule of $A\left[
X\right]  $). Now, $q_{1}p+r_{1}=f=q_{2}p+r_{2}$. Subtracting $q_{2}p+r_{1}$
from this equality, we obtain $q_{1}p-q_{2}p=r_{2}-r_{1}\in A\left[  X\right]
_{\leq n-1}$. Thus, $p\left(  q_{1}-q_{2}\right)  =\left(  q_{1}-q_{2}\right)
p=q_{1}p-q_{2}p\in A\left[  X\right]  _{\leq n-1}$. Proposition
\ref{prop.pol.q=0} (applied to $q=q_{1}-q_{2}$) thus shows that $q_{1}%
-q_{2}=0$. In other words, $q_{1}=q_{2}$. Now, $q_{2}p+r_{2}%
=f=\underbrace{q_{1}}_{=q_{2}}p+r_{1}=q_{2}p+r_{1}$. Subtracting $q_{2}p$ from
this equality, we obtain $r_{2}=r_{1}$, so that $r_{1}=r_{2}$. Now, $\left(
\underbrace{q_{1}}_{=q_{2}},\underbrace{r_{1}}_{=r_{2}}\right)  =\left(
q_{2},r_{2}\right)  $.

Now, let us forget that we fixed $\left(  q_{1},r_{1}\right)  $ and $\left(
q_{2},r_{2}\right)  $. We thus have shown that if $\left(  q_{1},r_{1}\right)
$ and $\left(  q_{2},r_{2}\right)  $ are two pairs $\left(  q,r\right)  \in
A\left[  X\right]  \times A\left[  X\right]  _{\leq n-1}$ such that $f=qp+r$,
then $\left(  q_{1},r_{1}\right)  =\left(  q_{2},r_{2}\right)  $. In other
words, there exists \textbf{at most one} pair $\left(  q,r\right)  \in
A\left[  X\right]  \times A\left[  X\right]  _{\leq n-1}$ such that $f=qp+r$.
This proves Lemma \ref{lem.pol.quorem.uni}.
\end{proof}
\end{verlong}

Now, let us state the existence part of Theorem \ref{thm.pol.quorem};
actually, let us make a slightly stronger claim:

\begin{lemma}
\label{lem.pol.quorem.exi}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$.
Let $p\in A\left[  X\right]  $ be a monic polynomial of degree $n$. Let
$d\in\left\{  -1,0,1,\ldots\right\}  $. Let $f\in A\left[  X\right]  _{\leq
d}$. Then, there exists \textbf{at least one} pair $\left(  q,r\right)  \in
A\left[  X\right]  _{\leq d-n}\times A\left[  X\right]  _{\leq n-1}$ such that
$f=qp+r$.
\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.pol.quorem.exi}.]We shall prove Lemma
\ref{lem.pol.quorem.exi} by induction on $d$:

\textit{Induction base:} Lemma \ref{lem.pol.quorem.exi} holds in the case when
$d=-1$ (because in this case, we have $f\in A\left[  X\right]  _{\leq
d}=A\left[  X\right]  _{\leq-1}=\left\{  0\right\}  $ and therefore $f=0$, so
that we can take $\left(  q,r\right)  =\left(  0,0\right)  $).

\textit{Induction step:} Let $D\in\mathbb{N}$. Assume that Lemma
\ref{lem.pol.quorem.exi} holds in the case when $d=D-1$. We must prove that
Lemma \ref{lem.pol.quorem.exi} holds in the case when $d=D$.

Let $A$, $n$ and $p$ be as in Lemma \ref{lem.pol.quorem.exi}. Let $f\in
A\left[  X\right]  _{\leq D}$. We are going to show the following claim:

\begin{statement}
\textit{Claim 1:} There exists \textbf{at least one} pair $\left(  q,r\right)
\in A\left[  X\right]  _{\leq D-n}\times A\left[  X\right]  _{\leq n-1}$ such
that $f=qp+r$.
\end{statement}

\textit{Proof of Claim 1:} If $D\leq n-1$, then $A\left[  X\right]  _{\leq
D}\subseteq A\left[  X\right]  _{\leq n-1}$ (by Lemma \ref{lem.pol.sub}).
Hence, if $D\leq n-1$, then Claim 1 holds (because we can just set $\left(
q,r\right)  =\left(  0,f\right)  $, using the fact that $f\in A\left[
X\right]  _{\leq D}\subseteq A\left[  X\right]  _{\leq n-1}$). Hence, we WLOG
assume that we don't have $D\leq n-1$. Thus, $D\geq n$ (since $D$ and $n$ are
integers), so that $D-n\in\mathbb{N}$.

Define an element $\alpha\in A$ by $\alpha=\left[  X^{D}\right]  f$. Then, the
polynomial $\alpha X^{D-n}\in A\left[  X\right]  $ is well-defined (since
$D-n\in\mathbb{N}$) and belongs to $A\left[  X\right]  _{\leq D-n}$. Hence,
Lemma \ref{lem.pol.timesmon} \textbf{(a)} (applied to $D-n$ and $\alpha
X^{D-n}$ instead of $g$ and $q$) yields $p\alpha X^{D-n}\in A\left[  X\right]
_{\leq\left(  D-n\right)  +n}=A\left[  X\right]  _{\leq D}$. Moreover, Lemma
\ref{lem.pol.timesmon} \textbf{(b)} (applied to $D-n$ and $\alpha X^{D-n}$
instead of $g$ and $q$) yields
\[
\left[  X^{\left(  D-n\right)  +n}\right]  \left(  p\alpha X^{D-n}\right)
=\left[  X^{D-n}\right]  \left(  \alpha X^{D-n}\right)  =\alpha.
\]
Since $\left(  D-n\right)  +n=D$, this rewrites as $\left[  X^{D}\right]
\left(  p\alpha X^{D-n}\right)  =\alpha$.

Both $f$ and $p\alpha X^{D-n}$ belong to $A\left[  X\right]  _{\leq D}$.
Hence, the difference $f-p\alpha X^{D-n}$ also belongs to $A\left[  X\right]
_{\leq D}$ (since $A\left[  X\right]  _{\leq D}$ is an $A$-submodule of
$A\left[  X\right]  $). In other words, $f-p\alpha X^{D-n}\in A\left[
X\right]  _{\leq D}$. Furthermore,%
\[
\left[  X^{D}\right]  \left(  f-p\alpha X^{D-n}\right)  =\underbrace{\left[
X^{D}\right]  f}_{=\alpha}-\underbrace{\left[  X^{D}\right]  \left(  p\alpha
X^{D-n}\right)  }_{=\alpha}=\alpha-\alpha=0.
\]
Hence, Lemma \ref{lem.pol.nn-1} (applied to $D$ and $f-p\alpha X^{D-n}$
instead of $n$ and $q$) shows that $f-p\alpha X^{D-n}\in A\left[  X\right]
_{\leq D-1}$. Therefore, we can apply Lemma \ref{lem.pol.quorem.exi} to $D-1$
and $f-p\alpha X^{D-n}$ instead of $d$ and $f$ (since we have assumed that
Lemma \ref{lem.pol.quorem.exi} holds in the case when $d=D-1$). We thus obtain
that there exists \textbf{at least one} pair $\left(  q,r\right)  \in A\left[
X\right]  _{\leq\left(  D-1\right)  -n}\times A\left[  X\right]  _{\leq n-1}$
such that $f-p\alpha X^{D-n}=qp+r$. Denote this pair $\left(  q,r\right)  $ by
$\left(  \widetilde{q},\widetilde{r}\right)  $. Thus, $\left(  \widetilde{q}%
,\widetilde{r}\right)  $ is a pair in $A\left[  X\right]  _{\leq\left(
D-1\right)  -n}\times A\left[  X\right]  _{\leq n-1}$ satisfying $f-p\alpha
X^{D-n}=\widetilde{q}p+\widetilde{r}$.

We have $\left(  \widetilde{q},\widetilde{r}\right)  \in A\left[  X\right]
_{\leq\left(  D-1\right)  -n}\times A\left[  X\right]  _{\leq n-1}$; in other
words, $\widetilde{q}\in A\left[  X\right]  _{\leq\left(  D-1\right)  -n}$ and
$\widetilde{r}\in A\left[  X\right]  _{\leq n-1}$. Now, $\widetilde{q}\in
A\left[  X\right]  _{\leq\left(  D-1\right)  -n}\subseteq A\left[  X\right]
_{\leq D-n}$ (since $\left(  D-1\right)  -n\leq D-n$). Hence, both
$\widetilde{q}$ and $\alpha X^{D-n}$ belong to $A\left[  X\right]  _{\leq
D-n}$ (since we know that $\alpha X^{D-n}\in A\left[  X\right]  _{\leq D-n}$).
Thus, the sum $\widetilde{q}+\alpha X^{D-n}$ also belongs to $A\left[
X\right]  _{\leq D-n}$ (since $A\left[  X\right]  _{\leq D-n}$ is an
$A$-submodule of $A\left[  X\right]  $). In other words, $\widetilde{q}+\alpha
X^{D-n}\in A\left[  X\right]  _{\leq D-n}$. Combining this with $\widetilde{r}%
\in A\left[  X\right]  _{\leq n-1}$, we obtain $\left(  \widetilde{q}+\alpha
X^{D-n},\widetilde{r}\right)  \in A\left[  X\right]  _{\leq D-n}\times
A\left[  X\right]  _{\leq n-1}$. Furthermore, from $f-p\alpha X^{D-n}%
=\widetilde{q}p+\widetilde{r}$, we obtain%
\[
f=\underbrace{p\alpha X^{D-n}+\widetilde{q}p}_{=\left(  \widetilde{q}+\alpha
X^{D-n}\right)  p}+\widetilde{r}=\left(  \widetilde{q}+\alpha X^{D-n}\right)
p+\widetilde{r}.
\]
Thus, there exists \textbf{at least one} pair $\left(  q,r\right)  \in
A\left[  X\right]  _{\leq D-n}\times A\left[  X\right]  _{\leq n-1}$ such that
$f=qp+r$ (namely, $\left(  q,r\right)  =\left(  \widetilde{q}+\alpha
X^{D-n},\widetilde{r}\right)  $). This proves Claim 1.

Now, forget that we have fixed $A$, $n$, $p$ and $f$. We thus have shown that
if $A$, $n$ and $p$ are as in Lemma \ref{lem.pol.quorem.exi}, and if $f\in
A\left[  X\right]  _{\leq D}$, then Claim 1 holds. In other words, Lemma
\ref{lem.pol.quorem.exi} holds in the case when $d=D$. This completes the
induction step. The induction proof of Lemma \ref{lem.pol.quorem.exi} is thus complete.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.pol.quorem.exi}.]We shall prove Lemma
\ref{lem.pol.quorem.exi} by induction on $d$:

\textit{Induction base:} Lemma \ref{lem.pol.quorem.exi} holds in the case when
$d=-1$\ \ \ \ \footnote{\textit{Proof.} Let $n$, $p$, $d$ and $f$ be as in
Lemma \ref{lem.pol.quorem.exi}. Assume that $d=-1$. We must prove that Lemma
\ref{lem.pol.quorem.exi} holds in this case.
\par
We have $f\in A\left[  X\right]  _{\leq d}=A\left[  X\right]  _{\leq-1}$
(since $d=-1$), so that $f\in A\left[  X\right]  _{\leq-1}=\left\{  0\right\}
$ (by Lemma \ref{lem.pol.0}). In other words, $f=0$.
\par
Now, $0\in A\left[  X\right]  _{\leq n-d}$ (since $A\left[  X\right]  _{\leq
n-d}$ is an $A$-submodule of $A\left[  X\right]  $) and $0\in A\left[
X\right]  _{\leq n-1}$ (since $A\left[  X\right]  _{\leq n-1}$ is an
$A$-submodule of $A\left[  X\right]  $). Hence, $\left(  0,0\right)  \in
A\left[  X\right]  _{\leq n-d}\times A\left[  X\right]  _{\leq n-1}$.
Moreover, $0p+0=0=f$, so that $f=0p+0$. Hence, there exists \textbf{at least
one} pair $\left(  q,r\right)  \in A\left[  X\right]  _{\leq n-d}\times
A\left[  X\right]  _{\leq n-1}$ such that $f=qp+r$ (namely, $\left(
q,p\right)  =\left(  0,0\right)  $). In other words, Lemma
\ref{lem.pol.quorem.exi} holds. Qed.}. This completes the induction base.

\textit{Induction step:} Let $D\in\left\{  -1,0,1,\ldots\right\}  $ be such
that $D>-1$. Assume that Lemma \ref{lem.pol.quorem.exi} holds in the case when
$d=D-1$. We must prove that Lemma \ref{lem.pol.quorem.exi} holds in the case
when $d=D$.

We have $D>-1$. Thus, $D\geq0$ (since $D$ is an integer), so that
$D\in\mathbb{N}$. Hence, $D-1\in\left\{  -1,0,1,\ldots\right\}  $.

We have assumed that Lemma \ref{lem.pol.quorem.exi} holds in the case when
$d=D-1$. In other words, the following fact holds:

\begin{statement}
\textit{Fact 1:} Let $A$ be a commutative ring. Let $n\in\mathbb{N}$. Let
$p\in A\left[  X\right]  $ be a monic polynomial of degree $n$. Let $f\in
A\left[  X\right]  _{\leq D-1}$. Then, there exists \textbf{at least one} pair
$\left(  q,r\right)  \in A\left[  X\right]  _{\leq\left(  D-1\right)
-n}\times A\left[  X\right]  _{\leq n-1}$ such that $f=qp+r$.
\end{statement}

Now, let $A$, $n$ and $p$ be as in Lemma \ref{lem.pol.quorem.exi}. Let $f\in
A\left[  X\right]  _{\leq D}$. We are going to show the following claim:

\begin{statement}
\textit{Claim 2:} There exists \textbf{at least one} pair $\left(  q,r\right)
\in A\left[  X\right]  _{\leq D-n}\times A\left[  X\right]  _{\leq n-1}$ such
that $f=qp+r$.
\end{statement}

\textit{Proof of Claim 2:} If $D\leq n-1$, then Claim 2
holds\footnote{\textit{Proof.} Assume that $D\leq n-1$. We must prove that
Claim 2 holds.
\par
We have $D\leq n-1$, and thus $A\left[  X\right]  _{\leq D}\subseteq A\left[
X\right]  _{\leq n-1}$ (by Lemma \ref{lem.pol.sub}, applied to $g=D$ and
$h=n-1$). Now, $f\in A\left[  X\right]  _{\leq D}\subseteq A\left[  X\right]
_{\leq n-1}$.
\par
But $0\in A\left[  X\right]  _{\leq D-n}$ (since $A\left[  X\right]  _{\leq
D-n}$ is an $A$-submodule of $A\left[  X\right]  $). Combining this with $f\in
A\left[  X\right]  _{\leq n-1}$, we obtain $\left(  0,f\right)  \in A\left[
X\right]  _{\leq D-n}\times A\left[  X\right]  _{\leq n-1}$. Moreover,
$\underbrace{0p}_{=0}+f=f$, so that $f=0p+f$. Hence, there exists \textbf{at
least one} pair $\left(  q,r\right)  \in A\left[  X\right]  _{\leq D-n}\times
A\left[  X\right]  _{\leq n-1}$ such that $f=qp+r$ (namely, $\left(
q,r\right)  =\left(  0,f\right)  $). In other words, Claim 2 holds. Qed.}.
Hence, for the rest of our proof of Claim 2, we can WLOG assume that we don't
have $D\leq n-1$. Assume this.

We have $D>n-1$ (since we don't have $D\leq n-1$). Thus, $D\geq\left(
n-1\right)  +1$ (since $D$ and $n-1$ are integers). Thus, $D\geq\left(
n-1\right)  +1=n$. Hence, $D-n\in\mathbb{N}$.

Define an element $\alpha\in A$ by $\alpha=\left[  X^{D}\right]  f$. Then, the
polynomial $\alpha X^{D-n}\in A\left[  X\right]  $ is well-defined (since
$D-n\in\mathbb{N}$). Clearly, $\alpha X^{D-n}\in A\left[  X\right]  _{\leq
D-n}$\ \ \ \ \footnote{\textit{Proof.} Let $m\in\mathbb{N}$ be such that
$m>D-n$. Then, $m\neq D-n$ (since $m>D-n$). Hence, Lemma \ref{lem.pol.XkXn}
\textbf{(b)} (applied to $D-n$ and $m$ instead of $n$ and $k$) yields $\left[
X^{m}\right]  \left(  X^{D-n}\right)  =0$. Now, $\left[  X^{m}\right]  \left(
\alpha X^{D-n}\right)  =\alpha\underbrace{\left[  X^{m}\right]  \left(
X^{D-n}\right)  }_{=0}=0$.
\par
Now, forget that we fixed $m$. We thus have shown that $\left[  X^{m}\right]
\left(  \alpha X^{D-n}\right)  =0$ for every $m\in\mathbb{N}$ satisfying
$m>D-n$. Hence, Lemma \ref{lem.pol.AXn} \textbf{(b)} (applied to $D-n$ and
$\alpha X^{D-n}$ instead of $n$ and $q$) yields $\alpha X^{D-n}\in A\left[
X\right]  _{\leq D-n}$. Qed.}. Hence, Lemma \ref{lem.pol.timesmon}
\textbf{(a)} (applied to $D-n$ and $\alpha X^{D-n}$ instead of $g$ and $q$)
yields $p\alpha X^{D-n}\in A\left[  X\right]  _{\leq\left(  D-n\right)
+n}=A\left[  X\right]  _{\leq D}$ (since $\left(  D-n\right)  +n=D$).
Moreover, Lemma \ref{lem.pol.timesmon} \textbf{(b)} (applied to $D-n$ and
$\alpha X^{D-n}$ instead of $g$ and $q$) yields
\[
\left[  X^{\left(  D-n\right)  +n}\right]  \left(  p\alpha X^{D-n}\right)
=\left[  X^{D-n}\right]  \left(  \alpha X^{D-n}\right)  =\alpha
\underbrace{\left[  X^{D-n}\right]  \left(  X^{D-n}\right)  }%
_{\substack{=1\\\text{(by Lemma \ref{lem.pol.XkXn} \textbf{(a)},
applied}\\\text{to }D-n\text{ instead of }n\text{)}}}=\alpha.
\]
Since $\left(  D-n\right)  +n=D$, this rewrites as $\left[  X^{D}\right]
\left(  p\alpha X^{D-n}\right)  =\alpha$.

Both $f$ and $p\alpha X^{D-n}$ belong to $A\left[  X\right]  _{\leq D}$.
Hence, the difference $f-p\alpha X^{D-n}$ also belongs to $A\left[  X\right]
_{\leq D}$ (since $A\left[  X\right]  _{\leq D}$ is an $A$-submodule of
$A\left[  X\right]  $). In other words, $f-p\alpha X^{D-n}\in A\left[
X\right]  _{\leq D}$. Furthermore,%
\[
\left[  X^{D}\right]  \left(  f-p\alpha X^{D-n}\right)  =\underbrace{\left[
X^{D}\right]  f}_{=\alpha}-\underbrace{\left[  X^{D}\right]  \left(  p\alpha
X^{D-n}\right)  }_{=\alpha}=\alpha-\alpha=0.
\]
Hence, Lemma \ref{lem.pol.nn-1} (applied to $D$ and $f-p\alpha X^{D-n}$
instead of $n$ and $q$) shows that $f-p\alpha X^{D-n}\in A\left[  X\right]
_{\leq D-1}$. Therefore, Fact 1 (applied to $f-p\alpha X^{D-n}$ instead of
$f$) shows that there exists \textbf{at least one} pair $\left(  q,r\right)
\in A\left[  X\right]  _{\leq\left(  D-1\right)  -n}\times A\left[  X\right]
_{\leq n-1}$ such that $f-p\alpha X^{D-n}=qp+r$. Denote this pair $\left(
q,r\right)  $ by $\left(  \widetilde{q},\widetilde{r}\right)  $. Thus,
$\left(  \widetilde{q},\widetilde{r}\right)  $ is a pair in $A\left[
X\right]  _{\leq\left(  D-1\right)  -n}\times A\left[  X\right]  _{\leq n-1}$
satisfying $f-p\alpha X^{D-n}=\widetilde{q}p+\widetilde{r}$.

Now, $\left(  D-1\right)  -n=D-n-1\leq D-n$ and thus $A\left[  X\right]
_{\leq\left(  D-1\right)  -n}\subseteq A\left[  X\right]  _{\leq D-n}$ (by
Lemma \ref{lem.pol.sub}, applied to $g=\left(  D-1\right)  -n$ and $h=D-n$).
But $\left(  \widetilde{q},\widetilde{r}\right)  \in A\left[  X\right]
_{\leq\left(  D-1\right)  -n}\times A\left[  X\right]  _{\leq n-1}$; in other
words, $\widetilde{q}\in A\left[  X\right]  _{\leq\left(  D-1\right)  -n}$ and
$\widetilde{r}\in A\left[  X\right]  _{\leq n-1}$. Now, $\widetilde{q}\in
A\left[  X\right]  _{\leq\left(  D-1\right)  -n}\subseteq A\left[  X\right]
_{\leq D-n}$. Hence, both $\widetilde{q}$ and $\alpha X^{D-n}$ belong to
$A\left[  X\right]  _{\leq D-n}$ (since we know that $\alpha X^{D-n}\in
A\left[  X\right]  _{\leq D-n}$). Thus, the sum $\widetilde{q}+\alpha X^{D-n}$
also belongs to $A\left[  X\right]  _{\leq D-n}$ (since $A\left[  X\right]
_{\leq D-n}$ is an $A$-submodule of $A\left[  X\right]  $). In other words,
$\widetilde{q}+\alpha X^{D-n}\in A\left[  X\right]  _{\leq D-n}$. Combining
this with $\widetilde{r}\in A\left[  X\right]  _{\leq n-1}$, we obtain
$\left(  \widetilde{q}+\alpha X^{D-n},\widetilde{r}\right)  \in A\left[
X\right]  _{\leq D-n}\times A\left[  X\right]  _{\leq n-1}$. Furthermore, from
$f-p\alpha X^{D-n}=\widetilde{q}p+\widetilde{r}$, we obtain%
\[
f=\underbrace{p\alpha X^{D-n}+\widetilde{q}p}_{=\widetilde{q}p+p\alpha
X^{D-n}}+\widetilde{r}=\underbrace{\widetilde{q}p+p\alpha X^{D-n}}_{=\left(
\widetilde{q}+\alpha X^{D-n}\right)  p}+\widetilde{r}=\left(  \widetilde{q}%
+\alpha X^{D-n}\right)  p+\widetilde{r}.
\]
Thus, there exists \textbf{at least one} pair $\left(  q,r\right)  \in
A\left[  X\right]  _{\leq D-n}\times A\left[  X\right]  _{\leq n-1}$ such that
$f=qp+r$ (namely, $\left(  q,r\right)  =\left(  \widetilde{q}+\alpha
X^{D-n},\widetilde{r}\right)  $). This proves Claim 2.

Now, Claim 2 is proven. In other words, we have proven that there exists
\textbf{at least one} pair $\left(  q,r\right)  \in A\left[  X\right]  _{\leq
D-n}\times A\left[  X\right]  _{\leq n-1}$ such that $f=qp+r$.

Now, forget that we have fixed $A$, $n$, $p$ and $f$. We thus have shown that
if $A$, $n$ and $p$ are as in Lemma \ref{lem.pol.quorem.exi}, and if $f\in
A\left[  X\right]  _{\leq D}$, then there exists \textbf{at least one} pair
$\left(  q,r\right)  \in A\left[  X\right]  _{\leq D-n}\times A\left[
X\right]  _{\leq n-1}$ such that $f=qp+r$. In other words, Lemma
\ref{lem.pol.quorem.exi} holds in the case when $d=D$. This completes the
induction step. The induction proof of Lemma \ref{lem.pol.quorem.exi} is thus complete.
\end{proof}
\end{verlong}

Finally, we can prove Theorem \ref{thm.pol.quorem}:

\begin{vershort}
\begin{proof}
[Proof of Theorem \ref{thm.pol.quorem}.]The existence of the pair $\left(
q,r\right)  $ follows from Lemma \ref{lem.pol.quorem.exi}\footnote{Here we are
using the fact that there exists some $d\in\mathbb{N}$ such that $f\in
A\left[  X\right]  _{\leq d}$. (But this follows immediately from Proposition
\ref{prop.pol.union}.)}; its uniqueness from Lemma \ref{lem.pol.quorem.uni}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Theorem \ref{thm.pol.quorem}.]Renaming the index $n$ as $d$ in
Proposition \ref{prop.pol.union}, we obtain $A\left[  X\right]  =\bigcup
_{d\in\mathbb{N}}A\left[  X\right]  _{\leq d}$.

Now, $f\in A\left[  X\right]  =\bigcup_{d\in\mathbb{N}}A\left[  X\right]
_{\leq d}$. Hence, there exists some $d\in\mathbb{N}$ such that $q\in A\left[
X\right]  _{\leq d}$. Consider this $d$. Clearly, $d\in\mathbb{N}%
\subseteq\left\{  -1,0,1,\ldots\right\}  $.

Lemma \ref{lem.pol.quorem.exi} shows that there exists \textbf{at least one}
pair $\left(  q,r\right)  \in A\left[  X\right]  _{\leq d-n}\times A\left[
X\right]  _{\leq n-1}$ such that $f=qp+r$. Denote this pair $\left(
q,r\right)  $ by $\left(  \widetilde{q},\widetilde{r}\right)  $. Thus,
$\left(  \widetilde{q},\widetilde{r}\right)  $ is a pair in $A\left[
X\right]  _{\leq d-n}\times A\left[  X\right]  _{\leq n-1}$ satisfying
$f=\widetilde{q}p+\widetilde{r}$.

We have $\left(  \widetilde{q},\widetilde{r}\right)  \in\underbrace{A\left[
X\right]  _{\leq d-n}}_{\subseteq A\left[  X\right]  }\times A\left[
X\right]  _{\leq n-1}\subseteq A\left[  X\right]  \times A\left[  X\right]
_{\leq n-1}$. Thus, there exists \textbf{at least one} pair $\left(
q,r\right)  \in A\left[  X\right]  \times A\left[  X\right]  _{\leq n-1}$ such
that $f=qp+r$ (namely, $\left(  q,r\right)  =\left(  \widetilde{q}%
,\widetilde{r}\right)  $). Since we also know that there exists \textbf{at
most one} such pair (indeed, this is exactly what Lemma
\ref{lem.pol.quorem.uni} says), we can therefore conclude that there exists a
unique pair $\left(  q,r\right)  \in A\left[  X\right]  \times A\left[
X\right]  _{\leq n-1}$ such that $f=qp+r$. This proves Theorem
\ref{thm.pol.quorem}.
\end{proof}
\end{verlong}

\subsection{$aR\cap pR=apR$ for monic $p$ and arbitrary $a$}

\begin{convention}
Here and in the following, we shall observe the following convention:
Multiplication (of elements of a ring, or of ideals of a ring, or of an
element of a ring with an ideal of a ring) precedes set-theoretical operations
such as $\cap$ and $\cup$. Thus, if $A$ is a ring, if $S$ and $T$ are two
ideals of $A$, and if $a$ is an element of $A$, then the expression
\textquotedblleft$aS\cap T$\textquotedblright\ means \textquotedblleft$\left(
aS\right)  \cap T$\textquotedblright\ (and not \textquotedblleft$a\left(
S\cap T\right)  $\textquotedblright). Similarly, if $U$, $V$ and $W$ are three
ideals of a ring, then the expression \textquotedblleft$U\cap VW$%
\textquotedblright\ means \textquotedblleft$U\cap\left(  VW\right)
$\textquotedblright\ (and not \textquotedblleft$\left(  U\cap V\right)
W$\textquotedblright).
\end{convention}

Now, let us prove a consequence of Theorem \ref{thm.pol.quorem} that will
reveal its use later:

\begin{corollary}
\label{cor.pol.intersect}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$.
Let $p\in A\left[  X\right]  $ be a monic polynomial of degree $n$. Let $a\in
A$. Set $R=A\left[  X\right]  $. Then, $aR\cap pR=apR$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.pol.intersect}.]Combining $a\underbrace{pR}%
_{\subseteq R}\subseteq aR$ with $\underbrace{ap}_{=pa}R=p\underbrace{aR}%
_{\subseteq R}\subseteq pR$, we obtain $apR\subseteq aR\cap pR$.

Now, let $u\in aR\cap pR$. Thus, $u\in aR\cap pR\subseteq aR$. In other words,
there exists some $f\in R$ such that $u=af$. Consider this $f$.

We have $f\in R=A\left[  X\right]  $. Hence, Theorem \ref{thm.pol.quorem}
shows that there exists a unique pair $\left(  q,r\right)  \in A\left[
X\right]  \times A\left[  X\right]  _{\leq n-1}$ such that $f=qp+r$. Consider
this $\left(  q,r\right)  $.

We have $\left(  q,r\right)  \in A\left[  X\right]  \times A\left[  X\right]
_{\leq n-1}$; in other words, $q\in A\left[  X\right]  $ and $r\in A\left[
X\right]  _{\leq n-1}$. Now, $u=a\underbrace{f}_{=qp+r}=a\left(  qp+r\right)
=aqp+ar$.

On the other hand, $u\in aR\cap pR\subseteq pR$. In other words, there exists
some $v\in R$ such that $u=pv$. Consider this $v$. We have $pv=u=aqp+ar$.
Solving this equation for $ar$, we obtain $ar=pv-aqp=pv-paq=p\left(
v-aq\right)  $.

But $r\in A\left[  X\right]  _{\leq n-1}$ and thus $ar\in A\left[  X\right]
_{\leq n-1}$ (since $a\in A$ and since $A\left[  X\right]  _{\leq n-1}$ is an
$A$-submodule of $A\left[  X\right]  $). Hence, $p\left(  v-aq\right)  =ar\in
A\left[  X\right]  _{\leq n-1}$. Thus, Proposition \ref{prop.pol.q=0} (applied
to $v-aq$ instead of $q$) yields $v-aq=0$. Hence, $v=aq$, so that
$u=p\underbrace{v}_{=aq}=paq=ap\underbrace{q}_{\in R}\in apR$.

Now, forget that we fixed $u$. We thus have shown that every $u\in aR\cap pR$
satisfies $u\in apR$. In other words, $aR\cap pR=apR$. Combined with
$apR\subseteq aR\cap pR$, this yields $aR\cap pR=apR$. This proves Corollary
\ref{cor.pol.intersect}.
\end{proof}

The claim of Corollary \ref{cor.pol.intersect} can be restated as follows: If
$A$, $n$, $p$, $a$ and $R$ as in Corollary \ref{cor.pol.intersect}, then every
polynomial in $R$ that is divisible by $a$ and by $p$ must be divisible by
$ap$. This is a \textquotedblleft sort of coprimality
statement\textquotedblright\ (not in the usual sense of the word
\textquotedblleft coprimality\textquotedblright, but more akin to the property
of coprime positive integers $m$ and $n$ to satisfy $\operatorname{lcm}\left(
m,n\right)  =mn$). Theorems \ref{thm.div.G} and \ref{thm.div.all} are also
statements of this kind, and ultimately we will use Corollary
\ref{cor.pol.intersect} as the first stepping stone in deriving these two theorems.

For the sake of (future) convenience, let us state a variant of Corollary
\ref{cor.pol.intersect} \textquotedblleft translated through a ring
isomorphism\textquotedblright:

\begin{corollary}
\label{cor.pol.intersect.iso}Let $A$ and $B$ be two commutative rings. Let
$f:A\left[  X\right]  \rightarrow B$ be a ring isomorphism. Let $n\in
\mathbb{N}$. Let $p\in B$ be such that $f^{-1}\left(  p\right)  $ is a monic
polynomial of degree $n$. Let $b\in B$ be such that $f^{-1}\left(  b\right)
\in A$. Then, $bB\cap pB=bpB$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.pol.intersect.iso}.]Set $R=A\left[  X\right]  $.
Corollary \ref{cor.pol.intersect} (applied to $f^{-1}\left(  b\right)  $ and
$f^{-1}\left(  p\right)  $ instead of $a$ and $p$) yields $f^{-1}\left(
b\right)  R\cap f^{-1}\left(  p\right)  R=f^{-1}\left(  b\right)
f^{-1}\left(  p\right)  R$.

\begin{vershort}
But $f$ is a ring isomorphism from $A\left[  X\right]  $ to $B$, thus from $R$
to $B$ (since $R=A\left[  X\right]  $). Hence, applying $f$ to both sides of
the equality $f^{-1}\left(  b\right)  R\cap f^{-1}\left(  p\right)
R=f^{-1}\left(  b\right)  f^{-1}\left(  p\right)  R$, we obtain $bB\cap
pB=bpB$. This proves Corollary \ref{cor.pol.intersect.iso}.
\end{vershort}

\begin{verlong}
But $f$ is a ring isomorphism from $A\left[  X\right]  $ to $B$. Hence, $f$ is
bijective, and thus injective and surjective. Now, $f\left(  \underbrace{R}%
_{=A\left[  X\right]  }\right)  =f\left(  A\left[  X\right]  \right)  =B$
(since $f$ is surjective). Also, any two subsets $U$ and $V$ of $A\left[
X\right]  $ satisfy $f\left(  U\cap V\right)  =f\left(  U\right)  \cap
f\left(  V\right)  $ (since $f$ is bijective). Applying this to $U=f^{-1}%
\left(  b\right)  R$ and $V=f^{-1}\left(  p\right)  R$, we obtain%
\begin{align*}
f\left(  f^{-1}\left(  b\right)  R\cap f^{-1}\left(  p\right)  R\right)   &
=\underbrace{f\left(  f^{-1}\left(  b\right)  R\right)  }_{\substack{=f\left(
f^{-1}\left(  b\right)  \right)  f\left(  R\right)  \\\text{(since }f\text{ is
a ring homomorphism)}}}\cap\underbrace{f\left(  f^{-1}\left(  p\right)
R\right)  }_{\substack{=f\left(  f^{-1}\left(  p\right)  \right)  f\left(
R\right)  \\\text{(since }f\text{ is a ring homomorphism)}}}\\
&  =\underbrace{f\left(  f^{-1}\left(  b\right)  \right)  }_{=b}%
\underbrace{f\left(  R\right)  }_{=B}\cap\underbrace{f\left(  f^{-1}\left(
p\right)  \right)  }_{=p}\underbrace{f\left(  R\right)  }_{=B}=bB\cap pB.
\end{align*}
Comparing this with%
\begin{align*}
f\left(  \underbrace{f^{-1}\left(  b\right)  R\cap f^{-1}\left(  p\right)
R}_{=f^{-1}\left(  b\right)  f^{-1}\left(  p\right)  R}\right)   &  =f\left(
f^{-1}\left(  b\right)  f^{-1}\left(  p\right)  R\right)
=\underbrace{f\left(  f^{-1}\left(  b\right)  \right)  }_{=b}%
\underbrace{f\left(  f^{-1}\left(  p\right)  \right)  }_{=p}%
\underbrace{f\left(  R\right)  }_{=B}\\
&  =bpB,
\end{align*}
we obtain $bB\cap pB=bpB$. This proves Corollary \ref{cor.pol.intersect.iso}.
\end{verlong}
\end{proof}

\section{\label{sect.mulpol}On multivariate polynomials}

\subsection{Notations and the isomorphisms $\rho_{i}$}

Next, we shall discuss some properties of multivariate polynomial rings of the
form $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $. First, we recall that such
rings can be obtained recursively by adjoining one variable after the other;
indeed, for each $n\in\mathbb{N}$ and each $i\in\left\{  1,2,\ldots,n\right\}
$, there is a canonical isomorphism%
\begin{equation}
A\left[  X_{1},X_{2},\ldots,X_{n}\right]  \cong\left(  A\left[  X_{1}%
,X_{2},\ldots,\widehat{X_{i}},\ldots,X_{n}\right]  \right)  \left[
X_{i}\right]  \label{eq.mulpol.iso}%
\end{equation}
(where the \textquotedblleft$\widehat{X_{i}}$\textquotedblright\ means that
the element $X_{i}$ is removed from the list), which is often used to identify
$A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ with $\left(  A\left[
X_{1},X_{2},\ldots,\widehat{X_{i}},\ldots,X_{n}\right]  \right)  \left[
X_{i}\right]  $ (though we shall not make such identification). Let us first
explain our notations:

\begin{convention}
Let $n\in\mathbb{N}$. Let $\left(  u_{1},u_{2},\ldots,u_{n}\right)  $ be a
list of $n$ arbitrary objects. Let $i\in\left\{  1,2,\ldots,n\right\}  $.
Then, $\left(  u_{1},u_{2},\ldots,\widehat{u_{i}},\ldots,u_{n}\right)  $ will
denote the list $\left(  u_{1},u_{2},\ldots,u_{i-1},u_{i+1},u_{i+2}%
,\ldots,u_{n}\right)  $ (this is a list of $n-1$ objects). Thus, the hat over
the symbol $u_{i}$ signifies that the $i$-th entry of the list is being
removed. (It does \textbf{not} mean that every object that happens to be equal
to $u_{i}$ is removed from the list; we only remove the $i$-th object. So, for
example, the list $\left(  \left(  -5\right)  ^{2},\left(  -4\right)
^{2},\ldots,\widehat{2^{2}},\ldots,5^{2}\right)  $ contains the entry $\left(
-2\right)  ^{2}$, even though this entry equals the removed entry $2^{2}$.)
\end{convention}

\begin{definition}
\label{def.mulpol.rho}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$. Let
$R$ denote the polynomial ring $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $.
For each $i\in\left\{  1,2,\ldots,n\right\}  $, we let $R_{i}$ denote the
polynomial ring $A\left[  X_{1},X_{2},\ldots,\widehat{X_{i}},\ldots
,X_{n}\right]  $ (a polynomial ring in $n-1$ indeterminates).

For each $i\in\left\{  1,2,\ldots,n\right\}  $, we regard $R_{i}$ as an
$A$-subalgebra of $R$. For each $i\in\left\{  1,2,\ldots,n\right\}  $, we let
$\rho_{i}:R_{i}\left[  X\right]  \rightarrow R$ be the $R_{i}$-algebra
homomorphism which sends every $p\in R_{i}\left[  X\right]  $ to $p\left(
X_{i}\right)  $. In other words, $\rho_{i}:\left(  A\left[  X_{1},X_{2}%
,\ldots,\widehat{X_{i}},\ldots,X_{n}\right]  \right)  \left[  X\right]
\rightarrow A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ is the $A$-algebra
homomorphism which satisfies%
\begin{equation}
\left(  \rho_{i}\left(  X_{j}\right)  =X_{j}\ \ \ \ \ \ \ \ \ \ \text{for
every }j\in\left\{  1,2,\ldots,\widehat{i},\ldots,n\right\}  \right)
\label{eq.def.mulpol.rho.rhoiXj}%
\end{equation}
an%
\begin{equation}
\rho_{i}\left(  X\right)  =X_{i}. \label{eq.def.mulpol.rho.rhoiX}%
\end{equation}
It is well-known that this $\rho_{i}$ is actually an $R_{i}$-algebra
isomorphism. Indeed, this $\rho_{i}$ is the isomorphism responsible for
(\ref{eq.mulpol.iso}) (at least if we rename the indeterminate $X$ in
$R_{i}\left[  X\right]  $ as $X_{i}$). Since the map $\rho_{i}$ is a $R_{i}%
$-algebra isomorphism, its inverse $\rho_{i}^{-1}$ is well-defined and also a
$R_{i}$-algebra isomorphism.

These notations $A$, $n$, $R$, $R_{i}$ and $\rho_{i}$ shall be in place for
the whole Section \ref{sect.mulpol}.
\end{definition}

\subsection{Regularity of $X_{i}-X_{j}$}

\begin{proposition}
\label{prop.mulpol.monic}Let $i$ and $j$ be two distinct elements of $\left\{
1,2,\ldots,n\right\}  $.

\textbf{(a)} The polynomial $\rho_{i}^{-1}\left(  X_{i}-X_{j}\right)  \in
R_{i}\left[  X\right]  $ is monic of degree $1$.

\textbf{(b)} The element $X_{i}-X_{j}$ of $R$ is regular.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.mulpol.monic}.]The definition of $R_{i}$
yields $R_{i}=A\left[  X_{1},X_{2},\ldots,\widehat{X_{i}},\ldots,X_{n}\right]
$.

Combining $j\in\left\{  1,2,\ldots,n\right\}  $ with $j\neq i$ (since $i$ and
$j$ are distinct), we obtain $j\in\left\{  1,2,\ldots,n\right\}
\setminus\left\{  i\right\}  =\left\{  1,2,\ldots,\widehat{i},\ldots
,n\right\}  $. Hence, $X_{j}\in A\left[  X_{1},X_{2},\ldots,\widehat{X_{i}%
},\ldots,X_{n}\right]  =R_{i}$. Hence, $X-X_{j}$ is a well-defined polynomial
in $R_{i}\left[  X\right]  $. Clearly, this polynomial $X-X_{j}\in
R_{i}\left[  X\right]  $ is monic of degree $1$.

