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\ihead{Errata to ``A Determinantal Formula for the Exterior Powers...'}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{A Determinantal Formula for the Exterior Powers of the Polynomial
Ring}

\textit{Dan Laksov \& Anders Thorup}

Indiana University Mathematics Journal,

Vol. 56, No. 2 (2007), pp. 825--845.

\url{http://dx.doi.org/10.1512/iumj.2007.56.2937}

\textbf{Errata and addenda by Darij Grinberg (version of
%TCIMACRO{\TeXButton{today}{\today}}%
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\maketoc


\section*{***}

I will refer to the results appearing in the paper \textquotedblleft\textbf{A
Determinantal Formula for the Exterior Powers of the Polynomial Ring}%
\textquotedblright\ by Dan Laksov and Anders Thorup\ by the numbers under
which they appear in this preprint.

\setcounter{section}{9}

\section{Errata}

\begin{itemize}
\item \textbf{page 828, \S 0.3:} It is worth saying that if $g=b_{N}%
T^{N}+b_{N-1}T^{N-1}+b_{N-2}T^{N-2}+\cdots$ is a Laurent series, and if $i$ is
an integer such that $i>N$, then you set $b_{i}=0$. (This is used, for
example, in the definition of $\operatorname*{Res}\left(  g_{1},\ldots
,g_{n}\right)  $, since otherwise the entries $b_{i,-j}$ of the determinant
are undefined if one of the series $g_{1},\ldots,g_{n}$ begins with a very low
negative power of $T$.)

\item \textbf{page 829, Theorem 0.1:} In part (3), I'd replace
\textquotedblleft exponents\textquotedblright\ by \textquotedblleft
nonnegative exponents\textquotedblright\ for clarity.

\item \textbf{page 829, Remark 0.5:} The reasoning for why \textquotedblleft%
(2) is a consequence of (3)\textquotedblright\ can be simplified. All you need
is the following: In order to prove (2) in general, it suffices to prove (2)
in the case when $f_{1},\ldots,f_{n}$ are monomials (since any polynomial is
an $A$-linear combination of monomials). In other words, it suffices to prove
(2) in the case when each $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies
$f_{i}=T^{q_{i}}$ for some $q_{i}\in\mathbb{N}$. So assume WLOG that we are in
this case. Then, note that both the left hand side and the right hand side of
(2) are alternating as functions in the parameters $q_{1},q_{2},\ldots,q_{n}$
(because the residue of $n$ polynomials $f_{1},f_{2},\ldots,f_{n}$ vanishes
when two of the polynomials are equal, and switches sign if two of the
polynomials are swapped). Thus, we can assume WLOG that $q_{1}>q_{2}%
>\cdots>q_{n}$. Assume this, and write the $q_{i}$ in the form $q_{i}%
=h_{i}+n-i$ for some $h_{i}\in\mathbb{N}$; then, the inequalities $q_{1}%
>q_{2}>\cdots>q_{n}\geq0$ lead to $h_{1}\geq h_{2}\geq\cdots\geq h_{n}\geq0$.
Thus, $f_{1}\left(  X\right)  \wedge\cdots\wedge f_{n}\left(  X\right)
=X^{h_{1}+n-1}\wedge X^{h_{2}+n-2}\wedge\cdots\wedge X^{h_{n}}$ (since each
$i\in\left\{  1,2,\ldots,n\right\}  $ satisfies $f_{i}=T^{q_{i}}=T^{h_{i}%
+n-i}$ (since $q_{i}=h_{i}+n-i$)) and%
\[
\operatorname*{Res}\left(  \dfrac{f_{1}}{P},\ldots,\dfrac{f_{n}}{P}\right)
=s_{h_{1},h_{2},\ldots,h_{n}}%
\]
(this has been proven in the previous paragraph). Thus, the claim of (2) is
precisely the assertion (3).

Note that we did not need to assume (1) to make this argument.

