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\ihead{Regular elements and lcm-coprimality}
\ohead{page \thepage}
\cfoot{}
\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 $i0$
(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 $-\inftyn$, 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, $\underbrace{m}_{\geq
g+n}-\underbrace{k}_{\leq n-1\left( g+n\right) -n=g\geq0$, 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,
$nn}-\underbrace{k}_{\leq n}>n-n=0$, 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 ng+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,
$g+n-\underbrace{k}_{\leq n-1g+n-n=g\geq0$, 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 $n-1$. 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 $\prod_{1\leq i0$. 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\ i0}}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 $%
\begin{cases}
x_{j-i}, & \text{if }j\geq i;\\
0, & \text{if }j0$ (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 $ij$. 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 ia_{2}>\cdots>a_{n}$, the quotient $\dfrac{\det F}{\prod_{1\leq ia_{2}>\cdots>a_{n}$, the
quotient $\dfrac{\det F}{\prod_{1\leq i0$, 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} %
\bibitem[BieLou89]{BieLou89}L. C. Biedenharn, J. D. Louck, \textit{A new class
of symmetric polynomials defined in terms of tableaux}, Advances in Applied
Mathematics, Volume 10, Issue 4, December 1989, pp. 396--438.\newline\url{https://doi.org/10.1016/0196-8858(89)90023-7}
\bibitem[BuMcNa14]{BuMcNa14}Daniel Bump, Peter J. McNamara and Maki Nakasuji,
\textit{Factorial Schur Functions and the Yang-Baxter Equation},
arXiv:1108.3087v3.\newline\url{https://arxiv.org/abs/1108.3087v3}
\bibitem[CheLou93]{CheLou93}William Y. C. Chen and James D. Louck, \textit{The
factorial Schur function}, J. Math. Phys. 34, 4144 (1993).\newline\url{http://dx.doi.org/10.1063/1.530032}
\bibitem[EGHLSVY11]{EGHLSVY11}Pavel Etingof, Oleg Golberg, Sebastian Hensel,
Tiankai Liu, Alex Schwendner, Dmitry Vaintrob, Elena Yudovina,
\textit{Introduction to Representation Theory}, Student Mathematical Library
\#59, AMS 2011.\newline A draft of this book can be found on the arXiv as
\href{http://arxiv.org/abs/0901.0827v5}{arXiv:0901.0827v5}.
\bibitem[EleRos12]{EleRos12}Georges Elencwajg, Julian Rosen and others,
\textit{Answers to math.stackexchange question \#187952 (\textquotedblleft A
non-nilpotent formal power series with nilpotent
coefficients\textquotedblright)}.\newline\url{https://math.stackexchange.com/q/187952}
\bibitem[Fields71]{Fields71}David E. Fields, \textit{Zero divisors and
nilpotent elements in power series rings}, Proc. Amer. Math. Soc. \textbf{27}
(1971), pp. 427--433. \url{https://doi.org/10.1090/S0002-9939-1971-0271100-6}
\bibitem[Garrett09]{Garrett09}Paul Garrett, \textit{Abstract Algebra}, lecture
notes, 2009.\newline\url{http://www.math.umn.edu/~garrett/m/algebra/notes/Whole.pdf}
\bibitem[Grinbe15]{detnotes}Darij Grinberg, \textit{Notes on the combinatorial
fundamentals of algebra}, 10 January 2019.\newline%
\url{http://www.cip.ifi.lmu.de/~grinberg/primes2015/sols.pdf} \newline The
numbering of theorems and formulas in this link might shift when the project
gets updated; for a \textquotedblleft frozen\textquotedblright\ version whose
numbering is guaranteed to match that in the citations above, see
\url{https://github.com/darijgr/detnotes/releases/tag/2019-01-10} .
\bibitem[GriRei18]{HopfComb}Darij Grinberg, Victor Reiner, \textit{Hopf
algebras in Combinatorics}, version of 11 May 2018,
\href{http://www.arxiv.org/abs/1409.8356v5}{\texttt{arXiv:1409.8356v5}}.
\newline See also
\url{http://www.cip.ifi.lmu.de/~grinberg/algebra/HopfComb-sols.pdf} for a
version that gets updated.
\bibitem[Knapp2016]{Knapp2016}Anthony W. Knapp, \textit{Basic Algebra},
Digital Second Edition, 2016.\newline\url{http://www.math.stonybrook.edu/~aknapp/download.html}
\bibitem[LLPT95]{LLPT95}D. Laksov, A. Lascoux, P. Pragacz, and A. Thorup,
\textit{The LLPT Notes}, edited by A. Thorup, 1995,\newline%
\url{http://www.math.ku.dk/~thorup/notes/sympol.pdf} .
\bibitem[Macdon95]{Macdonald}I. G. Macdonald, \textit{Symmetric Functions and
Hall Polynomials}, Oxford University Press, 2nd edition 1995.
\end{thebibliography}
\end{document}