Now, $\rho_{i}$ is an $R_{i}$-algebra homomorphism. Hence,%
\[
\rho_{i}\left(  X-X_{j}\right)  =\underbrace{\rho_{i}\left(  X\right)
}_{\substack{=X_{i}\\\text{(by (\ref{eq.def.mulpol.rho.rhoiX}))}%
}}-\underbrace{\rho_{i}\left(  X_{j}\right)  }_{\substack{=X_{j}\\\text{(by
(\ref{eq.def.mulpol.rho.rhoiXj}))}}}=X_{i}-X_{j}.
\]
Hence, $X-X_{j}=\rho_{i}^{-1}\left(  X_{i}-X_{j}\right)  $.

Now, recall that the polynomial $X-X_{j}\in R_{i}\left[  X\right]  $ is monic
of degree $1$. Since $X-X_{j}=\rho_{i}^{-1}\left(  X_{i}-X_{j}\right)  $, this
rewrites as follows: The polynomial $\rho_{i}^{-1}\left(  X_{i}-X_{j}\right)
\in R_{i}\left[  X\right]  $ is monic of degree $1$. This proves Proposition
\ref{prop.mulpol.monic} \textbf{(a)}.

\textbf{(b)} The map $\rho_{i}:R_{i}\left[  X\right]  \rightarrow R$ is an
$R_{i}$-algebra isomorphism, thus a ring isomorphism.

The polynomial $X-X_{j}\in R_{i}\left[  X\right]  $ is monic of degree $1$.
Thus, Corollary \ref{cor.pol.monic-reg} (applied to $1$, $R_{i}$ and $X-X_{j}$
instead of $n$, $A$ and $p$) shows that the element $X-X_{j}$ of $R_{i}\left[
X\right]  $ is regular. Hence, Proposition \ref{prop.reg.iso} (applied to
$R_{i}\left[  X\right]  $, $R$, $\rho_{i}$ and $X-X_{j}$ instead of $A$, $B$,
$f$ and $a$) shows that $\rho_{i}\left(  X-X_{j}\right)  $ is a regular
element of $R$ (since $\rho_{i}:R_{i}\left[  X\right]  \rightarrow R$ is a
ring isomorphism). In other words, $X_{i}-X_{j}$ is a regular element of $R$
(since $\rho_{i}\left(  X-X_{j}\right)  =X_{i}-X_{j}$). This proves
Proposition \ref{prop.mulpol.monic}.
\end{proof}

\begin{corollary}
\label{cor.mulpol.prodmonic}The polynomial $\prod_{1\leq i<j\leq n}\left(
X_{i}-X_{j}\right)  $ is a regular element of $R$.
\end{corollary}

\begin{vershort}
\begin{proof}
[Proof of Corollary \ref{cor.mulpol.prodmonic}.]Proposition
\ref{prop.mulpol.monic} \textbf{(b)} shows that $\prod_{1\leq i<j\leq
n}\left(  X_{i}-X_{j}\right)  $ is a product of finitely many regular
elements. Hence, $\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  $ is
itself regular (since Proposition \ref{prop.reg.prod} shows that a product of
finitely many regular elements must itself be regular).
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Corollary \ref{cor.mulpol.prodmonic}.]Let $G$ be the set $\left\{
\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}
$. Then, $G$ is a subset of $\left\{  1,2,\ldots,n\right\}  ^{2}$. Hence, $G$
is a finite set (since $\left\{  1,2,\ldots,n\right\}  ^{2}$ is a finite set).
Every element of $G$ has the form $\left(  i,j\right)  $ for some $\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$ (by the
definition of $G$).

For every $\left(  i,j\right)  \in G$, define an element $a_{\left(
i,j\right)  }$ of $R$ by $a_{\left(  i,j\right)  }=X_{i}-X_{j}$. Then, for
every $g\in G$, the element $a_{g}$ is a regular element of $R$%
\ \ \ \ \footnote{\textit{Proof.} Let $g\in G$. Then, $g\in G=\left\{  \left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}  $. In
other words, $g=\left(  i,j\right)  $ for some $\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$. Consider this
$\left(  i,j\right)  $.
\par
From $g=\left(  i,j\right)  $, we obtain $a_{g}=a_{\left(  i,j\right)  }%
=X_{i}-X_{j}$ (by the definition of $a_{\left(  i,j\right)  }$). But $i$ and
$j$ are two elements of $\left\{  1,2,\ldots,n\right\}  $ (since $\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$) and are distinct (since
$i<j$). Therefore, the element $X_{i}-X_{j}$ of $R$ is regular (by Proposition
\ref{prop.mulpol.monic} \textbf{(b)}). Since $a_{g}=X_{i}-X_{j}$, this
rewrites as follows: The element $a_{g}$ of $R$ is regular. Qed.}. Hence,
Proposition \ref{prop.reg.prod} (applied to $R$ instead of $A$) shows that the
element $\prod_{g\in G}a_{g}$ of $R$ is regular.

But every element of $G$ has the form $\left(  i,j\right)  $ for some $\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$. Hence,
we can rename the index $g$ as $\left(  i,j\right)  $ in the product
$\prod_{g\in G}a_{g}$. We thus obtain
\begin{align*}
\prod_{g\in G}a_{g}  &  =\underbrace{\prod_{\left(  i,j\right)  \in G}%
}_{\substack{=\prod_{\substack{\left(  i,j\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2};\\i<j}}\\\text{(since }G=\left\{  \left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}  \text{)}%
}}\underbrace{a_{\left(  i,j\right)  }}_{\substack{=X_{i}-X_{j}\\\text{(by
the}\\\text{definition of }a_{\left(  i,j\right)  }\text{)}}%
}=\underbrace{\prod_{\substack{\left(  i,j\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2};\\i<j}}}_{=\prod_{1\leq i<j\leq n}}\left(  X_{i}-X_{j}\right)
\\
&  =\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  .
\end{align*}


Now, recall that the element $\prod_{g\in G}a_{g}$ of $R$ is regular. In other
words, the element $\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  $ of
$R$ is regular (since $\prod_{g\in G}a_{g}=\prod_{1\leq i<j\leq n}\left(
X_{i}-X_{j}\right)  $). This proves Corollary \ref{cor.mulpol.prodmonic}.
\end{proof}
\end{verlong}

\subsection{Simultaneous multiples\ of $X_{p}-X_{q}$ and $X_{u}-X_{v}$}

\begin{proposition}
\label{prop.mulpol.RiRj}Let $H$ be the set $\left\{  \left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}  $. Let $\left(
p,q\right)  $ and $\left(  u,v\right)  $ be two distinct elements of $H$.
Then, either $X_{u}-X_{v}\in R_{p}$ or $X_{u}-X_{v}\in R_{q}$ (or both).
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.mulpol.RiRj}.]We have $\left(  p,q\right)  \in
H$; in other words, $\left(  p,q\right)  \in\left\{  1,2,\ldots,n\right\}
^{2}$ and $p<q$ (by the definition of $H$). Similarly, $\left(  u,v\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}$ and $u<v$.

We now observe that if $r$ is an element of $\left\{  1,2,\ldots,n\right\}  $
satisfying $r\notin\left\{  u,v\right\}  $, then
\begin{equation}
X_{u}-X_{v}\in R_{r} \label{pf.prop.mulpol.RiRj.short.Rr}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.mulpol.RiRj.short.Rr}):} Let $r$ be
an element of $\left\{  1,2,\ldots,n\right\}  $ satisfying $r\notin\left\{
u,v\right\}  $. From $r\notin\left\{  u,v\right\}  $, we obtain $r\neq u$ and
$r\neq v$. From $r\neq u$, we obtain $u\neq r$, so that $u\in\left\{
1,2,\ldots,n\right\}  \setminus\left\{  r\right\}  =\left\{  1,2,\ldots
,\widehat{r},\ldots,n\right\}  $. Hence, $X_{u}\in A\left[  X_{1},X_{2}%
,\ldots,\widehat{X_{r}},\ldots,X_{n}\right]  =R_{r}$ (since $R_{r}$ was
defined as $A\left[  X_{1},X_{2},\ldots,\widehat{X_{r}},\ldots,X_{n}\right]
$). Similarly, $X_{v}\in R_{r}$. Thus, both polynomials $X_{u}$ and $X_{v}$
belong to $R_{r}$; hence, so does their difference $X_{u}-X_{v}$. In other
words, $X_{u}-X_{v}\in R_{r}$. This proves (\ref{pf.prop.mulpol.RiRj.short.Rr}%
).}.

But it is impossible that both $p$ and $q$ belong to the set $\left\{
u,v\right\}  $\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Hence,
both $p$ and $q$ belong to the set $\left\{  u,v\right\}  $. Since $p<q$, this
shows that $p$ is the smaller of the two integers $u$ and $v$, whereas $q$ is
the larger of the two integers $u$ and $v$. But the smaller of the two
integers $u$ and $v$ is $u$ (since $u<v$). Thus, $p=u$ (since $p$ is the
smaller of the two integers $u$ and $v$); similarly, $q=v$. Hence, $\left(
\underbrace{p}_{=u},\underbrace{q}_{=v}\right)  =\left(  u,v\right)  $. This
contradicts the fact that $\left(  p,q\right)  $ and $\left(  u,v\right)  $
are distinct. This contradiction shows that our assumption was wrong, qed.}.
Hence, we are in one of the following two cases:

\textit{Case 1:} We have $p\notin\left\{  u,v\right\}  $.

\textit{Case 2:} We have $q\notin\left\{  u,v\right\}  $.

In Case 1, we have $p\notin\left\{  u,v\right\}  $ and therefore $X_{u}%
-X_{v}\in R_{p}$ (by (\ref{pf.prop.mulpol.RiRj.short.Rr}), applied to $r=p$).
Hence, Proposition \ref{prop.mulpol.RiRj} holds in Case 1.

In Case 2, we have $q\notin\left\{  u,v\right\}  $ and therefore $X_{u}%
-X_{v}\in R_{q}$ (by (\ref{pf.prop.mulpol.RiRj.short.Rr}), applied to $r=q$).
Hence, Proposition \ref{prop.mulpol.RiRj} holds in Case 2.

Thus, Proposition \ref{prop.mulpol.RiRj} holds in both Cases 1 and 2. Hence,
Proposition \ref{prop.mulpol.RiRj} holds always.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.mulpol.RiRj}.]We have $\left(  p,q\right)  \in
H=\left\{  \left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}%
\ \mid\ i<j\right\}  $ (by the definition of $H$). In other words, $\left(
p,q\right)  $ is an element $\left(  i,j\right)  $ of $\left\{  1,2,\ldots
,n\right\}  ^{2}$ satisfying $i<j$. In other words, $\left(  p,q\right)  $ is
an element of $\left\{  1,2,\ldots,n\right\}  ^{2}$ and satisfies $p<q$. From
$\left(  p,q\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$, we obtain
$p\in\left\{  1,2,\ldots,n\right\}  $ and $q\in\left\{  1,2,\ldots,n\right\}
$.

We have $\left(  u,v\right)  \in H=\left\{  \left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}  $ (by the definition of $H$).
In other words, $\left(  u,v\right)  $ is an element $\left(  i,j\right)  $ of
$\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$. In other words,
$\left(  u,v\right)  $ is an element of $\left\{  1,2,\ldots,n\right\}  ^{2}$
and satisfies $u<v$. From $\left(  u,v\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2}$, we obtain $u\in\left\{  1,2,\ldots,n\right\}  $ and
$v\in\left\{  1,2,\ldots,n\right\}  $.

We now observe that if $r$ is an element of $\left\{  1,2,\ldots,n\right\}  $
satisfying $r\notin\left\{  u,v\right\}  $, then
\begin{equation}
X_{u}-X_{v}\in R_{r} \label{pf.prop.mulpol.RiRj.Rr}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.mulpol.RiRj.Rr}):} Let $r$ be an
element of $\left\{  1,2,\ldots,n\right\}  $ satisfying $r\notin\left\{
u,v\right\}  $. From $r\notin\left\{  u,v\right\}  $, we obtain $r\neq u$ and
$r\neq v$. From $r\neq u$, we obtain $u\neq r$. Combining $u\in\left\{
1,2,\ldots,n\right\}  $ with $u\neq r$, we obtain $u\in\left\{  1,2,\ldots
,n\right\}  \setminus\left\{  r\right\}  =\left\{  1,2,\ldots,\widehat{r}%
,\ldots,n\right\}  $. Hence, $X_{u}\in A\left[  X_{1},X_{2},\ldots
,\widehat{X_{r}},\ldots,X_{n}\right]  =R_{r}$ (since $R_{r}=A\left[
X_{1},X_{2},\ldots,\widehat{X_{r}},\ldots,X_{n}\right]  $ (by the definition
of $R_{r}$)). Also, from $r\neq v$, we obtain $v\neq r$. Combining
$v\in\left\{  1,2,\ldots,n\right\}  $ with $v\neq r$, we obtain $v\in\left\{
1,2,\ldots,n\right\}  \setminus\left\{  r\right\}  =\left\{  1,2,\ldots
,\widehat{r},\ldots,n\right\}  $. Hence, $X_{v}\in A\left[  X_{1},X_{2}%
,\ldots,\widehat{X_{r}},\ldots,X_{n}\right]  =R_{r}$.
\par
But $R_{r}$ is an $A$-algebra. Since the elements $X_{u}$ and $X_{v}$ belong
to $R_{r}$, we thus conclude that their difference $X_{u}-X_{v}$ belongs to
$R_{r}$ as well. In other words, $X_{u}-X_{v}\in R_{r}$. This proves
(\ref{pf.prop.mulpol.RiRj.Rr}).}.

The elements $\left(  p,q\right)  $ and $\left(  u,v\right)  $ are distinct.
In other words, $\left(  p,q\right)  \neq\left(  u,v\right)  $. Also, recall
that $p<q$ and $u<v$. From this, we can easily conclude $\left\{  p,q\right\}
\not \subseteq \left\{  u,v\right\}  $\ \ \ \ \footnote{\textit{Proof.} Assume
the contrary. Thus, $\left\{  p,q\right\}  \subseteq\left\{  u,v\right\}  $.
\par
Now, $q\in\left\{  p,q\right\}  \subseteq\left\{  u,v\right\}  $. In other
words, either $q=u$ or $q=v$ (or both). Hence, we are in one of the following
cases:
\par
\textit{Case 1:} We have $q=u$.
\par
\textit{Case 2:} We have $q=v$.
\par
Let us consider Case 1 first. In this case, we have $q=u$. Now, $p<q=u$, so
that $p\neq u$. Combining $p\in\left\{  p,q\right\}  =\left\{  u,v\right\}  $
with $p\neq u$, we obtain $p\in\left\{  u,v\right\}  \setminus\left\{
u\right\}  \subseteq\left\{  v\right\}  $, so that $p=v$. Hence, $v=p<q=u$.
But this contradicts $u<v$. Hence, we have found a contradiction in Case 1.
\par
Let us now consider Case 2. In this case, we have $q=v$. Now, $p<q=v$, so that
$p\neq v$. Combining $p\in\left\{  p,q\right\}  =\left\{  u,v\right\}  $ with
$p\neq v$, we obtain $p\in\left\{  u,v\right\}  \setminus\left\{  v\right\}
\subseteq\left\{  u\right\}  $, so that $p=u$. Combining this with $q=v$, we
obtain $\left(  p,q\right)  =\left(  u,v\right)  $. This contradicts $\left(
p,q\right)  \neq\left(  u,v\right)  $. Hence, we have found a contradiction in
Case 2.
\par
We have now found a contradiction in each of the two Cases 1 and 2. Since
these two Cases cover all possibilities, this shows that we always have a
contradiction. Thus, our assumption was wrong, qed.}.

Now, we are in one of the following two cases:

\textit{Case 1:} We have $p\notin\left\{  u,v\right\}  $.

\textit{Case 2:} We have $p\in\left\{  u,v\right\}  $.

Let us first consider Case 1. In this case, we have $p\notin\left\{
u,v\right\}  $. Hence, $X_{u}-X_{v}\in R_{p}$ (by
(\ref{pf.prop.mulpol.RiRj.Rr}), applied to $r=p$). Thus, either $X_{u}%
-X_{v}\in R_{p}$ or $X_{u}-X_{v}\in R_{q}$ (or both). Hence, Proposition
\ref{prop.mulpol.RiRj} is proven in Case 1.

Let us now consider Case 2. In this case, we have $p\in\left\{  u,v\right\}
$. Thus, $q\notin\left\{  u,v\right\}  $\ \ \ \ \footnote{\textit{Proof.}
Assume the contrary. Thus, $q\in\left\{  u,v\right\}  $. Hence, $\left\{
q\right\}  \subseteq\left\{  u,v\right\}  $. Moreover, $\left\{  p\right\}
\subseteq\left\{  u,v\right\}  $ (since $p\in\left\{  u,v\right\}  $). Thus,
$\left\{  p,q\right\}  =\underbrace{\left\{  p\right\}  }_{\subseteq\left\{
u,v\right\}  }\cup\underbrace{\left\{  q\right\}  }_{\subseteq\left\{
u,v\right\}  }\subseteq\left\{  u,v\right\}  \cup\left\{  u,v\right\}
=\left\{  u,v\right\}  $. This contradicts $\left\{  p,q\right\}
\not \subseteq \left\{  u,v\right\}  $. This contradiction shows that our
assumption was wrong. Qed.}. Hence, $X_{u}-X_{v}\in R_{q}$ (by
(\ref{pf.prop.mulpol.RiRj.Rr}), applied to $r=q$). Thus, either $X_{u}%
-X_{v}\in R_{p}$ or $X_{u}-X_{v}\in R_{q}$ (or both). Hence, Proposition
\ref{prop.mulpol.RiRj} is proven in Case 2.

We have now proven Proposition \ref{prop.mulpol.RiRj} in each of the two Cases
1 and 2. Since these two Cases cover all possibilities, this shows that
Proposition \ref{prop.mulpol.RiRj} always holds.
\end{proof}
\end{verlong}

We now prove a \textquotedblleft coprimality statement\textquotedblright\ for
polynomials of the form $X_{i}-X_{j}$:

\begin{proposition}
\label{prop.mulpol.copri}Let $H$ be the set $\left\{  \left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}  $. For every
$\left(  i,j\right)  \in H$, define an element $a_{\left(  i,j\right)  }$ of
$R$ by $a_{\left(  i,j\right)  }=X_{i}-X_{j}$. Let $g$ and $h$ be two distinct
elements of $H$. Then, $a_{g}R\cap a_{h}R=a_{g}a_{h}R$.
\end{proposition}

Before we prove this, let us state a lemma (for the sake of convenience):

\begin{lemma}
\label{lem.mulpol.copri.2}Let $u$, $v$ and $r$ be three elements of $\left\{
1,2,\ldots,n\right\}  $ such that $u\neq v$. Assume that $X_{u}-X_{v}\in
R_{r}$.

Set $b=X_{u}-X_{v}$. Also, let $p\in R$ be such that $\rho_{r}^{-1}\left(
p\right)  \in R_{r}\left[  X\right]  $ is a monic polynomial of degree $1$.
Then, $bR\cap pR=bpR$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.mulpol.copri.2}.]Recall that $\rho_{i}:R_{i}\left[
X\right]  \rightarrow R$ is an $R_{i}$-algebra isomorphism for every
$i\in\left\{  1,2,\ldots,n\right\}  $. Applying this to $i=r$, we see that
$\rho_{r}:R_{r}\left[  X\right]  \rightarrow R$ is an $R_{r}$-algebra
isomorphism, thus a ring isomorphism.

We have $b=X_{u}-X_{v}\in R_{r}\subseteq R_{r}\left[  X\right]  $. Hence,
$\rho_{r}\left(  b\right)  $ is well-defined. Moreover, $b\in R_{r}$ and thus
$\rho_{r}\left(  b\right)  =b$ (since $\rho_{r}$ is an $R_{r}$-algebra
homomorphism); thus, $\rho_{r}^{-1}\left(  b\right)  =b\in R_{r}$. Thus,
Corollary \ref{cor.pol.intersect.iso} (applied to $R_{r}$, $R$, $\rho_{r}$ and
$1$ instead of $A$, $B$, $f$ and $n$) shows that $bR\cap pR=bpR$. This proves
Lemma \ref{lem.mulpol.copri.2}.
\end{proof}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.mulpol.copri}.]Write the elements $g$ and $h$
of $H$ in the forms $g=\left(  p,q\right)  $ and $h=\left(  u,v\right)  $.
Thus, $p<q$ and $u<v$.

The elements $g$ and $h$ are distinct. In other words, the elements $\left(
p,q\right)  $ and $\left(  u,v\right)  $ are distinct (since $g=\left(
p,q\right)  $ and $h=\left(  u,v\right)  $).

From $g=\left(  p,q\right)  $, we obtain $a_{g}=a_{\left(  p,q\right)  }%
=X_{p}-X_{q}$ (by the definition of $a_{\left(  p,q\right)  }$). Similarly,
$a_{h}=X_{u}-X_{v}$.

Proposition \ref{prop.mulpol.RiRj} shows that we have either $X_{u}-X_{v}\in
R_{p}$ or $X_{u}-X_{v}\in R_{q}$ (or both). In other words, we are in one of
the following two cases:

\textit{Case 1:} We have $X_{u}-X_{v}\in R_{p}$.

\textit{Case 2:} We have $X_{u}-X_{v}\in R_{q}$.

Let us first consider Case 1. In this case, we have $X_{u}-X_{v}\in R_{p}$.
Hence, $a_{h}=X_{u}-X_{v}\in R_{p}$.

But the elements $p$ and $q$ of $\left\{  1,2,\ldots,n\right\}  $ are distinct
(since $p<q$). Proposition \ref{prop.mulpol.monic} \textbf{(a)} (applied to
$i=p$ and $j=q$) thus yields that the polynomial $\rho_{p}^{-1}\left(
X_{p}-X_{q}\right)  \in R_{p}\left[  X\right]  $ is monic of degree $1$. In
other words, the polynomial $\rho_{p}^{-1}\left(  a_{g}\right)  \in
R_{p}\left[  X\right]  $ is monic of degree $1$ (since $a_{g}=X_{p}-X_{q}$).
Thus, Lemma \ref{lem.mulpol.copri.2} (applied to $p$, $a_{h}$ and $a_{g}$
instead of $r$, $b$ and $p$) yields $a_{h}R\cap a_{g}R=a_{h}a_{g}R$. Thus,%
\[
a_{g}R\cap a_{h}R=a_{h}R\cap a_{g}R=\underbrace{a_{h}a_{g}}_{=a_{g}a_{h}%
}R=a_{g}a_{h}R.
\]
Hence, Proposition \ref{prop.mulpol.copri} is proven in Case 1.

Let us now consider Case 2. In this case, we have $X_{u}-X_{v}\in R_{q}$.
Hence, $a_{h}=X_{u}-X_{v}\in R_{q}$.

But the elements $q$ and $p$ of $\left\{  1,2,\ldots,n\right\}  $ are distinct
(since $p<q$). Proposition \ref{prop.mulpol.monic} \textbf{(a)} (applied to
$i=q$ and $j=p$) thus yields that the polynomial $\rho_{q}^{-1}\left(
X_{q}-X_{p}\right)  \in R_{q}\left[  X\right]  $ is monic of degree $1$. In
other words, the polynomial $\rho_{q}^{-1}\left(  -a_{g}\right)  \in
R_{q}\left[  X\right]  $ is monic of degree $1$ (since $-\underbrace{a_{g}%
}_{=X_{p}-X_{q}}=-\left(  X_{p}-X_{q}\right)  =X_{q}-X_{p}$). Thus, Lemma
\ref{lem.mulpol.copri.2} (applied to $q$, $a_{h}$ and $-a_{g}$ instead of $r$,
$b$ and $p$) yields $a_{h}R\cap\left(  -a_{g}\right)  R=a_{h}\left(
-a_{g}\right)  R$. Now,
\begin{align*}
a_{g}R\cap a_{h}R  &  =a_{h}R\cap\underbrace{a_{g}R}_{=-a_{g}R=\left(
-a_{g}\right)  R}=a_{h}R\cap\left(  -a_{g}\right)  R=a_{h}\underbrace{\left(
-a_{g}\right)  R}_{=-a_{g}R=a_{g}R}\\
&  =\underbrace{a_{h}a_{g}}_{=a_{g}a_{h}}R=a_{g}a_{h}R.
\end{align*}
Hence, Proposition \ref{prop.mulpol.copri} is proven in Case 2.

We have now proven Proposition \ref{prop.mulpol.copri} in each of the two
Cases 1 and 2. Thus, Proposition \ref{prop.mulpol.copri} always holds.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.mulpol.copri}.]We have
\[
g\in H=\left\{  \left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}
^{2}\ \mid\ i<j\right\}  =\left\{  \left(  p,q\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}\ \mid\ p<q\right\}
\]
(here, we have renamed the index $\left(  i,j\right)  $ as $\left(
p,q\right)  $). Hence, $g=\left(  p,q\right)  $ for some $\left(  p,q\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying $p<q$. Consider this
$\left(  p,q\right)  $. We have $\left(  p,q\right)  =g\in H$. From $p<q$, we
obtain $p\neq q$. In other words, $q\neq p$. We have $\left(  p,q\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}$; in other words, $p\in\left\{
1,2,\ldots,n\right\}  $ and $q\in\left\{  1,2,\ldots,n\right\}  $.

We have
\[
h\in H=\left\{  \left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}
^{2}\ \mid\ i<j\right\}  =\left\{  \left(  u,v\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}\ \mid\ u<v\right\}
\]
(here, we have renamed the index $\left(  i,j\right)  $ as $\left(
u,v\right)  $). Hence, $h=\left(  u,v\right)  $ for some $\left(  u,v\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying $u<v$. Consider this
$\left(  u,v\right)  $. We have $\left(  u,v\right)  =h\in H$. From $u<v$, we
obtain $u\neq v$. We have $\left(  u,v\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2}$; in other words, $u\in\left\{  1,2,\ldots,n\right\}  $ and
$v\in\left\{  1,2,\ldots,n\right\}  $.

The elements $g$ and $h$ are distinct. In other words, the elements $\left(
p,q\right)  $ and $\left(  u,v\right)  $ are distinct (since $g=\left(
p,q\right)  $ and $h=\left(  u,v\right)  $).

From $g=\left(  p,q\right)  $, we obtain $a_{g}=a_{\left(  p,q\right)  }%
=X_{p}-X_{q}$ (by the definition of $a_{\left(  p,q\right)  }$).

From $h=\left(  u,v\right)  $, we obtain $a_{h}=a_{\left(  u,v\right)  }%
=X_{u}-X_{v}$ (by the definition of $a_{\left(  u,v\right)  }$).

Proposition \ref{prop.mulpol.RiRj} shows that we have either $X_{u}-X_{v}\in
R_{p}$ or $X_{u}-X_{v}\in R_{q}$ (or both). In other words, we are in one of
the following two cases:

\textit{Case 1:} We have $X_{u}-X_{v}\in R_{p}$.

\textit{Case 2:} We have $X_{u}-X_{v}\in R_{q}$.

Let us first consider Case 1. In this case, we have $X_{u}-X_{v}\in R_{p}$.
Hence, $a_{h}=X_{u}-X_{v}\in R_{p}$.

But the elements $p$ and $q$ of $\left\{  1,2,\ldots,n\right\}  $ are distinct
(since $p\neq q$). Proposition \ref{prop.mulpol.monic} \textbf{(a)} (applied
to $i=p$ and $j=q$) thus yields that the polynomial $\rho_{p}^{-1}\left(
X_{p}-X_{q}\right)  \in R_{p}\left[  X\right]  $ is monic of degree $1$. In
other words, the polynomial $\rho_{p}^{-1}\left(  a_{g}\right)  \in
R_{p}\left[  X\right]  $ is monic of degree $1$ (since $a_{g}=X_{p}-X_{q}$).
Thus, Lemma \ref{lem.mulpol.copri.2} (applied to $p$, $a_{h}$ and $a_{g}$
instead of $r$, $b$ and $p$) yields $a_{h}R\cap a_{g}R=a_{h}a_{g}R$. Thus,%
\[
a_{g}R\cap a_{h}R=a_{h}R\cap a_{g}R=\underbrace{a_{h}a_{g}}_{=a_{g}a_{h}%
}R=a_{g}a_{h}R.
\]
Hence, Proposition \ref{prop.mulpol.copri} is proven in Case 1.

Let us now consider Case 2. In this case, we have $X_{u}-X_{v}\in R_{q}$.
Hence, $a_{h}=X_{u}-X_{v}\in R_{q}$.

But the elements $q$ and $p$ of $\left\{  1,2,\ldots,n\right\}  $ are distinct
(since $q\neq p$). Proposition \ref{prop.mulpol.monic} \textbf{(a)} (applied
to $i=q$ and $j=p$) thus yields that the polynomial $\rho_{q}^{-1}\left(
X_{q}-X_{p}\right)  \in R_{q}\left[  X\right]  $ is monic of degree $1$. In
other words, the polynomial $\rho_{q}^{-1}\left(  -a_{g}\right)  \in
R_{q}\left[  X\right]  $ is monic of degree $1$ (since $-\underbrace{a_{g}%
}_{=X_{p}-X_{q}}=-\left(  X_{p}-X_{q}\right)  =X_{q}-X_{p}$). Thus, Lemma
\ref{lem.mulpol.copri.2} (applied to $q$, $a_{h}$ and $-a_{g}$ instead of $r$,
$b$ and $p$) yields $a_{h}R\cap\left(  -a_{g}\right)  R=a_{h}\left(
-a_{g}\right)  R$.

But $\left(  -a_{g}\right)  R=-a_{g}R=a_{g}R$ (since $a_{g}R$ is an
$A$-module). Now,
\begin{align*}
a_{g}R\cap a_{h}R  &  =a_{h}R\cap\underbrace{a_{g}R}_{\substack{=\left(
-a_{g}\right)  R\\\text{(since }\left(  -a_{g}\right)  R=a_{g}R\text{)}%
}}=a_{h}R\cap\left(  -a_{g}\right)  R=a_{h}\underbrace{\left(  -a_{g}\right)
R}_{=a_{g}R}\\
&  =\underbrace{a_{h}a_{g}}_{=a_{g}a_{h}}R=a_{g}a_{h}R.
\end{align*}
Hence, Proposition \ref{prop.mulpol.copri} is proven in Case 2.

We have now proven Proposition \ref{prop.mulpol.copri} in each of the two
Cases 1 and 2. Since these two Cases cover all possibilities, this shows that
Proposition \ref{prop.mulpol.copri} always holds.
\end{proof}
\end{verlong}

\section{\label{sect.lcm-copri}\textquotedblleft
lcm-coprimality\textquotedblright}

\subsection{A general fact about intersections of principal ideals}

One way to characterize coprime positive integers is the following: Two
positive integers $n$ and $m$ are coprime if and only if $\operatorname{lcm}%
\left(  n,m\right)  =nm$. In slightly more algebraic terms: Two positive
integers $n$ and $m$ are coprime if and only if $n\mathbb{Z}\cap
m\mathbb{Z}=nm\mathbb{Z}$. (This also holds for arbitrary nonzero integers $n$
and $m$, but fails if $n$ or $m$ is allowed to be zero.) In analogy to this,
we might wonder when two elements $b$ and $c$ of a commutative ring $R$
satisfy $bR\cap cR=bcR$. This is one possible way to generalize coprimality of
positive integers to ring elements\footnote{though not the only possible way
-- for example, another, non-equivalent generalization is to study elements
$b$ and $c$ of $R$ satisfying $bR+cR=R$}.

Our proof of Theorem \ref{thm.div.G} will use a combination of Proposition
\ref{prop.mulpol.copri} and the following general fact:

\begin{proposition}
\label{prop.copri.prod}Let $R$ be a commutative ring. Let $G$ be a finite set.
For every $g\in G$, let $a_{g}$ be a regular element of $R$. Assume that every
two distinct elements $g$ and $h$ of $G$ satisfy $a_{g}R\cap a_{h}R=a_{g}%
a_{h}R$. Then, $\bigcap_{g\in G}\left(  a_{g}R\right)  =\left(  \prod_{g\in
G}a_{g}\right)  R$.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.copri.prod}.]We have assumed that every two
distinct elements $g$ and $h$ of $G$ satisfy%
\begin{equation}
a_{g}R\cap a_{h}R=a_{g}a_{h}R. \label{pf.prop.copri.prod.short.ass-2}%
\end{equation}


Now, we shall prove that every subset $S$ of $G$ satisfies%
\begin{equation}
\bigcap_{g\in S}\left(  a_{g}R\right)  =\left(  \prod_{g\in S}a_{g}\right)  R.
\label{pf.prop.copri.prod.short.main}%
\end{equation}


\textit{Proof of (\ref{pf.prop.copri.prod.short.main}):} We will prove
(\ref{pf.prop.copri.prod.short.main}) by induction on $\left\vert S\right\vert
$:

\textit{Induction base:} The equality (\ref{pf.prop.copri.prod.short.main})
holds in the case when $\left\vert S\right\vert =0$\ \ \ \ \footnote{Indeed,
in the case when $\left\vert S\right\vert =0$, the set $S$ is empty; thus, the
left hand side of (\ref{pf.prop.copri.prod.short.main}) is the empty
intersection of subsets of $R$ (and thus equal to $R$), whereas the right hand
side of (\ref{pf.prop.copri.prod.short.main}) is $1R$ (because the product
$\prod_{g\in S}a_{g}$ is the empty product and thus equals $1$). So
(\ref{pf.prop.copri.prod.short.main}) becomes $R=1R$ in this case; but this
holds obviously.}.

\textit{Induction step:} Let $N$ be a positive integer. Assume (as our
induction hypothesis) that (\ref{pf.prop.copri.prod.short.main}) holds in the
case when $\left\vert S\right\vert =N-1$. We must prove that
(\ref{pf.prop.copri.prod.short.main}) holds in the case when $\left\vert
S\right\vert =N$.

Let $S$ be a subset of $G$ satisfying $\left\vert S\right\vert =N$.

We have $\left\vert S\right\vert =N>0$. Hence, there exists a $k\in S$. Pick
such a $k$.

Define an element $b\in R$ by $b=\prod_{g\in S\setminus\left\{  k\right\}
}a_{g}$.

From $k\in S$, we obtain $\left\vert S\setminus\left\{  k\right\}  \right\vert
=\underbrace{\left\vert S\right\vert }_{=N}-1=N-1$. Hence, we can apply
(\ref{pf.prop.copri.prod.short.main}) to $S\setminus\left\{  k\right\}  $
instead of $S$ (because of the induction hypothesis). We thus obtain%
\begin{equation}
\bigcap_{g\in S\setminus\left\{  k\right\}  }\left(  a_{g}R\right)
=\underbrace{\left(  \prod_{g\in S\setminus\left\{  k\right\}  }a_{g}\right)
}_{=b}R=bR. \label{pf.prop.copri.prod.short.indass}%
\end{equation}


But $k\in S$. Hence,%
\begin{equation}
\bigcap_{g\in S}\left(  a_{g}R\right)  =a_{k}R\cap\underbrace{\bigcap_{g\in
S\setminus\left\{  k\right\}  }\left(  a_{g}R\right)  }%
_{\substack{=bR\\\text{(by (\ref{pf.prop.copri.prod.short.indass}))}}%
}=a_{k}R\cap bR. \label{pf.prop.copri.prod.short.indass.3}%
\end{equation}


On the other hand, $k\in S$, so that%
\begin{equation}
\prod_{g\in S}a_{g}=a_{k}\underbrace{\prod_{g\in S\setminus\left\{  k\right\}
}a_{g}}_{=b}=a_{k}b. \label{pf.prop.copri.prod.short.prodag}%
\end{equation}


Combining $a_{k}\underbrace{bR}_{\subseteq R}\subseteq a_{k}R$ with
$\underbrace{a_{k}b}_{=ba_{k}}R=b\underbrace{a_{k}R}_{\subseteq R}\subseteq
bR$, we obtain $a_{k}bR\subseteq a_{k}R\cap bR$.

On the other hand, let $x\in a_{k}R\cap bR$. Then, $x\in a_{k}R\cap
bR=\bigcap_{g\in S}\left(  a_{g}R\right)  $ (by
(\ref{pf.prop.copri.prod.short.indass.3})). In other words,%
\begin{equation}
x\in a_{g}R\ \ \ \ \ \ \ \ \ \ \text{for every }g\in S.
\label{pf.prop.copri.prod.short.xin}%
\end{equation}


But $x\in a_{k}R\cap bR\subseteq a_{k}R$. In other words, there exists some
$z\in R$ such that $x=a_{k}z$. Consider this $z$. We have%
\begin{equation}
z\in a_{g}R\ \ \ \ \ \ \ \ \ \ \text{for every }g\in S\setminus\left\{
k\right\}  . \label{pf.prop.copri.prod.short.zin}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.copri.prod.short.zin}):} Let $g\in
S\setminus\left\{  k\right\}  $. We want to show that $z\in a_{g}R$.

We have $g\in S\setminus\left\{  k\right\}  $. In other words, $g\in S$ and
$g\neq k$. From $g\in S\subseteq G$ and $k\in S\subseteq G$, we conclude that
$g$ and $k$ are two elements of $G$. Furthermore, these two elements are
distinct (since $g\neq k$). Hence, (\ref{pf.prop.copri.prod.short.ass-2})
(applied to $h=k$) yields $a_{g}R\cap a_{k}R=a_{g}a_{k}R$.

But (\ref{pf.prop.copri.prod.short.xin}) yields $x\in a_{g}R$. Combining this
with $x\in a_{k}R$, we obtain $x\in a_{g}R\cap a_{k}R=a_{g}a_{k}R$. In other
words, there exists some $w\in R$ such that $x=a_{g}a_{k}w$. Consider this $w$.

We have $a_{k}\left(  z-a_{g}w\right)  =\underbrace{a_{k}z}_{=x}%
-\underbrace{a_{k}a_{g}}_{=a_{g}a_{k}}w=x-\underbrace{a_{g}a_{k}w}_{=x}%
=x-x=0$. But the element $a_{k}$ of $R$ is regular (indeed, $a_{h}$ is a
regular element of $R$ for each $h\in G$). Hence, from $a_{k}\left(
z-a_{g}w\right)  =0$, we obtain $z-a_{g}w=0$. Hence, $z=a_{g}\underbrace{w}%
_{\in R}\in a_{g}R$. This proves (\ref{pf.prop.copri.prod.short.zin}).]

From (\ref{pf.prop.copri.prod.short.zin}), we obtain $z\in\bigcap_{g\in
S\setminus\left\{  k\right\}  }\left(  a_{g}R\right)  =bR$ (by
(\ref{pf.prop.copri.prod.short.indass})). Hence, $x=a_{k}\underbrace{z}_{\in
bR}\in a_{k}bR$.

Now, forget that we fixed $x$. We thus have shown that every $x\in a_{k}R\cap
bR$ satisfies $x\in a_{k}bR$. In other words, $a_{k}R\cap bR\subseteq a_{k}%
bR$. Combining this with $a_{k}bR\subseteq a_{k}R\cap bR$, we obtain
$a_{k}R\cap bR=a_{k}bR$.

Now, (\ref{pf.prop.copri.prod.short.indass.3}) becomes%
\[
\bigcap_{g\in S}\left(  a_{g}R\right)  =a_{k}R\cap bR=\underbrace{a_{k}%
b}_{\substack{=\prod_{g\in S}a_{g}\\\text{(by
(\ref{pf.prop.copri.prod.short.prodag}))}}}R=\left(  \prod_{g\in S}%
a_{g}\right)  R.
\]


Now, forget that we fixed $S$. We thus have shown that every subset $S$ of $G$
satisfying $\left\vert S\right\vert =N$ satisfies $\bigcap_{g\in S}\left(
a_{g}R\right)  =\left(  \prod_{g\in S}a_{g}\right)  R$. In other words,
(\ref{pf.prop.copri.prod.short.main}) holds in the case when $\left\vert
S\right\vert =N$. This completes the induction proof of
(\ref{pf.prop.copri.prod.short.main}).

Now, Proposition \ref{prop.copri.prod} follows by applying
(\ref{pf.prop.copri.prod.short.main}) to $S=G$.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.copri.prod}.]We have assumed that every two
distinct elements $g$ and $h$ of $G$ satisfy%
\begin{equation}
a_{g}R\cap a_{h}R=a_{g}a_{h}R. \label{pf.prop.copri.prod.ass-2}%
\end{equation}


Moreover, if $k\in G$ and $y\in R$ are such that $a_{k}y=0$, then%
\begin{equation}
y=0 \label{pf.prop.copri.prod.ass-0}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.copri.prod.ass-0}):} Let $k\in G$ and
$y\in R$ be such that $a_{k}y=0$.
\par
For every $g\in G$, we know that $a_{g}$ is a regular element of $R$. Applying
this to $g=k$, we conclude that $a_{k}$ is a regular element of $R$.
\par
But the element $a_{k}$ of $R$ is regular if and only if every $x\in R$
satisfying $a_{k}x=0$ satisfies $x=0$ (by the definition of \textquotedblleft
regular\textquotedblright). Hence,%
\[
\text{every }x\in R\text{ satisfying }a_{k}x=0\text{ satisfies }x=0
\]
(since the element $a_{k}$ of $R$ is regular). Applying this to $x=y$, we
conclude that $y=0$ (since $a_{k}y=0$). This proves
(\ref{pf.prop.copri.prod.ass-0}).}.