\item \textbf{page 831, proof of Lemma 1.1:} Before \textquotedblleft To prove
assertion (3)\textquotedblright, add \textquotedblleft Now assume $I$ to be
equipped with a total order.\textquotedblright.

\item \textbf{page 832, \S 1.3:} When defining $S$ here, you should perhaps
say that this $S$ is not the same $S$ that was defined in \S 0.1, but rather a
generalization thereof.

\item \textbf{page 832, proof of Proposition 1.3:} Remove the four sentences
that begin with \textquotedblleft Then we have an equality\textquotedblright%
\ and end with \textquotedblleft and hence $fy$ is in the
kernel\textquotedblright. (These four sentences merely repeat the four
preceding sentences.)

\item \textbf{page 832, proof of Proposition 1.3:} \textquotedblleft Assume
that $M$ is $A$ free\textquotedblright\ $\rightarrow$ \textquotedblleft Assume
that $M$ is $A$-free\textquotedblright.

\item \textbf{page 833, \S 2.1:} In \textquotedblleft and it is clear that the
action of $S$ on $A\left[  X_{1},\ldots,X_{n}\right]  $ is determined by these
equations\textquotedblright, replace \textquotedblleft$A\left[  X_{1}%
,\ldots,X_{n}\right]  $\textquotedblright\ by \textquotedblleft$\wedge_{A}%
^{n}A\left[  X\right]  $\textquotedblright. (Or perhaps remove these words
altogether, since you don't seem to use them anywhere.)

\item \textbf{page 833, proof of Proposition 2.1:} Replace \textquotedblleft
of (2.2)\textquotedblright\ by \textquotedblleft of (2.1)\textquotedblright.

\item \textbf{page 833, proof of Proposition 2.1:} This proof is not complete.
You implicitly build up an involution for each $k\in\left\{  1,2,\ldots
,n-1\right\}  $ that pairs up cancelling addends on the right hand side of
(2.1); but why don't these involutions for different values of $k$
\textquotedblleft snatch away\textquotedblright\ addends from one another? An
addend may be paired up by more than one of the involutions.

I suggest replacing the proof by a cleaner argument, which I show in the
Appendix to these errata (Section \ref{sect.pf.2.1}):

\item \textbf{page 834, proof of Corollary 2.2:} I find this proof somewhat
harrowing to read; the combinatorics requires too much handwaving. I present a
different proof (longer, but a lot less reliant on mental acrobatics) in the
Appendix to these errata (Section \ref{sect.pf.2.2}).

\item \textbf{page 834, proof of Corollary 2.2:} In the proof of Corollary 2.2
(your proof, not mine), replace \textquotedblleft into $n+1$
intervals\textquotedblright\ by \textquotedblleft into $m+1$
intervals\textquotedblright.

\item \textbf{page 835, \S 2.2:} \textquotedblleft Note that the Schur
function $s_{h_{1}h_{2}\ldots h_{m}0\ldots0}$\textquotedblright\ $\rightarrow$
\textquotedblleft Note that the Schur function $s_{h_{1},h_{2},\ldots
,h_{m},0,\ldots,0}$\textquotedblright.

\item \textbf{page 835, \S 2.2:} The sentence \textquotedblleft For $m=1$
Gatto's formula clearly holds\textquotedblright\ should be moved
\textbf{after} the next sentence (\textquotedblleft We prove Gatto's formula
...\textquotedblright), since at its current position it is unclear what the
\textquotedblleft$m$\textquotedblright\ stands for.

\item \textbf{page 835, \S 2.2:} \textquotedblleft on the positive integers
$m$\textquotedblright\ $\rightarrow$ \textquotedblleft on the smallest
positive integer $m$\textquotedblright.

\item \textbf{page 835, \S 2.2:} Before \textquotedblleft Development of the
determinant\textquotedblright, insert \textquotedblleft Now assume that
$\left(  h_{1},h_{2},\ldots,h_{n}\right)  \in\mathbb{N}^{n}$ has
$h_{m+1}=h_{m+2}=\cdots=h_{n}=0$ but $h_{m}>0$\textquotedblright.

\item \textbf{page 835, \S 2.2:} In the first displayed equation of \S 2.2,
replace the subscript \textquotedblleft$h_{1},\ldots,h_{i-1},h_{i+1}%
,\ldots,h_{m-1},0,\ldots,0$\textquotedblright\ by \textquotedblleft%
$h_{1},\ldots,h_{i-1},h_{i+1}-1,\ldots,h_{m}-1,0,\ldots,0$\textquotedblright.

\item \textbf{page 835, proof of Lemma 2.3:} I cannot follow this proof at the
point where you argue that \textquotedblleft$fX^{n-1}\wedge X^{n-2}%
\wedge\cdots\wedge X^{0}$ contains the term $X^{h_{1}+n-1}\wedge X^{h_{2}%
+n-2}\wedge\cdots\wedge X^{h_{n}}$\textquotedblright. A slightly different
(but cleaner) proof of Lemma 2.3 is shown in the Appendix to these errata
(Section \ref{sect.pf.2.3}).

\item \textbf{page 836, \S 3:} On the first line of \S 3, I would replace
\textquotedblleft of Section 1\textquotedblright\ by \textquotedblleft of
(1.2)\textquotedblright, just to be a bit more specific.

\item \textbf{page 836, \S 3.1:} I suggest replacing \textquotedblleft free
$A$-module of rank $1$ over $S$\textquotedblright\ by \textquotedblleft free
$S$-module of rank $1$\textquotedblright.

\item \textbf{page 836, \S 3.1:} Replace \textquotedblleft$\operatorname*{alt}%
\left(  X_{1}^{h_{1}+n-1}\cdots X_{n}^{0}\right)  $\textquotedblright\ by
\textquotedblleft$\operatorname*{alt}\left(  X_{1}^{h_{1}+n-1}\cdots
X_{n}^{h_{n}}\right)  $\textquotedblright.

\item \textbf{page 836, \S 3.1:} I would simplify this whole paragraph,
avoiding the first reference to [22], as follows:

\textquotedblleft Let $\Delta$ be the Vandermonde determinant
\[
\det\left(  \left(  X_{j}^{n-i}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}\right)  =\operatorname*{alt}\left(  X_{1}^{n-1}X_{2}^{n-2}\cdots
X_{n}^{n-n}\right)  .
\]
Note that $\Delta=\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  $.

The Jacobi-Trudi formula ([22, I.3, (3.4) on p. 41]) shows that
\begin{equation}
s_{h_{1},h_{2},\ldots,h_{n}}=\operatorname*{alt}\left(  X_{1}^{h_{1}+n-1}%
X_{2}^{h_{2}+n-2}\cdots X_{n}^{h_{n}+n-n}\right)  /\Delta\label{p836.3.1.2}%
\end{equation}
for any $\left(  h_{1},h_{2},\ldots,h_{n}\right)  \in\mathbb{N}^{n}$
satisfying $h_{1}\geq h_{2}\geq\cdots\geq h_{n}$. (Keep in mind that what is
called $s_{\lambda}$ in [22, \S I.3] corresponds to our \newline%
$\operatorname*{alt}\left(  X_{1}^{h_{1}+n-1}X_{2}^{h_{2}+n-2}\cdots
X_{n}^{h_{n}+n-n}\right)  /\Delta$, where $\left(  h_{1},h_{2},\ldots
,h_{n}\right)  =\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)  $;
meanwhile, what we call $s_{h_{1},h_{2},\ldots,h_{n}}$ corresponds to the
determinant $\det\left(  h_{\lambda_{i}-i+j}\right)  _{1\leq i,j\leq n}$ in
[22, \S I.3].)