Every subset $S$ of $G$ is finite (since $G$ is finite). Hence, for every
subset $S$ of $G$, the cardinality $\left\vert S\right\vert $ is a nonnegative integer.

Now, we shall prove that every subset $S$ of $G$ satisfies%
\begin{equation}
\bigcap_{g\in S}\left(  a_{g}R\right)  =\left(  \prod_{g\in S}a_{g}\right)  R.
\label{pf.prop.copri.prod.main}%
\end{equation}


\textit{Proof of (\ref{pf.prop.copri.prod.main}):} We will prove
(\ref{pf.prop.copri.prod.main}) by induction on $\left\vert S\right\vert $:

\textit{Induction base:} The equality (\ref{pf.prop.copri.prod.main}) holds in
the case when $\left\vert S\right\vert =0$\ \ \ \ \footnote{\textit{Proof.}
Assume that $\left\vert S\right\vert =0$. We must prove that the equality
(\ref{pf.prop.copri.prod.main}) holds.
\par
We have $\left\vert S\right\vert =0$, so that $S=\varnothing$. Hence,
\[
\bigcap_{g\in S}\left(  a_{g}R\right)  =\bigcap_{g\in\varnothing}\left(
a_{g}R\right)  =\left(  \text{an empty intersection of subsets of }R\right)
=R.
\]
Comparing this with%
\begin{align*}
\left(  \prod_{g\in S}a_{g}\right)  R  &  =\underbrace{\left(  \prod
_{g\in\varnothing}a_{g}\right)  }_{\substack{=\left(  \text{an empty product
of elements of }R\right)  \\=1}}R\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}S=\varnothing\right) \\
&  =1R=R,
\end{align*}
we obtain $\bigcap_{g\in S}\left(  a_{g}R\right)  =\left(  \prod_{g\in S}%
a_{g}\right)  R$. In other words, the equality (\ref{pf.prop.copri.prod.main})
holds. Qed.}. This completes the induction base.

\textit{Induction step:} Let $N$ be a positive integer. Assume that the
equality (\ref{pf.prop.copri.prod.main}) holds in the case when $\left\vert
S\right\vert =N-1$. We must prove that the equality
(\ref{pf.prop.copri.prod.main}) holds in the case when $\left\vert
S\right\vert =N$.

We have assumed that the equality (\ref{pf.prop.copri.prod.main}) holds in the
case when $\left\vert S\right\vert =N-1$. In other words, the following fact holds:

\begin{statement}
\textit{Fact 1:} Every subset $S$ of $G$ satisfying $\left\vert S\right\vert
=N-1$ satisfies
\[
\bigcap_{g\in S}\left(  a_{g}R\right)  =\left(  \prod_{g\in S}a_{g}\right)
R.
\]

\end{statement}

Now, let $S$ be a subset of $G$ satisfying $\left\vert S\right\vert =N$. We
shall prove that $\bigcap_{g\in S}\left(  a_{g}R\right)  =\left(  \prod_{g\in
S}a_{g}\right)  R$.

We have $\left\vert S\right\vert =N>0$ (since $N$ is a positive integer).
Hence, the set $S$ is nonempty. In other words, there exists a $k\in S$.
Consider this $k$.

Define an element $b\in R$ by $b=\prod_{g\in S\setminus\left\{  k\right\}
}a_{g}$.

From $k\in S$, we obtain $\left\vert S\setminus\left\{  k\right\}  \right\vert
=\underbrace{\left\vert S\right\vert }_{=N}-1=N-1$. Hence, Fact 1 (applied to
$S\setminus\left\{  k\right\}  $ instead of $S$) yields%
\begin{equation}
\bigcap_{g\in S\setminus\left\{  k\right\}  }\left(  a_{g}R\right)
=\underbrace{\left(  \prod_{g\in S\setminus\left\{  k\right\}  }a_{g}\right)
}_{=b}R=bR. \label{pf.prop.copri.prod.indass}%
\end{equation}


But $k\in S$. Hence, we can split off the term for $g=k$ from the intersection
$\bigcap_{g\in S}\left(  a_{g}R\right)  $. We thus obtain
\begin{equation}
\bigcap_{g\in S}\left(  a_{g}R\right)  =a_{k}R\cap\underbrace{\bigcap_{g\in
S\setminus\left\{  k\right\}  }\left(  a_{g}R\right)  }%
_{\substack{=bR\\\text{(by (\ref{pf.prop.copri.prod.indass}))}}}=a_{k}R\cap
bR. \label{pf.prop.copri.prod.indass.3}%
\end{equation}


On the other hand, $k\in S$. Hence, we can split off the factor for $g=k$ from
the product $\prod_{g\in S}a_{g}$. We thus obtain
\begin{equation}
\prod_{g\in S}a_{g}=a_{k}\underbrace{\prod_{g\in S\setminus\left\{  k\right\}
}a_{g}}_{=b}=a_{k}b. \label{pf.prop.copri.prod.prodag}%
\end{equation}


Combining $a_{k}\underbrace{bR}_{\subseteq R}\subseteq a_{k}R$ with
$\underbrace{a_{k}b}_{=ba_{k}}R=b\underbrace{a_{k}R}_{\subseteq R}\subseteq
bR$, we obtain $a_{k}bR\subseteq a_{k}R\cap bR$.

On the other hand, let $x\in a_{k}R\cap bR$. Then, $x\in a_{k}R\cap
bR=\bigcap_{g\in S}\left(  a_{g}R\right)  $ (by
(\ref{pf.prop.copri.prod.indass.3})). In other words,%
\begin{equation}
x\in a_{g}R\ \ \ \ \ \ \ \ \ \ \text{for every }g\in S.
\label{pf.prop.copri.prod.xin}%
\end{equation}


But $x\in a_{k}R\cap bR\subseteq a_{k}R$. In other words, there exists some
$z\in R$ such that $x=a_{k}z$. Consider this $z$. We have%
\begin{equation}
z\in a_{g}R\ \ \ \ \ \ \ \ \ \ \text{for every }g\in S\setminus\left\{
k\right\}  \label{pf.prop.copri.prod.zin}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.copri.prod.zin}):} Let $g\in
S\setminus\left\{  k\right\}  $. We want to show that $z\in a_{g}R$.
\par
We have $g\in S\setminus\left\{  k\right\}  $. In other words, $g\in S$ and
$g\neq k$. From $g\in S\subseteq G$ and $k\in S\subseteq G$, we conclude that
$g$ and $k$ are two elements of $G$. Furthermore, these two elements are
distinct (since $g\neq k$). Hence, (\ref{pf.prop.copri.prod.ass-2}) (applied
to $h=k$) yields $a_{g}R\cap a_{k}R=a_{g}a_{k}R$.
\par
But (\ref{pf.prop.copri.prod.xin}) yields $x\in a_{g}R$. Combining this with
$x\in a_{k}R$, we obtain $x\in a_{g}R\cap a_{k}R=a_{g}a_{k}R$. In other words,
there exists some $w\in R$ such that $x=a_{g}a_{k}w$. Consider this $w$.
\par
We have $a_{k}\left(  z-a_{g}w\right)  =\underbrace{a_{k}z}_{=x}%
-\underbrace{a_{k}a_{g}}_{=a_{g}a_{k}}w=x-\underbrace{a_{g}a_{k}w}_{=x}%
=x-x=0$. Hence, (\ref{pf.prop.copri.prod.ass-0}) (applied to $y=z-a_{g}w$)
yields $z-a_{g}w=0$. Hence, $z=a_{g}\underbrace{w}_{\in R}\in a_{g}R$. This
proves (\ref{pf.prop.copri.prod.zin}).}. In other words, $z\in\bigcap_{g\in
S\setminus\left\{  k\right\}  }\left(  a_{g}R\right)  $. In view of
(\ref{pf.prop.copri.prod.indass}), this rewrites as $z\in bR$. Hence,
$x=a_{k}\underbrace{z}_{\in bR}\in a_{k}bR$.

Now, forget that we fixed $x$. We thus have shown that every $x\in a_{k}R\cap
bR$ satisfies $x\in a_{k}bR$. In other words, $a_{k}R\cap bR\subseteq a_{k}%
bR$. Combining this with $a_{k}bR\subseteq a_{k}R\cap bR$, we obtain
$a_{k}R\cap bR=a_{k}bR$.

Now, (\ref{pf.prop.copri.prod.indass.3}) becomes%
\[
\bigcap_{g\in S}\left(  a_{g}R\right)  =a_{k}R\cap bR=\underbrace{a_{k}%
b}_{\substack{=\prod_{g\in S}a_{g}\\\text{(by (\ref{pf.prop.copri.prod.prodag}%
))}}}R=\left(  \prod_{g\in S}a_{g}\right)  R.
\]


Now, forget that we fixed $S$. We thus have shown that every subset $S$ of $G$
satisfying $\left\vert S\right\vert =N$ satisfies $\bigcap_{g\in S}\left(
a_{g}R\right)  =\left(  \prod_{g\in S}a_{g}\right)  R$. In other words, the
equality (\ref{pf.prop.copri.prod.main}) holds in the case when $\left\vert
S\right\vert =N$. This completes the induction proof of
(\ref{pf.prop.copri.prod.main}).

Now, we can apply (\ref{pf.prop.copri.prod.main}) to $S=G$. We thus obtain
$\bigcap_{g\in G}\left(  a_{g}R\right)  =\left(  \prod_{g\in G}a_{g}\right)
R$. This proves Proposition \ref{prop.copri.prod}.
\end{proof}
\end{verlong}

\subsection{Application to the polynomials $X_{i}-X_{j}$}

\begin{corollary}
\label{cor.mulpol.copri.prod}Let $A$ be a commutative ring with unity. Let
$n\in\mathbb{N}$. Let $G$ be a subset of the set $\left\{  \left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}  $. Let $R$ be the
polynomial ring $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $. Then,%
\[
\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}-X_{j}\right)
R\right)  =\left(  \prod_{\left(  i,j\right)  \in G}\left(  X_{i}%
-X_{j}\right)  \right)  R.
\]

\end{corollary}

\begin{vershort}
\begin{proof}
[Proof of Corollary \ref{cor.mulpol.copri.prod}.]Let $H$ be the set $\left\{
\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}
$. Thus, $G\subseteq H\subseteq\left\{  1,2,\ldots,n\right\}  ^{2}$.

For every $\left(  i,j\right)  \in H$, define an element $a_{\left(
i,j\right)  }$ of $R$ by $a_{\left(  i,j\right)  }=X_{i}-X_{j}$. Hence, an
element $a_{g}$ of $R$ is defined for each $g\in G$ (since $G\subseteq H$).

For every $g\in G$, the element $a_{g}$ is a regular element of $R$%
\ \ \ \ \footnote{\textit{Proof.} Let $g\in G$. Then, $g\in G\subseteq\left\{
\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}
$. In other words, $g=\left(  i,j\right)  $ for some $\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$. Consider this
$\left(  i,j\right)  $. From $g=\left(  i,j\right)  $, we obtain
$a_{g}=a_{\left(  i,j\right)  }=X_{i}-X_{j}$ (by the definition of $a_{\left(
i,j\right)  }$).
\par
Clearly, $i$ and $j$ are elements of $\left\{  1,2,\ldots,n\right\}  $ (since
$\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$) and are distinct
(since $i<j$). Therefore, Proposition \ref{prop.mulpol.monic} \textbf{(b)}
shows that the element $X_{i}-X_{j}$ of $R$ is regular. In other words, the
element $a_{g}$ of $R$ is regular (since $a_{g}=X_{i}-X_{j}$). Qed.}. Every
two distinct elements $g$ and $h$ of $G$ satisfy $a_{g}R\cap a_{h}R=a_{g}%
a_{h}R$\ \ \ \ \footnote{\textit{Proof.} Let $g$ and $h$ be two distinct
elements of $G$. Then, $g\in G\subseteq H$ and $h\in G\subseteq H$. Hence, $g$
and $h$ are two elements of $H$. Thus, Proposition \ref{prop.mulpol.copri}
shows that $a_{g}R\cap a_{h}R=a_{g}a_{h}R$. Qed.}. Hence, Proposition
\ref{prop.copri.prod} shows that
\[
\bigcap_{g\in G}\left(  a_{g}R\right)  =\left(  \prod_{g\in G}a_{g}\right)
R.
\]
Renaming the index $g$ as $\left(  i,j\right)  $ on both sides of this
equality, we obtain%
\[
\bigcap_{\left(  i,j\right)  \in G}\left(  a_{\left(  i,j\right)  }R\right)
=\left(  \prod_{\left(  i,j\right)  \in G}a_{\left(  i,j\right)  }\right)  R.
\]
Since $a_{\left(  i,j\right)  }=X_{i}-X_{j}$ for every $\left(  i,j\right)
\in G$, this rewrites as%
\[
\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}-X_{j}\right)
R\right)  =\left(  \prod_{\left(  i,j\right)  \in G}\left(  X_{i}%
-X_{j}\right)  \right)  R.
\]
This proves Corollary \ref{cor.mulpol.copri.prod}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Corollary \ref{cor.mulpol.copri.prod}.]Let $H$ be the set $\left\{
\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}
$. Thus, every element of $H$ has the form $\left(  i,j\right)  $ for some
$\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$.
Clearly, $H\subseteq\left\{  1,2,\ldots,n\right\}  ^{2}$. Hence, the set $H$
is finite (since the set $\left\{  1,2,\ldots,n\right\}  ^{2}$ is finite).

We have $G\subseteq\left\{  \left(  i,j\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2}\ \mid\ i<j\right\}  =H$. Hence, the set $G$ is finite (since
the set $H$ is finite).

For every $\left(  i,j\right)  \in H$, define an element $a_{\left(
i,j\right)  }$ of $R$ by $a_{\left(  i,j\right)  }=X_{i}-X_{j}$. Hence, an
element $a_{h}$ of $R$ is defined for each $h\in H$. Thus, an element $a_{g}$
of $R$ is defined for each $g\in G$\ \ \ \ \footnote{\textit{Proof.} Let $g\in
G$. Then, $g\in G\subseteq H$. Now, recall that an element $a_{h}$ of $R$ is
defined for each $h\in H$. Applying this to $h=g$, we conclude that an element
$a_{g}$ of $R$ is defined. Qed.}.

For every $g\in G$, the element $a_{g}$ is a regular element of $R$%
\ \ \ \ \footnote{\textit{Proof.} Let $g\in G$. Then, $g\in G\subseteq\left\{
\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}
$. In other words, $g=\left(  i,j\right)  $ for some $\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$. Consider this
$\left(  i,j\right)  $. From $g=\left(  i,j\right)  $, we obtain
$a_{g}=a_{\left(  i,j\right)  }=X_{i}-X_{j}$ (by the definition of $a_{\left(
i,j\right)  }$).
\par
Clearly, $i$ and $j$ are elements of $\left\{  1,2,\ldots,n\right\}  $ (since
$\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$) and are distinct
(since $i<j$). Therefore, Proposition \ref{prop.mulpol.monic} \textbf{(b)}
shows that the element $X_{i}-X_{j}$ of $R$ is regular. In other words, the
element $a_{g}$ of $R$ is regular (since $a_{g}=X_{i}-X_{j}$). In other words,
the element $a_{g}$ is a regular element of $R$. Qed.}. Every two distinct
elements $g$ and $h$ of $G$ satisfy $a_{g}R\cap a_{h}R=a_{g}a_{h}%
R$\ \ \ \ \footnote{\textit{Proof.} Let $g$ and $h$ be two distinct elements
of $G$. Then, $g\in G\subseteq H$ and $h\in G\subseteq H$. Hence, $g$ and $h$
are two elements of $H$. Thus, Proposition \ref{prop.mulpol.copri} shows that
$a_{g}R\cap a_{h}R=a_{g}a_{h}R$. Qed.}. Hence, Proposition
\ref{prop.copri.prod} shows that
\begin{equation}
\bigcap_{g\in G}\left(  a_{g}R\right)  =\left(  \prod_{g\in G}a_{g}\right)  R.
\label{pf.cor.mulpol.copri.prod.0}%
\end{equation}


Now, recall that every element of $H$ has the form $\left(  i,j\right)  $ for
some $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying
$i<j$. Hence, we can rename the index $g$ as $\left(  i,j\right)  $ in the
intersection $\bigcap_{g\in G}\left(  a_{g}R\right)  $. We thus obtain%
\begin{equation}
\bigcap_{g\in G}\left(  a_{g}R\right)  =\bigcap_{\left(  i,j\right)  \in
G}\left(  \underbrace{a_{\left(  i,j\right)  }}_{\substack{=X_{i}%
-X_{j}\\\text{(by the definition of }a_{\left(  i,j\right)  }\text{)}%
}}R\right)  =\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}%
-X_{j}\right)  R\right)  . \label{pf.cor.mulpol.copri.prod.1}%
\end{equation}
On the other hand, every element of $H$ has the form $\left(  i,j\right)  $
for some $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$
satisfying $i<j$. Hence, we can rename the index $g$ as $\left(  i,j\right)  $
in the product $\prod_{g\in G}a_{g}$. We thus obtain%
\[
\prod_{g\in G}a_{g}=\prod_{\left(  i,j\right)  \in G}\underbrace{a_{\left(
i,j\right)  }}_{\substack{=X_{i}-X_{j}\\\text{(by the definition of
}a_{\left(  i,j\right)  }\text{)}}}=\prod_{\left(  i,j\right)  \in G}\left(
X_{i}-X_{j}\right)  .
\]
Now, comparing (\ref{pf.cor.mulpol.copri.prod.1}) with
(\ref{pf.cor.mulpol.copri.prod.0}), we obtain%
\[
\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}-X_{j}\right)
R\right)  =\underbrace{\left(  \prod_{g\in G}a_{g}\right)  }_{=\prod_{\left(
i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  }R=\left(  \prod_{\left(
i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  \right)  R.
\]
This proves Corollary \ref{cor.mulpol.copri.prod}.
\end{proof}
\end{verlong}

\subsection{Proving Theorems \ref{thm.div.G} and \ref{thm.div.all}}

We can now prove Theorems \ref{thm.div.G} and \ref{thm.div.all}:

\begin{proof}
[Proof of Theorem \ref{thm.div.G}.]Let $R$ be the polynomial ring $A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $.

\begin{vershort}
We know that $f$ is divisible by $X_{i}-X_{j}$ for every $\left(  i,j\right)
\in G$. In other words, $f\in\left(  X_{i}-X_{j}\right)  R$ for every $\left(
i,j\right)  \in G$. In other words, $f\in\bigcap_{\left(  i,j\right)  \in
G}\left(  \left(  X_{i}-X_{j}\right)  R\right)  $.
\end{vershort}

\begin{verlong}
We know that
\begin{equation}
f\text{ is divisible by }X_{i}-X_{j}\text{ for every }\left(  i,j\right)  \in
G. \label{pf.thm.div.G.1}%
\end{equation}
Thus, $f\in\left(  X_{i}-X_{j}\right)  R$ for every $\left(  i,j\right)  \in
G$\ \ \ \ \footnote{\textit{Proof.} Let $\left(  i,j\right)  \in G$. Then, $f$
is divisible by $X_{i}-X_{j}$ (by (\ref{pf.thm.div.G.1})). This is a
divisibility in the ring $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $. In
other words, it is a divisibility in the ring $R$ (since $R=A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $). Thus, $f$ is divisible by $X_{i}-X_{j}$
in the ring $R$. In other words, $f\in\left(  X_{i}-X_{j}\right)  R$. Qed.}.
In other words, $f\in\bigcap_{\left(  i,j\right)  \in G}\left(  \left(
X_{i}-X_{j}\right)  R\right)  $.
\end{verlong}

\begin{vershort}
Now,%
\[
f\in\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}-X_{j}\right)
R\right)  =\left(  \prod_{\left(  i,j\right)  \in G}\left(  X_{i}%
-X_{j}\right)  \right)  R\ \ \ \ \ \ \ \ \ \ \left(  \text{by Corollary
\ref{cor.mulpol.copri.prod}}\right)  .
\]
In other words, $f$ is divisible by $\prod_{\left(  i,j\right)  \in G}\left(
X_{i}-X_{j}\right)  $ in the ring $R$. This proves Theorem \ref{thm.div.G}.
\end{vershort}

\begin{verlong}
Now,%
\[
f\in\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}-X_{j}\right)
R\right)  =\left(  \prod_{\left(  i,j\right)  \in G}\left(  X_{i}%
-X_{j}\right)  \right)  R\ \ \ \ \ \ \ \ \ \ \left(  \text{by Corollary
\ref{cor.mulpol.copri.prod}}\right)  .
\]
In other words, $f$ is divisible by $\prod_{\left(  i,j\right)  \in G}\left(
X_{i}-X_{j}\right)  $ in the ring $R$. In other words, $f$ is divisible by
$\prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  $ in the ring
$A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ (since $R=A\left[  X_{1}%
,X_{2},\ldots,X_{n}\right]  $). This proves Theorem \ref{thm.div.G}.
\end{verlong}
\end{proof}

\begin{vershort}
\begin{proof}
[Proof of Theorem \ref{thm.div.all}.]Theorem \ref{thm.div.all} follows from
Theorem \ref{thm.div.G} if we set \newline$G=\left\{  \left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}  $.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Theorem \ref{thm.div.all}.]We have assumed that
\begin{equation}
f\text{ is divisible by }X_{i}-X_{j} \label{pf.thm.div.all.1}%
\end{equation}
for every $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$
satisfying $i<j$.

Let $G$ be the set $\left\{  \left(  i,j\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2}\ \mid\ i<j\right\}  $. Then, $G$ is a subset of
\newline$\left\{  \left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}
^{2}\ \mid\ i<j\right\}  $. Clearly,%
\[
G=\left\{  \left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}%
\ \mid\ i<j\right\}  =\left\{  \left(  u,v\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2}\ \mid\ u<v\right\}
\]
(here, we have renamed the index $\left(  i,j\right)  $ as $\left(
u,v\right)  $).

Furthermore, $f$ is divisible by $X_{i}-X_{j}$ for every $\left(  i,j\right)
\in G$\ \ \ \ \footnote{\textit{Proof.} Let $\left(  i,j\right)  \in G$. Then,
$\left(  i,j\right)  \in G=\left\{  \left(  u,v\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}\ \mid\ u<v\right\}  $. In other words, $\left(
i,j\right)  $ is an element $\left(  u,v\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2}$ satisfying $u<v$. In other words, $\left(  i,j\right)  $ is
an element of $\left\{  1,2,\ldots,n\right\}  ^{2}$ and satisfies $i<j$.
Hence, $f$ is divisible by $X_{i}-X_{j}$ (by (\ref{pf.thm.div.all.1})). Qed.}.
Theorem \ref{thm.div.G} thus shows that $f$ is divisible by $\prod_{\left(
i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  $. In other words, $f$ is
divisible by $\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  $ (since%
\begin{align*}
\underbrace{\prod_{\left(  i,j\right)  \in G}}_{\substack{=\prod
_{\substack{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}%
;\\i<j}}\\\text{(since }G=\left\{  \left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}  \text{)}}}\left(  X_{i}%
-X_{j}\right)   &  =\underbrace{\prod_{\substack{\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2};\\i<j}}}_{=\prod_{1\leq i<j\leq n}%
}\left(  X_{i}-X_{j}\right) \\
&  =\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)
\end{align*}
). This proves Theorem \ref{thm.div.all}.
\end{proof}
\end{verlong}

\section{\label{sect.pow}Analogues for power series}

\subsection{When is $X-a\in A\left[  \left[  X\right]  \right]  $ regular?}

Let us now try to extend some of the above results from polynomials to power series.

Recall that if $A$ is a commutative ring, then $A\left[  \left[  X\right]
\right]  $ denotes the ring of formal power series in one indeterminate $X$
over $A$. This ring contains $A\left[  X\right]  $ as a subring. First, I
shall show two facts that can be viewed as partial analogues of Corollary
\ref{cor.pol.monic-reg}:

\begin{proposition}
\label{prop.pow.X-a-reg.nil}Let $A$ be a commutative ring. Let $a$ be a
nilpotent element of $A$. Then, the element $X-a$ of $A\left[  \left[
X\right]  \right]  $ is regular.
\end{proposition}

\begin{proposition}
\label{prop.pow.X-a-reg.reg}Let $A$ be a commutative ring. Let $a$ be a
regular element of $A$. Then, the element $X-a$ of $A\left[  \left[  X\right]
\right]  $ is regular.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.pow.X-a-reg.nil}.]Let $x\in A\left[  \left[
X\right]  \right]  $ be such that $\left(  X-a\right)  x=0$. We shall prove
that $x=0$.

Write the power series $x$ in the form $x=\sum_{i\in\mathbb{N}}c_{i}X^{i}$ for
some sequence $\left(  c_{0},c_{1},c_{2},\ldots\right)  \in A^{\mathbb{N}}$.

\begin{vershort}
Extend the sequence $\left(  c_{0},c_{1},c_{2},\ldots\right)  \in
A^{\mathbb{N}}$ to a sequence $\left(  c_{-1},c_{0},c_{1},c_{2},\ldots\right)
\in A^{\left\{  -1,0,1,\ldots\right\}  }$ by setting $c_{-1}=0$. Then,%
\begin{align}
\sum_{i\in\mathbb{N}}c_{i-1}X^{i}  &  =\underbrace{c_{-1}}_{=0}X^{0}%
+\sum_{\substack{i\in\mathbb{N};\\i>0}}c_{i-1}X^{i}=\underbrace{0X^{0}}%
_{=0}+\sum_{\substack{i\in\mathbb{N};\\i>0}}c_{i-1}X^{i}=\sum_{\substack{i\in
\mathbb{N};\\i>0}}c_{i-1}X^{i}\nonumber\\
&  =\sum_{i\in\mathbb{N}}c_{i}\underbrace{X^{i+1}}_{=XX^{i}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }i\text{ for
}i-1\text{ in the sum}\right) \nonumber\\
&  =\sum_{i\in\mathbb{N}}c_{i}XX^{i}=X\underbrace{\sum_{i\in\mathbb{N}}%
c_{i}X^{i}}_{=x}=Xx. \label{pf.prop.pow.X-a-reg.nil.short.2}%
\end{align}
On the other hand, $\left(  X-a\right)  x=0$. Thus,%
\begin{align*}
0  &  =\left(  X-a\right)  x=\underbrace{Xx}_{\substack{=\sum_{i\in\mathbb{N}%
}c_{i-1}X^{i}\\\text{(by (\ref{pf.prop.pow.X-a-reg.nil.short.2}))}%
}}-a\underbrace{x}_{=\sum_{i\in\mathbb{N}}c_{i}X^{i}}=\sum_{i\in\mathbb{N}%
}c_{i-1}X^{i}-\underbrace{a\sum_{i\in\mathbb{N}}c_{i}X^{i}}_{=\sum
_{i\in\mathbb{N}}ac_{i}X^{i}}\\
&  =\sum_{i\in\mathbb{N}}c_{i-1}X^{i}-\sum_{i\in\mathbb{N}}ac_{i}X^{i}.
\end{align*}
In other words,%
\[
\sum_{i\in\mathbb{N}}c_{i-1}X^{i}=\sum_{i\in\mathbb{N}}ac_{i}X^{i}.
\]
Comparing coefficients on both sides of this equality, we obtain%
\begin{equation}
c_{i-1}=ac_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\mathbb{N}.
\label{pf.prop.pow.X-a-reg.nil.short.5}%
\end{equation}


Now, we can easily see that every $i\in\left\{  -1,0,1,\ldots\right\}  $ and
$j\in\mathbb{N}$ satisfy%
\begin{equation}
c_{i}=a^{j}c_{i+j} \label{pf.prop.pow.X-a-reg.nil.short.7}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.pow.X-a-reg.nil.short.7}):} Fix
$i\in\left\{  -1,0,1,\ldots\right\}  $. We shall prove
(\ref{pf.prop.pow.X-a-reg.nil.short.7}) by induction over $j$:
\par
\textit{Induction base:} We have $\underbrace{a^{0}}_{=1}\underbrace{c_{i+0}%
}_{=c_{i}}=c_{i}$. Thus, $c_{i}=a^{0}c_{i+0}$. In other words,
(\ref{pf.prop.pow.X-a-reg.nil.short.7}) holds for $j=0$. This completes the
induction base.
\par
\textit{Induction step:} Let $J\in\mathbb{N}$ be positive. Assume that
(\ref{pf.prop.pow.X-a-reg.nil.short.7}) holds for $j=J-1$. We must now show
that (\ref{pf.prop.pow.X-a-reg.nil.short.7}) holds for $j=J$.
\par
We have assumed that (\ref{pf.prop.pow.X-a-reg.nil.short.7}) holds for
$j=J-1$. In other words, we have $c_{i}=a^{J-1}c_{i+J-1}$. Now,%
\[
c_{i}=a^{J-1}\underbrace{c_{i+J-1}}_{\substack{=ac_{i+J}\\\text{(by
(\ref{pf.prop.pow.X-a-reg.nil.short.5}) (applied to }i+J\\\text{instead of
}i\text{))}}}=\underbrace{a^{J-1}a}_{=a^{J}}c_{i+J}=a^{J}c_{i+J}.
\]
In other words, (\ref{pf.prop.pow.X-a-reg.nil.short.7}) holds for $j=J$. Thus,
the induction step is complete. Hence, (\ref{pf.prop.pow.X-a-reg.nil.short.7})
is proven by induction.}.

But the element $a$ of $A$ is nilpotent. In other words, there exists some
$k\in\mathbb{N}$ such that $a^{k}=0$ (by the definition of \textquotedblleft
nilpotent\textquotedblright). Consider this $k$. Every $i\in\mathbb{N}$
satisfies%
\begin{align}
c_{i}  &  =\underbrace{a^{k}}_{=0}c_{i+k}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.pow.X-a-reg.nil.short.7}) (applied to }j=k\text{)}\right)
\nonumber\\
&  =0. \label{pf.prop.pow.X-a-reg.nil.short.9}%
\end{align}
Now, $x=\sum_{i\in\mathbb{N}}\underbrace{c_{i}}_{\substack{=0\\\text{(by
(\ref{pf.prop.pow.X-a-reg.nil.short.9}))}}}X^{i}=\sum_{i\in\mathbb{N}}%
0X^{i}=0$.
\end{vershort}

\begin{verlong}
For every $i\in\mathbb{N}$, the element $%
\begin{cases}
0, & \text{if }i=0;\\
c_{i-1}, & \text{if }i\neq0
\end{cases}
$ of $A$ is well-defined\footnote{\textit{Proof.} Fix $i\in\mathbb{N}$. We
must prove that the element $%
\begin{cases}
0, & \text{if }i=0;\\
c_{i-1}, & \text{if }i\neq0
\end{cases}
$ of $A$ is well-defined.
\par
We are in one of the following two cases:
\par
\textit{Case 1:} We have $i=0$.
\par
\textit{Case 2:} We have $i\neq0$.
\par
Let us first consider Case 1. In this case, we have $i=0$. Hence, $%
\begin{cases}
0, & \text{if }i=0;\\
c_{i-1}, & \text{if }i\neq0
\end{cases}
=0$. Thus, the element $%
\begin{cases}
0, & \text{if }i=0;\\
c_{i-1}, & \text{if }i\neq0
\end{cases}
$ of $A$ is well-defined (since the element $0$ of $A$ is well-defined).
Hence, we have shown that the element $%
\begin{cases}
0, & \text{if }i=0;\\
c_{i-1}, & \text{if }i\neq0
\end{cases}
$ of $A$ is well-defined in Case 1.
\par
Let us now consider Case 2. In this case, we have $i\neq0$. Combining
$i\in\mathbb{N}$ with $i\neq0$, we obtain $i\in\mathbb{N}\setminus\left\{
0\right\}  =\left\{  1,2,3,\ldots\right\}  $. Thus, $i-1\in\left\{
0,1,2,\ldots\right\}  =\mathbb{N}$. Therefore, the element $c_{i-1}$ of $A$ is
well-defined (since $\left(  c_{0},c_{1},c_{2},\ldots\right)  \in
A^{\mathbb{N}}$). But recall that $i\neq0$. Hence, $%
\begin{cases}
0, & \text{if }i=0;\\
c_{i-1}, & \text{if }i\neq0
\end{cases}
=c_{i-1}$. Thus, the element $%
\begin{cases}
0, & \text{if }i=0;\\
c_{i-1}, & \text{if }i\neq0
\end{cases}
$ of $A$ is well-defined (since the element $c_{i-1}$ of $A$ is well-defined).
Hence, we have shown that the element $%
\begin{cases}
0, & \text{if }i=0;\\
c_{i-1}, & \text{if }i\neq0
\end{cases}
$ of $A$ is well-defined in Case 2.
\par
We now have shown that the element $%
\begin{cases}
0, & \text{if }i=0;\\
c_{i-1}, & \text{if }i\neq0
\end{cases}
$ of $A$ is well-defined in each of the two Cases 1 and 2. Since these two
Cases cover all possibilities, we thus conclude that the element $%
\begin{cases}
0, & \text{if }i=0;\\
c_{i-1}, & \text{if }i\neq0
\end{cases}
$ of $A$ is always well-defined. Qed.}. Thus, we can define a sequence
$\left(  d_{0},d_{1},d_{2},\ldots\right)  \in A^{\mathbb{N}}$ by%
\[
\left(  d_{i}=%
\begin{cases}
0, & \text{if }i=0;\\
c_{i-1}, & \text{if }i\neq0
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\mathbb{N}\right)  .
\]
Then, the definition of $d_{0}$ yields%
\begin{equation}
d_{0}=%
\begin{cases}
0, & \text{if }0=0;\\
c_{0-1}, & \text{if }0\neq0
\end{cases}
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }0=0\right)  .
\label{pf.prop.pow.X-a-reg.nil.0}%
\end{equation}
Also, for every $i\in\mathbb{N}$ satisfying $i\neq0$, we have%
\begin{equation}
d_{i}=%
\begin{cases}
0, & \text{if }i=0;\\
c_{i-1}, & \text{if }i\neq0
\end{cases}
=c_{i-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\neq0\right)  .
\label{pf.prop.pow.X-a-reg.nil.1}%
\end{equation}


Now,%
\begin{align}
\sum_{i\in\mathbb{N}}d_{i}X^{i}  &  =\underbrace{d_{0}}_{=0}X^{0}%
+\sum_{\substack{i\in\mathbb{N};\\i\neq0}}\underbrace{d_{i}}%
_{\substack{=c_{i-1}\\\text{(by (\ref{pf.prop.pow.X-a-reg.nil.1}))}}%
}X^{i}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the}\\
\text{addend for }i=0\text{ from the sum}%
\end{array}
\right) \nonumber\\
&  =\underbrace{0X^{0}}_{=0}+\sum_{\substack{i\in\mathbb{N};\\i\neq0}%
}c_{i-1}X^{i}=\underbrace{\sum_{\substack{i\in\mathbb{N};\\i\neq0}}}%
_{=\sum_{i\in\mathbb{N}\setminus\left\{  0\right\}  }=\sum_{i\in\left\{
1,2,3,\ldots\right\}  }}c_{i-1}X^{i}=\sum_{i\in\left\{  1,2,3,\ldots\right\}
}c_{i-1}X^{i}\nonumber\\
&  =\sum_{i\in\mathbb{N}}\underbrace{c_{\left(  i+1\right)  -1}}%
_{\substack{=c_{i}\\\text{(since }\left(  i+1\right)  -1=i\text{)}%
}}\underbrace{X^{i+1}}_{=XX^{i}}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we
have substituted }i+1\text{ for }i\text{ in the sum}\right) \nonumber\\
&  =\sum_{i\in\mathbb{N}}c_{i}XX^{i}=X\underbrace{\sum_{i\in\mathbb{N}}%
c_{i}X^{i}}_{\substack{=x\\\text{(since }x=\sum_{i\in\mathbb{N}}c_{i}%
X^{i}\text{)}}}=Xx. \label{pf.prop.pow.X-a-reg.nil.2}%
\end{align}


On the other hand, $\left(  X-a\right)  x=0$. Thus,%
\begin{align*}
0  &  =\left(  X-a\right)  x=\underbrace{Xx}_{\substack{=\sum_{i\in\mathbb{N}%
}d_{i}X^{i}\\\text{(by (\ref{pf.prop.pow.X-a-reg.nil.2}))}}}-a\underbrace{x}%
_{=\sum_{i\in\mathbb{N}}c_{i}X^{i}}=\sum_{i\in\mathbb{N}}d_{i}X^{i}%
-\underbrace{a\sum_{i\in\mathbb{N}}c_{i}X^{i}}_{=\sum_{i\in\mathbb{N}}%
ac_{i}X^{i}}\\
&  =\sum_{i\in\mathbb{N}}d_{i}X^{i}-\sum_{i\in\mathbb{N}}ac_{i}X^{i}.
\end{align*}
In other words,%
\[
\sum_{i\in\mathbb{N}}d_{i}X^{i}=\sum_{i\in\mathbb{N}}ac_{i}X^{i}.
\]
Comparing coefficients on both sides of this equality, we obtain%
\begin{equation}
d_{i}=ac_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\mathbb{N}.
\label{pf.prop.pow.X-a-reg.nil.5}%
\end{equation}


Thus, every $i\in\mathbb{N}$ satisfies%
\begin{equation}
c_{i}=ac_{i+1} \label{pf.prop.pow.X-a-reg.nil.6}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.pow.X-a-reg.nil.6}):} Let
$i\in\mathbb{N}$. Then, $i+1\in\mathbb{N}$. Hence,
(\ref{pf.prop.pow.X-a-reg.nil.5}) (applied to $i+1$ instead of $i$) yields
$d_{i+1}=ac_{i+1}$.
\par
But $i\in\mathbb{N}$, hence $i\geq0$ and thus $\underbrace{i}_{\geq0}%
+1\geq1>0$. Therefore, $i+1\neq0$. Hence, (\ref{pf.prop.pow.X-a-reg.nil.1})
(applied to $i+1$ instead of $i$) yields $d_{i+1}=c_{\left(  i+1\right)
-1}=c_{i}$ (since $\left(  i+1\right)  -1=i$). Comparing this with
$d_{i+1}=ac_{i+1}$, we obtain $c_{i}=ac_{i+1}$. This proves
(\ref{pf.prop.pow.X-a-reg.nil.6}).}.

Now, we can easily see that every $i\in\mathbb{N}$ and $j\in\mathbb{N}$
satisfy%
\begin{equation}
c_{i}=a^{j}c_{i+j} \label{pf.prop.pow.X-a-reg.nil.7}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.pow.X-a-reg.nil.7}):} Fix
$i\in\mathbb{N}$. We shall prove (\ref{pf.prop.pow.X-a-reg.nil.7}) by
induction over $j$:
\par
\textit{Induction base:} We have $\underbrace{a^{0}}_{=1}c_{i+0}=c_{i+0}%
=c_{i}$ (since $i+0=i$). Thus, $c_{i}=a^{0}c_{i+0}$. In other words,
(\ref{pf.prop.pow.X-a-reg.nil.7}) holds for $j=0$. This completes the
induction base.
\par
\textit{Induction step:} Let $J\in\mathbb{N}$. Assume that
(\ref{pf.prop.pow.X-a-reg.nil.7}) holds for $j=J$. We must now show that
(\ref{pf.prop.pow.X-a-reg.nil.7}) holds for $j=J+1$.
\par
We have assumed that (\ref{pf.prop.pow.X-a-reg.nil.7}) holds for $j=J$. In
other words, we have $c_{i}=a^{J}c_{i+J}$. Now,%
\[
c_{i}=a^{J}\underbrace{c_{i+J}}_{\substack{=ac_{i+J+1}\\\text{(by
(\ref{pf.prop.pow.X-a-reg.nil.6}) (applied to }i+J\\\text{instead of
}i\text{))}}}=\underbrace{a^{J}a}_{=a^{J+1}}\underbrace{c_{i+J+1}%
}_{=c_{i+\left(  J+1\right)  }}=a^{J+1}c_{i+\left(  J+1\right)  }.
\]
In other words, (\ref{pf.prop.pow.X-a-reg.nil.7}) holds for $j=J+1$. Thus, the
induction step is complete. Hence, (\ref{pf.prop.pow.X-a-reg.nil.7}) is proven
by induction.}.