Let $\gamma$ denote the $S$-linear isomorphism $\bigwedge\nolimits_{A}%
^{n}A\left[  X\right]  \rightarrow A\left[  X_{1},\ldots,X_{n}\right]
^{\operatorname*{alt}}$ constructed in Proposition 3.1. Then, $\gamma$ is
bijective (since $\gamma$ is an isomorphism) and thus injective. Now, for any
$\left(  h_{1},h_{2},\ldots,h_{n}\right)  \in\mathbb{N}^{n}$ satisfying
$h_{1}\geq h_{2}\geq\cdots\geq h_{n}$, we have%
\begin{align*}
&  \gamma\left(  X^{h_{1}+n-1}\wedge X^{h_{2}+n-2}\wedge\cdots\wedge
X^{h_{n}+n-n}\right) \\
&  =\operatorname*{alt}\left(  X_{1}^{h_{1}+n-1}X_{2}^{h_{2}+n-2}\cdots
X_{n}^{h_{n}+n-n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition
of }\gamma\right) \\
&  =s_{h_{1},h_{2},\ldots,h_{n}}\Delta\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{p836.3.1.2})}\right) \\
&  =s_{h_{1},h_{2},\ldots,h_{n}}\underbrace{\operatorname*{alt}\left(
X_{1}^{n-1}X_{2}^{n-2}\cdots X_{n}^{n-n}\right)  }_{=\gamma\left(
X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge X^{n-n}\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta=\operatorname*{alt}\left(
X_{1}^{n-1}X_{2}^{n-2}\cdots X_{n}^{n-n}\right)  \right) \\
&  =s_{h_{1},h_{2},\ldots,h_{n}}\gamma\left(  X^{n-1}\wedge X^{n-2}%
\wedge\cdots\wedge X^{n-n}\right) \\
&  =\gamma\left(  s_{h_{1},h_{2},\ldots,h_{n}}\cdot X^{n-1}\wedge
X^{n-2}\wedge\cdots\wedge X^{n-n}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\gamma\text{ is }S\text{-linear}\right)
\end{align*}
and therefore%
\begin{equation}
X^{h_{1}+n-1}\wedge X^{h_{2}+n-2}\wedge\cdots\wedge X^{h_{n}+n-n}%
=s_{h_{1},h_{2},\ldots,h_{n}}\cdot X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge
X^{n-n}. \label{p836.3.1.5}%
\end{equation}
This proves part (3) of the Main Theorem. As we know from Remark 0.5, this
entails that part (2) of the Main Theorem also holds. It remains to prove part (1).

The family%
\[
\left(  X^{i_{1}}\wedge X^{i_{2}}\wedge\cdots\wedge X^{i_{n}}\right)
_{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathbb{N}^{n};\ i_{1}%
>i_{2}>\cdots>i_{n}}%
\]
and the family%
\[
\left(  X^{h_{1}+n-1}\wedge X^{h_{2}+n-2}\wedge\cdots\wedge X^{h_{n}%
+n-n}\right)  _{\left(  h_{1},h_{2},\ldots,h_{n}\right)  \in\mathbb{N}%
^{n};\ h_{1}\geq h_{2}\geq\cdots\geq h_{n}}%
\]
can be obtained from one another by relabelling\footnote{Indeed, there is a
bijection from the set $\left\{  \left(  i_{1},i_{2},\ldots,i_{n}\right)
\in\mathbb{N}^{n}\ \mid\ i_{1}>i_{2}>\cdots>i_{n}\right\}  $ to the set
$\left\{  \left(  h_{1},h_{2},\ldots,h_{n}\right)  \in\mathbb{N}^{n}%
\ \mid\ h_{1}\geq h_{2}\geq\cdots\geq h_{n}\right\}  $; this bijection sends
each $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ to $\left(  i_{1}%
-n+1,i_{2}-n+2,\ldots,i_{n}-n+n\right)  $. This bijection has the property
that if it sends some $n$-tuple $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ to
an $n$-tuple $\left(  h_{1},h_{2},\ldots,h_{n}\right)  $, then $X^{h_{1}%
+n-1}\wedge X^{h_{2}+n-2}\wedge\cdots\wedge X^{h_{n}+n-n}=X^{i_{1}}\wedge
X^{i_{2}}\wedge\cdots\wedge X^{i_{n}}$. Therefore, if we relabel the first of
our two families using this bijection, then we obtain the second family.}.
Hence, these two families have the same span. Since the first family spans the
$A$-module $\bigwedge\nolimits_{A}^{n}A\left[  X\right]  $ (because the
$A$-module $A\left[  X\right]  $ is spanned by $X^{0},X^{1},X^{2},\ldots$), we
thus conclude that the second family spans the $A$-module $\bigwedge
\nolimits_{A}^{n}A\left[  X\right]  $ as well.

For any $\left(  h_{1},h_{2},\ldots,h_{n}\right)  \in\mathbb{N}^{n}$
satisfying $h_{1}\geq h_{2}\geq\cdots\geq h_{n}$, we have%
\begin{align*}
&  X^{h_{1}+n-1}\wedge X^{h_{2}+n-2}\wedge\cdots\wedge X^{h_{n}+n-n}\\
&  =\underbrace{s_{h_{1},h_{2},\ldots,h_{n}}}_{\in S}\cdot X^{n-1}\wedge
X^{n-2}\wedge\cdots\wedge X^{n-n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{p836.3.1.5})}\right) \\
&  \in S\cdot\left(  X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge X^{n-n}\right)  .
\end{align*}
In other words, $f\in S\cdot\left(  X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge
X^{n-n}\right)  $ whenever $f$ is an element of the family $\left(
X^{h_{1}+n-1}\wedge X^{h_{2}+n-2}\wedge\cdots\wedge X^{h_{n}+n-n}\right)
_{\left(  h_{1},h_{2},\ldots,h_{n}\right)  \in\mathbb{N}^{n};\ h_{1}\geq
h_{2}\geq\cdots\geq h_{n}}$. Hence, $f\in S\cdot\left(  X^{n-1}\wedge
X^{n-2}\wedge\cdots\wedge X^{n-n}\right)  $ holds for each $f\in
\bigwedge\nolimits_{A}^{n}A\left[  X\right]  $ (since the family $\left(
X^{h_{1}+n-1}\wedge X^{h_{2}+n-2}\wedge\cdots\wedge X^{h_{n}+n-n}\right)
_{\left(  h_{1},h_{2},\ldots,h_{n}\right)  \in\mathbb{N}^{n};\ h_{1}\geq
h_{2}\geq\cdots\geq h_{n}}$ spans the $A$-module $\bigwedge\nolimits_{A}%
^{n}A\left[  X\right]  $). In other words,
\[
\bigwedge\nolimits_{A}^{n}A\left[  X\right]  \subseteq S\cdot\left(
X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge X^{n-n}\right)  .
\]
Combining this with $S\cdot\left(  X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge
X^{n-n}\right)  \subseteq\bigwedge\nolimits_{A}^{n}A\left[  X\right]  $ (which
is obvious), we obtain%
\[
\bigwedge\nolimits_{A}^{n}A\left[  X\right]  =S\cdot\left(  X^{n-1}\wedge
X^{n-2}\wedge\cdots\wedge X^{n-n}\right)  .
\]
Hence, the $n$-vector $X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge X^{n-n}$
generates the $S$-module $\bigwedge\nolimits_{A}^{n}A\left[  X\right]  $.
Since the annihilator of this $n$-vector is zero (by Lemma 2.3), we thus
conclude that the $1$-tuple $\left(  X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge
X^{n-n}\right)  $ is a basis of the $S$-module $\bigwedge\nolimits_{A}%
^{n}A\left[  X\right]  $. In other words, the $S$-module $\bigwedge
\nolimits_{A}^{n}A\left[  X\right]  $ is free of rank $1$, generated by the
$n$-vector $X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge X^{n-n}$. This proves part
(1) of the Main Theorem. Thus, the Main Theorem is completely proven.