But the element $a$ of $A$ is nilpotent. In other words, there exists some
$k\in\mathbb{N}$ such that $a^{k}=0$ (by the definition of \textquotedblleft
nilpotent\textquotedblright). Consider this $k$. Every $i\in\mathbb{N}$
satisfies%
\begin{align}
c_{i}  &  =\underbrace{a^{k}}_{=0}c_{i+k}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.pow.X-a-reg.nil.7}) (applied to }j=k\text{)}\right) \nonumber\\
&  =0. \label{pf.prop.pow.X-a-reg.nil.9}%
\end{align}
Now, $x=\sum_{i\in\mathbb{N}}\underbrace{c_{i}}_{\substack{=0\\\text{(by
(\ref{pf.prop.pow.X-a-reg.nil.9}))}}}X^{i}=\sum_{i\in\mathbb{N}}0X^{i}=0$.
\end{verlong}

Now, forget that we fixed $x$. We thus have shown that every $x\in A\left[
\left[  X\right]  \right]  $ satisfying $\left(  X-a\right)  x=0$ satisfies
$x=0$.

But the element $X-a$ of $A\left[  \left[  X\right]  \right]  $ is regular if
and only if every $x\in A\left[  \left[  X\right]  \right]  $ satisfying
$\left(  X-a\right)  x=0$ satisfies $x=0$ (by the definition of
\textquotedblleft regular\textquotedblright). Hence, the element $X-a$ of
$A\left[  \left[  X\right]  \right]  $ is regular (since every $x\in A\left[
\left[  X\right]  \right]  $ satisfying $\left(  X-a\right)  x=0$ satisfies
$x=0$). This proves Proposition \ref{prop.pow.X-a-reg.nil}.
\end{proof}

We shall generalize Proposition \ref{prop.pow.X-a-reg.nil} in Subsection
\ref{subsect.pow.more.regps} (and show a second proof for it).

\begin{proof}
[Proof of Proposition \ref{prop.pow.X-a-reg.reg}.]The element $a$ of $A$ is
regular if and only if every $x\in A$ satisfying $ax=0$ satisfies $x=0$ (by
the definition of \textquotedblleft regular\textquotedblright). Hence,%
\begin{equation}
\text{every }x\in A\text{ satisfying }ax=0\text{ satisfies }x=0
\label{pf.prop.pow.X-a-reg.reg.a-reg}%
\end{equation}
(since the element $a$ of $A$ is regular).

Let $x\in A\left[  \left[  X\right]  \right]  $ be such that $\left(
X-a\right)  x=0$. We shall prove that $x=0$.

Write the power series $x$ in the form $x=\sum_{i\in\mathbb{N}}c_{i}X^{i}$ for
some sequence $\left(  c_{0},c_{1},c_{2},\ldots\right)  \in A^{\mathbb{N}}$.

\begin{vershort}
Extend the sequence $\left(  c_{0},c_{1},c_{2},\ldots\right)  \in
A^{\mathbb{N}}$ to a sequence $\left(  c_{-1},c_{0},c_{1},c_{2},\ldots\right)
\in A^{\left\{  -1,0,1,\ldots\right\}  }$ by setting $c_{-1}=0$. Then, the
equality (\ref{pf.prop.pow.X-a-reg.nil.short.5}) holds (for the same reasons
that we explained in the proof of Proposition \ref{prop.pow.X-a-reg.nil}).
Using this equality and the regularity of $a$, we can easily find that%
\begin{equation}
c_{i}=0\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  -1,0,1,\ldots
\right\}  \label{pf.prop.pow.X-a-reg.reg.short.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.pow.X-a-reg.reg.short.5}):} We shall
prove (\ref{pf.prop.pow.X-a-reg.reg.short.5}) by induction over $i$:
\par
\textit{Induction base:} We have $c_{-1}=0$. In other words,
(\ref{pf.prop.pow.X-a-reg.reg.short.5}) holds for $i=-1$. This completes the
induction base.
\par
\textit{Induction step:} Let $j\in\mathbb{N}$. Assume that
(\ref{pf.prop.pow.X-a-reg.reg.short.5}) holds for $i=j-1$. We must prove that
(\ref{pf.prop.pow.X-a-reg.reg.short.5}) holds for $i=j$.
\par
We have assumed that (\ref{pf.prop.pow.X-a-reg.reg.short.5}) holds for
$i=j-1$. In other words, we have $c_{j-1}=0$. Now,
(\ref{pf.prop.pow.X-a-reg.nil.short.5}) (applied to $i=j$) yields
$c_{j-1}=ac_{j}$. Hence, $ac_{j}=c_{j-1}=0$. Thus, $c_{j}=0$ (by
(\ref{pf.prop.pow.X-a-reg.reg.a-reg}) (applied to $c_{j}$ instead of $x$)). In
other words, (\ref{pf.prop.pow.X-a-reg.reg.short.5}) holds for $i=j$. This
completes the induction step. Hence, the induction proof of
(\ref{pf.prop.pow.X-a-reg.reg.short.5}) is complete.}.

Now, $x=\sum_{i\in\mathbb{N}}\underbrace{c_{i}}_{\substack{=0\\\text{(by
(\ref{pf.prop.pow.X-a-reg.reg.short.5}))}}}X^{i}=\sum_{i\in\mathbb{N}}%
0X^{i}=0$.
\end{vershort}

\begin{verlong}
Define a sequence $\left(  d_{0},d_{1},d_{2},\ldots\right)  \in A^{\mathbb{N}%
}$ as in the proof of Proposition \ref{prop.pow.X-a-reg.nil}. Then, $d_{0}=0$.
(Indeed, this can be proven in the same way as we proved
(\ref{pf.prop.pow.X-a-reg.nil.0}) during our proof of Proposition
\ref{prop.pow.X-a-reg.nil}.) Furthermore, we have%
\begin{equation}
d_{i}=ac_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\mathbb{N}.
\label{pf.prop.pow.X-a-reg.reg.1}%
\end{equation}
(Indeed, this can be proven in the same way as we proved
(\ref{pf.prop.pow.X-a-reg.nil.5}) during our proof of Proposition
\ref{prop.pow.X-a-reg.nil}.) Moreover, every $i\in\mathbb{N}$ satisfies%
\begin{equation}
c_{i}=ac_{i+1}. \label{pf.prop.pow.X-a-reg.reg.2}%
\end{equation}
(Indeed, this can be proven in the same way as we proved
(\ref{pf.prop.pow.X-a-reg.nil.6}) during our proof of Proposition
\ref{prop.pow.X-a-reg.nil}.)

Now, we observe that%
\begin{equation}
c_{i}=0\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\mathbb{N}.
\label{pf.prop.pow.X-a-reg.reg.5}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.pow.X-a-reg.reg.5}):} We shall prove
(\ref{pf.prop.pow.X-a-reg.reg.5}) by induction over $i$:

\textit{Induction base:} The equality (\ref{pf.prop.pow.X-a-reg.reg.1})
(applied to $i=0$) yields $d_{0}=ac_{0}$. Comparing this with $d_{0}=0$, we
obtain $ac_{0}=0$. Hence, $c_{0}=0$ (by (\ref{pf.prop.pow.X-a-reg.reg.a-reg})
(applied to $c_{0}$ instead of $x$)). In other words,
(\ref{pf.prop.pow.X-a-reg.reg.5}) holds for $i=0$. This completes the
induction base.

\textit{Induction step:} Let $j\in\mathbb{N}$. Assume that
(\ref{pf.prop.pow.X-a-reg.reg.5}) holds for $i=j$. We must prove that
(\ref{pf.prop.pow.X-a-reg.reg.5}) holds for $i=j+1$.

We have assumed that (\ref{pf.prop.pow.X-a-reg.reg.5}) holds for $i=j$. In
other words, we have $c_{j}=0$. Now, (\ref{pf.prop.pow.X-a-reg.reg.2})
(applied to $i=j$) yields $c_{j}=ac_{j+1}$. Hence, $ac_{j+1}=c_{j}=0$. Thus,
$c_{j+1}=0$ (by (\ref{pf.prop.pow.X-a-reg.reg.a-reg}) (applied to $c_{j+1}$
instead of $x$)). In other words, (\ref{pf.prop.pow.X-a-reg.reg.5}) holds for
$i=j+1$. This completes the induction step. Hence, the induction proof of
(\ref{pf.prop.pow.X-a-reg.reg.5}) is complete.]

Now, $x=\sum_{i\in\mathbb{N}}\underbrace{c_{i}}_{\substack{=0\\\text{(by
(\ref{pf.prop.pow.X-a-reg.reg.5}))}}}X^{i}=\sum_{i\in\mathbb{N}}0X^{i}=0$.
\end{verlong}

Now, forget that we fixed $x$. We thus have shown that every $x\in A\left[
\left[  X\right]  \right]  $ satisfying $\left(  X-a\right)  x=0$ satisfies
$x=0$.

But the element $X-a$ of $A\left[  \left[  X\right]  \right]  $ is regular if
and only if every $x\in A\left[  \left[  X\right]  \right]  $ satisfying
$\left(  X-a\right)  x=0$ satisfies $x=0$ (by the definition of
\textquotedblleft regular\textquotedblright). Hence, the element $X-a$ of
$A\left[  \left[  X\right]  \right]  $ is regular (since every $x\in A\left[
\left[  X\right]  \right]  $ satisfying $\left(  X-a\right)  x=0$ satisfies
$x=0$). This proves Proposition \ref{prop.pow.X-a-reg.reg}.
\end{proof}

Proposition \ref{prop.pow.X-a-reg.nil} and Proposition
\ref{prop.pow.X-a-reg.reg} give two sufficient criteria for a power series of
the form $X-a$ to be a regular element of $A\left[  \left[  X\right]  \right]
$. These criteria are not necessary (not even in combination); for example,
the power series $X-\overline{2}\in\left(  \mathbb{Z}/6\mathbb{Z}\right)
\left[  \left[  X\right]  \right]  $ (where $\overline{2}$ denotes the residue
class of $2\in\mathbb{Z}$ modulo $6$) is regular, even though the element
$\overline{2}$ of $\mathbb{Z}/6\mathbb{Z}$ is neither nilpotent nor regular.
More generally, the element $X-a$ of $A\left[  \left[  X\right]  \right]  $ is
regular whenever the ring $A$ is Noetherian\footnote{Actually, a much stronger
statement holds: If $A$ is Noetherian, then a power series $f \in A \left[
\left[  X\right]  \right]  $ is regular in $A \left[  \left[  X\right]
\right]  $ if and only if every $b \in A$ satisfying $bf = 0$ must satisfy $b
= 0$. See \cite[Theorem 5]{Fields71}.}; it is also regular whenever there
exists a $k\in\mathbb{N}$ satisfying $a^{k}A=a^{k+1}A$. We leave the proofs of
these facts to the reader. Let me observe that \textbf{some} condition on $A$
and $a$ is needed to guarantee the regularity of $X-a$; in full generality the
claim would not be true, as the following example shows:

\begin{example}
\label{exa.pow.X-a-reg.not}Let $K$ be a field. Let $A$ be the commutative
$K$-algebra with generators $a,x_{-1},x_{0},x_{1},\ldots$ and relations
\[
x_{-1}=0\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ x_{i}=ax_{i+1}\text{
for all }i\geq-1.
\]
Then, the power series $X-a\in A\left[  \left[  X\right]  \right]  $ is
\textbf{not} regular.
\end{example}

\begin{proof}
[Proof of Example \ref{exa.pow.X-a-reg.not} (sketched).]Define a power series
$x\in A\left[  \left[  X\right]  \right]  $ by $x=\sum_{i\in\mathbb{N}}%
x_{i}X^{i}$. Then, it is easy to see that $\left(  X-a\right)  x=0$.

Let us now prove that $x_{0}\neq0$. Indeed, let us assume the contrary. Thus,
$x_{0}=0$.

Let $\mathbf{A}$ be the polynomial algebra over $K$ in the indeterminates
$\mathbf{a},\mathbf{x}_{-1},\mathbf{x}_{0},\mathbf{x}_{1},\mathbf{x}%
_{2},\ldots$. Then, there is a unique ring homomorphism $\pi:\mathbf{A}%
\rightarrow A$ sending $\mathbf{a}$ to $a$ and sending each $\mathbf{x}_{i}$
to the corresponding $x_{i}$. This ring homomorphism $\pi$ is surjective, and
its kernel is the ideal $\mathbf{I}$ of $\mathbf{A}$ generated by
$\mathbf{x}_{-1}$ and $\mathbf{x}_{i}-\mathbf{ax}_{i+1}$ for all $i\geq-1$.
Since $\pi\left(  \mathbf{x}_{0}\right)  =x_{0}=0$, we see that the polynomial
$\mathbf{x}_{0}\in\mathbf{A}$ lies in this ideal $\mathbf{I}$. In other words,
$\mathbf{x}_{0}$ is an $\mathbf{A}$-linear combination of $\mathbf{x}_{-1}$
and $\mathbf{x}_{i}-\mathbf{ax}_{i+1}$ for all $i\geq-1$. In other words,
there exists some $s\in\left\{  -1,0,1,\ldots\right\}  $, some $\mathbf{u}%
\in\mathbf{A}$ and some $\mathbf{v}_{-1},\mathbf{v}_{0},\ldots,\mathbf{v}%
_{s-1}\in\mathbf{A}$ such that%
\begin{equation}
\mathbf{x}_{0}=\mathbf{u}\cdot\mathbf{x}_{-1}+\sum_{i=-1}^{s-1}\mathbf{v}%
_{i}\cdot\left(  \mathbf{x}_{i}-\mathbf{ax}_{i+1}\right)  .
\label{pf.exa.pow.X-a-reg.not.1}%
\end{equation}
Consider these $s$, $\mathbf{u}$ and $\mathbf{v}_{-1},\mathbf{v}_{0}%
,\ldots,\mathbf{v}_{s-1}$.

Now, consider the polynomial ring $K\left[  T\right]  $. The substitution%
\[
\mathbf{a}\mapsto T,\ \ \ \ \ \ \ \ \ \ \mathbf{x}_{i}\mapsto%
\begin{cases}
T^{s-i}, & \text{if }i\leq s;\\
0, & \text{if }i>s
\end{cases}
\]
defines a $K$-algebra homomorphism $\mathbf{A}\rightarrow K\left[  T\right]
$. Let $\widetilde{\mathbf{u}}$ and $\widetilde{\mathbf{v}}_{i}$ denote the
images of $\mathbf{u}$ and $\mathbf{v}_{i}$ under this substitution. If we
apply this substitution to both sides of the equality
(\ref{pf.exa.pow.X-a-reg.not.1}), then we obtain%
\[
T^{s}=\widetilde{\mathbf{u}}\cdot T^{s+1}+\sum_{i=-1}^{s-1}%
\widetilde{\mathbf{v}}_{i}\cdot\underbrace{\left(  T^{s-i}-TT^{s-\left(
i+1\right)  }\right)  }_{=0}=\widetilde{\mathbf{u}}\cdot T^{s+1}.
\]
Hence, $T^{s}$ is divisible by $T^{s+1}$ in $K\left[  T\right]  $. But this is
clearly absurd. Hence, we have obtained a contradiction.

Thus, $x_{0}\neq0$ is proven. Since $x_{0}$ is the constant term of the power
series $x$, this entails that $x\neq0$. Hence, $X-a$ is not regular (because
if $X-a$ were regular, then $\left(  X-a\right)  x=0$ would yield $x=0$, which
would contradict $x\neq0$). This completes the proof of Example
\ref{exa.pow.X-a-reg.not}.

(Notice that this proof mostly begs the question how the $K$-algebra $A$ looks
like -- e.g., whether it has an explicit combinatorial basis. It only shows
that the element $x_{0}$ of $A$ is nonzero; as we saw, this was enough for our
purposes.\footnote{An explicit combinatorial basis can nevertheless be found.
Namely,%
\[
\left(  1,a,a^{2},a^{3},\ldots,x_{0},x_{1},x_{2},\ldots\right)
\]
is a basis of the $K$-vector space $A$. Indeed, to prove its linear
independence, argue using substitutions such as the one above (for varying
$s$). In order to prove that it spans $A$, argue that any product of two (or
more) $x_{i}$'s is $0$ (because $x_{i}\underbrace{x_{j}}_{=a^{i+1}x_{j+i+1}%
}=\underbrace{a^{i+1}x_{i}}_{=x_{-1}=0}x_{j+i+1}=0$), and that any product of
the form $a^{i}x_{j}$ can be rewritten as $%
\begin{cases}
x_{j-i}, & \text{if }j\geq i;\\
0, & \text{if }j<i
\end{cases}
$.})
\end{proof}

More general criteria for regularity of power series in $A\left[  \left[
X\right]  \right]  $ will be discussed in Section \ref{sect.pow.more}.

\subsection{Notations and the isomorphisms $\rho_{i}$}

We can now prove an analogue of Corollary \ref{cor.mulpol.prodmonic} for power
series, whose proof will be more or less the same as the proof of Corollary
\ref{cor.mulpol.prodmonic} itself. Let us first build up some notations.

Rings of formal power series (of the form $A\left[  \left[  X_{1},X_{2}%
,\ldots,X_{n}\right]  \right]  $) can be obtained recursively by adjoining one
variable after the other (similarly to polynomial rings $A\left[  X_{1}%
,X_{2},\ldots,X_{n}\right]  $); indeed, for each $n\in\mathbb{N}$ and each
$i\in\left\{  1,2,\ldots,n\right\}  $, there is a canonical ring isomorphism%
\begin{equation}
A\left[  \left[  X_{1},X_{2},\ldots,X_{n}\right]  \right]  \cong\left(
A\left[  \left[  X_{1},X_{2},\ldots,\widehat{X_{i}},\ldots,X_{n}\right]
\right]  \right)  \left[  \left[  X_{i}\right]  \right]  \label{eq.mulpow.iso}%
\end{equation}
(where the \textquotedblleft$\widehat{X_{i}}$\textquotedblright\ again means
that the element $X_{i}$ is removed from the list).\footnote{Let me make some
comments about this isomorphism:
\par
Rings of formal power series are topological rings; the topology ensures that
\textquotedblleft reasonable\textquotedblright\ infinite sums such as
$\sum_{i\in\mathbb{N}}X^{i}$ or $\sum_{\left(  i,j\right)  \in\mathbb{N}^{2}%
}X^{i}Y^{j}$ converge. This topology is defined as follows:
\par
Let $B$ be a topological ring. Then, for any $m\in\mathbb{N}$ and any $m$
indeterminates $Y_{1},Y_{2},\ldots,Y_{m}$, we endow the ring $B\left[  \left[
Y_{1},Y_{2},\ldots,Y_{m}\right]  \right]  $ with the following topology: The
$B$-module $B^{\mathbb{N}^{m}}$ is canonically equipped with a product
topology (since $B$ itself has a topology). Use the $B$-module isomorphism
\begin{align*}
B\left[  \left[  Y_{1},Y_{2},\ldots,Y_{m}\right]  \right]   &  \rightarrow
B^{\mathbb{N}^{m}},\\
f  &  \mapsto\left(  \text{the coefficient of }Y_{1}^{k_{1}}Y_{2}^{k_{2}%
}\cdots Y_{m}^{k_{m}}\text{ in }f\right)  _{\left(  k_{1},k_{2},\ldots
,k_{m}\right)  \in\mathbb{N}^{m}}%
\end{align*}
to transport the product topology from $B^{\mathbb{N}^{m}}$ to $B\left[
\left[  Y_{1},Y_{2},\ldots,Y_{m}\right]  \right]  $. The resulting topology on
$B\left[  \left[  Y_{1},Y_{2},\ldots,Y_{m}\right]  \right]  $ makes the
$B$-algebra $B\left[  \left[  Y_{1},Y_{2},\ldots,Y_{m}\right]  \right]  $ into
a topological $B$-algebra; this is the topology we want. Notice that it
depends on the topology on $B$. Explicitly, it is characterized by the
following property: A net $\left(  f_{s}\right)  _{s\in S}\in\left(  B\left[
\left[  Y_{1},Y_{2},\ldots,Y_{m}\right]  \right]  \right)  ^{S}$ of power
series in $B\left[  \left[  Y_{1},Y_{2},\ldots,Y_{m}\right]  \right]  $
converges to a power series $f\in B\left[  \left[  Y_{1},Y_{2},\ldots
,Y_{m}\right]  \right]  $ if and only if for each $\left(  k_{1},k_{2}%
,\ldots,k_{m}\right)  \in\mathbb{N}^{m}$, the net $\left(  \text{the
coefficient of }Y_{1}^{k_{1}}Y_{2}^{k_{2}}\cdots Y_{m}^{k_{m}}\text{ in }%
f_{s}\right)  _{s\in S}\in B^{S}$ converges to $\left(  \text{the coefficient
of }Y_{1}^{k_{1}}Y_{2}^{k_{2}}\cdots Y_{m}^{k_{m}}\text{ in }f\right)  $ in
$B$.
\par
If $B$ is just a ring (not a topological ring), then the preceding definition
still applies, provided that we regard $B$ as a topological ring by equipping
it with the discrete topology.
\par
Thus, the commutative ring $A$ becomes a topological ring (using the discrete
topology), and therefore both sides of the isomorphism (\ref{eq.mulpow.iso})
become topological rings. The isomorphism (\ref{eq.mulpow.iso}) becomes an
isomorphism of topological rings.
\par
Let me once again reiterate that the topology on $B\left[  \left[  Y_{1}%
,Y_{2},\ldots,Y_{m}\right]  \right]  $ is constructed using the topology on
$B$. In particular, the topology on $\left(  A\left[  \left[  X_{1}%
,X_{2},\ldots,\widehat{X_{i}},\ldots,X_{n}\right]  \right]  \right)  \left[
\left[  X_{i}\right]  \right]  $ is constructed using the topology on
$A\left[  \left[  X_{1},X_{2},\ldots,\widehat{X_{i}},\ldots,X_{n}\right]
\right]  $, which is not (in general) the discrete topology!} This isomorphism
is rather useful; let me introduce a notation for it:

\begin{definition}
\label{def.mulpow.rho}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$. Let
$R$ denote the ring $A\left[  \left[  X_{1},X_{2},\ldots,X_{n}\right]
\right]  $ (a ring of formal power series in $n$ indeterminates). For each
$i\in\left\{  1,2,\ldots,n\right\}  $, we let $R_{i}$ denote the ring
$A\left[  \left[  X_{1},X_{2},\ldots,\widehat{X_{i}},\ldots,X_{n}\right]
\right]  $ (a ring of formal power series in $n-1$ indeterminates).

For each $i\in\left\{  1,2,\ldots,n\right\}  $, we regard $R_{i}$ as an
$A$-subalgebra of $R$. For each $i\in\left\{  1,2,\ldots,n\right\}  $, we let
$\rho_{i}:R_{i}\left[  \left[  X\right]  \right]  \rightarrow R$ be the
continuous $R_{i}$-algebra homomorphism which sends every $p\in R_{i}\left[
\left[  X\right]  \right]  $ to $p\left(  X_{i}\right)  $. In other words,
$\rho_{i}:\left(  A\left[  \left[  X_{1},X_{2},\ldots,\widehat{X_{i}}%
,\ldots,X_{n}\right]  \right]  \right)  \left[  \left[  X\right]  \right]
\rightarrow A\left[  \left[  X_{1},X_{2},\ldots,X_{n}\right]  \right]  $ is
the continuous $A$-algebra homomorphism which satisfies%
\begin{equation}
\left(  \rho_{i}\left(  X_{j}\right)  =X_{j}\ \ \ \ \ \ \ \ \ \ \text{for
every }j\in\left\{  1,2,\ldots,\widehat{i},\ldots,n\right\}  \right)
\label{eq.def.mulpow.rho.rhoiXj}%
\end{equation}
an%
\begin{equation}
\rho_{i}\left(  X\right)  =X_{i}. \label{eq.def.mulpow.rho.rhoiX}%
\end{equation}
It is well-known that this $\rho_{i}$ is actually an $R_{i}$-algebra
isomorphism. Indeed, this $\rho_{i}$ is the isomorphism responsible for
(\ref{eq.mulpow.iso}) (at least if we rename the indeterminate $X$ in
$R_{i}\left[  \left[  X\right]  \right]  $ as $X_{i}$). Since the map
$\rho_{i}$ is a $R_{i}$-algebra isomorphism, its inverse $\rho_{i}^{-1}$ is
well-defined and also a $R_{i}$-algebra isomorphism.

These notations $A$, $n$, $R$, $R_{i}$ and $\rho_{i}$ shall be in place for
the rest of Section \ref{sect.pow}.
\end{definition}

\subsection{Regularity of $X_{i}$ and $X_{i}-X_{j}$}

\begin{proposition}
\label{prop.mulpow.reg1}Let $i\in\left\{  1,2,\ldots,n\right\}  $. Then, the
element $X_{i}$ of $R$ is regular.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.mulpow.reg1}.]The map $\rho_{i}:R_{i}\left[
\left[  X\right]  \right]  \rightarrow R$ is an $R_{i}$-algebra isomorphism,
thus a ring isomorphism.

The element $0$ of $R_{i}$ is nilpotent (since $0^{1}=0$). Hence, Proposition
\ref{prop.pow.X-a-reg.nil} (applied to $R_{i}$ and $0$ instead of $A$ and $a$)
yields that the element $X-0$ of $R_{i}\left[  \left[  X\right]  \right]  $ is
regular. In other words, the element $X$ of $R_{i}\left[  \left[  X\right]
\right]  $ is regular (since $X-0=X$). Hence, Proposition \ref{prop.reg.iso}
(applied to $R_{i}\left[  \left[  X\right]  \right]  $, $R$, $\rho_{i}$ and
$X$ instead of $A$, $B$, $f$ and $a$) shows that $\rho_{i}\left(  X\right)  $
is a regular element of $R$ (since $\rho_{i}:R_{i}\left[  \left[  X\right]
\right]  \rightarrow R$ is a ring isomorphism). In other words, $X_{i}$ is a
regular element of $R$ (since $\rho_{i}\left(  X\right)  =X_{i}$). This proves
Proposition \ref{prop.mulpow.reg1}.
\end{proof}

\begin{proposition}
\label{prop.mulpow.reg2}Let $i$ and $j$ be two distinct elements of $\left\{
1,2,\ldots,n\right\}  $. The element $X_{i}-X_{j}$ of $R$ is regular.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.mulpow.reg2}.]The definition of $R_{i}$ yields
$R_{i}=A\left[  \left[  X_{1},X_{2},\ldots,\widehat{X_{i}},\ldots
,X_{n}\right]  \right]  $. Recall that $R_{i}$ is an $A$-subalgebra of $R$,
thus a subring of $R$.

Combining $j\in\left\{  1,2,\ldots,n\right\}  $ with $j\neq i$ (since $i$ and
$j$ are distinct), we obtain $j\in\left\{  1,2,\ldots,n\right\}
\setminus\left\{  i\right\}  =\left\{  1,2,\ldots,\widehat{i},\ldots
,n\right\}  $. Hence, $X_{j}\in A\left[  \left[  X_{1},X_{2},\ldots
,\widehat{X_{i}},\ldots,X_{n}\right]  \right]  =R_{i}$. So we have proven that
$X_{j}$ is an element of $R_{i}$.

Now, $\rho_{i}$ is an $R_{i}$-algebra homomorphism. Hence,%
\[
\rho_{i}\left(  X-X_{j}\right)  =\underbrace{\rho_{i}\left(  X\right)
}_{\substack{=X_{i}\\\text{(by (\ref{eq.def.mulpow.rho.rhoiX}))}%
}}-\underbrace{\rho_{i}\left(  X_{j}\right)  }_{\substack{=X_{j}\\\text{(by
(\ref{eq.def.mulpow.rho.rhoiXj}))}}}=X_{i}-X_{j}.
\]


Proposition \ref{prop.mulpow.reg1} (applied to $j$ instead of $i$) shows that
the element $X_{j}$ of $R$ is regular. Hence, Proposition \ref{prop.reg.sub}
(applied to $R_{i}$, $R$ and $X_{j}$ instead of $A$, $B$ and $a$) shows that
$X_{j}$ is a regular element of $R_{i}$ (since $R_{i}$ is a subring of $R$,
and since $X_{j}$ is an element of $R_{i}$). Hence, Proposition
\ref{prop.pow.X-a-reg.reg} (applied to $R_{i}$ and $X_{j}$ instead of $A$ and
$a$) yields that the element $X-X_{j}$ of $R_{i}\left[  \left[  X\right]
\right]  $ is regular. Hence, Proposition \ref{prop.reg.iso} (applied to
$R_{i}\left[  \left[  X\right]  \right]  $, $R$, $\rho_{i}$ and $X-X_{j}$
instead of $A$, $B$, $f$ and $a$) shows that $\rho_{i}\left(  X-X_{j}\right)
$ is a regular element of $R$ (since $\rho_{i}:R_{i}\left[  \left[  X\right]
\right]  \rightarrow R$ is a ring isomorphism). In other words, $X_{i}-X_{j}$
is a regular element of $R$ (since $\rho_{i}\left(  X-X_{j}\right)
=X_{i}-X_{j}$). This proves Proposition \ref{prop.mulpow.reg2}.
\end{proof}

We can now state an analogue of Corollary \ref{cor.mulpol.prodmonic}
(actually, a generalization):

\begin{corollary}
\label{cor.mulpow.prodmonic}The polynomial $\prod_{1\leq i<j\leq n}\left(
X_{i}-X_{j}\right)  $ is a regular element of $R$.
\end{corollary}

\begin{vershort}
\begin{proof}
[Proof of Corollary \ref{cor.mulpow.prodmonic}.]Analogous to the proof of
Corollary \ref{cor.mulpol.prodmonic} (but using Proposition
\ref{prop.mulpow.reg2} instead of Proposition \ref{prop.mulpol.monic}
\textbf{(b)}).
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Corollary \ref{cor.mulpow.prodmonic}.]Let $G$ be the set $\left\{
\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}
$. Then, $G$ is a subset of $\left\{  1,2,\ldots,n\right\}  ^{2}$. Hence, $G$
is a finite set (since $\left\{  1,2,\ldots,n\right\}  ^{2}$ is a finite set).
Every element of $G$ has the form $\left(  i,j\right)  $ for some $\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$ (by the
definition of $G$).

For every $\left(  i,j\right)  \in G$, define an element $a_{\left(
i,j\right)  }$ of $R$ by $a_{\left(  i,j\right)  }=X_{i}-X_{j}$. Then, for
every $g\in G$, the element $a_{g}$ is a regular element of $R$%
\ \ \ \ \footnote{\textit{Proof.} Let $g\in G$. Then, $g\in G=\left\{  \left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}  $. In
other words, $g=\left(  i,j\right)  $ for some $\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$. Consider this
$\left(  i,j\right)  $.
\par
From $g=\left(  i,j\right)  $, we obtain $a_{g}=a_{\left(  i,j\right)  }%
=X_{i}-X_{j}$ (by the definition of $a_{\left(  i,j\right)  }$). But $i$ and
$j$ are two elements of $\left\{  1,2,\ldots,n\right\}  $ (since $\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$) and are distinct (since
$i<j$). Therefore, the element $X_{i}-X_{j}$ of $R$ is regular (by Proposition
\ref{prop.mulpow.reg2}). Since $a_{g}=X_{i}-X_{j}$, this rewrites as follows:
The element $a_{g}$ of $R$ is regular. Qed.}. Hence, Proposition
\ref{prop.reg.prod} (applied to $R$ instead of $A$) shows that the element
$\prod_{g\in G}a_{g}$ of $R$ is regular.

But every element of $G$ has the form $\left(  i,j\right)  $ for some $\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$. Hence,
we can rename the index $g$ as $\left(  i,j\right)  $ in the product
$\prod_{g\in G}a_{g}$. We thus obtain
\begin{align*}
\prod_{g\in G}a_{g}  &  =\underbrace{\prod_{\left(  i,j\right)  \in G}%
}_{\substack{=\prod_{\substack{\left(  i,j\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2};\\i<j}}\\\text{(since }G=\left\{  \left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}  \text{)}%
}}\underbrace{a_{\left(  i,j\right)  }}_{\substack{=X_{i}-X_{j}\\\text{(by
the}\\\text{definition of }a_{\left(  i,j\right)  }\text{)}}%
}=\underbrace{\prod_{\substack{\left(  i,j\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2};\\i<j}}}_{=\prod_{1\leq i<j\leq n}}\left(  X_{i}-X_{j}\right)
\\
&  =\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  .
\end{align*}


Now, recall that the element $\prod_{g\in G}a_{g}$ of $R$ is regular. In other
words, the element $\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  $ of
$R$ is regular (since $\prod_{g\in G}a_{g}=\prod_{1\leq i<j\leq n}\left(
X_{i}-X_{j}\right)  $). This proves Corollary \ref{cor.mulpow.prodmonic}.
\end{proof}
\end{verlong}

\subsection{Analogues of other properties of polynomials}

Analogues of Theorem \ref{thm.div.G}, Theorem \ref{thm.div.all} and Corollary
\ref{cor.mulpol.copri.prod} for power series can also be stated:

\begin{theorem}
\label{thm.mulpow.div.G}Let $G$ be a subset of the set $\left\{  \left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}  $.

Let $f\in R$ be a formal power series in the $n$ indeterminates $X_{1}%
,X_{2},\ldots,X_{n}$ over $A$. Assume that $f$ is divisible by $X_{i}-X_{j}$
for every $\left(  i,j\right)  \in G$. Then, $f$ is divisible by
$\prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  $.
\end{theorem}

\begin{theorem}
\label{thm.mulpow.div.all}Let $f\in R$ be a formal power series in the $n$
indeterminates $X_{1},X_{2},\ldots,X_{n}$ over $A$. Assume that $f$ is
divisible by $X_{i}-X_{j}$ for every $\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$. Then, $f$ is divisible by
$\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  $.

(Here, the symbol \textquotedblleft$\prod_{1\leq i<j\leq n}$\textquotedblright%
\ is to be understood as in Theorem \ref{thm.div.all}.)
\end{theorem}

\begin{corollary}
\label{cor.mulpow.copri.prod}Let $G$ be a subset of the set $\left\{  \left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}  $.
Then,%
\[
\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}-X_{j}\right)
R\right)  =\left(  \prod_{\left(  i,j\right)  \in G}\left(  X_{i}%
-X_{j}\right)  \right)  R.
\]

\end{corollary}

Theorem \ref{thm.mulpow.div.G}, Theorem \ref{thm.mulpow.div.all} and Corollary
\ref{cor.mulpow.copri.prod} are analogues of Theorem \ref{thm.div.G}, Theorem
\ref{thm.div.all} and Corollary \ref{cor.mulpol.copri.prod}, respectively. I
am not sure whether they can be proven in a similar manner to the proofs given
above. However, they can be derived from their polynomial counterparts using a
fairly straightforward homogeneity argument (indeed, the power series
$X_{i}-X_{j}$ are actually homogeneous polynomials, and therefore so are their
products; hence, the divisibility of a power series $f$ by a product of such
power series can be decided by separately studying each homogeneous component
of $f$). We shall give these proofs in Subsection \ref{subsect.pow.hg}.

Let us next discuss some properties of the polynomial ring $A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $ that do \textbf{not} directly carry over to
the ring $R=A\left[  \left[  X_{1},X_{2},\ldots,X_{n}\right]  \right]  $ of
formal power series.

\begin{question}
What conditions would make an analogue of Corollary \ref{cor.pol.intersect}
for power series true? If $A$ is a ring, and $a$ and $b$ are two elements of
$A$, then what should be required on $A$, $a$ and $b$ in order to guarantee
that
\begin{equation}
bA\left[  \left[  X\right]  \right]  \cap\left(  X-a\right)  A\left[  \left[
X\right]  \right]  =b\left(  X-a\right)  A\left[  \left[  X\right]  \right]
\label{eq.quest.pow.intersect.when}%
\end{equation}
? It can be shown that requiring $a$ to be nilpotent suffices. However,
requiring that $a$ be regular does \textbf{not} suffice, as Example
\ref{exa.pow.intersect.not} below shows.
\end{question}

\begin{example}
\label{exa.pow.intersect.not}Let $K$ be a field. Let $A$ be the commutative
$K$-algebra with generators
\begin{equation}
a,\ \underbrace{x_{-1},x_{0},x_{1},\ldots}_{\text{countably many}%
},\ b,\ \underbrace{z_{-1},z_{0},z_{1},\ldots}_{\text{countably many}}
\label{eq.exa.pow.intersect.not.gens}%
\end{equation}
and relations
\[
x_{-1}=0\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ x_{i}-ax_{i+1}%
=bz_{i}\text{ for all }i\geq-1.
\]
Then, the element $a\in A$ is regular, but we have%
\begin{equation}
bA\left[  \left[  X\right]  \right]  \cap\left(  X-a\right)  A\left[  \left[
X\right]  \right]  \neq b\left(  X-a\right)  A\left[  \left[  X\right]
\right]  . \label{eq.exa.pow.intersect.not.neq}%
\end{equation}

\end{example}

\begin{proof}
[Proof of Example \ref{exa.pow.intersect.not} (sketched).]\textit{1st step:}
Let $\mathbf{A}$ be the polynomial algebra over $K$ in the indeterminates%
\begin{equation}
\mathbf{a},\ \underbrace{\mathbf{x}_{-1},\mathbf{x}_{0},\mathbf{x}_{1},\ldots
}_{\text{countably many}},\ \mathbf{b},\ \underbrace{\mathbf{z}_{-1}%
,\mathbf{z}_{0},\mathbf{z}_{1},\ldots}_{\text{countably many}}.
\label{pf.exa.pow.intersect.not.gensA}%
\end{equation}
Then, there is a unique ring homomorphism $\pi:\mathbf{A}\rightarrow A$
sending each of the indeterminates listed in
(\ref{pf.exa.pow.intersect.not.gensA}) to the corresponding generator in
(\ref{eq.exa.pow.intersect.not.gens}). This ring homomorphism $\pi$ is
surjective, and its kernel is the ideal $\mathbf{I}$ of $\mathbf{A}$ generated
by $\mathbf{x}_{-1}$ and $\mathbf{x}_{i}-\mathbf{ax}_{i+1}-\mathbf{bz}_{i}$
for all $i\geq-1$.

\textit{2nd step:} We shall now prove that the element $a\in A$ is regular.

Indeed, let $p\in A$ be such that $ap=0$. We must prove that $p=0$.

Since $\pi$ is surjective, we can write $p$ in the form $p=\pi\left(
\mathbf{p}\right)  $ for some $\mathbf{p}\in\mathbf{A}$. Consider this
$\mathbf{p}$.

From $a=\pi\left(  \mathbf{a}\right)  $ and $p=\pi\left(  \mathbf{p}\right)
$, we obtain $ap=\pi\left(  \mathbf{a}\right)  \pi\left(  \mathbf{p}\right)
=\pi\left(  \mathbf{ap}\right)  $, so that $\pi\left(  \mathbf{ap}\right)
=ap=0$ and therefore $\mathbf{ap}\in\mathbf{I}$. In other words, $\mathbf{ap}$
is an $\mathbf{A}$-linear combination of $\mathbf{x}_{-1}$ and $\mathbf{x}%
_{i}-\mathbf{ax}_{i+1}-\mathbf{bz}_{i}$ for all $i\geq-1$. In other words,
there exists some $s\in\mathbb{N}$, some $\mathbf{u}\in\mathbf{A}$ and some
$\mathbf{v}_{-1},\mathbf{v}_{0},\ldots,\mathbf{v}_{s-1}\in\mathbf{A}$ such
that%
\begin{equation}
\mathbf{ap}=\mathbf{u}\cdot\mathbf{x}_{-1}+\sum_{i=-1}^{s-1}\mathbf{v}%
_{i}\cdot\left(  \mathbf{x}_{i}-\mathbf{ax}_{i+1}-\mathbf{bz}_{i}\right)  .
\label{pf.exa.pow.intersect.not.step2.ap=}%
\end{equation}
Consider these $s$, $\mathbf{u}$ and $\mathbf{v}_{-1},\mathbf{v}_{0}%
,\ldots,\mathbf{v}_{s-1}$. We WLOG assume that $s$ is large enough that only
the indeterminates%
\begin{equation}
\mathbf{a},\ \mathbf{x}_{-1},\mathbf{x}_{0},\mathbf{x}_{1},\ldots
,\mathbf{x}_{s},\ \mathbf{b},\ \mathbf{z}_{-1},\mathbf{z}_{0},\mathbf{z}%
_{1},\ldots,\mathbf{z}_{s} \label{pf.exa.pow.intersect.not.step2.gensAs}%
\end{equation}
appear in the polynomials $\mathbf{p},\mathbf{u,}\mathbf{v}_{-1}%
,\mathbf{v}_{0},\ldots,\mathbf{v}_{s-1}$ (since we can always increase $s$).

Let $\mathbf{A}_{s}$ be the polynomial algebra over $K$ in the indeterminates
listed in (\ref{pf.exa.pow.intersect.not.step2.gensAs}). Clearly,
$\mathbf{A}_{s}$ is a subalgebra of $\mathbf{A}$. Furthermore, all of the
polynomials $\mathbf{p},\mathbf{u,}\mathbf{v}_{-1},\mathbf{v}_{0}%
,\ldots,\mathbf{v}_{s-1}$ as well as $\mathbf{a}$, $\mathbf{x}_{-1}$ and the
differences $\mathbf{x}_{i}-\mathbf{ax}_{i+1}-\mathbf{bz}_{i}$ for all
$i\in\left\{  -1,0,1,\ldots,s-1\right\}  $ belong to $\mathbf{A}_{s}$. Hence,
we can regard (\ref{pf.exa.pow.intersect.not.step2.ap=}) as an equality inside
$\mathbf{A}_{s}$.