\item \textbf{page 836, \S 4.1:} Before \textquotedblleft The residue algebra
$S\left[  T\right]  /P$ is freely generated\textquotedblright, add
\textquotedblleft The polynomial $P\in S\left[  T\right]  $ is monic of degree
$n$. Thus,\textquotedblright.

\item \textbf{page 836, \S 4.1:} After \textquotedblleft well-known to be an
isomorphism\textquotedblright, add \textquotedblleft(but this latter fact will
not be used)\textquotedblright.

\item \textbf{page 836, \S 4.1:} After \textquotedblleft and it is free of
rank $1$\textquotedblright, add \textquotedblleft with a basis consisting of
the single element $\xi^{n-1}\wedge\xi^{n-2}\wedge\cdots\wedge\xi^{0}$ (since
$S\left[  \xi\right]  $ is a free $S$-module with basis $\xi^{n-1},\xi
^{n-2},\ldots,\xi^{0}$)\textquotedblright.

\item \textbf{page 837, \S 4.1:} \textquotedblleft exterior
product\textquotedblright\ $\rightarrow$ \textquotedblleft exterior
power\textquotedblright.

\item \textbf{page 837, proof of Lemma 4.1:} \textquotedblleft from the
equation $0=P\left(  \xi\right)  =\xi^{n-1}-c_{1}\xi^{n-2}+\cdots+\left(
-1\right)  ^{n}c_{n}$\textquotedblright\ $\rightarrow$ \textquotedblleft from
the equation $0=P\left(  \xi\right)  =\xi^{n}-c_{1}\xi^{n-1}+\cdots+\left(
-1\right)  ^{n}c_{n}$\textquotedblright.

\item \textbf{page 838, proof of Theorem 4.2:} \textquotedblleft natural
injection of $S$-algebras\textquotedblright\ $\rightarrow$ \textquotedblleft
natural homomorphism of $S$-algebras\textquotedblright. (It is true that this
homomorphism is an injection, but this is not obvious at this point, and not
needed for your argument; it is thus only distraction.)

\item \textbf{page 838, proof of Theorem 4.2:} After \textquotedblleft and
equal to the identity on $S$\textquotedblright, add \textquotedblleft(by the
universal property of the residue algebra $S\left[  \xi\right]  =S\left[
T\right]  /P$, since $P\left(  X_{i}\right)  =0$)\textquotedblright.

\item \textbf{page 838, proof of Theorem 4.2:} \textquotedblleft with the
natural surjection $A\left[  X_{1},\ldots,A_{n}\right]  \rightarrow
\bigwedge\nolimits_{A}^{n}A\left[  X\right]  $\textquotedblright%
\ $\rightarrow$ \textquotedblleft with the natural surjection $A\left[
X_{1},\ldots,X_{n}\right]  =\bigotimes\nolimits_{A}^{n}A\left[  X\right]
\rightarrow\bigwedge\nolimits_{A}^{n}A\left[  X\right]  $\textquotedblright.

\item \textbf{page 838, proof of Theorem 4.2:} I suggest explaining somewhere
what you mean by \textquotedblleft alternating\textquotedblright\ when talking
about maps out of a tensor power. (Namely, you say that an $A$-linear map
$f:\bigotimes\nolimits_{A}^{n}V\rightarrow W$ (for two $A$-modules $V$ and
$W$) is \textit{alternating} if and only if the map
\begin{align*}
V^{n}  &  \rightarrow W,\\
\left(  v_{1},v_{2},\ldots,v_{n}\right)   &  \mapsto f\left(  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{n}\right)
\end{align*}
is alternating; equivalently, you say that $f:\bigotimes\nolimits_{A}%
^{n}V\rightarrow W$ is alternating if and only if $f$ factors through the
canonical projection $\bigotimes\nolimits_{A}^{n}V\rightarrow\bigwedge
\nolimits_{A}^{n}W$.)