Let $\mathbf{B}_{s}$ be the polynomial algebra over $K$ in the indeterminates
\begin{equation}
\mathbf{a},\ \widetilde{\mathbf{x}}_{s},\ \mathbf{b},\ \mathbf{z}%
_{-1},\mathbf{z}_{0},\mathbf{z}_{1},\ldots,\mathbf{z}_{s}.
\label{pf.exa.pow.intersect.not.step2.gensBs}%
\end{equation}
(You are reading right: The indeterminate $\widetilde{\mathbf{x}}_{s}$ is not
$\mathbf{x}_{s}$, although it is closely related.) Inside $\mathbf{B}_{s}$,
define new elements $\widetilde{\mathbf{x}}_{s-1},\widetilde{\mathbf{x}}%
_{s-2},\ldots,\widetilde{\mathbf{x}}_{-1}$ recursively by setting%
\begin{equation}
\widetilde{\mathbf{x}}_{s-i}=\mathbf{a}\widetilde{\mathbf{x}}_{s-\left(
i-1\right)  }+\mathbf{bz}_{s-i}\ \ \ \ \ \ \ \ \ \ \text{for each }%
i\in\left\{  1,2,\ldots,s+1\right\}  .
\label{pf.exa.pow.intersect.not.step2.xs-i=}%
\end{equation}
Then, it is easy to prove (by induction over $i$) that%
\[
\widetilde{\mathbf{x}}_{s-i}=\mathbf{a}^{i}\widetilde{\mathbf{x}}_{s}%
+\sum_{k=1}^{i}\mathbf{a}^{i-k}\mathbf{bz}_{s-k}\ \ \ \ \ \ \ \ \ \ \text{for
each }i\in\left\{  0,1,\ldots,s+1\right\}  .
\]
Applying this to $i=s+1$, we obtain%
\begin{equation}
\widetilde{\mathbf{x}}_{-1}=\mathbf{a}^{s+1}\widetilde{\mathbf{x}}_{s}%
+\sum_{k=1}^{s+1}\mathbf{a}^{s+1-k}\mathbf{bz}_{s-k}.
\label{pf.exa.pow.intersect.not.step2.x-1=}%
\end{equation}


For each $i\in\left\{  -1,0,\ldots,s-1\right\}  $, we have%
\begin{equation}
\widetilde{\mathbf{x}}_{i}=\mathbf{a}\widetilde{\mathbf{x}}_{i+1}%
+\mathbf{bz}_{i} \label{pf.exa.pow.intersect.not.step2.xi=}%
\end{equation}
(by (\ref{pf.exa.pow.intersect.not.step2.xs-i=}), applied to $s-i$ instead of
$i$).

We can regard $\widetilde{\mathbf{x}}_{-1}\in\mathbf{B}_{s}$ as a polynomial
in the indeterminate $\mathbf{a}$ over the unique factorization domain
$K\left[  \widetilde{\mathbf{x}}_{s},\ \mathbf{b},\ \mathbf{z}_{-1}%
,\mathbf{z}_{0},\mathbf{z}_{1},\ldots,\mathbf{z}_{s}\right]  $. Then,
(\ref{pf.exa.pow.intersect.not.step2.x-1=}) shows that this polynomial
$\widetilde{\mathbf{x}}_{-1}$ has leading coefficient $\widetilde{\mathbf{x}%
}_{s}$, and that the prime $\mathbf{b}$ of the unique factorization domain
$K\left[  \widetilde{\mathbf{x}}_{s},\ \mathbf{b},\ \mathbf{z}_{-1}%
,\mathbf{z}_{0},\mathbf{z}_{1},\ldots,\mathbf{z}_{s}\right]  $ divides all
coefficients of $\widetilde{\mathbf{x}}_{-1}$ except for its leading
coefficient (which is $\widetilde{\mathbf{x}}_{s}$), and that $\mathbf{b}^{2}$
does not divide the constant term of $\widetilde{\mathbf{x}}_{-1}$ (because
the constant term of $\widetilde{\mathbf{x}}_{-1}$ is $\mathbf{bz}_{-1}$).
Therefore, Eisenstein's Irreducibility Criterion (see, e.g., \cite[Corollary
8.22]{Knapp2016}) shows that this polynomial $\widetilde{\mathbf{x}}_{-1}$ is
irreducible over the fraction field of $K\left[  \widetilde{\mathbf{x}}%
_{s},\ \mathbf{b},\ \mathbf{z}_{-1},\mathbf{z}_{0},\mathbf{z}_{1}%
,\ldots,\mathbf{z}_{s}\right]  $. Moreover, the coefficients of this
polynomial $\widetilde{\mathbf{x}}_{-1}$ have no common divisor except for
units\footnote{Indeed, this is clear from observing that the leading
coefficient is $\widetilde{\mathbf{x}}_{s}$, whereas the constant term is
$\mathbf{bz}_{-1}$.}; therefore, $\widetilde{\mathbf{x}}_{-1}$ is irreducible
over the ring $K\left[  \widetilde{\mathbf{x}}_{s},\ \mathbf{b},\ \mathbf{z}%
_{-1},\mathbf{z}_{0},\mathbf{z}_{1},\ldots,\mathbf{z}_{s}\right]  $ as well.
In other words, $\widetilde{\mathbf{x}}_{-1}$ is a prime element of the unique
factorization domain $\mathbf{B}_{s}$.

The substitution%
\[
\mathbf{a}\mapsto\mathbf{a},\ \ \ \ \ \ \ \ \ \ \mathbf{x}_{i}\mapsto
\widetilde{\mathbf{x}}_{i},\ \ \ \ \ \ \ \ \ \ \mathbf{b}\mapsto
\mathbf{b},\ \ \ \ \ \ \ \ \ \ \mathbf{z}_{i}\mapsto\mathbf{z}_{i}%
\]
defines a $K$-algebra homomorphism $\alpha:\mathbf{A}_{s}\rightarrow
\mathbf{B}_{s}$. Let $\widetilde{\mathbf{p}}$, $\widetilde{\mathbf{u}}$ and
$\widetilde{\mathbf{v}}_{i}$ denote the images of $\mathbf{p}$, $\mathbf{u}$
and $\mathbf{v}_{i}$ under this substitution. If we apply this substitution to
both sides of the equality (\ref{pf.exa.pow.intersect.not.step2.ap=}), then we
obtain%
\[
\mathbf{a}\widetilde{\mathbf{p}}=\widetilde{\mathbf{u}}\cdot
\widetilde{\mathbf{x}}_{-1}+\sum_{i=-1}^{s-1}\widetilde{\mathbf{v}}_{i}%
\cdot\underbrace{\left(  \widetilde{\mathbf{x}}_{i}-\mathbf{a}%
\widetilde{\mathbf{x}}_{i+1}-\mathbf{bz}_{i}\right)  }%
_{\substack{=0\\\text{(by (\ref{pf.exa.pow.intersect.not.step2.xi=}))}%
}}=\widetilde{\mathbf{u}}\cdot\widetilde{\mathbf{x}}_{-1}.
\]
Hence, $\mathbf{a}\widetilde{\mathbf{p}}$ is divisible by
$\widetilde{\mathbf{x}}_{-1}$ in $\mathbf{B}_{s}$. Since
$\widetilde{\mathbf{x}}_{-1}$ is a prime element of the unique factorization
domain $\mathbf{B}_{s}$, we thus conclude that either $\mathbf{a}$ or
$\widetilde{\mathbf{p}}$ is divisible by $\widetilde{\mathbf{x}}_{-1}$ in
$\mathbf{B}_{s}$. But since $\mathbf{a}$ is not divisible by
$\widetilde{\mathbf{x}}_{-1}$ in $\mathbf{B}_{s}$ (indeed, the constant term
of $\widetilde{\mathbf{x}}_{-1}$, when regarded as a polynomial in
$\mathbf{a}$, is $\mathbf{bz}_{-1}$, which has no common divisors with
$\mathbf{a}$ except for units), this shows that $\widetilde{\mathbf{p}}$ is
divisible by $\widetilde{\mathbf{x}}_{-1}$ in $\mathbf{B}_{s}$. In other
words, $\widetilde{\mathbf{p}}=\widetilde{\mathbf{x}}_{-1}%
\widetilde{\mathbf{q}}$ for some $\widetilde{\mathbf{q}}\in\mathbf{B}_{s}$.
Consider this $\widetilde{\mathbf{q}}$.

Let $\mathbf{J}_{s}$ be the ideal of $\mathbf{A}_{s}$ generated by the
polynomials $\mathbf{x}_{i}-\mathbf{ax}_{i+1}-\mathbf{bz}_{i}$ with
$i\in\left\{  -1,0,1,\ldots,s-1\right\}  $. The homomorphism $\alpha
:\mathbf{A}_{s}\rightarrow\mathbf{B}_{s}$ sends the generators of
$\mathbf{J}_{s}$ to $0$; therefore, it factors through $\mathbf{A}%
_{s}/\mathbf{J}_{s}$, yielding a $K$-algebra homomorphism $\alpha^{\prime
}:\mathbf{A}_{s}/\mathbf{J}_{s}\rightarrow\mathbf{B}_{s}$. This homomorphism
$\alpha^{\prime}$ is invertible\footnote{\textit{Proof.} The substitution%
\[
\mathbf{a}\mapsto\mathbf{a},\ \ \ \ \ \ \ \ \ \ \widetilde{\mathbf{x}}%
_{s}\mapsto\mathbf{x}_{s},\ \ \ \ \ \ \ \ \ \ \mathbf{b}\mapsto\mathbf{b}%
,\ \ \ \ \ \ \ \ \ \ \mathbf{z}_{i}\mapsto\mathbf{z}_{i}%
\]
defines a $K$-algebra homomorphism $\beta:\mathbf{B}_{s}\rightarrow
\mathbf{A}_{s}$. Composing this $K$-algebra homomorphism $\beta:\mathbf{B}%
_{s}\rightarrow\mathbf{A}_{s}$ with the canonical projection $\mathbf{A}%
_{s}\rightarrow\mathbf{A}_{s}/\mathbf{J}_{s}$ produces a $K$-algebra
homomorphism $\beta^{\prime}:\mathbf{B}_{s}\rightarrow\mathbf{A}%
_{s}/\mathbf{J}_{s}$. Using (\ref{pf.exa.pow.intersect.not.step2.xs-i=}), we
can show (by induction over $i$) that this homomorphism $\beta^{\prime}$ sends
$\widetilde{\mathbf{x}}_{s-i}$ to (the remainder class of) $\mathbf{x}_{s-i}$
for each $i\in\left\{  0,1,\ldots,s+1\right\}  $. In other words, $\beta$
sends $\widetilde{\mathbf{x}}_{i}$ to (the remainder class of) $\mathbf{x}%
_{i}$ for each $i\in\left\{  -1,0,1,\ldots,s-1\right\}  $. Now, the two
$K$-algebra homomorphisms $\alpha^{\prime}\circ\beta^{\prime}$ and
$\beta^{\prime}\circ\alpha^{\prime}$ both preserve the generators listed in
(\ref{pf.exa.pow.intersect.not.step2.gensBs}) and
(\ref{pf.exa.pow.intersect.not.step2.gensAs}), respectively (or, rather, their
remainder classes), and therefore are the identity maps. In other words, the
homomorphisms $\alpha^{\prime}$ and $\beta^{\prime}$ are mutually inverse.
Hence, $\alpha^{\prime}$ is invertible. Qed.}, and therefore injective. In
other words, $\operatorname*{Ker}\alpha=\mathbf{J}_{s}$. Also, $\alpha$ is
surjective (since $\alpha^{\prime}$ is invertible), and thus there exists some
$\mathbf{q}\in\mathbf{A}_{s}$ such that $\widetilde{\mathbf{q}}=\alpha\left(
\mathbf{q}\right)  $. Consider this $\mathbf{q}$.

Now, $\alpha\left(  \mathbf{p}\right)  =\widetilde{\mathbf{p}}%
=\underbrace{\widetilde{\mathbf{x}}_{-1}}_{=\alpha\left(  \mathbf{x}%
_{-1}\right)  }\underbrace{\widetilde{\mathbf{q}}}_{=\alpha\left(
\mathbf{q}\right)  }=\alpha\left(  \mathbf{x}_{-1}\right)  \alpha\left(
\mathbf{q}\right)  =\alpha\left(  \mathbf{x}_{-1}\mathbf{q}\right)  $, so that
$\mathbf{p}-\mathbf{x}_{-1}\mathbf{q}\in\operatorname*{Ker}\alpha
=\mathbf{J}_{s}$.

But the homomorphism $\pi$ sends $\mathbf{J}_{s}$ to $0$. Thus, from
$\mathbf{p}-\mathbf{x}_{-1}\mathbf{q}\in\mathbf{J}_{s}$, we obtain $\pi\left(
\mathbf{p}-\mathbf{x}_{-1}\mathbf{q}\right)  =0$. In other words, $\pi\left(
\mathbf{p}\right)  =\pi\left(  \mathbf{x}_{-1}\mathbf{q}\right)
=\underbrace{\pi\left(  \mathbf{x}_{-1}\right)  }_{=0}\pi\left(
\mathbf{q}\right)  =0$. Thus, $p=\pi\left(  \mathbf{p}\right)  =0$. This
completes our 2nd step.

\textit{3rd step:} It remains to prove (\ref{eq.exa.pow.intersect.not.neq}).

Define a power series $x\in A\left[  \left[  X\right]  \right]  $ by
$x=\sum_{i\in\mathbb{N}}x_{i}X^{i}$. Then, it is easy to see that
\[
\left(  X-a\right)  x=\sum_{i\in\mathbb{N}}\underbrace{\left(  x_{i-1}%
-ax_{i}\right)  }_{\substack{=bz_{i-1}\\\text{(by one of the}\\\text{defining
relations of }A\text{)}}}X^{i}=\sum_{i\in\mathbb{N}}bz_{i-1}X^{i}=b\sum
_{i\in\mathbb{N}}z_{i-1}X^{i}\in bA\left[  \left[  X\right]  \right]  .
\]
Combining this with $\left(  X-a\right)  x\in\left(  X-a\right)  A\left[
\left[  X\right]  \right]  $ (which is obvious), we obtain%
\[
\left(  X-a\right)  x\in bA\left[  \left[  X\right]  \right]  \cap\left(
X-a\right)  A\left[  \left[  X\right]  \right]  .
\]
If we now can prove that%
\begin{equation}
\left(  X-a\right)  x\notin b\left(  X-a\right)  A\left[  \left[  X\right]
\right]  , \label{pf.exa.pow.intersect.not.step3.notin}%
\end{equation}
then (\ref{eq.exa.pow.intersect.not.neq}) will follow, and thus we will be
done. Hence, it remains to prove (\ref{pf.exa.pow.intersect.not.step3.notin}).

\textit{4rd step:} Let us prove (\ref{pf.exa.pow.intersect.not.step3.notin}).
Indeed, assume the contrary. Thus,%
\[
\left(  X-a\right)  x\in b\left(  X-a\right)  A\left[  \left[  X\right]
\right]  .
\]
In other words,
\begin{equation}
\left(  X-a\right)  x=b\left(  X-a\right)  y
\label{pf.exa.pow.intersect.not.step4.=}%
\end{equation}
for some $y\in A\left[  \left[  X\right]  \right]  $. Consider this $y$.

Let $y_{0}$ be the constant term of the power series $y$. Recall that $x_{0}$
is the constant term of the power series $x$. Comparing the constant terms on
both sides of (\ref{pf.exa.pow.intersect.not.step4.=}), we thus obtain
$-ax_{0}=-bay_{0}$. Thus, $ax_{0}=bay_{0}$. Since the element $a$ of $A$ is
regular, we can cancel $a$ from this equality, and therefore obtain
$x_{0}=by_{0}$.

Now, let $A^{\prime}$, $a^{\prime}$ and $x_{i}^{\prime}$ be what was denoted
by $A$, $a$ and $x_{i}$ in Example \ref{exa.pow.X-a-reg.not}. Then,
$x_{0}^{\prime}\neq0$. (Indeed, this is exactly the statement
\textquotedblleft$x_{0}\neq0$\textquotedblright\ that we proved in our proof
of Example \ref{exa.pow.X-a-reg.not}.)

Let $J$ be the ideal of $A$ generated by $b,\ \underbrace{z_{-1},z_{0}%
,z_{1},\ldots}_{\text{countably many}}$. Then, there is a canonical
isomorphism $A/J\rightarrow A^{\prime}$ sending the projections of
$a,\ \underbrace{x_{-1},x_{0},x_{1},\ldots}_{\text{countably many}}$ onto
$A/J$ to the generators $a^{\prime},\ \underbrace{x_{-1}^{\prime}%
,x_{0}^{\prime},x_{1}^{\prime},\ldots}_{\text{countably many}}$ of $A^{\prime
}$ (indeed, just compare the definitions of $A$ and $A^{\prime}$, and observe
that the former becomes the latter if $b,\ \underbrace{z_{-1},z_{0}%
,z_{1},\ldots}_{\text{countably many}}$ are set to $0$).

The canonical projection $A\rightarrow A/J$ sends $x_{0}$ to $x_{0}^{\prime}$
and sends $b$ to $0$. Hence, projecting both sides of the equality
$x_{0}=by_{0}$ onto $A/J$, we obtain $x_{0}^{\prime}=0y_{0}^{\prime}$, where
$y_{0}^{\prime}$ is the image of $y_{0}$. Therefore, $x_{0}^{\prime}%
=0y_{0}^{\prime}=0$, contradicting $x_{0}^{\prime}\neq0$. This contradiction
completes our proof of (\ref{pf.exa.pow.intersect.not.step3.notin}), and this
in turn finishes our proof of Example \ref{exa.pow.intersect.not}.
\end{proof}

In Section \ref{sect.pow.more}, we shall see some sufficient criteria for
analogues of Corollary \ref{cor.pol.intersect} to hold in $A\left[  \left[
X\right]  \right]  $.

\subsection{\label{subsect.pow.hg}Appendix: The graded component trick}

In this subsection, we shall prove Theorem \ref{thm.mulpow.div.G}, Theorem
\ref{thm.mulpow.div.all} and Corollary \ref{cor.mulpow.copri.prod}. Their
proofs will all rely on the notion of homogeneous components. Let us recall
how this notion is defined and introduce a notation for it:

\begin{definition}
\label{def.mulpow.hgc}Let $f\in R$ be a formal power series. Let
$d\in\mathbb{Z}$. Then, $f_{\deg=d}$ shall mean the $d$-th homogeneous
component of $f$; this is defined as follows: Write the formal power series
$f$ in the form $f=\sum_{\left(  i_{1},i_{2},\ldots,i_{n}\right)
\in\mathbb{N}^{n}}a_{\left(  i_{1},i_{2},\ldots,i_{n}\right)  }X_{1}^{i_{1}%
}X_{2}^{i_{2}}\cdots X_{n}^{i_{n}}$, where $a_{\left(  i_{1},i_{2}%
,\ldots,i_{n}\right)  }\in A$ are its coefficients. Then, the $d$-th
homogeneous component of $f$ is defined to be the finite sum%
\[
\sum_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathbb{N}%
^{n};\\i_{1}+i_{2}+\cdots+i_{n}=d}}a_{\left(  i_{1},i_{2},\ldots,i_{n}\right)
}X_{1}^{i_{1}}X_{2}^{i_{2}}\cdots X_{n}^{i_{n}}.
\]


Thus,%
\begin{equation}
f_{\deg=d}=\sum_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)
\in\mathbb{N}^{n};\\i_{1}+i_{2}+\cdots+i_{n}=d}}a_{\left(  i_{1},i_{2}%
,\ldots,i_{n}\right)  }X_{1}^{i_{1}}X_{2}^{i_{2}}\cdots X_{n}^{i_{n}}.
\label{eq.def.mulpow.hgc.sum}%
\end{equation}
Hence,%
\begin{equation}
f_{\deg=d}\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]
\label{eq.def.mulpow.hgc.poly}%
\end{equation}
(since the right hand side of (\ref{eq.def.mulpow.hgc.sum}) is a
\textbf{finite} sum of monomials), and
\begin{equation}
f_{\deg=d}\text{ is a homogeneous polynomial of degree }d
\label{eq.def.mulpow.hgc.sumhgp}%
\end{equation}
(because all the monomials $X_{1}^{i_{1}}X_{2}^{i_{2}}\cdots X_{n}^{i_{n}}$
appearing on the right hand side of (\ref{eq.def.mulpow.hgc.sum}) are of
degree $d$).
\end{definition}

If $f\in R$ and if $d$ is a negative integer, then $f_{\deg=d}=0$ (because the
sum on the right hand side of (\ref{eq.def.mulpow.hgc.sum}) is empty in this
case). Furthermore, it is clear that every formal power series $f\in R$
satisfies%
\begin{equation}
f=\sum_{d\in\mathbb{N}}f_{\deg=d}=\sum_{d\in\mathbb{Z}}f_{\deg=d}.
\label{eq.def.mulpow.hgc.f=sum}%
\end{equation}
Note that the sums $\sum_{d\in\mathbb{Z}}f_{\deg=d}$ and $\sum_{d\in
\mathbb{N}}f_{\deg=d}$ in this equality differ only in the presence (or
absence) of the addends $f_{\deg=d}$ for negative integers $d$; but all these
addends are $0$. Thus, the difference is insubstantial.

The polynomial ring $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ is a graded
ring (since every polynomial $f\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $
is a \textbf{finite} sum of homogeneous polynomials), unlike the ring
$R=A\left[  \left[  X_{1},X_{2},\ldots,X_{n}\right]  \right]  $ of formal
power series. However, $R$, too, has the following property:

\begin{lemma}
\label{lem.mulpow.hg1}Let $\left(  p_{d}\right)  _{d\in\mathbb{Z}}\in
R^{\mathbb{Z}}$ be a family of formal power series. Assume that for each
$d\in\mathbb{Z}$,%
\begin{equation}
\text{the power series }p_{d}\text{ is a homogeneous polynomial of degree
}d\text{.} \label{eq.lem.mulpow.hg1.ass}%
\end{equation}
Then:

\textbf{(a)} The infinite sum $\sum_{d\in\mathbb{Z}}p_{d}$ is well-defined
(i.e., it converges with respect to the topology on $R$).

\textbf{(b)} We have%
\[
\left(  \sum_{d\in\mathbb{Z}}p_{d}\right)  _{\deg=e}=p_{e}%
\ \ \ \ \ \ \ \ \ \ \text{for all }e\in\mathbb{Z}.
\]

\end{lemma}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.mulpow.hg1}.]For each $d\in\mathbb{Z}$, the power
series $p_{d}$ is a homogeneous polynomial of degree $d$ (by
(\ref{eq.lem.mulpow.hg1.ass})), and thus can be written in the form%
\begin{equation}
p_{d}=\sum_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in
\mathbb{N}^{n};\\i_{1}+i_{2}+\cdots+i_{n}=d}}a_{d,\left(  i_{1},i_{2}%
,\ldots,i_{n}\right)  }X_{1}^{i_{1}}X_{2}^{i_{2}}\cdots X_{n}^{i_{n}}
\label{pf.lem.mulpow.hg1.1}%
\end{equation}
for some coefficients $a_{d,\left(  i_{1},i_{2},\ldots,i_{n}\right)  }\in A$.
Consider these coefficients $a_{d,\left(  i_{1},i_{2},\ldots,i_{n}\right)  }$.

\textbf{(a)} The sum $\sum_{\left(  i_{1},i_{2},\ldots,i_{n}\right)
\in\mathbb{N}^{n}}a_{i_{1}+i_{2}+\cdots+i_{n},\left(  i_{1},i_{2},\ldots
,i_{n}\right)  }X_{1}^{i_{1}}X_{2}^{i_{2}}\cdots X_{n}^{i_{n}}$ is clearly
well-defined in $R$ (since it is a sum of distinct monomials with
coefficients). Now, we can rewrite this sum as follows:%
\begin{align}
&  \underbrace{\sum_{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in
\mathbb{N}^{n}}}_{=\sum_{d\in\mathbb{Z}}\sum_{\substack{\left(  i_{1}%
,i_{2},\ldots,i_{n}\right)  \in\mathbb{N}^{n};\\i_{1}+i_{2}+\cdots+i_{n}=d}%
}}a_{i_{1}+i_{2}+\cdots+i_{n},\left(  i_{1},i_{2},\ldots,i_{n}\right)  }%
X_{1}^{i_{1}}X_{2}^{i_{2}}\cdots X_{n}^{i_{n}}\nonumber\\
&  =\sum_{d\in\mathbb{Z}}\sum_{\substack{\left(  i_{1},i_{2},\ldots
,i_{n}\right)  \in\mathbb{N}^{n};\\i_{1}+i_{2}+\cdots+i_{n}=d}%
}\underbrace{a_{i_{1}+i_{2}+\cdots+i_{n},\left(  i_{1},i_{2},\ldots
,i_{n}\right)  }}_{\substack{=a_{d,\left(  i_{1},i_{2},\ldots,i_{n}\right)
}\\\text{(since }i_{1}+i_{2}+\cdots+i_{n}=d\text{)}}}X_{1}^{i_{1}}X_{2}%
^{i_{2}}\cdots X_{n}^{i_{n}}\nonumber\\
&  =\sum_{d\in\mathbb{Z}}\underbrace{\sum_{\substack{\left(  i_{1}%
,i_{2},\ldots,i_{n}\right)  \in\mathbb{N}^{n};\\i_{1}+i_{2}+\cdots+i_{n}%
=d}}a_{d,\left(  i_{1},i_{2},\ldots,i_{n}\right)  }X_{1}^{i_{1}}X_{2}^{i_{2}%
}\cdots X_{n}^{i_{n}}}_{\substack{=p_{d}\\\text{(by (\ref{pf.lem.mulpow.hg1.1}%
))}}}=\sum_{d\in\mathbb{Z}}p_{d}. \label{pf.lem.mulpow.hg1.2}%
\end{align}
Hence, the sum $\sum_{d\in\mathbb{Z}}p_{d}$ is well-defined as well. This
proves Lemma \ref{lem.mulpow.hg1} \textbf{(a)}.

\textbf{(b)} From (\ref{pf.lem.mulpow.hg1.2}), we obtain%
\[
\sum_{d\in\mathbb{Z}}p_{d}=\sum_{\left(  i_{1},i_{2},\ldots,i_{n}\right)
\in\mathbb{N}^{n}}a_{i_{1}+i_{2}+\cdots+i_{n},\left(  i_{1},i_{2},\ldots
,i_{n}\right)  }X_{1}^{i_{1}}X_{2}^{i_{2}}\cdots X_{n}^{i_{n}}.
\]
Hence, the definition of $\left(  \sum_{d\in\mathbb{Z}}p_{d}\right)  _{\deg
=e}$ yields%
\begin{align*}
\left(  \sum_{d\in\mathbb{Z}}p_{d}\right)  _{\deg=e}  &  =\sum
_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathbb{N}^{n}%
;\\i_{1}+i_{2}+\cdots+i_{n}=e}}\underbrace{a_{i_{1}+i_{2}+\cdots+i_{n},\left(
i_{1},i_{2},\ldots,i_{n}\right)  }}_{\substack{=a_{e,\left(  i_{1}%
,i_{2},\ldots,i_{n}\right)  }\\\text{(since }i_{1}+i_{2}+\cdots+i_{n}%
=e\text{)}}}X_{1}^{i_{1}}X_{2}^{i_{2}}\cdots X_{n}^{i_{n}}\\
&  =\sum_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathbb{N}%
^{n};\\i_{1}+i_{2}+\cdots+i_{n}=e}}a_{e,\left(  i_{1},i_{2},\ldots
,i_{n}\right)  }X_{1}^{i_{1}}X_{2}^{i_{2}}\cdots X_{n}^{i_{n}}.
\end{align*}
Comparing this with%
\[
p_{e}=\sum_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in
\mathbb{N}^{n};\\i_{1}+i_{2}+\cdots+i_{n}=e}}a_{e,\left(  i_{1},i_{2}%
,\ldots,i_{n}\right)  }X_{1}^{i_{1}}X_{2}^{i_{2}}\cdots X_{n}^{i_{n}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.mulpow.hg1.1}), applied to
}d=e\right)  ,
\]
we obtain $\left(  \sum_{d\in\mathbb{Z}}p_{d}\right)  _{\deg=e}=p_{e}$. This
proves Lemma \ref{lem.mulpow.hg1} \textbf{(b)}.
\end{proof}
\end{verlong}

Using Lemma \ref{lem.mulpow.hg1}, we can show another basic property of
homogeneous components:

\begin{lemma}
\label{lem.mulpow.hg2}Let $c\in\mathbb{Z}$. Let $p\in A\left[  X_{1}%
,X_{2},\ldots,X_{n}\right]  $ be a homogeneous polynomial of degree $c$. Let
$g\in R$ be any formal power series. Then,%
\[
\left(  pg\right)  _{\deg=e}=p\cdot g_{\deg=e-c}\ \ \ \ \ \ \ \ \ \ \text{for
all }e\in\mathbb{Z}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.mulpow.hg2}.]Applying (\ref{eq.def.mulpow.hgc.f=sum})
to $f=g$, we obtain%
\[
g=\sum_{d\in\mathbb{Z}}g_{\deg=d}=\sum_{d\in\mathbb{Z}}g_{\deg=d-c}%
\]
(here, we have substituted $d-c$ for $d$ in the sum). Hence,%
\begin{equation}
p\underbrace{g}_{=\sum_{d\in\mathbb{Z}}g_{\deg=d-c}}=p\cdot\sum_{d\in
\mathbb{Z}}g_{\deg=d-c}=\sum_{d\in\mathbb{Z}}p\cdot g_{\deg=d-c}.
\label{pf.lem.mulpow.hg2.1}%
\end{equation}


\begin{vershort}
Now, if $d\in\mathbb{Z}$ is arbitrary, then $p\cdot g_{\deg=d-c}$ is a
homogeneous polynomial of degree $d$\ \ \ \ \footnote{\textit{Proof.} Let
$d\in\mathbb{Z}$ be arbitrary. Then, (\ref{eq.def.mulpow.hgc.poly}) (applied
to $g$ and $d-c$ instead of $f$ and $d$) shows that $g_{\deg=d-c}\in A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $. Also, (\ref{eq.def.mulpow.hgc.sumhgp})
(applied to $g$ and $d-c$ instead of $f$ and $d$) shows that $g_{\deg=d-c}$ is
a homogeneous polynomial of degree $d-c$. On the other hand, $p$ is a
homogeneous polynomial in $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ of
degree $c$ (since $p\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $).
\par
But $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ is a graded ring. Hence, the
product of two homogeneous elements of $A\left[  X_{1},X_{2},\ldots
,X_{n}\right]  $ is again homogeneous, and its degree is the sum of the
degrees of the two factors. Applying this to $p$ and $g_{\deg=d-c}$, we
conclude that the product $p\cdot g_{\deg=d-c}$ is a homogeneous polynomial of
degree $c+\left(  d-c\right)  $ (since $p$ is a homogeneous polynomial in
$A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ of degree $c$, and since
$g_{\deg=d-c}$ is a homogeneous polynomial in $A\left[  X_{1},X_{2}%
,\ldots,X_{n}\right]  $ of degree $d-c$). In other words, $p\cdot g_{\deg
=d-c}$ is a homogeneous polynomial of degree $d$ (since $c+\left(  d-c\right)
=d$). Qed.}.
\end{vershort}

\begin{verlong}
Now, if $d\in\mathbb{Z}$ is arbitrary, then $p\cdot g_{\deg=d-c}$ is a
homogeneous polynomial of degree $d$\ \ \ \ \footnote{\textit{Proof.} Let
$d\in\mathbb{Z}$ be arbitrary. Then, (\ref{eq.def.mulpow.hgc.poly}) (applied
to $g$ and $d-c$ instead of $f$ and $d$) shows that $g_{\deg=d-c}\in A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $. Also, (\ref{eq.def.mulpow.hgc.sumhgp})
(applied to $g$ and $d-c$ instead of $f$ and $d$) shows that $g_{\deg=d-c}$ is
a homogeneous polynomial of degree $d-c$. On the other hand, $p$ is a
homogeneous polynomial in $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ of
degree $c$ (since $p\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $).
\par
But $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ is a graded ring. Hence, the
product of two homogeneous elements of $A\left[  X_{1},X_{2},\ldots
,X_{n}\right]  $ is again homogeneous. More precisely: If $u\in\mathbb{N}$ and
$v\in\mathbb{N}$, and if $\alpha\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]
$ is a homogeneous polynomial of degree $u$, and if $\beta\in A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $ is a homogeneous polynomial of degree $v$,
then $\alpha\cdot\beta$ is a homogeneous polynomial of degree $u+v$. We can
apply this to $u=c$, $v=d-c$, $\alpha=p$ and $\beta=g_{\deg=d-c}$ (since we
know that $p$ is a homogeneous polynomial in $A\left[  X_{1},X_{2}%
,\ldots,X_{n}\right]  $ of degree $c$, and since we know that $g_{\deg=d-c}$
is a homogeneous polynomial in $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ of
degree $d-c$). We thus conclude that $p\cdot g_{\deg=d-c}$ is a homogeneous
polynomial of degree $c+\left(  d-c\right)  $. In other words, $p\cdot
g_{\deg=d-c}$ is a homogeneous polynomial of degree $d$ (since $c+\left(
d-c\right)  =d$). Qed.}.
\end{verlong}

Hence, Lemma \ref{lem.mulpow.hg1} \textbf{(a)} (applied to $\left(
p_{d}\right)  _{d\in\mathbb{Z}}=\left(  p\cdot g_{\deg=d-c}\right)
_{d\in\mathbb{Z}}$) yields that the infinite sum $\sum_{d\in\mathbb{Z}}p\cdot
g_{\deg=d-c}$ is well-defined. Moreover, Lemma \ref{lem.mulpow.hg1}
\textbf{(b)} (applied to $\left(  p_{d}\right)  _{d\in\mathbb{Z}}=\left(
p\cdot g_{\deg=d-c}\right)  _{d\in\mathbb{Z}}$) yields that%
\[
\left(  \sum_{d\in\mathbb{Z}}p\cdot g_{\deg=d-c}\right)  _{\deg=e}=p\cdot
g_{\deg=e-c}\ \ \ \ \ \ \ \ \ \ \text{for all }e\in\mathbb{Z}.
\]
But (\ref{pf.lem.mulpow.hg2.1}) yields $pg=\sum_{d\in\mathbb{Z}}p\cdot
g_{\deg=d-c}$. Hence,
\[
\left(  pg\right)  _{\deg=e}=\left(  \sum_{d\in\mathbb{Z}}p\cdot g_{\deg
=d-c}\right)  _{\deg=e}=p\cdot g_{\deg=e-c}\ \ \ \ \ \ \ \ \ \ \text{for all
}e\in\mathbb{Z}.
\]
This proves Lemma \ref{lem.mulpow.hg2}.
\end{proof}

Using Lemma \ref{lem.mulpow.hg2}, we can now prove the following fact, which
helps reduce divisibilities in $R$ to divisibilities of polynomials:

\begin{lemma}
\label{lem.mulpow.hg3}Let $p\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ be
a homogeneous polynomial. Let $f\in R$ be any formal power series. Assume
that
\begin{equation}
p\mid f_{\deg=e}\text{ (in the ring }R\text{)}\ \ \ \ \ \ \ \ \ \ \text{for
each }e\in\mathbb{Z}. \label{eq.lem.mulpow.hg3.ass}%
\end{equation}
Then, $p\mid f$ in the ring $R$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.mulpow.hg3}.]The polynomial $p$ is homogeneous. In
other words, there exists a $c\in\mathbb{Z}$ such that $p$ is homogeneous of
degree $c$. Consider this $c$.

Let $e\in\mathbb{Z}$. Then, (\ref{eq.lem.mulpow.hg3.ass}) shows that $p\mid
f_{\deg=e}$ (in the ring $R$). In other words, there exists an element
$g_{e}\in R$ such that $f_{\deg=e}=pg_{e}$. Consider this $g_{e}$.

At this point it would be tempting to finish the proof by saying that%
\[
f=\sum_{e\in\mathbb{Z}}\underbrace{f_{\deg=e}}_{=pg_{e}}=\sum_{e\in\mathbb{Z}%
}pg_{e}=p\sum_{e\in\mathbb{Z}}g_{e}.
\]
However, this is not entirely correct, since the sum $\sum_{e\in\mathbb{Z}%
}g_{e}$ might not be well-defined (it is an infinite sum, and nothing
guarantees that it converges with respect to the topology on $R$). Thus, we
need to be somewhat more careful.

We first observe an obvious fact: If $u\in A\left[  X_{1},X_{2},\ldots
,X_{n}\right]  $ is a homogeneous polynomial of degree $e$, then%
\begin{equation}
u_{\deg=e}=u \label{pf.lem.mulpow.hg3.udege=u}%
\end{equation}
(since $u_{\deg=e}$ is the $e$-th homogeneous component of $u$, but $u$ is
already homogeneous of degree $e$).

Applying (\ref{eq.def.mulpow.hgc.poly}) to $d=e$, we see that $f_{\deg=e}\in
A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $. Applying
(\ref{eq.def.mulpow.hgc.sumhgp}) to $d=e$, we see that $f_{\deg=e}$ is a
homogeneous polynomial of degree $e$. Hence, (\ref{pf.lem.mulpow.hg3.udege=u})
(applied to $u=f_{\deg=e}$) yields $\left(  f_{\deg=e}\right)  _{\deg
=e}=f_{\deg=e}$. Hence,%
\begin{equation}
f_{\deg=e}=\left(  \underbrace{f_{\deg=e}}_{=pg_{e}}\right)  _{\deg=e}=\left(
pg_{e}\right)  _{\deg=e}=p\cdot\left(  g_{e}\right)  _{\deg=e-c}
\label{pf.lem.mulpow.hg3.2}%
\end{equation}
(by Lemma \ref{lem.mulpow.hg2}, applied to $g_{e}$ instead of $g$). Note that
$\left(  g_{e}\right)  _{\deg=e-c}$ is a homogeneous polynomial of degree
$e-c$ (by (\ref{eq.def.mulpow.hgc.sumhgp}), applied to $g_{e}$ and $e-c$
instead of $f$ and $d$).

Now, forget that we fixed $e$. Thus, for each $e\in\mathbb{Z}$, we have
constructed a $g_{e}\in R$ such that (\ref{pf.lem.mulpow.hg3.2}) holds, and
such that
\begin{equation}
\left(  g_{e}\right)  _{\deg=e-c}\text{ is a homogeneous polynomial of degree
}e-c. \label{pf.lem.mulpow.hg3.3}%
\end{equation}


Now, for each $d\in\mathbb{Z}$, the power series $\left(  g_{d+c}\right)
_{\deg=d}$ is a homogeneous polynomial of degree $d$%
\ \ \ \ \footnote{\textit{Proof.} Let $d\in\mathbb{Z}$. Then,
(\ref{pf.lem.mulpow.hg3.3}) (applied to $e=d+c$) shows that $\left(
g_{d+c}\right)  _{\deg=\left(  d+c\right)  -c}$ is a homogeneous polynomial of
degree $\left(  d+c\right)  -c$. In other words, $\left(  g_{d+c}\right)
_{\deg=d}$ is a homogeneous polynomial of degree $d$ (since $\left(
d+c\right)  -c=d$). Qed.}. Hence, Lemma \ref{lem.mulpow.hg1} \textbf{(a)}
(applied to $\left(  p_{d}\right)  _{d\in\mathbb{Z}}=\left(  \left(
g_{d+c}\right)  _{\deg=d}\right)  _{d\in\mathbb{Z}}$) shows that the infinite
sum $\sum_{d\in\mathbb{Z}}\left(  g_{d+c}\right)  _{\deg=d}$ is well-defined.
Hence,
\begin{align*}
p\cdot\sum_{d\in\mathbb{Z}}\left(  g_{d+c}\right)  _{\deg=d}  &  =\sum
_{d\in\mathbb{Z}}p\cdot\left(  g_{d+c}\right)  _{\deg=d}=\sum_{e\in\mathbb{Z}%
}p\cdot\underbrace{\left(  g_{\left(  e-c\right)  +c}\right)  _{\deg=e-c}%
}_{\substack{=\left(  g_{e}\right)  _{\deg=e-c}\\\text{(since }\left(
e-c\right)  +c=e\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }e-c\text{ for
}d\text{ in the sum}\right) \\
&  =\sum_{e\in\mathbb{Z}}\underbrace{p\cdot\left(  g_{e}\right)  _{\deg=e-c}%
}_{\substack{=f_{\deg=e}\\\text{(by (\ref{pf.lem.mulpow.hg3.2}))}}}=\sum
_{e\in\mathbb{Z}}f_{\deg=e}=\sum_{d\in\mathbb{Z}}f_{\deg=d}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}e\text{ as }d\right) \\
&  =f\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.def.mulpow.hgc.f=sum}%
)}\right)  .
\end{align*}
Hence, $f=p\cdot\sum_{d\in\mathbb{Z}}\left(  g_{d+c}\right)  _{\deg=d}$. Thus,
$p\mid f$ in the ring $R$. This proves Lemma \ref{lem.mulpow.hg3}.
\end{proof}

We can now prove Theorem \ref{thm.mulpow.div.G}:

\begin{proof}
[Proof of Theorem \ref{thm.mulpow.div.G}.]We have assumed that
\begin{equation}
f\text{ is divisible by }X_{i}-X_{j}\text{ for every }\left(  i,j\right)  \in
G. \label{pf.thm.mulpow.div.G.ass}%
\end{equation}


\begin{vershort}
The product $\prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  $ is
clearly a homogeneous polynomial of degree $\left\vert G\right\vert $.
\end{vershort}

\begin{verlong}
Recall that the polynomial ring $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $
is a graded ring. Hence, any finite product of homogeneous polynomials in
$A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ is again a homogeneous
polynomial. Thus, the product $\prod_{\left(  i,j\right)  \in G}\left(
X_{i}-X_{j}\right)  $ is a homogeneous polynomial (since it is the product of
the homogeneous polynomials $X_{i}-X_{j}$ for $\left(  i,j\right)  \in G$).
\end{verlong}

Now, let $e\in\mathbb{Z}$. Then, (\ref{eq.def.mulpow.hgc.poly}) (applied to
$d=e$) shows that $f_{\deg=e}\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $.

Let $\left(  i,j\right)  \in G$. Then, $f$ is divisible by $X_{i}-X_{j}$ (by
(\ref{pf.thm.mulpow.div.G.ass})). In other words, there exists a formal power
series $g\in R$ such that $f=\left(  X_{i}-X_{j}\right)  \cdot g$. Consider
this $g$. Note that (\ref{eq.def.mulpow.hgc.poly}) (applied to $g$ and $e-1$
instead of $f$ and $d$) shows that $g_{\deg=e-1}\in A\left[  X_{1}%
,X_{2},\ldots,X_{n}\right]  $.

Clearly, $X_{i}-X_{j}$ is a homogeneous polynomial of degree $1$, and thus
belongs to $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $. Hence, Lemma
\ref{lem.mulpow.hg2} (applied to $c=1$ and $p=X_{i}-X_{j}$) yields%
\[
\left(  \left(  X_{i}-X_{j}\right)  \cdot g\right)  _{\deg=e}=\left(
X_{i}-X_{j}\right)  \cdot g_{\deg=e-1}.
\]
In view of $f=\left(  X_{i}-X_{j}\right)  \cdot g$, this rewrites as
\[
f_{\deg=e}=\left(  X_{i}-X_{j}\right)  \cdot g_{\deg=e-1}.
\]
Hence, $f_{\deg=e}$ is divisible by $X_{i}-X_{j}$ in the ring $A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $ (since $g_{\deg=e-1}\in A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $).

Now, forget that we fixed $\left(  i,j\right)  $. Thus, we have shown that
$f_{\deg=e}$ is divisible by $X_{i}-X_{j}$ in the ring $A\left[  X_{1}%
,X_{2},\ldots,X_{n}\right]  $ for every $\left(  i,j\right)  \in G$. Hence,
Theorem \ref{thm.div.G} (applied to $f_{\deg=e}$ instead of $f$) yields that
$f_{\deg=e}$ is divisible by $\prod_{\left(  i,j\right)  \in G}\left(
X_{i}-X_{j}\right)  $ in the ring $A\left[  X_{1},X_{2},\ldots,X_{n}\right]
$. In other words, $\prod_{\left(  i,j\right)  \in G}\left(  X_{i}%
-X_{j}\right)  \mid f_{\deg=e}$ in the ring $A\left[  X_{1},X_{2},\ldots
,X_{n}\right]  $. Therefore, $\prod_{\left(  i,j\right)  \in G}\left(
X_{i}-X_{j}\right)  \mid f_{\deg=e}$ in the ring $R$ (since $A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $ is a subring of $R$).

Now, forget that we fixed $e$. We thus have shown that $\prod_{\left(
i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  \mid f_{\deg=e}$ (in the ring
$R$) for each $e\in\mathbb{Z}$. Hence, Lemma \ref{lem.mulpow.hg3} (applied to
$p=\prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  $) shows that
$\prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  \mid f$ in the
ring $R$ (since $\prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)
$ is a homogeneous polynomial). In other words, $f$ is divisible by
$\prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  $. This proves
Theorem \ref{thm.mulpow.div.G}.
\end{proof}

\begin{vershort}
\begin{proof}
[Proof of Theorem \ref{thm.mulpow.div.all}.]Theorem \ref{thm.mulpow.div.all}
follows from Theorem \ref{thm.mulpow.div.G} if we set \newline$G=\left\{
\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j\right\}
$.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Theorem \ref{thm.mulpow.div.all}.]Theorem \ref{thm.mulpow.div.all}
can be derived from Theorem \ref{thm.mulpow.div.G} in the same way as we
derived Theorem \ref{thm.div.all} from Theorem \ref{thm.div.G}.
\end{proof}
\end{verlong}

\begin{vershort}
\begin{proof}
[Proof of Corollary \ref{cor.mulpow.copri.prod}.]It is easy to see that%
\[
\left(  \prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  \right)
R\subseteq\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}%
-X_{j}\right)  R\right)
\]
(since every element of $\left(  \prod_{\left(  i,j\right)  \in G}\left(
X_{i}-X_{j}\right)  \right)  R$ is a multiple of the product $\prod_{\left(
i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  $, and therefore a multiple of
each factor $X_{i}-X_{j}$ of this product). But Theorem \ref{thm.mulpow.div.G}
says that
\[
\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}-X_{j}\right)
R\right)  \subseteq\left(  \prod_{\left(  i,j\right)  \in G}\left(
X_{i}-X_{j}\right)  \right)  R
\]
(since $\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}-X_{j}\right)
R\right)  $ is the set of all $f\in R$ that are divisible by all $X_{i}-X_{j}$
with $\left(  i,j\right)  \in G$, whereas $\left(  \prod_{\left(  i,j\right)
\in G}\left(  X_{i}-X_{j}\right)  \right)  R$ is the set of all $f\in R$ that
are divisible by $\prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)
$). Combining these two relations, we obtain%
\[
\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}-X_{j}\right)
R\right)  =\left(  \prod_{\left(  i,j\right)  \in G}\left(  X_{i}%
-X_{j}\right)  \right)  R.
\]
This proves Corollary \ref{cor.mulpow.copri.prod}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Corollary \ref{cor.mulpow.copri.prod}.]It is easy to see that
\[
\left(  \prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  \right)
R\subseteq\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}%
-X_{j}\right)  R\right)
\]
\ \footnote{\textit{Proof.} Let $a\in\left(  \prod_{\left(  i,j\right)  \in
G}\left(  X_{i}-X_{j}\right)  \right)  R$. We shall show that $a\in
\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}-X_{j}\right)
R\right)  $.
\par
Indeed, we have
\[
a\in\left(  \prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)
\right)  R=\left(  \prod_{\left(  u,v\right)  \in G}\left(  X_{u}%
-X_{v}\right)  \right)  R
\]
(here, we have renamed the index $\left(  i,j\right)  $ as $\left(
u,v\right)  $ in the product). In other words, $a=\left(  \prod_{\left(
u,v\right)  \in G}\left(  X_{u}-X_{v}\right)  \right)  r$ for some $r\in R$.
Consider this $r$.
\par
Now, let $\left(  i,j\right)  \in G$. Then,
\[
\prod_{\left(  u,v\right)  \in G}\left(  X_{u}-X_{v}\right)  =\left(
X_{i}-X_{j}\right)  \cdot\prod_{\substack{\left(  u,v\right)  \in G;\\\left(
u,v\right)  \neq\left(  i,j\right)  }}\left(  X_{u}-X_{v}\right)
\]
(here, we have split off the factor for $\left(  u,v\right)  =\left(
i,j\right)  $ from the product). Now,%
\[
a=\underbrace{\left(  \prod_{\left(  u,v\right)  \in G}\left(  X_{u}%
-X_{v}\right)  \right)  }_{=\left(  X_{i}-X_{j}\right)  \cdot\prod
_{\substack{\left(  u,v\right)  \in G;\\\left(  u,v\right)  \neq\left(
i,j\right)  }}\left(  X_{u}-X_{v}\right)  }r=\left(  X_{i}-X_{j}\right)
\cdot\underbrace{\left(  \prod_{\substack{\left(  u,v\right)  \in G;\\\left(
u,v\right)  \neq\left(  i,j\right)  }}\left(  X_{u}-X_{v}\right)  \right)
\cdot r}_{\in R}\in\left(  X_{i}-X_{j}\right)  R.
\]
\par
Now, forget that we fixed $\left(  i,j\right)  $. We thus have shown that
$a\in\left(  X_{i}-X_{j}\right)  R$ for each $\left(  i,j\right)  \in G$. In
other words, $a\in\bigcap_{\left(  i,j\right)  \in G}\left(  \left(
X_{i}-X_{j}\right)  R\right)  $.
\par
Now, forget that we fixed $a$. We thus have proven that $a\in\bigcap_{\left(
i,j\right)  \in G}\left(  \left(  X_{i}-X_{j}\right)  R\right)  $ for each
$a\in\left(  \prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)
\right)  R$. In other words, $\left(  \prod_{\left(  i,j\right)  \in G}\left(
X_{i}-X_{j}\right)  \right)  R\subseteq\bigcap_{\left(  i,j\right)  \in
G}\left(  \left(  X_{i}-X_{j}\right)  R\right)  $.}. But from Theorem
\ref{thm.mulpow.div.G}, we obtain%
\[
\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}-X_{j}\right)
R\right)  \subseteq\left(  \prod_{\left(  i,j\right)  \in G}\left(
X_{i}-X_{j}\right)  \right)  R
\]
\footnote{\textit{Proof.} Let $f\in\bigcap_{\left(  i,j\right)  \in G}\left(
\left(  X_{i}-X_{j}\right)  R\right)  $. We shall show that $f\in\left(
\prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  \right)  R$.
\par
We have $f\in\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}%
-X_{j}\right)  R\right)  $. In other words, we have $f\in\left(  X_{i}%
-X_{j}\right)  R$ for each $\left(  i,j\right)  \in G$. Thus, the formal power
series $f$ is divisible by $X_{i}-X_{j}$ for each $\left(  i,j\right)  \in G$
(because this is what $f\in\left(  X_{i}-X_{j}\right)  R$ means). So we have
shown that $f$ is divisible by $X_{i}-X_{j}$ for every $\left(  i,j\right)
\in G$. Hence, Theorem \ref{thm.mulpow.div.G} yields that $f$ is divisible by
$\prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  $. In other
words, $f\in\left(  \prod_{\left(  i,j\right)  \in G}\left(  X_{i}%
-X_{j}\right)  \right)  R$.
\par
Now, forget that we fixed $f$. We thus have shown that $f\in\left(
\prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  \right)  R$ for
each $f\in\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}%
-X_{j}\right)  R\right)  $. In other words, $\bigcap_{\left(  i,j\right)  \in
G}\left(  \left(  X_{i}-X_{j}\right)  R\right)  \subseteq\left(
\prod_{\left(  i,j\right)  \in G}\left(  X_{i}-X_{j}\right)  \right)  R$.}.
Combining these two relations, we obtain%
\[
\bigcap_{\left(  i,j\right)  \in G}\left(  \left(  X_{i}-X_{j}\right)
R\right)  =\left(  \prod_{\left(  i,j\right)  \in G}\left(  X_{i}%
-X_{j}\right)  \right)  R.
\]
This proves Corollary \ref{cor.mulpow.copri.prod}.
\end{proof}
\end{verlong}

\section{\label{sect.lcm-coprim-reg}lcm-coprimality meets regularity}

\subsection{Connecting lcm-coprimality to regularity in $A/pA$}

Let us take one more look at \textquotedblleft
lcm-coprimality\textquotedblright\ -- i.e., relations of the form $bR\cap
cR=bcR$ for two elements $b$ and $c$ of a commutative ring $R$ (as they
appear, for example, in Proposition \ref{prop.mulpol.copri}). Let us first
connect such relations with regular elements:

\begin{proposition}
\label{prop.lcm-copri-reg.equiv}Let $A$ be a commutative ring. Let $p$ and $d$
be two elements of $A$. For each $a\in A$, let $\overline{a}$ denote the
residue class of $a$ modulo the ideal $pA$; this is an element of the quotient
ring $A/pA$.

\textbf{(a)} If the element $\overline{d}$ of $A/pA$ is regular, then $dA\cap
pA=dpA$.

\textbf{(b)} Assume that the element $d$ of $A$ is regular. If $dA\cap
pA=dpA$, then the element $\overline{d}$ of $A/pA$ is regular.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.lcm-copri-reg.equiv}.]\textbf{(a)} Assume that
the element $\overline{d}$ of $A/pA$ is regular. We must show that $dA\cap
pA=dpA$. We have%
\begin{equation}
dpA=dpA\cap\underbrace{dp}_{=pd}A=d\underbrace{pA}_{\subseteq A}\cap
p\underbrace{dA}_{\subseteq A}\subseteq dA\cap pA.
\label{pf.prop.lcm-copri-reg.equiv.a.obvi}%
\end{equation}


The element $\overline{d}$ of $A/pA$ is regular if and only if every $x\in
A/pA$ satisfying $\overline{d}x=0$ satisfies $x=0$ (by the definition of
\textquotedblleft regular\textquotedblright). Hence,%
\begin{equation}
\text{every }x\in A/pA\text{ satisfying }\overline{d}x=0\text{ satisfies }x=0
\label{pf.prop.lcm-copri-reg.equiv.a.reg}%
\end{equation}
(since the element $\overline{d}$ of $A/pA$ is regular).

Let $y\in dA\cap pA$. Then, $y\in dA\cap pA\subseteq pA$. Thus, $\overline
{y}=0$ (since $\overline{y}$ is the residue class of $y$ modulo the ideal $pA$).

We have $y\in dA\cap pA\subseteq dA$. In other words, there exists some $z\in
A$ such that $y=dz$. Consider this $z$. We have $y=dz$ and thus $\overline
{y}=\overline{dz}=\overline{d}\cdot\overline{z}$. Comparing this with
$\overline{y}=0$, we obtain $\overline{d}\cdot\overline{z}=0$. Thus,
(\ref{pf.prop.lcm-copri-reg.equiv.a.reg}) (applied to $x=\overline{z}$) yields
$\overline{z}=0$. In other words, $z\in pA$ (since $\overline{z}$ is the
residue class of $z$ modulo the ideal $pA$). Now, $y=d\underbrace{z}_{\in
pA}\in dpA$.

Now, forget that we fixed $y$. We thus have shown that $y\in dpA$ for every
$y\in dA\cap pA$. In other words, $dA\cap pA\subseteq dpA$. Combining this
with (\ref{pf.prop.lcm-copri-reg.equiv.a.obvi}), we obtain $dA\cap pA=dpA$.
This proves Proposition \ref{prop.lcm-copri-reg.equiv} \textbf{(a)}.

\textbf{(b)} The element $d$ of $A$ is regular if and only if every $x\in A$
satisfying $dx=0$ satisfies $x=0$ (by the definition of \textquotedblleft
regular\textquotedblright). Hence,%
\begin{equation}
\text{every }x\in A\text{ satisfying }dx=0\text{ satisfies }x=0
\label{pf.prop.lcm-copri-reg.equiv.b.reg1}%
\end{equation}
(since the element $d$ of $A$ is regular).

Assume that $dA\cap pA=dpA$. Let $x\in A/pA$ be such that $\overline{d}x=0$.
We shall show that $x=0$.

We have $x\in A/pA$. In other words, $x=\overline{y}$ for some $y\in A$.
Consider this $y$. We have $x=\overline{y}$ and thus $\overline{d}%
\underbrace{x}_{=\overline{y}}=\overline{d}\cdot\overline{y}=\overline{dy}$.
Thus, $\overline{dy}=\overline{d}x=0$. In other words, $dy\in pA$ (since
$\overline{dy}$ is the residue class of $dy$ modulo the ideal $pA$). Combining
$d\underbrace{y}_{\in A}\in dA$ with $dy\in pA$, we obtain $dy\in dA\cap
pA=dpA$. In other words, there exists some $z\in A$ such that $dy=dpz$.
Consider this $z$. We have $d\left(  y-pz\right)  =dy-dpz=0$ (since $dy=dpz$).
Hence, (\ref{pf.prop.lcm-copri-reg.equiv.b.reg1}) (applied to $y-pz$ instead
of $x$) yields $y-pz=0$. Thus, $y=p\underbrace{z}_{\in A}\in pA$, so that
$\overline{y}=0$ (since $\overline{y}$ is the residue class of $y$ modulo the
ideal $pA$). Hence, $x=\overline{y}=0$.

Now, forget that we fixed $x$. We thus have shown that%
\begin{equation}
\text{every }x\in A/pA\text{ satisfying }\overline{d}x=0\text{ satisfies }x=0.
\label{pf.prop.lcm-copri-reg.equiv.b.reg2}%
\end{equation}


But the element $\overline{d}$ of $A/pA$ is regular if and only if every $x\in
A/pA$ satisfying $\overline{d}x=0$ satisfies $x=0$ (by the definition of
\textquotedblleft regular\textquotedblright). Hence, the element $\overline
{d}$ of $A/pA$ is regular (since every $x\in A/pA$ satisfying $\overline
{d}x=0$ satisfies $x=0$). This proves Proposition
\ref{prop.lcm-copri-reg.equiv} \textbf{(b)}.
\end{proof}

Using Proposition \ref{prop.lcm-copri-reg.equiv}, we can show the following
counterpart to Proposition \ref{prop.mulpol.copri}:

\begin{proposition}
\label{prop.lcm-copri-reg.prod}Let $R$ be a commutative ring. Let $G$ be a
finite set. For every $g\in G$, let $a_{g}$ be a regular element of $R$. Let
$p\in R$. Assume that every $g\in G$ satisfies $pR\cap a_{g}R=pa_{g}R$. Let
$b=\prod_{g\in G}a_{g}$. Then, $pR\cap bR=pbR$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.lcm-copri-reg.prod}.]For each $a\in R$, let
$\overline{a}$ denote the residue class of $a$ modulo the ideal $pR$; this is
an element of the quotient ring $R/pR$. Notice that%
\[
\prod_{g\in G}\overline{a_{g}}=\overline{\prod_{g\in G}a_{g}}=\overline{b}%
\]
(since $\prod_{g\in G}a_{g}=b$).

We have assumed that every $g\in G$ satisfies%
\begin{equation}
pR\cap a_{g}R=pa_{g}R. \label{pf.prop.lcm-copri-reg.prod.1}%
\end{equation}


Now, let $g\in G$. Then, $a_{g}$ is a regular element of $R$. Also,
\begin{align*}
a_{g}R\cap pR  &  =pR\cap a_{g}R=\underbrace{pa_{g}}_{=a_{g}p}%
R\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.lcm-copri-reg.prod.1}%
)}\right) \\
&  =a_{g}pR.
\end{align*}
Hence, Proposition \ref{prop.lcm-copri-reg.equiv} \textbf{(b)} (applied to
$A=R$ and $d=a_{g}$) yields that the element $\overline{a_{g}}$ of $R/pR$ is regular.

Now, forget that we fixed $g$. We thus have proven that the element
$\overline{a_{g}}$ of $R/pR$ is regular for each $g\in G$. Thus, Proposition
\ref{prop.reg.prod} (applied to $R/pR$ and $\overline{a_{g}}$ instead of $A$
and $a_{g}$) yields that the element $\prod_{g\in G}\overline{a_{g}}$ of
$R/pR$ is regular. In other words, the element $\overline{b}$ of $R/pR$ is
regular (since $\prod_{g\in G}\overline{a_{g}}=\overline{b}$). Hence,
Proposition \ref{prop.lcm-copri-reg.equiv} \textbf{(a)} (applied to $A=R$ and
$d=b$) yields that $bR\cap pR=\underbrace{bp}_{=pb}R=pbR$. Thus, $pR\cap
bR=bR\cap pR=pbR$. This proves Proposition \ref{prop.lcm-copri-reg.prod}.
\end{proof}

\subsection{A second proof of Proposition \ref{prop.copri.prod}}

Proposition \ref{prop.lcm-copri-reg.prod} leads to an alternative proof of
Proposition \ref{prop.copri.prod} (or, more precisely, allows us to simplify
our proof above):

\begin{vershort}
\begin{proof}
[Second proof of Proposition \ref{prop.copri.prod}.]We proceed precisely as in
the above proof of Proposition \ref{prop.copri.prod}, until we prove
(\ref{pf.prop.copri.prod.short.prodag}) (in the induction step). From there,
Proposition \ref{prop.lcm-copri-reg.prod} allows us to take the following shortcut:

The element $a_{g}$ is a regular element of $R$ for every $g\in S\setminus
\left\{  k\right\}  $. Moreover, every $g\in S\setminus\left\{  k\right\}  $
satisfies $a_{k}R\cap a_{g}R=a_{k}a_{g}R$ (by
(\ref{pf.prop.copri.prod.short.ass-2}), applied to $k$ and $g$ instead of $g$
and $h$). Hence, Proposition \ref{prop.lcm-copri-reg.prod} (applied to
$S\setminus\left\{  k\right\}  $ and $a_{k}$ instead of $G$ and $p$) yields
$a_{k}R\cap bR=a_{k}bR$. Hence, (\ref{pf.prop.copri.prod.short.indass.3})
becomes
\[
\bigcap_{g\in S}\left(  a_{g}R\right)  =a_{k}R\cap bR=\underbrace{a_{k}%
b}_{\substack{=\prod_{g\in S}a_{g}\\\text{(by
(\ref{pf.prop.copri.prod.short.prodag}))}}}R=\left(  \prod_{g\in S}%
a_{g}\right)  R.
\]
Once again, this completes the induction step, and thus Proposition
\ref{prop.copri.prod} is proven again.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Second proof of Proposition \ref{prop.copri.prod}.]We have assumed that every
two distinct elements $g$ and $h$ of $G$ satisfy%
\begin{equation}
a_{g}R\cap a_{h}R=a_{g}a_{h}R. \label{pf.prop.copri.prod.pf2.ass-2}%
\end{equation}
Also, we have assumed that%
\begin{equation}
a_{g}\text{ is a regular element of }R \label{pf.prop.copri.prod.pf2.ass-3}%
\end{equation}
for every $g\in G$.

Every subset $S$ of $G$ is finite (since $G$ is finite). Hence, for every
subset $S$ of $G$, the cardinality $\left\vert S\right\vert $ is a nonnegative integer.

Now, we shall prove that every subset $S$ of $G$ satisfies%
\begin{equation}
\bigcap_{g\in S}\left(  a_{g}R\right)  =\left(  \prod_{g\in S}a_{g}\right)  R.
\label{pf.prop.copri.prod.pf2.main}%
\end{equation}


\textit{Proof of (\ref{pf.prop.copri.prod.pf2.main}):} We will prove
(\ref{pf.prop.copri.prod.pf2.main}) by induction on $\left\vert S\right\vert $:

\textit{Induction base:} The equality (\ref{pf.prop.copri.prod.pf2.main})
holds in the case when $\left\vert S\right\vert =0$%
\ \ \ \ \footnote{\textit{Proof.} Assume that $\left\vert S\right\vert =0$. We
must prove that the equality (\ref{pf.prop.copri.prod.pf2.main}) holds.
\par
We have $\left\vert S\right\vert =0$, so that $S=\varnothing$. Hence,
\[
\bigcap_{g\in S}\left(  a_{g}R\right)  =\bigcap_{g\in\varnothing}\left(
a_{g}R\right)  =\left(  \text{an empty intersection of subsets of }R\right)
=R.
\]
Comparing this with%
\begin{align*}
\left(  \prod_{g\in S}a_{g}\right)  R  &  =\underbrace{\left(  \prod
_{g\in\varnothing}a_{g}\right)  }_{\substack{=\left(  \text{an empty product
of elements of }R\right)  \\=1}}R\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}S=\varnothing\right) \\
&  =1R=R,
\end{align*}
we obtain $\bigcap_{g\in S}\left(  a_{g}R\right)  =\left(  \prod_{g\in S}%
a_{g}\right)  R$. In other words, the equality
(\ref{pf.prop.copri.prod.pf2.main}) holds. Qed.}. This completes the induction base.

\textit{Induction step:} Let $N$ be a positive integer. Assume that the
equality (\ref{pf.prop.copri.prod.pf2.main}) holds in the case when
$\left\vert S\right\vert =N-1$. We must prove that the equality
(\ref{pf.prop.copri.prod.pf2.main}) holds in the case when $\left\vert
S\right\vert =N$.

We have assumed that the equality (\ref{pf.prop.copri.prod.pf2.main}) holds in
the case when $\left\vert S\right\vert =N-1$. In other words, the following
fact holds:

\begin{statement}
\textit{Fact 1:} Every subset $S$ of $G$ satisfying $\left\vert S\right\vert
=N-1$ satisfies
\[
\bigcap_{g\in S}\left(  a_{g}R\right)  =\left(  \prod_{g\in S}a_{g}\right)
R.
\]

\end{statement}

Now, let $S$ be a subset of $G$ satisfying $\left\vert S\right\vert =N$. We
shall prove that $\bigcap_{g\in S}\left(  a_{g}R\right)  =\left(  \prod_{g\in
S}a_{g}\right)  R$.

We have $\left\vert S\right\vert =N>0$ (since $N$ is a positive integer).
Hence, the set $S$ is nonempty. In other words, there exists a $k\in S$.
Consider this $k$.

Define an element $b\in R$ by $b=\prod_{g\in S\setminus\left\{  k\right\}
}a_{g}$.

The element $a_{g}$ is a regular element of $R$ for every $g\in S\setminus
\left\{  k\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let $g\in
S\setminus\left\{  k\right\}  $. Then, $g\in S\setminus\left\{  k\right\}
\subseteq S\subseteq G$. Hence, $a_{g}$ is a regular element of $R$ (by
(\ref{pf.prop.copri.prod.pf2.ass-3})). Qed.}. Every $g\in S\setminus\left\{
k\right\}  $ satisfies%
\begin{equation}
a_{k}R\cap a_{g}R=a_{k}a_{g}R \label{pf.prop.copri.prod.pf2.everyg}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.copri.prod.pf2.everyg}):} Let $g\in
S\setminus\left\{  k\right\}  $. Thus, $g\in S$ and $g\neq k$. Now, $g$ and
$k$ are two elements of $G$ (since $g\in S\subseteq G$ and $k\in S\subseteq
G$), and are distinct (since $g\neq k$). Thus, $g$ and $k$ are two distinct
elements of $G$. In other words, $k$ and $g$ are two distinct elements of $G$.
Thus, (\ref{pf.prop.copri.prod.pf2.ass-2}) (applied to $k$ and $g$ instead of
$g$ and $k$) yields $a_{k}R\cap a_{g}R=a_{k}a_{g}R$. This proves
(\ref{pf.prop.copri.prod.pf2.everyg}).}. Hence, Proposition
\ref{prop.lcm-copri-reg.prod} (applied to $S\setminus\left\{  k\right\}  $ and
$a_{k}$ instead of $G$ and $p$) yields
\begin{equation}
a_{k}R\cap bR=a_{k}bR. \label{pf.prop.copri.prod.pf2.1}%
\end{equation}


From $k\in S$, we obtain $\left\vert S\setminus\left\{  k\right\}  \right\vert
=\underbrace{\left\vert S\right\vert }_{=N}-1=N-1$. Hence, Fact 1 (applied to
$S\setminus\left\{  k\right\}  $ instead of $S$) yields%
\begin{equation}
\bigcap_{g\in S\setminus\left\{  k\right\}  }\left(  a_{g}R\right)
=\underbrace{\left(  \prod_{g\in S\setminus\left\{  k\right\}  }a_{g}\right)
}_{=b}R=bR. \label{pf.prop.copri.prod.pf2.indass}%
\end{equation}


But $k\in S$. Hence, we can split off the term for $g=k$ from the intersection
$\bigcap_{g\in S}\left(  a_{g}R\right)  $. We thus obtain
\begin{align}
\bigcap_{g\in S}\left(  a_{g}R\right)   &  =a_{k}R\cap\underbrace{\bigcap
_{g\in S\setminus\left\{  k\right\}  }\left(  a_{g}R\right)  }%
_{\substack{=bR\\\text{(by (\ref{pf.prop.copri.prod.pf2.indass}))}}%
}=a_{k}R\cap bR\nonumber\\
&  =a_{k}bR\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.copri.prod.pf2.1})}\right)  . \label{pf.prop.copri.prod.pf2.2}%
\end{align}


On the other hand, $k\in S$. Hence, we can split off the factor for $g=k$ from
the product $\prod_{g\in S}a_{g}$. We thus obtain
\begin{equation}
\prod_{g\in S}a_{g}=a_{k}\underbrace{\prod_{g\in S\setminus\left\{  k\right\}
}a_{g}}_{=b}=a_{k}b. \label{pf.prop.copri.prod.pf2.prodag}%
\end{equation}
Now, (\ref{pf.prop.copri.prod.pf2.2}) becomes%
\[
\bigcap_{g\in S}\left(  a_{g}R\right)  =\underbrace{a_{k}b}_{\substack{=\prod
_{g\in S}a_{g}\\\text{(by (\ref{pf.prop.copri.prod.pf2.prodag}))}}}R=\left(
\prod_{g\in S}a_{g}\right)  R.
\]


Now, forget that we fixed $S$. We thus have shown that every subset $S$ of $G$
satisfying $\left\vert S\right\vert =N$ satisfies $\bigcap_{g\in S}\left(
a_{g}R\right)  =\left(  \prod_{g\in S}a_{g}\right)  R$. In other words, the
equality (\ref{pf.prop.copri.prod.pf2.main}) holds in the case when
$\left\vert S\right\vert =N$. This completes the induction proof of
(\ref{pf.prop.copri.prod.pf2.main}).

Now, we can apply (\ref{pf.prop.copri.prod.pf2.main}) to $S=G$. We thus obtain
$\bigcap_{g\in G}\left(  a_{g}R\right)  =\left(  \prod_{g\in G}a_{g}\right)
R$. This proves Proposition \ref{prop.copri.prod}.
\end{proof}
\end{verlong}

\begin{remark}
Proposition \ref{prop.copri.prod} \textbf{cannot} be generalized by lifting
the condition that the $a_{g}$ be regular. Here is a counterexample:

Let $K$ be a field. Let $R$ be the commutative $K$-algebra given by generators
$x,y,z$ and relations $yz=zx=xy$. Notice that $R$ is a quotient of the
polynomial ring $K\left[  \mathbf{x},\mathbf{y},\mathbf{z}\right]  $ by the
homogeneous ideal generated by $\mathbf{yz}-\mathbf{zx}$ and $\mathbf{zx}%
-\mathbf{xy}$; thus, computations inside $R$ can easily be done on a computer.
(A Gr\"{o}bner basis of said ideal with respect to the lexicographic order is
$\left(  \mathbf{xy}-\mathbf{yz},\mathbf{xz}-\mathbf{yz},\mathbf{y}%
^{2}\mathbf{z}-\mathbf{yz}^{2}\right)  $. A basis of the $K$-vector space $R$
is the family $\left(  x^{i}\right)  _{i\geq0}\cup\left(  yz^{i}\right)
_{i\geq0}\cup\left(  z^{i}\right)  _{i\geq1}$.)

Let $G=\left\{  1,2,3\right\}  $, and set $a_{1}=x$, $a_{2}=y$ and $a_{3}=z$.
(Of course, none of the three elements $a_{1},a_{2},a_{3}$ of $R$ is regular.)
It is easy to see that every two distinct elements $g$ and $h$ of $G$ satisfy
$a_{g}R\cap a_{h}R=a_{g}a_{h}R$. However, it is not true that $\bigcap_{g\in
G}\left(  a_{g}R\right)  =\left(  \prod_{g\in G}a_{g}\right)  R$ (since the
element $yz=zx=xy$ of $R$ belongs to $\bigcap_{g\in G}\left(  a_{g}R\right)  $
but not to $\left(  \prod_{g\in G}a_{g}\right)  R$).

Proposition \ref{prop.lcm-copri-reg.prod} also fails if we lift the condition
that the $a_{g}$ be regular. A counterexample can be obtained from the same
setting, using $G=\left\{  1,2\right\}  $ and $p=a_{3}$ this time.
\end{remark}

\section{\label{sect.subst}Some words on substitutions}

We are next going to restate Theorem \ref{thm.div.all} in a different way (one
which is often easier to apply in practical use cases):

\begin{corollary}
\label{cor.div.substituted}Let $A$ be a commutative ring. Let $n\in\mathbb{N}%
$. Let $f\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ be a polynomial in
the $n$ indeterminates $X_{1},X_{2},\ldots,X_{n}$ over $A$. Assume that $f$
satisfies the following property:

\begin{statement}
\textit{Property 1:} For every $\left(  i,j\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2}$ satisfying $i<j$, the result of substituting $X_{j}$ for
$X_{i}$ in $f$ is $0$.
\end{statement}

Then, $f$ is divisible by $\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)
$.
\end{corollary}

In order to derive this corollary from Theorem \ref{thm.div.all}, we shall
need the following lemma:

\begin{lemma}
\label{lem.subst.factor-out}Let $A$ be a commutative ring. Let $n\in
\mathbb{N}$. Let $f\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ be a
polynomial in the $n$ indeterminates $X_{1},X_{2},\ldots,X_{n}$ over $A$. Let
$\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$ be such that
$i<j$.

Assume that the result of substituting $X_{j}$ for $X_{i}$ in $f$ is $0$.
Then, $f$ is divisible by $X_{i}-X_{j}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.subst.factor-out}.]Let us use the notations $R$,
$R_{i}$ and $\rho_{i}$ introduced in Definition \ref{def.mulpol.rho}. Thus,
$f\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  =R$.

Let $\kappa$ be the map from $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ to
$A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ which sends each $p\in A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $ to the result of substituting $X_{j}$ for
$X_{i}$ in $p$. This map $\kappa$ is an $A$-algebra homomorphism (in fact, it
is an evaluation homomorphism); it satisfies%
\begin{equation}
\kappa\left(  X_{u}\right)  =%
\begin{cases}
X_{u}, & \text{if }u\neq i;\\
X_{j}, & \text{if }u=i
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for each }u\in\left\{  1,2,\ldots,n\right\}  .
\label{pf.lem.subst.factor-out.1}%
\end{equation}


The map $\kappa$ is an $A$-algebra homomorphism from $A\left[  X_{1}%
,X_{2},\ldots,X_{n}\right]  $ to $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $.
Since $R=A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $, this rewrites as
follows: The map $\kappa$ is an $A$-algebra homomorphism from $R$ to $R$.

Let $u$ be the element $X_{i}-X_{j}$ of $R$. Clearly, $uR$ is an ideal of $R$;
thus, $R/uR$ is a quotient ring of $R$. Let $\pi$ be the canonical projection
$R\rightarrow R/uR$. Then, $\pi$ is an $A$-algebra homomorphism satisfying
$\operatorname*{Ker}\pi=uR$. Thus, $\pi\circ\kappa:R\rightarrow R/uR$ is an
$A$-algebra homomorphism (since it is the composition of the two $A$-algebra
homomorphisms $\pi:R\rightarrow R/uR$ and $\kappa:R\rightarrow R$).

We have $\left(  \pi\circ\kappa\right)  \left(  X_{u}\right)  =\pi\left(
X_{u}\right)  $ for each $u\in\left\{  1,2,\ldots,n\right\}  $%
\ \ \ \ \footnote{\textit{Proof.} Let $u\in\left\{  1,2,\ldots,n\right\}  $.
We must prove that $\left(  \pi\circ\kappa\right)  \left(  X_{u}\right)
=\pi\left(  X_{u}\right)  $.
\par
We are in one of the following two cases:
\par
\textit{Case 1:} We have $u=i$.
\par
\textit{Case 2:} We have $u\neq i$.
\par
Let us first consider Case 1. In this case, we have $u=i$. Now,
(\ref{pf.lem.subst.factor-out.1}) yields $\kappa\left(  X_{u}\right)  =%
\begin{cases}
X_{u}, & \text{if }u\neq i;\\
X_{j}, & \text{if }u=i
\end{cases}
=X_{j}$ (since $u=i$). Now,%
\[
\left(  \pi\circ\kappa\right)  \left(  X_{u}\right)  =\pi\left(
\underbrace{\kappa\left(  X_{u}\right)  }_{=X_{j}}\right)  =\pi\left(
X_{j}\right)  .
\]
Hence,%
\begin{align*}
\pi\left(  \underbrace{X_{u}}_{\substack{=X_{i}\\\text{(since }u=i\text{)}%
}}\right)  -\underbrace{\left(  \pi\circ\kappa\right)  \left(  X_{u}\right)
}_{=\pi\left(  X_{j}\right)  }  &  =\pi\left(  X_{i}\right)  -\pi\left(
X_{j}\right) \\
&  =\pi\left(  X_{i}-X_{j}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\pi\text{ is an }A\text{-algebra homomorphism}\right) \\
&  =0
\end{align*}
(since $X_{i}-X_{j}=u\in uR=\operatorname*{Ker}\pi$). Thus, $\left(  \pi
\circ\kappa\right)  \left(  X_{u}\right)  =\pi\left(  X_{u}\right)  $. Hence,
$\left(  \pi\circ\kappa\right)  \left(  X_{u}\right)  =\pi\left(
X_{u}\right)  $ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $u\neq i$. The equality
(\ref{pf.lem.subst.factor-out.1}) yields $\kappa\left(  X_{u}\right)  =%
\begin{cases}
X_{u}, & \text{if }u\neq i;\\
X_{j}, & \text{if }u=i
\end{cases}
=X_{u}$ (since $u\neq i$). Now,%
\[
\left(  \pi\circ\kappa\right)  \left(  X_{u}\right)  =\pi\left(
\underbrace{\kappa\left(  X_{u}\right)  }_{=X_{u}}\right)  =\pi\left(
X_{u}\right)  .
\]
Hence, $\left(  \pi\circ\kappa\right)  \left(  X_{u}\right)  =\pi\left(
X_{u}\right)  $ is proven in Case 2.
\par
We thus have proven $\left(  \pi\circ\kappa\right)  \left(  X_{u}\right)
=\pi\left(  X_{u}\right)  $ in both Cases 1 and 2. Since these two Cases cover
all possibilities, this shows that $\left(  \pi\circ\kappa\right)  \left(
X_{u}\right)  =\pi\left(  X_{u}\right)  $ always holds.}. In other words, the
two $A$-algebra homomorphisms $\pi\circ\kappa$ and $\pi$ are equal to each
other on each of the elements $X_{1},X_{2},\ldots,X_{n}$.

But $R=A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $. Hence, the elements
$X_{1},X_{2},\ldots,X_{n}$ generate the $A$-algebra $R$. Therefore, if two
$A$-algebra homomorphisms $\phi:R\rightarrow S$ and $\psi:R\rightarrow S$
(where $S$ is any $A$-algebra) are equal to each other on each of the elements
$X_{1},X_{2},\ldots,X_{n}$, then these two $A$-algebra homomorphisms $\phi$
and $\psi$ must be identical. Applying this to $S=R/uR$, $\phi=\pi\circ\kappa$
and $\psi=\pi$, we conclude that the two $A$-algebra homomorphisms $\pi
\circ\kappa$ and $\pi$ must be identical (since $\pi\circ\kappa$ and $\pi$ are
equal to each other on each of the elements $X_{1},X_{2},\ldots,X_{n}$). In
other words, $\pi\circ\kappa=\pi$.