\item \textbf{page 838, \S 4.3:} After \textquotedblleft is upper triangular
with $1$'s on the diagonal\textquotedblright, add \textquotedblleft(by the
equality $T^{h}/P\left(  T\right)  =\sum_{j=n-h}^{\infty}s_{h+j-n}T^{-j}$ in
Remark 0.5, and because $s_{0}=1$)\textquotedblright.

\item \textbf{page 838, \S 4.3:} You write: \textquotedblleft Therefore, since
$\bigwedge\nolimits_{S}^{n}S\left[  \xi\right]  $ is free of rank $1$ over $S$
with generator $\xi^{n-1}\wedge\cdots\wedge\xi^{0}$, equation (4.5) is
general\textquotedblright. In my opinion, this can be explained better.
Indeed, if $f_{1},f_{2},\ldots,f_{n}\in S\left[  T\right]  $ are $n$
polynomials, then the residue $\operatorname*{Res}\left(  \dfrac{f_{1}}%
{P},\dfrac{f_{2}}{P},\ldots,\dfrac{f_{n}}{P}\right)  $ depends only on the
residue classes of the polynomials $f_{1},f_{2},\ldots,f_{n}$ modulo $P$ (but
not on these polynomials themselves)\footnote{This is because the residue
$\operatorname*{Res}\left(  \dfrac{f_{1}}{P},\dfrac{f_{2}}{P},\ldots
,\dfrac{f_{n}}{P}\right)  $ does not change when a multiple of $P$ is added to
one of $f_{1},f_{2},\ldots,f_{n}$ (since this residue is $S$-multilinear in
$f_{1},f_{2},\ldots,f_{n}$ and vanishes when one of $f_{1},f_{2},\ldots,f_{n}$
is divisible by $P$).}. In other words, this residue can be depends only on
the values $f_{1}\left(  \xi\right)  ,f_{2}\left(  \xi\right)  ,\ldots
,f_{n}\left(  \xi\right)  $ (since these values encode the same information as
the residue classes of the polynomials $f_{1},f_{2},\ldots,f_{n}$ modulo $P$).
Hence, the map
\begin{align*}
\alpha:\left(  S\left[  \xi\right]  \right)  ^{n}  &  \rightarrow
\bigwedge\nolimits_{S}^{n}S\left[  \xi\right]  ,\\
\left(  f_{1}\left(  \xi\right)  ,f_{2}\left(  \xi\right)  ,\ldots
,f_{n}\left(  \xi\right)  \right)   &  \mapsto\operatorname*{Res}\left(
\dfrac{f_{1}}{P},\dfrac{f_{2}}{P},\ldots,\dfrac{f_{n}}{P}\right)  \xi
^{n-1}\wedge\xi^{n-2}\wedge\cdots\wedge\xi^{0}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{for }f_{1},f_{2},\ldots,f_{n}\in S\left[
T\right]  \right)
\end{align*}
is well-defined. This map $\alpha$ is furthermore $S$-multilinear (since the
residue is $S$-multilinear) and alternating (since the residue is
alternating). Hence, it induces an $S$-linear map%
\begin{align*}
\alpha^{\prime}:\bigwedge\nolimits_{S}^{n}S\left[  \xi\right]   &
\rightarrow\bigwedge\nolimits_{S}^{n}S\left[  \xi\right]  ,\\
f_{1}\left(  \xi\right)  \wedge f_{2}\left(  \xi\right)  \wedge\cdots\wedge