Now, the definition of $\kappa$ shows that $\kappa\left(  f\right)  $ is the
result of substituting $X_{j}$ for $X_{i}$ in $f$. In other words,
$\kappa\left(  f\right)  $ is $0$ (since the result of substituting $X_{j}$
for $X_{i}$ in $f$ is $0$). Thus, $\kappa\left(  f\right)  =0$. Hence,
$\left(  \pi\circ\kappa\right)  \left(  f\right)  =\pi\left(
\underbrace{\kappa\left(  f\right)  }_{=0}\right)  =\pi\left(  0\right)  =0$
(since $\pi$ is an $A$-algebra homomorphism). Since $\pi\circ\kappa=\pi$, this
rewrites as $\pi\left(  f\right)  =0$. In other words, $f\in
\operatorname*{Ker}\pi=uR$. In other words, $f$ is divisible by $u$. In other
words, $f$ is divisible by $X_{i}-X_{j}$ (since $u=X_{i}-X_{j}$). This proves
Lemma \ref{lem.subst.factor-out}.
\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor.div.substituted}.]The polynomial $f$ is divisible
by $X_{i}-X_{j}$ for every $\left(  i,j\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2}$ satisfying $i<j$\ \ \ \ \footnote{\textit{Proof.} Fix any
$\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$.
We must prove that $f$ is divisible by $X_{i}-X_{j}$.
\par
Property 1 (in the statement of Corollary \ref{cor.div.substituted}) shows
that the result of substituting $X_{j}$ for $X_{i}$ in $f$ is $0$. Thus, Lemma
\ref{lem.subst.factor-out} shows that $f$ is divisible by $X_{i}-X_{j}$.
Qed.}. Hence, Theorem \ref{thm.div.all} shows that $f$ is divisible by
$\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  $. This proves Corollary
\ref{cor.div.substituted}.
\end{proof}

\section{\label{sect.symmetric}A consequence on symmetric polynomials}

\subsection{Dividing by regular elements}

From Corollary \ref{cor.div.substituted}, we can furthermore derive a fact
about symmetric polynomials. Let us first prepare by showing two trivial facts
and making some definitions:

\begin{lemma}
\label{lem.regular.cancel}Let $A$ be a commutative ring. Let $b$ be a regular
element of $A$. Let $z_{1}$ and $z_{2}$ be two elements of $A$ such that
$bz_{1}=bz_{2}$. Then, $z_{1}=z_{2}$.
\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.regular.cancel}.]We have $b\left(  z_{1}%
-z_{2}\right)  =bz_{1}-bz_{2}=0$ (since $bz_{1}=bz_{2}$). Since $b$ is
regular, this results in $z_{1}-z_{2}=0$. In other words, $z_{1}=z_{2}$. This
proves Lemma \ref{lem.regular.cancel}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.regular.cancel}.]We have $b\left(  z_{1}%
-z_{2}\right)  =bz_{1}-bz_{2}=0$ (since $bz_{1}=bz_{2}$). But the element $b$
is regular if and only if every $x\in A$ satisfying $bx=0$ satisfies $x=0$ (by
the definition of \textquotedblleft regular\textquotedblright). Hence, every
$x\in A$ satisfying $bx=0$ satisfies $x=0$ (since the element $b$ is regular).
Applying this to $x=z_{1}-z_{2}$, we obtain $z_{1}-z_{2}=0$ (since $b\left(
z_{1}-z_{2}\right)  =0$). In other words, $z_{1}=z_{2}$. This proves Lemma
\ref{lem.regular.cancel}.
\end{proof}
\end{verlong}

\begin{proposition}
\label{prop.regular.quotient}Let $A$ be a commutative ring. Let $b$ be a
regular element of $A$. Let $c\in A$. If $c\in bA$, then there exists a
\textbf{unique} element $x\in A$ satisfying $c=bx$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.regular.quotient}.]Assume that $c\in bA$.

If $z_{1}$ and $z_{2}$ are two elements $x\in A$ satisfying $c=bx$, then
$z_{1}=z_{2}$\ \ \ \ \footnote{\textit{Proof.} Let $z_{1}$ and $z_{2}$ be two
elements $x\in A$ satisfying $c=bx$. Thus, $z_{1}$ and $z_{2}$ are two
elements of $A$ satisfying $c=bz_{1}$ and $c=bz_{2}$. Now, $bz_{1}=c=bz_{2}$.
Hence, Lemma \ref{lem.regular.cancel} shows that $z_{1}=z_{2}$. Qed.}. In
other words, there exists \textbf{at most one} element $x\in A$ satisfying
$c=bx$. On the other hand, there exists some element $x\in A$ satisfying
$c=bx$ (because $c\in bA$). Combining the preceding two sentences, we conclude
that there exists a \textbf{unique} element $x\in A$ satisfying $c=bx$. This
proves Proposition \ref{prop.regular.quotient}.
\end{proof}

\begin{definition}
Let $A$ be a commutative ring. Let $b$ be a regular element of $A$. Let $c\in
A$. If $c\in bA$, then there exists a \textbf{unique} element $x\in A$
satisfying $c=bx$ (according to Proposition \ref{prop.regular.quotient}). This
element $x$ is denoted by $\dfrac{c}{b}$ or by $c/b$. This notation
generalizes the standard notation for quotients of (e.g.) rational numbers; it
also satisfies analogous rules (for example, if $b\in A$ and $b^{\prime}\in A$
are any two regular elements, and if $c\in bA$ and $c^{\prime}\in b^{\prime}%
A$, then $\dfrac{c}{b}+\dfrac{c^{\prime}}{b^{\prime}}=\dfrac{cb^{\prime
}+bc^{\prime}}{bb^{\prime}}$ and $\dfrac{c}{b}\cdot\dfrac{c^{\prime}%
}{b^{\prime}}=\dfrac{cc^{\prime}}{bb^{\prime}}$).
\end{definition}

\subsection{Symmetric polynomials}

\begin{definition}
Let $n\in\mathbb{N}$. Then, $S_{n}$ denotes the group of permutations of the
set $\left\{  1,2,\ldots,n\right\}  $. This group is called the $n$\textit{-th
symmetric group}.

The sign of a permutation $\sigma\in S_{n}$ shall be denoted by $\left(
-1\right)  ^{\sigma}$.
\end{definition}

\begin{definition}
Let $A$ be a commutative ring. Let $n\in\mathbb{N}$. Let $g\in A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $ be a polynomial in the $n$ indeterminates
$X_{1},X_{2},\ldots,X_{n}$ over $A$. The polynomial $g$ is said to be
\textit{symmetric} if and only if each $\sigma\in S_{n}$ satisfies%
\[
g\left(  X_{\sigma\left(  1\right)  },X_{\sigma\left(  2\right)  }%
,\ldots,X_{\sigma\left(  n\right)  }\right)  =g\left(  X_{1},X_{2}%
,\ldots,X_{n}\right)  .
\]
(Of course, we have $g\left(  X_{1},X_{2},\ldots,X_{n}\right)  =g$.)
\end{definition}

For example, if $n=3$, then the polynomials $X_{1}+X_{2}+X_{3}-7$ and $\left(
X_{1}+5\right)  \left(  X_{2}+5\right)  +\left(  X_{1}+5\right)  \left(
X_{3}+5\right)  +\left(  X_{2}+5\right)  \left(  X_{3}+5\right)  $ are
symmetric, but the polynomial $X_{1}+X_{3}$ is not (unless the ring $A$ is trivial).

\subsection{Symmetric polynomials from dividing by $\prod_{1\leq i<j\leq
n}\left(  X_{i}-X_{j}\right)  $}

Now, we claim the following:

\begin{proposition}
\label{prop.symmetric.from-antisym}Let $A$ be a commutative ring. Let
$n\in\mathbb{N}$. Let $f\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ be a
polynomial in the $n$ indeterminates $X_{1},X_{2},\ldots,X_{n}$ over $A$.
Assume that $f$ satisfies the following two properties:

\begin{statement}
\textit{Property 1:} For every $\left(  i,j\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2}$ satisfying $i<j$, the result of substituting $X_{j}$ for
$X_{i}$ in $f$ is $0$.
\end{statement}

\begin{statement}
\textit{Property 2:} Each $\sigma\in S_{n}$ satisfies%
\begin{equation}
f\left(  X_{\sigma\left(  1\right)  },X_{\sigma\left(  2\right)  }%
,\ldots,X_{\sigma\left(  n\right)  }\right)  =\left(  -1\right)  ^{\sigma
}f\left(  X_{1},X_{2},\ldots,X_{n}\right)  .
\label{eq.prop.symmetric.from-antisym.p2}%
\end{equation}

\end{statement}

Then, $f$ is divisible by $\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)
$, and the quotient $\dfrac{f}{\prod_{1\leq i<j\leq n}\left(  X_{i}%
-X_{j}\right)  }\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ is a symmetric polynomial.
\end{proposition}

Note that Proposition \ref{prop.symmetric.from-antisym} is essentially the
\textquotedblleft(iii) $\Longrightarrow$ (ii)\textquotedblright\ part of
\cite[Chapter SYM, Lemma (6.1)]{LLPT95}.\footnote{\textquotedblleft
Essentially\textquotedblright\ because in \cite[Chapter SYM, Lemma
(6.1)]{LLPT95}, the result of substituting $X_{j}$ for $X_{i}$ in $f$ is
considered not for $i<j$ but for $i>j$. But this makes little difference (it
merely boils down to renaming the indeterminates).}

For the proof of Proposition \ref{prop.symmetric.from-antisym}, we shall use
the following well-known fact:

\begin{proposition}
\label{prop.symmetric.perm-vand}Let $n\in\mathbb{N}$. Let $\sigma\in S_{n}$.
Let $R$ be a commutative ring. If $x_{1},x_{2},\ldots,x_{n}$ are $n$ elements
of $R$, then%
\[
\prod_{1\leq i<j\leq n}\left(  x_{\sigma\left(  i\right)  }-x_{\sigma\left(
j\right)  }\right)  =\left(  -1\right)  ^{\sigma}\cdot\prod_{1\leq i<j\leq
n}\left(  x_{i}-x_{j}\right)  .
\]

\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.symmetric.perm-vand}.]In the case when
$R=\mathbb{C}$ (so that $x_{1},x_{2},\ldots,x_{n}$ are $n$ complex numbers),
the statement of Proposition \ref{prop.symmetric.perm-vand} is precisely the
claim of \cite[Exercise 5.13 \textbf{(a)}]{detnotes}. The proof given in
\cite[solution to Exercise 5.13 \textbf{(a)}]{detnotes} works just as well in
the general case.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.symmetric.from-antisym}.]Let $R$ be the
commutative ring $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $. Thus,
$X_{1},X_{2},\ldots,X_{n}$ are $n$ elements of $R$. Hence, for each $\sigma\in
S_{n}$, we have%
\begin{equation}
\prod_{1\leq i<j\leq n}\left(  X_{\sigma\left(  i\right)  }-X_{\sigma\left(
j\right)  }\right)  =\left(  -1\right)  ^{\sigma}\cdot\prod_{1\leq i<j\leq
n}\left(  X_{i}-X_{j}\right)  \label{pf.prop.symmetric.from-antisym.1}%
\end{equation}
(by Proposition \ref{prop.symmetric.perm-vand} (applied to $x_{i}=X_{i}$)).

Corollary \ref{cor.div.substituted} shows that $f$ is divisible by
$\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  $. In other words,
$f\in\left(  \prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  \right)  R$.
Furthermore, the polynomial $\prod_{1\leq i<j\leq n}\left(  X_{i}%
-X_{j}\right)  $ is a regular element of $R$ (by Corollary
\ref{cor.mulpol.prodmonic}). Hence, the quotient $\dfrac{f}{\prod_{1\leq
i<j\leq n}\left(  X_{i}-X_{j}\right)  }\in A\left[  X_{1},X_{2},\ldots
,X_{n}\right]  $ is well-defined.

Denote this quotient $\dfrac{f}{\prod_{1\leq i<j\leq n}\left(  X_{i}%
-X_{j}\right)  }\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ by $g$. Thus,
\begin{equation}
g\cdot\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  =f.
\label{pf.prop.symmetric.from-antisym.3}%
\end{equation}


Now, fix any permutation $\sigma\in S_{n}$. The element $\left(  -1\right)
^{\sigma}$ satisfies $\left(  -1\right)  ^{\sigma}\cdot\left(  -1\right)
^{\sigma}=\left(  \left(  -1\right)  ^{\sigma}\right)  ^{2}=1$ (since $\left(
-1\right)  ^{\sigma}\in\left\{  1,-1\right\}  $). In particular, $\left(
-1\right)  ^{\sigma}$ is invertible (as an element of $R$).

Substituting $X_{\sigma\left(  1\right)  },X_{\sigma\left(  2\right)  }%
,\ldots,X_{\sigma\left(  n\right)  }$ for $X_{1},X_{2},\ldots,X_{n}$ on both
sides of the equality (\ref{pf.prop.symmetric.from-antisym.3}), we obtain%
\[
g\left(  X_{\sigma\left(  1\right)  },X_{\sigma\left(  2\right)  }%
,\ldots,X_{\sigma\left(  n\right)  }\right)  \cdot\prod_{1\leq i<j\leq
n}\left(  X_{\sigma\left(  i\right)  }-X_{\sigma\left(  j\right)  }\right)
=f\left(  X_{\sigma\left(  1\right)  },X_{\sigma\left(  2\right)  }%
,\ldots,X_{\sigma\left(  n\right)  }\right)  .
\]
Hence,%
\begin{align*}
f\left(  X_{\sigma\left(  1\right)  },X_{\sigma\left(  2\right)  }%
,\ldots,X_{\sigma\left(  n\right)  }\right)   &  =g\left(  X_{\sigma\left(
1\right)  },X_{\sigma\left(  2\right)  },\ldots,X_{\sigma\left(  n\right)
}\right)  \cdot\underbrace{\prod_{1\leq i<j\leq n}\left(  X_{\sigma\left(
i\right)  }-X_{\sigma\left(  j\right)  }\right)  }_{\substack{=\left(
-1\right)  ^{\sigma}\cdot\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)
\\\text{(by (\ref{pf.prop.symmetric.from-antisym.1}))}}}\\
&  =g\left(  X_{\sigma\left(  1\right)  },X_{\sigma\left(  2\right)  }%
,\ldots,X_{\sigma\left(  n\right)  }\right)  \cdot\left(  -1\right)  ^{\sigma
}\cdot\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right) \\
&  =\left(  -1\right)  ^{\sigma}\cdot\left(  \prod_{1\leq i<j\leq n}\left(
X_{i}-X_{j}\right)  \right)  \cdot g\left(  X_{\sigma\left(  1\right)
},X_{\sigma\left(  2\right)  },\ldots,X_{\sigma\left(  n\right)  }\right)  .
\end{align*}
Comparing this with (\ref{eq.prop.symmetric.from-antisym.p2}), we obtain%
\[
\left(  -1\right)  ^{\sigma}\cdot\left(  \prod_{1\leq i<j\leq n}\left(
X_{i}-X_{j}\right)  \right)  \cdot g\left(  X_{\sigma\left(  1\right)
},X_{\sigma\left(  2\right)  },\ldots,X_{\sigma\left(  n\right)  }\right)
=\left(  -1\right)  ^{\sigma}f\left(  X_{1},X_{2},\ldots,X_{n}\right)  .
\]
We can cancel the factor $\left(  -1\right)  ^{\sigma}$ from both sides of
this equality (since $\left(  -1\right)  ^{\sigma}$ is invertible as an
element of $R$). As a result, we obtain%
\begin{align*}
&  \left(  \prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  \right)  \cdot
g\left(  X_{\sigma\left(  1\right)  },X_{\sigma\left(  2\right)  }%
,\ldots,X_{\sigma\left(  n\right)  }\right) \\
&  =f\left(  X_{1},X_{2},\ldots,X_{n}\right)  =f=g\cdot\prod_{1\leq i<j\leq
n}\left(  X_{i}-X_{j}\right)  =\left(  \prod_{1\leq i<j\leq n}\left(
X_{i}-X_{j}\right)  \right)  \cdot g.
\end{align*}
Hence, Lemma \ref{lem.regular.cancel} (applied to $R$, $\prod_{1\leq i<j\leq
n}\left(  X_{i}-X_{j}\right)  $, $g\left(  X_{\sigma\left(  1\right)
},X_{\sigma\left(  2\right)  },\ldots,X_{\sigma\left(  n\right)  }\right)  $
and $g$ instead of $A$, $b$, $z_{1}$ and $z_{2}$) shows that $g\left(
X_{\sigma\left(  1\right)  },X_{\sigma\left(  2\right)  },\ldots
,X_{\sigma\left(  n\right)  }\right)  =g$ (since $\prod_{1\leq i<j\leq
n}\left(  X_{i}-X_{j}\right)  $ is a regular element of $R$). Hence, $g\left(
X_{\sigma\left(  1\right)  },X_{\sigma\left(  2\right)  },\ldots
,X_{\sigma\left(  n\right)  }\right)  =g=g\left(  X_{1},X_{2},\ldots
,X_{n}\right)  $.

Now, forget that we fixed $\sigma$. We thus have shown that each $\sigma\in
S_{n}$ satisfies $g\left(  X_{\sigma\left(  1\right)  },X_{\sigma\left(
2\right)  },\ldots,X_{\sigma\left(  n\right)  }\right)  =g\left(  X_{1}%
,X_{2},\ldots,X_{n}\right)  $. In other words, the polynomial $g$ is
symmetric. In other words, the polynomial $\dfrac{f}{\prod_{1\leq i<j\leq
n}\left(  X_{i}-X_{j}\right)  }$ is symmetric (since $g=\dfrac{f}{\prod_{1\leq
i<j\leq n}\left(  X_{i}-X_{j}\right)  }$). In other words, the quotient
$\dfrac{f}{\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  }\in A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $ is a symmetric polynomial. This completes
the proof of Proposition \ref{prop.symmetric.from-antisym}.
\end{proof}

\subsection{Schur and factorial Schur polynomials}

As an application of Proposition \ref{prop.symmetric.from-antisym}, let us
construct the Schur polynomials (as quotients of determinants) and the
factorial Schur polynomials. We begin with a standard piece of notation:

\begin{definition}
Let $A$ be a commutative ring. Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let
$a_{i,j}$ be an element of $A$ for each $i\in\left\{  1,2,\ldots,n\right\}  $
and $j\in\left\{  1,2,\ldots,m\right\}  $. Then, $\left(  a_{i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq m}$ shall denote the $n\times m$-matrix over $A$
whose $\left(  i,j\right)  $-th entry is $a_{i,j}$ for all $i\in\left\{
1,2,\ldots,n\right\}  $ and $j\in\left\{  1,2,\ldots,m\right\}  $.
\end{definition}

For example, the matrix $\left(
\begin{array}
[c]{ccc}%
2 & 3 & 4\\
3 & 4 & 5
\end{array}
\right)  $ can be rewritten as $\left(  i+j\right)  _{1\leq i\leq2,\ 1\leq
j\leq3}$ using this notation.

Now, we can define the Schur polynomials:

\begin{corollary}
\label{cor.symmetric.schur}Let $A$ be a commutative ring. Let $n\in\mathbb{N}%
$. Consider the polynomial ring $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $
in the $n$ indeterminates $X_{1},X_{2},\ldots,X_{n}$ over $A$. Let
$a_{1},a_{2},\ldots,a_{n}$ be $n$ nonnegative integers.

Let $F$ be the $n\times n$-matrix $\left(  \left(  X_{i}\right)  ^{a_{j}%
}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$ over $A\left[  X_{1},X_{2}%
,\ldots,X_{n}\right]  $. Then, the polynomial $\det F$ is divisible by
$\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  $, and the quotient
$\dfrac{\det F}{\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  }\in
A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ is a symmetric polynomial.
\end{corollary}

We shall prove Corollary \ref{cor.symmetric.schur} later. When the integers
$a_{1},a_{2},\ldots,a_{n}$ in Corollary \ref{cor.symmetric.schur} satisfy
$a_{1}>a_{2}>\cdots>a_{n}$, the quotient $\dfrac{\det F}{\prod_{1\leq i<j\leq
n}\left(  X_{i}-X_{j}\right)  }$ is known as a \textit{Schur polynomial} (see,
e.g., \cite[Corollary 2.6.6]{HopfComb} or \cite[Chapter I, (3.1)]{Macdonald}).

\begin{example}
For this example, set $n=3$ and $a_{1}=4$ and $a_{2}=2$ and $a_{3}=0$ in
Corollary \ref{cor.symmetric.schur}. Then, the matrix $F$ is
\[
F=\left(
\begin{array}
[c]{ccc}%
X_{1}^{4} & X_{1}^{2} & X_{1}^{0}\\
X_{2}^{4} & X_{2}^{2} & X_{2}^{0}\\
X_{3}^{4} & X_{3}^{2} & X_{3}^{0}%
\end{array}
\right)  =\left(
\begin{array}
[c]{ccc}%
X_{1}^{4} & X_{1}^{2} & 1\\
X_{2}^{4} & X_{2}^{2} & 1\\
X_{3}^{4} & X_{3}^{2} & 1
\end{array}
\right)  ,
\]
and its determinant is%
\begin{align*}
\det F  &  =\det\left(
\begin{array}
[c]{ccc}%
X_{1}^{4} & X_{1}^{2} & 1\\
X_{2}^{4} & X_{2}^{2} & 1\\
X_{3}^{4} & X_{3}^{2} & 1
\end{array}
\right)  =X_{1}^{4}X_{2}^{2}-X_{1}^{4}X_{3}^{2}-X_{1}^{2}X_{2}^{4}+X_{1}%
^{2}X_{3}^{4}+X_{2}^{4}X_{3}^{2}-X_{2}^{2}X_{3}^{4}\\
&  =\left(  X_{1}-X_{2}\right)  \left(  X_{1}-X_{3}\right)  \left(
X_{2}-X_{3}\right)  \left(  X_{1}+X_{2}\right)  \left(  X_{1}+X_{3}\right)
\left(  X_{2}+X_{3}\right)  .
\end{align*}
Thus, $\det F$ is clearly divisible by $\left(  X_{1}-X_{2}\right)  \left(
X_{1}-X_{3}\right)  \left(  X_{2}-X_{3}\right)  =\prod_{1\leq i<j\leq
n}\left(  X_{i}-X_{j}\right)  $, and moreover the quotient
\[
\dfrac{\det F}{\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  }=\left(
X_{1}+X_{2}\right)  \left(  X_{1}+X_{3}\right)  \left(  X_{2}+X_{3}\right)
\]
is a symmetric polynomial. (This is a somewhat unusual example, since $F$ is a
Vandermonde determinant in this case: for general choices of $a_{1}%
,a_{2},a_{3}$, you should not expect the quotient $\dfrac{\det F}{\prod_{1\leq
i<j\leq n}\left(  X_{i}-X_{j}\right)  }$ to have a nice factorization.)
\end{example}

To construct the factorial Schur polynomials, we need another notation:

\begin{definition}
Let $A$ be a commutative ring. Let $a\in A$ and $k\in\mathbb{N}$. Then,
$a^{\underline{k}}$ shall denote the element $a\left(  a-1\right)  \left(
a-2\right)  \cdots\left(  a-k+1\right)  =\prod_{i=0}^{k-1}\left(  a-i\right)
$ of $A$. (This element is called the $k$\textit{-th lower factorial} or the
$k$\textit{-th falling factorial} of $a$.)
\end{definition}

For example, if $A$ is any commutative ring and $a\in A$, then%
\[
a^{\underline{0}}=1,\ \ \ \ \ \ \ \ \ \ a^{\underline{1}}%
=a,\ \ \ \ \ \ \ \ \ \ a^{\underline{2}}=a\left(  a-1\right)
,\ \ \ \ \ \ \ \ \ \ a^{\underline{3}}=a\left(  a-1\right)  \left(
a-2\right)  .
\]


\begin{corollary}
\label{cor.symmetric.factorialschur}Let $A$ be a commutative ring. Let
$n\in\mathbb{N}$. Consider the polynomial ring $A\left[  X_{1},X_{2}%
,\ldots,X_{n}\right]  $ in the $n$ indeterminates $X_{1},X_{2},\ldots,X_{n}$
over $A$. Let $a_{1},a_{2},\ldots,a_{n}$ be $n$ nonnegative integers.

Let $F$ be the $n\times n$-matrix $\left(  \left(  X_{i}\right)
^{\underline{a_{j}}}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$ over $A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $. Then, the polynomial $\det F$ is divisible
by $\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  $, and the quotient
$\dfrac{\det F}{\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  }\in
A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ is a symmetric polynomial.
\end{corollary}

\begin{example}
For this example, set $n=3$ and $a_{1}=3$ and $a_{2}=2$ and $a_{3}=0$ in
Corollary \ref{cor.symmetric.factorialschur}. Then, the matrix $F$ is%
\[
F=\left(
\begin{array}
[c]{ccc}%
X_{1}^{\underline{3}} & X_{1}^{\underline{2}} & X_{1}^{\underline{0}}\\
X_{2}^{\underline{3}} & X_{2}^{\underline{2}} & X_{2}^{\underline{0}}\\
X_{3}^{\underline{3}} & X_{3}^{\underline{2}} & X_{3}^{\underline{0}}%
\end{array}
\right)  =\left(
\begin{array}
[c]{ccc}%
X_{1}\left(  X_{1}-1\right)  \left(  X_{1}-2\right)   & X_{1}\left(
X_{1}-1\right)   & 1\\
X_{2}\left(  X_{2}-1\right)  \left(  X_{2}-2\right)   & X_{2}\left(
X_{2}-1\right)   & 1\\
X_{3}\left(  X_{3}-1\right)  \left(  X_{3}-2\right)   & X_{3}\left(
X_{3}-1\right)   & 1
\end{array}
\right)  ,
\]
and its determinant is%
\begin{align*}
\det F &  =\det\left(
\begin{array}
[c]{ccc}%
X_{1}\left(  X_{1}-1\right)  \left(  X_{1}-2\right)   & X_{1}\left(
X_{1}-1\right)   & 1\\
X_{2}\left(  X_{2}-1\right)  \left(  X_{2}-2\right)   & X_{2}\left(
X_{2}-1\right)   & 1\\
X_{3}\left(  X_{3}-1\right)  \left(  X_{3}-2\right)   & X_{3}\left(
X_{3}-1\right)   & 1
\end{array}
\right)  \\
&  =\left(  X_{1}-X_{2}\right)  \left(  X_{1}-X_{3}\right)  \left(
X_{2}-X_{3}\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left(  X_{1}X_{2}+X_{1}X_{3}+X_{2}X_{3}-X_{1}%
-X_{2}-X_{3}+1\right)  .
\end{align*}
This, again, is a polynomial that is divisible by $\left(  X_{1}-X_{2}\right)
\left(  X_{1}-X_{3}\right)  \left(  X_{2}-X_{3}\right)  =\prod_{1\leq i<j\leq
n}\left(  X_{i}-X_{j}\right)  $, and the quotient ($X_{1}X_{2}+X_{1}%
X_{3}+X_{2}X_{3}-X_{1}-X_{2}-X_{3}+1$) is again symmetric.
\end{example}

When the integers $a_{1},a_{2},\ldots,a_{n}$ in Corollary
\ref{cor.symmetric.factorialschur} satisfy $a_{1}>a_{2}>\cdots>a_{n}$, the
quotient $\dfrac{\det F}{\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  }$
is known as a \textit{factorial Schur polynomial} (see, e.g., \cite[Theorem
3.2]{CheLou93}\footnote{Note that the lower factorial $a^{\underline{k}}$ is
denoted by $\left(  a\right)  _{k}$ in \cite{CheLou93}.}; for a more general
result, see \cite[Theorem 2]{BuMcNa14}).

Instead of proving Corollary \ref{cor.symmetric.schur} and Corollary
\ref{cor.symmetric.factorialschur} separately, let us show a fact that
generalizes them both:

\begin{corollary}
\label{cor.symmetric.factorialschur-gen}Let $A$ be a commutative ring. Let
$n\in\mathbb{N}$. Consider the polynomial ring $A\left[  X_{1},X_{2}%
,\ldots,X_{n}\right]  $ in the $n$ indeterminates $X_{1},X_{2},\ldots,X_{n}$
over $A$. Let $P_{1},P_{2},\ldots,P_{n}$ be $n$ polynomials in a single
variable $T$ over the ring $A$.

Let $F$ be the $n\times n$-matrix $\left(  P_{j}\left(  X_{i}\right)  \right)
_{1\leq i\leq n,\ 1\leq j\leq n}$ over $A\left[  X_{1},X_{2},\ldots
,X_{n}\right]  $. Then, the polynomial $\det F$ is divisible by $\prod_{1\leq
i<j\leq n}\left(  X_{i}-X_{j}\right)  $, and the quotient $\dfrac{\det
F}{\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  }\in A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $ is a symmetric polynomial.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.symmetric.factorialschur-gen}.]Define a
polynomial $f\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ by $f=\det F$.
(This is clearly well-defined, since the entries of the $n\times n$-matrix $F$
belong to $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $.) Note that
\[
F=\left(  P_{j}\left(  X_{i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq
n}=\left(  P_{v}\left(  X_{u}\right)  \right)  _{1\leq u\leq n,\ 1\leq v\leq
n}%
\]
(here, we have renamed the index $\left(  i,j\right)  $ as $\left(
u,v\right)  $).

We claim that the polynomial $f$ satisfies the following two properties:

\begin{statement}
\textit{Property 1:} For every $\left(  i,j\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2}$ satisfying $i<j$, the result of substituting $X_{j}$ for
$X_{i}$ in $f$ is $0$.
\end{statement}

\begin{statement}
\textit{Property 2:} Each $\sigma\in S_{n}$ satisfies%
\[
f\left(  X_{\sigma\left(  1\right)  },X_{\sigma\left(  2\right)  }%
,\ldots,X_{\sigma\left(  n\right)  }\right)  =\left(  -1\right)  ^{\sigma
}f\left(  X_{1},X_{2},\ldots,X_{n}\right)  .
\]

\end{statement}

[\textit{Proof of Property 1:} Fix $\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$.

The matrix $F$ satisfies $F=\left(  P_{v}\left(  X_{u}\right)  \right)
_{1\leq u\leq n,\ 1\leq v\leq n}$.

Hence, the $i$-th row of $F$ is $\left(  P_{1}\left(  X_{i}\right)
,P_{2}\left(  X_{i}\right)  ,\ldots,P_{n}\left(  X_{i}\right)  \right)  $,
whereas the $j$-th row of $F$ is $\left(  P_{1}\left(  X_{j}\right)
,P_{2}\left(  X_{j}\right)  ,\ldots,P_{n}\left(  X_{j}\right)  \right)  $. If
we substitute $X_{j}$ for $X_{i}$, then these two rows become equal (because
$X_{i}$ and $X_{j}$ become equal upon this substitution). In other words, we
have%
\begin{align*}
&  \left(  \text{the result of substituting }X_{j}\text{ for }X_{i}\text{ in
the }i\text{-th row of }F\right) \\
&  =\left(  \text{the result of substituting }X_{j}\text{ for }X_{i}\text{ in
the }j\text{-th row of }F\right)  .
\end{align*}


Let $F^{\prime}$ be the matrix obtained by substituting $X_{j}$ for $X_{i}$ in
the matrix $F$. Thus,%
\begin{align*}
&  \left(  \text{the }i\text{-th row of }F^{\prime}\right) \\
&  =\left(  \text{the result of substituting }X_{j}\text{ for }X_{i}\text{ in
the }i\text{-th row of }F\right) \\
&  =\left(  \text{the result of substituting }X_{j}\text{ for }X_{i}\text{ in
the }j\text{-th row of }F\right)
\end{align*}
and%
\begin{align*}
&  \left(  \text{the }j\text{-th row of }F^{\prime}\right) \\
&  =\left(  \text{the result of substituting }X_{j}\text{ for }X_{i}\text{ in
the }j\text{-th row of }F\right)
\end{align*}
(since the matrix $F^{\prime}$ is obtained by substituting $X_{j}$ for $X_{i}$
in the matrix $F$). Comparing these two equalities, we conclude that $\left(
\text{the }i\text{-th row of }F^{\prime}\right)  =\left(  \text{the
}j\text{-th row of }F^{\prime}\right)  $. Hence, the matrix $F^{\prime}$ has
two equal rows (since $i<j$). Therefore, the determinant of $F^{\prime}$ is
$0$. In other words, $\det\left(  F^{\prime}\right)  =0$.

But recall that $f=\det F$. Hence,%
\begin{align*}
&  \left(  \text{the result of substituting }X_{j}\text{ for }X_{i}\text{ in
}f\right) \\
&  =\left(  \text{the result of substituting }X_{j}\text{ for }X_{i}\text{ in
}\det F\right) \\
&  =\det\underbrace{\left(  \text{the matrix obtained by substituting }%
X_{j}\text{ for }X_{i}\text{ in the matrix }F\right)  }_{\substack{=F^{\prime
}\\\text{(because this is how }F^{\prime}\text{ was defined)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because when a substitution is applied to all entries of the matrix,}\\
\text{the determinant of the matrix undergoes the same substitution}%
\end{array}
\right) \\
&  =\det\left(  F^{\prime}\right)  =0.
\end{align*}
In other words, the result of substituting $X_{j}$ for $X_{i}$ in $f$ is $0$.
This proves Property 1.]

[\textit{Proof of Property 2:} Let $\sigma\in S_{n}$.

It is well-known that if we permute the rows of an $n\times n$-matrix
according to the permutation $\sigma$, then the determinant of this matrix is
multiplied by $\left(  -1\right)  ^{\sigma}$. In other words, any $n\times
n$-matrix $\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$ satisfies%
\[
\det\left(  \left(  b_{\sigma\left(  i\right)  ,j}\right)  _{1\leq i\leq
n,\ 1\leq j\leq n}\right)  =\left(  -1\right)  ^{\sigma}\det\left(  \left(
b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\right)
\]
\footnote{See, for example, \cite[Lemma 6.17 \textbf{(a)}]{detnotes} (applied
to $\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$, $\sigma$ and
$\left(  b_{\sigma\left(  i\right)  ,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}$ instead of $B$, $\kappa$ and $B_{\kappa}$) for a proof of this fact.}.
Applying this to $\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}=\left(  P_{j}\left(  X_{i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq
n}$, we conclude that%
\[
\det\left(  \left(  P_{j}\left(  X_{\sigma\left(  i\right)  }\right)  \right)
_{1\leq i\leq n,\ 1\leq j\leq n}\right)  =\left(  -1\right)  ^{\sigma}%
\det\left(  \left(  P_{j}\left(  X_{i}\right)  \right)  _{1\leq i\leq
n,\ 1\leq j\leq n}\right)  .
\]


But $f=\det F=\det\left(  \left(  P_{j}\left(  X_{i}\right)  \right)  _{1\leq
i\leq n,\ 1\leq j\leq n}\right)  $ (since $F=\left(  P_{j}\left(
X_{i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq n}$). Substituting
$X_{\sigma\left(  1\right)  },X_{\sigma\left(  2\right)  },\ldots
,X_{\sigma\left(  n\right)  }$ for $X_{1},X_{2},\ldots,X_{n}$ in this
equality, we obtain%
\begin{align*}
f\left(  X_{\sigma\left(  1\right)  },X_{\sigma\left(  2\right)  }%
,\ldots,X_{\sigma\left(  n\right)  }\right)   &  =\det\left(  \left(
P_{j}\left(  X_{\sigma\left(  i\right)  }\right)  \right)  _{1\leq i\leq
n,\ 1\leq j\leq n}\right) \\
&  =\left(  -1\right)  ^{\sigma}\underbrace{\det\left(  \left(  P_{j}\left(
X_{i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq n}\right)  }_{=f=f\left(
X_{1},X_{2},\ldots,X_{n}\right)  }\\
&  =\left(  -1\right)  ^{\sigma}f\left(  X_{1},X_{2},\ldots,X_{n}\right)  .
\end{align*}
This proves Property 2.]

We have now proven that $f$ satisfies both Properties 1 and 2. Hence,
Proposition \ref{prop.symmetric.from-antisym} shows that $f$ is divisible by
$\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  $, and that the quotient
$\dfrac{f}{\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  }\in A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $ is a symmetric polynomial. In view of
$f=\det F$, this rewrites as follows: $\det F$ is divisible by $\prod_{1\leq
i<j\leq n}\left(  X_{i}-X_{j}\right)  $, and the quotient $\dfrac{\det
F}{\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  }\in A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $ is a symmetric polynomial. This proves
Corollary \ref{cor.symmetric.factorialschur-gen}.
\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor.symmetric.schur}.]Consider the polynomial ring
$A\left[  T\right]  $ in a single variable $T$ over the ring $A$. For each
$j\in\left\{  1,2,\ldots,n\right\}  $, define a polynomial $P_{j}\in A\left[
T\right]  $ by $P_{j}=T^{a_{j}}$. Thus, $P_{1},P_{2},\ldots,P_{n}$ are $n$
polynomials in a single variable $T$ over the ring $A$. Moreover, for each
$i\in\left\{  1,2,\ldots,n\right\}  $ and $j\in\left\{  1,2,\ldots,n\right\}
$, we have%
\[
P_{j}\left(  X_{i}\right)  =\left(  X_{i}\right)  ^{a_{j}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }P_{j}=T^{a_{j}}\right)  .
\]
Thus, we have $\left(  P_{j}\left(  X_{i}\right)  \right)  _{1\leq i\leq
n,\ 1\leq j\leq n}=\left(  \left(  X_{i}\right)  ^{a_{j}}\right)  _{1\leq
i\leq n,\ 1\leq j\leq n}$. But \newline$F=\left(  \left(  X_{i}\right)
^{a_{j}}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$ (by the definition of $F$).
Comparing these two equalities, we obtain $F=\left(  P_{j}\left(
X_{i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq n}$. Thus, $F$ is the
$n\times n$-matrix $\left(  P_{j}\left(  X_{i}\right)  \right)  _{1\leq i\leq
n,\ 1\leq j\leq n}$ over $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $. Hence,
Corollary \ref{cor.symmetric.factorialschur-gen} shows that the polynomial
$\det F$ is divisible by $\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)
$, and that the quotient $\dfrac{\det F}{\prod_{1\leq i<j\leq n}\left(
X_{i}-X_{j}\right)  }\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ is a
symmetric polynomial. This proves Corollary \ref{cor.symmetric.schur}.
\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor.symmetric.factorialschur}.]Consider the
polynomial ring $A\left[  T\right]  $ in a single variable $T$ over the ring
$A$. For each $j\in\left\{  1,2,\ldots,n\right\}  $, define a polynomial
$P_{j}\in A\left[  T\right]  $ by $P_{j}=T^{\underline{a_{j}}}$. Thus,
$P_{1},P_{2},\ldots,P_{n}$ are $n$ polynomials in a single variable $T$ over
the ring $A$. Moreover, for each $i\in\left\{  1,2,\ldots,n\right\}  $ and
$j\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{align*}
P_{j}\left(  X_{i}\right)   &  =X_{i}\left(  X_{i}-1\right)  \left(
X_{i}-2\right)  \cdots\left(  X_{i}-a_{j}+1\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }P_{j}=T^{\underline{a_{j}}}=T\left(  T-1\right)  \left(
T-2\right)  \cdots\left(  T-a_{j}+1\right) \\
\text{(by the definition of }T^{\underline{a_{j}}}\text{)}%
\end{array}
\right) \\
&  =\left(  X_{i}\right)  ^{\underline{a_{j}}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\left(  X_{i}\right)  ^{\underline{a_{j}}}=X_{i}\left(
X_{i}-1\right)  \left(  X_{i}-2\right)  \cdots\left(  X_{i}-a_{j}+1\right) \\
\text{(by the definition of }\left(  X_{i}\right)  ^{\underline{a_{j}}%
}\text{)}%
\end{array}
\right)  .
\end{align*}
Thus, we have $\left(  P_{j}\left(  X_{i}\right)  \right)  _{1\leq i\leq
n,\ 1\leq j\leq n}=\left(  \left(  X_{i}\right)  ^{\underline{a_{j}}}\right)
_{1\leq i\leq n,\ 1\leq j\leq n}$. But \newline$F=\left(  \left(
X_{i}\right)  ^{\underline{a_{j}}}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$
(by the definition of $F$). Comparing these two equalities, we obtain
$F=\left(  P_{j}\left(  X_{i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq
n}$. Thus, $F$ is the $n\times n$-matrix $\left(  P_{j}\left(  X_{i}\right)
\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$ over $A\left[  X_{1},X_{2}%
,\ldots,X_{n}\right]  $. Hence, Corollary
\ref{cor.symmetric.factorialschur-gen} shows that the polynomial $\det F$ is
divisible by $\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  $, and that
the quotient $\dfrac{\det F}{\prod_{1\leq i<j\leq n}\left(  X_{i}%
-X_{j}\right)  }\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ is a symmetric
polynomial. This proves Corollary \ref{cor.symmetric.factorialschur}.
\end{proof}

\section{\label{sect.pow.more}More about power series}

In this Section, we return to the study of the ring $A\left[  \left[
X\right]  \right]  $ of formal power series.

\subsection{\label{subsect.pow.more.reg}Regular and nilpotent elements in
general}

We begin with proving some general properties of nilpotency and regularity in
commutative rings. We begin with fairly obvious ones:

\begin{corollary}
\label{cor.reg.power}Let $A$ be a commutative ring. Let $r$ be a regular
element of $A$. Let $n\in\mathbb{N}$. Then, the element $r^{n}$ of $A$ is regular.
\end{corollary}

\begin{vershort}
\begin{proof}
[Proof of Corollary \ref{cor.reg.power}.]Apply Proposition \ref{prop.reg.prod}
to $G=\left\{  1,2,\ldots,n\right\}  $ and $a_{g}=r$.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Corollary \ref{cor.reg.power}.]Proposition \ref{prop.reg.prod}
(applied to $G=\left\{  1,2,\ldots,n\right\}  $ and $a_{g}=r$) shows that the
element $\prod_{g\in\left\{  1,2,\ldots,n\right\}  }r$ of $A$ is regular. In
view of%
\[
\prod_{g\in\left\{  1,2,\ldots,n\right\}  }r=r^{\left\vert \left\{
1,2,\ldots,n\right\}  \right\vert }=r^{n}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\left\vert \left\{  1,2,\ldots,n\right\}  \right\vert =n\right)
,
\]
this rewrites as follows: The element $r^{n}$ of $A$ is regular. This proves
Corollary \ref{cor.reg.power}.
\end{proof}
\end{verlong}

\begin{proposition}
\label{prop.reg.ab-to-a}Let $A$ be a commutative ring. Let $a\in A$ and $b\in
A$ be two elements of $A$ such that the element $ab$ of $A$ is regular. Then,
$a\in A$ is regular.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.reg.ab-to-a}.]Every $x\in A$ satisfying $ax=0$
satisfies $\left(  ab\right)  x=b\underbrace{ax}_{=0}=0$, thus $x=0$ (since
$ab$ is regular). In other words, $a$ is regular. This proves Proposition
\ref{prop.reg.ab-to-a}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.reg.ab-to-a}.]The element $ab$ is regular if
and only if every $x\in A$ satisfying $\left(  ab\right)  x=0$ satisfies $x=0$
(by the definition of \textquotedblleft regular\textquotedblright). Hence,%
\begin{equation}
\text{every }x\in A\text{ satisfying }\left(  ab\right)  x=0\text{ satisfies
}x=0 \label{pf.prop.reg.ab-to-a.reg-ab}%
\end{equation}
(since the element $ab$ is regular).

Now, every $x\in A$ satisfying $ax=0$ satisfies $x=0$%
\ \ \ \ \footnote{\textit{Proof.} Let $x\in A$ be such that $ax=0$. We must
prove that $x=0$.
\par
We have $\left(  ab\right)  x=abx=b\underbrace{ax}_{=0}=0$. Hence, $x=0$ (by
(\ref{pf.prop.reg.ab-to-a.reg-ab})). Qed.}.

But the element $a$ is regular if and only if every $x\in A$ satisfying $ax=0$
satisfies $x=0$ (by the definition of \textquotedblleft
regular\textquotedblright). Hence, the element $a$ is regular (since every
$x\in A$ satisfying $ax=0$ satisfies $x=0$). This proves Proposition
\ref{prop.reg.ab-to-a}.
\end{proof}
\end{verlong}

Next, let us recall a basic identity: If $A$ is a commutative ring, and if $a$
and $b$ are any elements of $A$, and if $m\in\mathbb{N}$, then%
\begin{equation}
\left(  a-b\right)  \sum_{i=0}^{m-1}a^{i}b^{m-1-i}=a^{m}-b^{m}.
\label{eq.sum.geometric-sum}%
\end{equation}


\begin{verlong}
\noindent(See \cite[(18)]{detnotes} for a proof of (\ref{eq.sum.geometric-sum}%
) in the case when $A=\mathbb{Z}$ or $A=\mathbb{Q}$ or $A=\mathbb{R}$ or
$A=\mathbb{C}$; the same proof applies in the general case.)
\end{verlong}

Now, we shall prove a surprisingly general property of regularity:

\begin{proposition}
\label{prop.reg.reg-nil}Let $A$ be a commutative ring. Let $r$ be a regular
element of $A$. Let $a$ be a nilpotent element of $A$. Then, the element $r-a$
of $A$ is regular.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.reg.reg-nil}.]The element $a$ of $A$ is
nilpotent. In other words, there exists some $m\in\mathbb{N}$ such that
$a^{m}=0$ (by the definition of \textquotedblleft nilpotent\textquotedblright%
). Consider this $m$.

Corollary \ref{cor.reg.power} (applied to $n=m$) shows that the element
$r^{m}$ of $A$ is regular.

But (\ref{eq.sum.geometric-sum}) (applied to $r$ and $a$ instead of $a$ and
$b$) yields%
\[
\left(  r-a\right)  \sum_{i=0}^{m-1}r^{i}b^{m-1-i}=r^{m}-\underbrace{a^{m}%
}_{=0}=r^{m}.
\]
Hence, the element $\left(  r-a\right)  \sum_{i=0}^{m-1}r^{i}b^{m-1-i}$ is
regular (since the element $r^{m}$ is regular). Thus, Proposition
\ref{prop.reg.ab-to-a} (applied to $r-a$ and $\sum_{i=0}^{m-1}r^{i}b^{m-1-i}$
instead of $a$ and $b$) shows that $r-a\in A$ is regular. This proves
Proposition \ref{prop.reg.reg-nil}.
\end{proof}

\subsection{\label{subsect.pow.more.regps}More regular power series}

Let us now resume studying formal power series in $A\left[  \left[  X\right]
\right]  $. We begin with an essentially trivial fact:

\begin{proposition}
\label{prop.pow.Xn-reg}Let $A$ be a commutative ring. Then:

\textbf{(a)} The element $X$ of $A\left[  \left[  X\right]  \right]  $ is regular.

\textbf{(b)} Let $n\in\mathbb{N}$. Then, the element $X^{n}$ of $A\left[
\left[  X\right]  \right]  $ is regular.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.pow.Xn-reg}.]\textbf{(a)} The quickest way to
prove this is by applying Proposition \ref{prop.pow.X-a-reg.nil} to $a=0$
(since $0$ is nilpotent). Alternatively, you can prove this directly: If $x\in
A\left[  \left[  X\right]  \right]  $ is such that $Xx=0$, then it is easily
seen that $x=0$ (since the coefficients of $Xx$ are precisely the coefficients
of $x$, just shifted by one degree); in other words, $X$ is regular. Either
way, Proposition \ref{prop.pow.Xn-reg} \textbf{(a)} is proven.

\textbf{(b)} Proposition \ref{prop.pow.Xn-reg} \textbf{(a)} shows that the
element $X$ of $A\left[  \left[  X\right]  \right]  $ is regular. Hence,
Corollary \ref{cor.reg.power} (applied to $A\left[  \left[  X\right]  \right]
$ and $X$ instead of $A$ and $r$) yields that the element $X^{n}$ of $A\left[
\left[  X\right]  \right]  $ is regular. This proves Proposition
\ref{prop.pow.Xn-reg} \textbf{(b)}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.pow.Xn-reg}.]\textbf{(a)} The quickest way to
prove this is the following: The element $0$ of $A$ is nilpotent (since
$0^{1}=0$). Hence, Proposition \ref{prop.pow.X-a-reg.nil} (applied to $a=0$)
shows that the element $X-0$ of $A\left[  \left[  X\right]  \right]  $ is
regular. In other words, the element $X$ of $A\left[  \left[  X\right]
\right]  $ is regular. This proves Proposition \ref{prop.pow.Xn-reg}
\textbf{(a)}.

However, let me also give an alternative proof of Proposition
\ref{prop.pow.Xn-reg} \textbf{(a)}, which avoids the use of Proposition
\ref{prop.pow.X-a-reg.nil} :

Let $x\in A\left[  \left[  X\right]  \right]  $ be such that $Xx=0$. We shall
prove that $x=0$.

Write the power series $x$ in the form $x=\sum_{i\in\mathbb{N}}c_{i}X^{i}$ for
some sequence $\left(  c_{0},c_{1},c_{2},\ldots\right)  \in A^{\mathbb{N}}$.

Define a sequence $\left(  d_{0},d_{1},d_{2},\ldots\right)  \in A^{\mathbb{N}%
}$ as in the proof of Proposition \ref{prop.pow.X-a-reg.nil}. Then, for every
$i\in\mathbb{N}$ satisfying $i\neq0$, we have%
\begin{equation}
d_{i}=c_{i-1}. \label{pf.prop.pow.Xn-reg.1}%
\end{equation}
(Indeed, this can be proven in the same way as we proved
(\ref{pf.prop.pow.X-a-reg.nil.1}) during our proof of Proposition
\ref{prop.pow.X-a-reg.nil}.) Furthermore, we have%
\begin{equation}
\sum_{i\in\mathbb{N}}d_{i}X^{i}=Xx. \label{pf.prop.pow.Xn-reg.2}%
\end{equation}
(Indeed, this can be proven in the same way as we proved
(\ref{pf.prop.pow.X-a-reg.nil.2}) during our proof of Proposition
\ref{prop.pow.X-a-reg.nil}.) Now,%
\[
\sum_{i\in\mathbb{N}}d_{i}X^{i}=Xx=0=\sum_{i\in\mathbb{N}}0X^{i}.
\]
Comparing coefficients on both sides of this equality, we obtain%
\begin{equation}
d_{i}=0\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\mathbb{N}.
\label{pf.prop.pow.Xn-reg.di=0}%
\end{equation}


Now, let $i\in\mathbb{N}$. Then, $i+1\in\mathbb{N}$ and $i+1\geq1>0$, so that
$i+1\neq0$. Hence, (\ref{pf.prop.pow.Xn-reg.1}) (applied to $i+1$ instead of
$i$) yields $d_{i+1}=c_{\left(  i+1\right)  -i}=c_{i}$, hence $c_{i}%
=d_{i+1}=0$ (by (\ref{pf.prop.pow.Xn-reg.di=0}), applied to $i+1$ instead of
$i$).

Now, forget that we fixed $i$. We thus have shown that $c_{i}=0$ for each
$i\in\mathbb{N}$. Hence, $\sum_{i\in\mathbb{N}}\underbrace{c_{i}}_{=0}%
X^{i}=\sum_{i\in\mathbb{N}}0X^{i}=0$. Thus, $x=\sum_{i\in\mathbb{N}}c_{i}%
X^{i}=0$.

Now, forget that we fixed $x$. We thus have shown that every $x\in A\left[
\left[  X\right]  \right]  $ satisfying $Xx=0$ satisfies $x=0$.

But the element $X$ of $A\left[  \left[  X\right]  \right]  $ is regular if
and only if every $x\in A\left[  \left[  X\right]  \right]  $ satisfying
$Xx=0$ satisfies $x=0$ (by the definition of \textquotedblleft
regular\textquotedblright). Hence, the element $X$ of $A\left[  \left[
X\right]  \right]  $ is regular (since every $x\in A\left[  \left[  X\right]
\right]  $ satisfying $Xx=0$ satisfies $x=0$). This proves Proposition
\ref{prop.pow.Xn-reg} \textbf{(a)} again.

\textbf{(b)} Proposition \ref{prop.pow.Xn-reg} \textbf{(a)} shows that the
element $X$ of $A\left[  \left[  X\right]  \right]  $ is regular. Hence,
Corollary \ref{cor.reg.power} (applied to $A\left[  \left[  X\right]  \right]
$ and $X$ instead of $A$ and $r$) yields that the element $X^{n}$ of $A\left[
\left[  X\right]  \right]  $ is regular. This proves Proposition
\ref{prop.pow.Xn-reg} \textbf{(b)}.
\end{proof}
\end{verlong}

We can now apply Proposition \ref{prop.reg.reg-nil} to rings of power series:

\begin{corollary}
\label{cor.pow.Xn-g-reg-nil}Let $A$ be a commutative ring. Let $n\in
\mathbb{N}$. Let $p\in A\left[  \left[  X\right]  \right]  $ be a formal power
series such that $X^{n}-p$ is nilpotent. Then, $p\in A\left[  \left[
X\right]  \right]  $ is regular.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.pow.Xn-g-reg-nil}.]Proposition
\ref{prop.pow.Xn-reg} \textbf{(b)} yields that the element $X^{n}$ of
$A\left[  \left[  X\right]  \right]  $ is regular. Meanwhile, the element
$X^{n}-p$ of $A\left[  \left[  X\right]  \right]  $ is nilpotent (by
assumption). Hence, Proposition \ref{prop.reg.reg-nil} (applied to $A\left[
\left[  X\right]  \right]  $, $X^{n}$ and $X^{n}-p$ instead of $A$, $r$ and
$a$) yields that the element $X^{n}-\left(  X^{n}-p\right)  $ of $A\left[
\left[  X\right]  \right]  $ is regular. In other words, the element $p$ of
$A\left[  \left[  X\right]  \right]  $ is regular (since $X^{n}-\left(
X^{n}-p\right)  =p$). This proves Corollary \ref{cor.pow.Xn-g-reg-nil}.
\end{proof}

We can now re-prove Proposition \ref{prop.pow.X-a-reg.nil}:

\begin{proof}
[Second proof of Proposition \ref{prop.pow.X-a-reg.nil}.]Recall that $A$ is a
subring of $A\left[  \left[  X\right]  \right]  $. We know that $a$ is
nilpotent. In other words, $X^{1}-\left(  X-a\right)  $ is nilpotent (since
$\underbrace{X^{1}}_{=X}-\left(  X-a\right)  =X-\left(  X-a\right)  =a$).
Thus, Corollary \ref{cor.pow.Xn-g-reg-nil} (applied to $n=1$ and $p=X-a$)
yields that $X-a\in A\left[  \left[  X\right]  \right]  $ is regular. This
proves Proposition \ref{prop.pow.X-a-reg.nil} again. (Of course, this second
proof of Proposition \ref{prop.pow.X-a-reg.nil} is circular if you have used
Proposition \ref{prop.pow.X-a-reg.nil} in proving Proposition
\ref{prop.pow.Xn-reg}; thus, for it to be valid, you need a proof of
Proposition \ref{prop.pow.Xn-reg} that is independent on Proposition
\ref{prop.pow.X-a-reg.nil}.)
\end{proof}

\subsection{Intermezzo on sums of nilpotents}

We next prove a standard fact about nilpotent elements in commutative rings:

\begin{proposition}
\label{prop.nil.a+b}Let $A$ be a commutative ring. Let $a$ and $b$ be two
nilpotent elements of $A$. Then, the element $a+b$ of $A$ is nilpotent.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.nil.a+b}.]The element $a$ of $A$ is nilpotent.
In other words, there exists some $p\in\mathbb{N}$ such that $a^{p}=0$ (by the
definition of \textquotedblleft nilpotent\textquotedblright). Consider this
$p$.

The element $b$ of $A$ is nilpotent. In other words, there exists some
$q\in\mathbb{N}$ such that $b^{q}=0$ (by the definition of \textquotedblleft
nilpotent\textquotedblright). Consider this $q$.

Every $k\in\left\{  0,1,\ldots,p\right\}  $ satisfies%
\begin{equation}
b^{p+q-k}=0. \label{pf.prop.nil.a+b.bq-k}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.nil.a+b.bq-k}):} Let $k\in\left\{
0,1,\ldots,p\right\}  $. Then, $k\leq p$, so that $p-k\geq0$. Hence,
$b^{q+\left(  p-k\right)  }=\underbrace{b^{q}}_{=0}b^{p-k}=0$. In view of
$q+\left(  p-k\right)  =p+q-k$, this rewrites as $b^{p+q-k}=0$. This proves
(\ref{pf.prop.nil.a+b.bq-k}).]

Every $k\in\left\{  p+1,p+2,\ldots,p+q\right\}  $ satisfies%
\begin{equation}
a^{k}=0. \label{pf.prop.nil.a+b.ak}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.nil.a+b.ak}):} Let $k\in\left\{
p+1,p+2,\ldots,p+q\right\}  $. Thus, $k\geq p+1\geq p$ and therefore
$k-p\geq0$. Hence, $a^{p+\left(  k-p\right)  }=\underbrace{a^{p}}_{=0}%
a^{k-p}=0$. In view of $p+\left(  k-p\right)  =k$, this rewrites as $a^{k}=0$.
This proves (\ref{pf.prop.nil.a+b.ak}).]

Recall that the ring $A$ is commutative. Hence, the binomial formula yields%
\begin{align*}
\left(  a+b\right)  ^{p+q}  &  =\sum_{k=0}^{p+q}\dbinom{p+q}{k}a^{k}%
b^{p+q-k}\\
&  =\sum_{k=0}^{p}\dbinom{p+q}{k}a^{k}\underbrace{b^{p+q-k}}%
_{\substack{=0\\\text{(by (\ref{pf.prop.nil.a+b.bq-k}))}}}+\sum_{k=p+1}%
^{p+q}\dbinom{p+q}{k}\underbrace{a^{k}}_{\substack{=0\\\text{(by
(\ref{pf.prop.nil.a+b.ak}))}}}b^{p+q-k}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split the sum at
}k=p\text{, since }0\leq p\leq p+q\right) \\
&  =\underbrace{\sum_{k=0}^{p}\dbinom{p+q}{k}a^{k}\cdot0}_{=0}%
+\underbrace{\sum_{k=p+1}^{p+q}\dbinom{p+q}{k}\cdot0b^{p+q-k}}_{=0}=0+0=0.
\end{align*}
Hence, there exists a $k\in\mathbb{N}$ such that $\left(  a+b\right)  ^{k}=0$
(namely, $k=p+q$). In other words, the element $a+b$ of $A$ is nilpotent (by
the definition of \textquotedblleft nilpotent\textquotedblright). This proves
Proposition \ref{prop.nil.a+b}.
\end{proof}

\begin{corollary}
\label{cor.nil.a-b}Let $A$ be a commutative ring. Let $a$ and $b$ be two
nilpotent elements of $A$. Then, the element $a-b$ of $A$ is nilpotent.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.nil.a-b}.]The element $b$ of $A$ is nilpotent. In
other words, there exists some $q\in\mathbb{N}$ such that $b^{q}=0$ (by the
definition of \textquotedblleft nilpotent\textquotedblright). Consider this
$q$. Then, $\left(  -b\right)  ^{q}=\left(  -1\right)  ^{q}\underbrace{b^{q}%
}_{=0}=0$. Hence, there exists a $k\in\mathbb{N}$ such that $\left(
-b\right)  ^{k}=0$ (namely, $k=q$). In other words, the element $-b$ of $A$ is
nilpotent (by the definition of \textquotedblleft nilpotent\textquotedblright%
). Hence, Proposition \ref{prop.nil.a+b} (applied to $-b$ instead of $b$)
shows that the element $a+\left(  -b\right)  $ of $A$ is nilpotent (since $a$
is nilpotent). In other words, the element $a-b$ of $A$ is nilpotent (since
$a+\left(  -b\right)  =a-b$). This proves Corollary \ref{cor.nil.a-b}.
\end{proof}

\subsection{Nilpotent power series have nilpotent coefficients}

In Definition \ref{def.pol.coeff} \textbf{(a)}, we have introduced the
notation $\left[  X^{n}\right]  p$ for the coefficient of $X^{n}$ in a
polynomial $p\in A\left[  X\right]  $. We can extend this notation to formal
power series $p\in A\left[  \left[  X\right]  \right]  $ in the obvious way:

\begin{definition}
\label{def.pow.coeff}Let $A$ be a commutative ring.

If $p\in A\left[  \left[  X\right]  \right]  $ is a formal power series in
some indeterminate $X$ over $A$, and if $n\in\mathbb{N}$, then $\left[
X^{n}\right]  p$ will denote the coefficient of $X^{n}$ in $p$. For example,%
\begin{align*}
\left[  X^{3}\right]  \left(  1+X+X^{2}+X^{3}+\cdots\right)   &  =1;\\
\left[  X^{4}\right]  \left(  1+2X+3X^{2}+4X^{3}+\cdots\right)   &  =5.
\end{align*}


Clearly, every formal power series $p\in A\left[  \left[  X\right]  \right]  $
satisfies $p=\sum_{n\in\mathbb{N}}\left(  \left[  X^{n}\right]  p\right)
X^{n}$. (Here, $\sum_{n\in\mathbb{N}}\left(  \left[  X^{n}\right]  p\right)
X^{n}$ is a well-defined infinite sum.)
\end{definition}

Clearly, Definition \ref{def.pow.coeff} extends Definition \ref{def.pol.coeff}
\textbf{(a)}.

Our next goal is a necessary criterion for the nilpotency of a formal power series:

\begin{theorem}
\label{thm.pow.nil-coeff}Let $A$ be a commutative ring. Let $g\in A\left[
\left[  X\right]  \right]  $ be nilpotent. Then, $\left[  X^{n}\right]  g$ is
nilpotent for each $n\in\mathbb{N}$.
\end{theorem}

In other words, each coefficient of a nilpotent formal power series in
$A\left[  \left[  X\right]  \right]  $ must be nilpotent.

Note that the converse of Theorem \ref{thm.pow.nil-coeff} does not hold in
general. (See \cite{EleRos12} and \cite{Fields71} for counterexamples.)

Before we prove Theorem \ref{thm.pow.nil-coeff}, we state a few simple lemmas.
First, we recall the rules for adding and multiplying formal power series:

\begin{proposition}
\label{prop.pow.+*}Let $A$ be a commutative ring. Let $n\in\mathbb{N}$.

\textbf{(a)} Every $p\in A\left[  \left[  X\right]  \right]  $ and $q\in
A\left[  \left[  X\right]  \right]  $ satisfy $\left[  X^{n}\right]  \left(
p+q\right)  =\left[  X^{n}\right]  p+\left[  X^{n}\right]  q$.

\textbf{(b)} Every $\lambda\in A$ and $p\in A\left[  \left[  X\right]
\right]  $ satisfy $\left[  X^{n}\right]  \left(  \lambda p\right)
=\lambda\left[  X^{n}\right]  p$.

\textbf{(c)} Every $p\in A\left[  \left[  X\right]  \right]  $ and $q\in
A\left[  \left[  X\right]  \right]  $ satisfy $\left[  X^{n}\right]  \left(
pq\right)  =\sum_{k=0}^{n}\left(  \left[  X^{k}\right]  p\right)  \cdot\left(
\left[  X^{n-k}\right]  q\right)  $.
\end{proposition}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.pow.+*}.]This is proven exactly as Proposition
\ref{prop.pol.+*} was proven.
\end{proof}
\end{verlong}

\begin{corollary}
\label{cor.pow.Xn0}Let $A$ be a commutative ring.

\textbf{(a)} Every $p\in A\left[  \left[  X\right]  \right]  $ and $q\in
A\left[  \left[  X\right]  \right]  $ satisfy $\left[  X^{0}\right]  \left(
pq\right)  =\left(  \left[  X^{0}\right]  p\right)  \cdot\left(  \left[
X^{0}\right]  q\right)  $.

\textbf{(b)} Every $p\in A\left[  \left[  X\right]  \right]  $ and
$n\in\mathbb{N}$ satisfy $\left[  X^{0}\right]  \left(  p^{n}\right)  =\left(
\left[  X^{0}\right]  p\right)  ^{n}$.
\end{corollary}

\begin{vershort}
\begin{proof}
[Proof of Corollary \ref{cor.pow.Xn0}.]\textbf{(a)} This follows easily from
Proposition \ref{prop.pow.+*} \textbf{(c)}.

\textbf{(b)} This follows by induction on $n$ (using Corollary
\ref{cor.pow.Xn0} \textbf{(a)} in the induction step).
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Corollary \ref{cor.pow.Xn0}.]\textbf{(a)} Let $p\in A\left[  \left[
X\right]  \right]  $ and $q\in A\left[  \left[  X\right]  \right]  $. Then,
Proposition \ref{prop.pow.+*} \textbf{(c)} (applied to $n=0$) yields%
\begin{align*}
\left[  X^{0}\right]  \left(  pq\right)   &  =\sum_{k=0}^{0}\left(  \left[
X^{k}\right]  p\right)  \cdot\left(  \left[  X^{0-k}\right]  q\right)
=\left(  \left[  X^{0}\right]  p\right)  \cdot\left(  \left[  X^{0-0}\right]
q\right) \\
&  =\left(  \left[  X^{0}\right]  p\right)  \cdot\left(  \left[  X^{0}\right]
q\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }0-0=0\right)  .
\end{align*}
This proves Corollary \ref{cor.pow.Xn0} \textbf{(a)}.

\textbf{(b)} We shall prove Corollary \ref{cor.pow.Xn0} \textbf{(b)} by
induction on $n$:

\textit{Induction base:} Every $p\in A\left[  \left[  X\right]  \right]  $
satisfies%
\[
\left[  X^{0}\right]  \left(  \underbrace{p^{0}}_{=1}\right)  =\left[
X^{0}\right]  \left(  1\right)  =1=\left(  \left[  X^{0}\right]  p\right)
^{0}%
\]
(since $\left(  \left[  X^{0}\right]  p\right)  ^{0}=1$). In other words,
Corollary \ref{cor.pow.Xn0} \textbf{(b)} holds for $n=0$. This completes the
induction base.

\textit{Induction step:} Let $m\in\mathbb{N}$. Assume that Corollary
\ref{cor.pow.Xn0} \textbf{(b)} holds for $n=m$. We must prove that Corollary
\ref{cor.pow.Xn0} \textbf{(b)} holds for $n=m+1$.

Let $p\in A\left[  \left[  X\right]  \right]  $. Then, we can apply Corollary
\ref{cor.pow.Xn0} \textbf{(b)} to $n=m$ (since Corollary \ref{cor.pow.Xn0}
\textbf{(b)} holds for $n=m$); thus, we obtain $\left[  X^{0}\right]  \left(
p^{m}\right)  =\left(  \left[  X^{0}\right]  p\right)  ^{m}$. Now,%
\begin{align*}
\left[  X^{0}\right]  \left(  \underbrace{p^{m+1}}_{=pp^{m}}\right)   &
=\left[  X^{0}\right]  \left(  pp^{m}\right)  =\left(  \left[  X^{0}\right]
p\right)  \cdot\underbrace{\left(  \left[  X^{0}\right]  \left(  p^{m}\right)
\right)  }_{=\left(  \left[  X^{0}\right]  p\right)  ^{m}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Corollary \ref{cor.pow.Xn0}
\textbf{(a)}, applied to }q=p^{m}\right) \\
&  =\left(  \left[  X^{0}\right]  p\right)  \cdot\left(  \left[  X^{0}\right]
p\right)  ^{m}=\left(  \left[  X^{0}\right]  p\right)  ^{m+1}.
\end{align*}


Now, forget that we fixed $p$. We thus have proven that Corollary
\ref{cor.pow.Xn0} \textbf{(b)} to $n=m+1$. This completes the induction step.
Thus, Corollary \ref{cor.pow.Xn0} \textbf{(b)} is proven.
\end{proof}
\end{verlong}

Now we can easily prove the particular case of Theorem \ref{thm.pow.nil-coeff}
for $n=0$:

\begin{lemma}
\label{lem.pow.nil-coeff-0}Let $A$ be a commutative ring. Let $g\in A\left[
\left[  X\right]  \right]  $ be nilpotent. Then, $\left[  X^{0}\right]  g$ is nilpotent.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.pow.nil-coeff-0}.]We have assumed that $g$ is
nilpotent. In other words, there exists an $n\in\mathbb{N}$ such that
$g^{n}=0$ (by the definition of \textquotedblleft nilpotent\textquotedblright%
). Consider this $n$. Now, Corollary \ref{cor.pow.Xn0} \textbf{(b)} (applied
to $p=g$) yields $\left[  X^{0}\right]  \left(  g^{n}\right)  =\left(  \left[
X^{0}\right]  g\right)  ^{n}$. Hence, $\left(  \left[  X^{0}\right]  g\right)
^{n}=\left[  X^{0}\right]  \underbrace{\left(  g^{n}\right)  }_{=0}=\left[
X^{0}\right]  0=0$. Thus, there exists a $k\in\mathbb{N}$ such that $\left(
\left[  X^{0}\right]  g\right)  ^{k}=0$ (namely, $k=n$). In other words,
$\left[  X^{0}\right]  g$ is nilpotent (by the definition of \textquotedblleft
nilpotent\textquotedblright). This proves Lemma \ref{lem.pow.nil-coeff-0}.
\end{proof}

In order to get a grip on the other coefficients of a nilpotent formal power
series, we need a few more basic results. We begin with a general property of
regular and nilpotent elements:

\begin{lemma}
\label{lem.nil.reg-div}Let $A$ be a commutative ring. Let $r\in A$ and $a\in
A$ be such that $ra$ is nilpotent and $r$ is regular. Then, $a$ is nilpotent.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.nil.reg-div}.]We have assumed that $ra$ is nilpotent.
In other words, there exists an $n\in\mathbb{N}$ such that $\left(  ra\right)
^{n}=0$ (by the definition of \textquotedblleft nilpotent\textquotedblright).
Consider this $n$. Then, $r^{n}a^{n}=\left(  ra\right)  ^{n}=0$.

\begin{vershort}
Corollary \ref{cor.reg.power} shows that the element $r^{n}$ of $A$ is
regular. In other words, every $x\in A$ satisfying $r^{n}x=0$ satisfies $x=0$.
Applying this to $x=a^{n}$, we obtain $a^{n}=0$ (since $r^{n}a^{n}=0$). Hence,
$a$ is nilpotent.
\end{vershort}

\begin{verlong}
Corollary \ref{cor.reg.power} shows that the element $r^{n}$ of $A$ is
regular. But the element $r^{n}$ is regular if and only if every $x\in A$
satisfying $r^{n}x=0$ satisfies $x=0$ (by the definition of \textquotedblleft
regular\textquotedblright). Hence, every $x\in A$ satisfying $r^{n}x=0$
satisfies $x=0$ (since the element $r^{n}$ is regular). Applying this to
$x=a^{n}$, we obtain $a^{n}=0$ (since $r^{n}a^{n}=0$). Hence, there exists a
$k\in\mathbb{N}$ such that $a^{k}=0$. In other words, $a$ is nilpotent (by the
definition of \textquotedblleft nilpotent\textquotedblright).
\end{verlong}

This proves Lemma \ref{lem.nil.reg-div}.
\end{proof}

\begin{corollary}
\label{cor.pow.nil-reg-div}Let $A$ be a commutative ring. Let $g\in A\left[
\left[  X\right]  \right]  $ be such that $Xg$ is nilpotent. Then, $g$ is nilpotent.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.pow.nil-reg-div}.]Proposition
\ref{prop.pow.Xn-reg} \textbf{(a)} shows that the element $X$ of $A\left[
\left[  X\right]  \right]  $ is regular. Hence, Lemma \ref{lem.nil.reg-div}
(applied to $A\left[  \left[  X\right]  \right]  $, $X$ and $g$ instead of
$A$, $r$ and $a$) yields that $g$ is nilpotent. This proves Corollary
\ref{cor.pow.nil-reg-div}.
\end{proof}

\begin{corollary}
\label{cor.pow.split-const-term}Let $A$ be a commutative ring. Let $g\in
A\left[  \left[  X\right]  \right]  $. Define a formal power series
$\widetilde{g}\in A\left[  \left[  X\right]  \right]  $ by $\widetilde{g}%
=\sum_{n\in\mathbb{N}}\left(  \left[  X^{n+1}\right]  g\right)  \cdot X^{n}$. Then:

\textbf{(a)} We have $g=\left[  X^{0}\right]  g+X\widetilde{g}$.

\textbf{(b)} We have $\left[  X^{k}\right]  g=\left[  X^{k-1}\right]
\widetilde{g}$ for each positive integer $k$.

\textbf{(c)} If $g$ is nilpotent, then $\widetilde{g}$ is nilpotent.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.pow.split-const-term}.]\textbf{(a)} Every formal
power series $f\in A\left[  \left[  X\right]  \right]  $ satisfies
$f=\sum_{n\in\mathbb{N}}\left(  \left[  X^{n}\right]  f\right)  \cdot X^{n}$
(since $\left[  X^{0}\right]  f,\left[  X^{1}\right]  f,\left[  X^{2}\right]
f,\ldots$ are the coefficients of $f$). Applying this to $f=g$, we obtain%
\begin{align*}
g  &  =\sum_{n\in\mathbb{N}}\left(  \left[  X^{n}\right]  g\right)  \cdot
X^{n}=\left(  \left[  X^{0}\right]  g\right)  \cdot\underbrace{X^{0}}%
_{=1}+\underbrace{\sum_{n\in\left\{  1,2,3,\ldots\right\}  }\left(  \left[
X^{n}\right]  f\right)  \cdot X^{n}}_{\substack{=\sum_{n\in\mathbb{N}}\left(
\left[  X^{n+1}\right]  g\right)  \cdot X^{n+1}\\\text{(here, we have
substituted }n+1\\\text{for }n\text{ in the sum)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}n=0\text{ from the sum}\right) \\
&  =\left[  X^{0}\right]  g+\sum_{n\in\mathbb{N}}\left(  \left[
X^{n+1}\right]  g\right)  \cdot\underbrace{X^{n+1}}_{=XX^{n}}=\left[
X^{0}\right]  g+\underbrace{\sum_{n\in\mathbb{N}}\left(  \left[
X^{n+1}\right]  g\right)  \cdot XX^{n}}_{=X\sum_{n\in\mathbb{N}}\left(
\left[  X^{n+1}\right]  g\right)  \cdot X^{n}}\\
&  =\left[  X^{0}\right]  g+X\underbrace{\sum_{n\in\mathbb{N}}\left(  \left[
X^{n+1}\right]  g\right)  \cdot X^{n}}_{=\widetilde{g}}=\left[  X^{0}\right]
g+X\widetilde{g}.
\end{align*}
This proves Corollary \ref{cor.pow.split-const-term} \textbf{(a)}.

\begin{vershort}
\textbf{(b)} We have $\widetilde{g}=\sum_{n\in\mathbb{N}}\left(  \left[
X^{n+1}\right]  g\right)  \cdot X^{n}$. Thus, the coefficients of the power
series $\widetilde{g}$ are $\left[  X^{1}\right]  g,\left[  X^{2}\right]
g,\left[  X^{3}\right]  g,\ldots$. In other words, for each $n\in\mathbb{N}$,
we have $\left[  X^{n}\right]  \widetilde{g}=\left[  X^{n+1}\right]  g$.
Substituting $k-1$ for $n$ in this result, we obtain the following: For each
$k\in\left\{  1,2,3,\ldots\right\}  $, we have $\left[  X^{k-1}\right]
\widetilde{g}=\left[  X^{k}\right]  g$. This proves Corollary
\ref{cor.pow.split-const-term} \textbf{(b)}.
\end{vershort}

\begin{verlong}
\textbf{(b)} We have $\widetilde{g}=\sum_{n\in\mathbb{N}}\left(  \left[
X^{n+1}\right]  g\right)  \cdot X^{n}$, and the coefficients $\left[
X^{1}\right]  g,\left[  X^{2}\right]  g,\left[  X^{3}\right]  g,\ldots$ on the
right hand side of this equality are elements of $A$. Thus, this equality
shows that the formal power series $\widetilde{g}$ has coefficients $\left[
X^{1}\right]  g,\left[  X^{2}\right]  g,\left[  X^{3}\right]  g,\ldots$. In
other words, for each $n\in\mathbb{N}$, we have%
\[
\left(  \text{the coefficient of }X^{n}\text{ in }\widetilde{g}\right)
=\left[  X^{n+1}\right]  g.
\]
Hence, for each $n\in\mathbb{N}$, we have%
\begin{align}
\left[  X^{n}\right]  \widetilde{g}  &  =\left(  \text{the coefficient of
}X^{n}\text{ in }\widetilde{g}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\left[  X^{n}\right]  \widetilde{g}\right) \nonumber\\
&  =\left[  X^{n+1}\right]  g. \label{pf.cor.pow.split-const-term.b.2}%
\end{align}


Now, let $k$ be a positive integer. Thus, $k\geq1$, so that $k-1\geq0$ and
thus $k-1\in\mathbb{N}$. Hence, (\ref{pf.cor.pow.split-const-term.b.2})
(applied to $n=k-1$) yields $\left[  X^{k-1}\right]  \widetilde{g}=\left[
X^{\left(  k-1\right)  +1}\right]  g=\left[  X^{k}\right]  g$ (since $\left(
k-1\right)  +1=k$). That is, $\left[  X^{k}\right]  g=\left[  X^{k-1}\right]
\widetilde{g}$. This proves Corollary \ref{cor.pow.split-const-term}
\textbf{(b)}.
\end{verlong}

\textbf{(c)} Assume that $g$ is nilpotent. Thus, Lemma
\ref{lem.pow.nil-coeff-0} shows that $\left[  X^{0}\right]  g$ is nilpotent.
Hence, Corollary \ref{cor.nil.a-b} (applied to $A\left[  \left[  X\right]
\right]  $, $g$ and $\left[  X^{0}\right]  g$ instead of $A$, $a$ and $b$)
shows that the element $g-\left[  X^{0}\right]  g$ of $A\left[  \left[
X\right]  \right]  $ is nilpotent.

But Corollary \ref{cor.pow.split-const-term} \textbf{(a)} yields $g=\left[
X^{0}\right]  g+X\widetilde{g}$; thus, $g-\left[  X^{0}\right]
g=X\widetilde{g}$. Hence, $X\widetilde{g}$ is nilpotent (since $g-\left[
X^{0}\right]  g$ is nilpotent). Therefore, Corollary \ref{cor.pow.nil-reg-div}
(applied to $\widetilde{g}$ instead of $g$) yields that $\widetilde{g}$ is
nilpotent. This proves Corollary \ref{cor.pow.split-const-term} \textbf{(c)}.
\end{proof}

We are now ready to prove Theorem \ref{thm.pow.nil-coeff}:

\begin{proof}
[Proof of Theorem \ref{thm.pow.nil-coeff}.]We shall prove Theorem
\ref{thm.pow.nil-coeff} by induction on $n$:

\textit{Induction base:} If $A$ is a commutative ring, and if $g\in A\left[
\left[  X\right]  \right]  $ is nilpotent, then $\left[  X^{0}\right]  g$ is
nilpotent (by Lemma \ref{lem.pow.nil-coeff-0}). In other words, Theorem
\ref{thm.pow.nil-coeff} holds for $n=0$. This completes the induction base.

\textit{Induction step:} Let $k$ be a positive integer. Assume that Theorem
\ref{thm.pow.nil-coeff} holds for $n=k-1$. We must prove that Theorem
\ref{thm.pow.nil-coeff} holds for $n=k$.

We have assumed that Theorem \ref{thm.pow.nil-coeff} holds for $n=k-1$. In
other words, the following statement holds:

\begin{statement}
\textit{Statement 1:} Let $A$ be a commutative ring. Let $g\in A\left[
\left[  X\right]  \right]  $ be nilpotent. Then, $\left[  X^{k-1}\right]  g$
is nilpotent.
\end{statement}

Now, we must prove that Theorem \ref{thm.pow.nil-coeff} holds for $n=k$. In
other words, we must prove the following statement:

\begin{statement}
\textit{Statement 2:} Let $A$ be a commutative ring. Let $g\in A\left[
\left[  X\right]  \right]  $ be nilpotent. Then, $\left[  X^{k}\right]  g$ is nilpotent.
\end{statement}

[\textit{Proof of Statement 2:} Define a formal power series $\widetilde{g}\in
A\left[  \left[  X\right]  \right]  $ by $\widetilde{g}=\sum_{n\in\mathbb{N}%
}\left(  \left[  X^{n+1}\right]  g\right)  \cdot X^{n}$. Then, Corollary
\ref{cor.pow.split-const-term} \textbf{(c)} shows that $\widetilde{g}$ is
nilpotent. Hence, Statement 1 (applied to $\widetilde{g}$ instead of $g$)
yields that $\left[  X^{k-1}\right]  \widetilde{g}$ is nilpotent. But
Corollary \ref{cor.pow.split-const-term} \textbf{(b)} yields $\left[
X^{k}\right]  g=\left[  X^{k-1}\right]  \widetilde{g}$. Hence, $\left[
X^{k}\right]  g$ is nilpotent (since $\left[  X^{k-1}\right]  \widetilde{g}$
is nilpotent). This proves Statement 2.]

So we have proven Statement 2. In other words, we have proven that Theorem
\ref{thm.pow.nil-coeff} holds for $n=k$. This completes the induction step.
Thus, Theorem \ref{thm.pow.nil-coeff} is proven.
\end{proof}

TODO!

\begin{noncompile}%
%TCIMACRO{\TEXTsymbol{\backslash}}%
%BeginExpansion
$\backslash$%
%EndExpansion
/%
%TCIMACRO{\TEXTsymbol{\backslash}}%
%BeginExpansion
$\backslash$%
%EndExpansion
/ Do not use the notion of degree.%

%TCIMACRO{\TEXTsymbol{\backslash}}%
%BeginExpansion
$\backslash$%
%EndExpansion
/%
%TCIMACRO{\TEXTsymbol{\backslash}}%
%BeginExpansion
$\backslash$%
%EndExpansion
/ Do not use the naked version of "monic" (without degree).

TODO: Division by $X-a$ when $a$ is nilpotent works in $A\left[  \left[
X\right]  \right]  $.

TODO: More generally: If $P\in A\left[  X\right]  $ is a monic polynomial of
degree $n$ such that $P=X^{n}+\left(  \text{nilpotent}\right)  $, then
division with remainder by $P$ works in $A\left[  \left[  X\right]  \right]  $.

TODO: Thus, coprimality (Corollary \ref{cor.pol.intersect}) does work when $a$
is nilpotent.

TODO: What about the case when $P=X^{n}+\left(  \text{nilpotent}\right)  $ but
$P$ is not necessarily monic? Do we still have division with remainder? At
least uniqueness holds because $P$ is regular.
\end{noncompile}

\begin{thebibliography}{999999999}                                                                                        %


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\end{thebibliography}


\end{document}