f_{n}\left(  \xi\right)   &  \mapsto\operatorname*{Res}\left(  \dfrac{f_{1}%
}{P},\dfrac{f_{2}}{P},\ldots,\dfrac{f_{n}}{P}\right)  \xi^{n-1}\wedge\xi
^{n-2}\wedge\cdots\wedge\xi^{0}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{for }f_{1},f_{2},\ldots,f_{n}\in S\left[
T\right]  \right)  .
\end{align*}
But you have proved the equality (4.5) in the case when $f_{i}\left(
T\right)  =T^{n-i}$ for all $i\in\left\{  1,2,\ldots,n\right\}  $. In other
words, you have shown that%
\[
\xi^{n-1}\wedge\xi^{n-2}\wedge\cdots\wedge\xi^{n-n}=\operatorname*{Res}\left(
\dfrac{T^{n-1}}{P},\dfrac{T^{n-2}}{P},\ldots,\dfrac{T^{n-n}}{P}\right)
\xi^{n-1}\wedge\xi^{n-2}\wedge\cdots\wedge\xi^{0}.
\]
But the definition of $\alpha^{\prime}$ (applied to the polynomials
$T^{n-1},T^{n-2},\ldots,T^{n-n}$ instead of $f_{1},f_{2},\ldots,f_{n}$) yields%
\begin{align*}
&  \alpha^{\prime}\left(  \xi^{n-1}\wedge\xi^{n-2}\wedge\cdots\wedge\xi
^{n-n}\right) \\
&  =\operatorname*{Res}\left(  \dfrac{T^{n-1}}{P},\dfrac{T^{n-2}}{P}%
,\ldots,\dfrac{T^{n-n}}{P}\right)  \xi^{n-1}\wedge\xi^{n-2}\wedge\cdots
\wedge\xi^{0}.
\end{align*}
Comparing these two equalities, we find%
\begin{align*}
\alpha^{\prime}\left(  \xi^{n-1}\wedge\xi^{n-2}\wedge\cdots\wedge\xi
^{n-n}\right)   &  =\xi^{n-1}\wedge\xi^{n-2}\wedge\cdots\wedge\xi^{n-n}\\
&  =\operatorname*{id}\left(  \xi^{n-1}\wedge\xi^{n-2}\wedge\cdots\wedge
\xi^{n-n}\right)  .
\end{align*}
Hence, the two $S$-linear maps $\alpha^{\prime}:\bigwedge\nolimits_{S}%
^{n}S\left[  \xi\right]  \rightarrow\bigwedge\nolimits_{S}^{n}S\left[
\xi\right]  $ and $\operatorname*{id}:\bigwedge\nolimits_{S}^{n}S\left[
\xi\right]  \rightarrow\bigwedge\nolimits_{S}^{n}S\left[  \xi\right]  $ are
equal to each other on the element $\xi^{n-1}\wedge\xi^{n-2}\wedge\cdots
\wedge\xi^{n-n}$. Since this element $\xi^{n-1}\wedge\xi^{n-2}\wedge
\cdots\wedge\xi^{n-n}$ generates the $S$-module $\bigwedge\nolimits_{S}%
^{n}S\left[  \xi\right]  $, we can thus conclude that the two maps
$\alpha^{\prime}$ and $\operatorname*{id}$ are identical (because if two
$S$-linear maps are equal to each other on a given generating set of their
domain, then they must be identical). In other words, $\alpha^{\prime
}=\operatorname*{id}$. Hence, every $f_{1},f_{2},\ldots,f_{n}\in S\left[
T\right]  $ satisfy%
\begin{align*}
&  \underbrace{\alpha^{\prime}}_{=\operatorname*{id}}\left(  f_{1}\left(
\xi\right)  \wedge f_{2}\left(  \xi\right)  \wedge\cdots\wedge f_{n}\left(
\xi\right)  \right) \\
&  =\operatorname*{id}\left(  f_{1}\left(  \xi\right)  \wedge f_{2}\left(
\xi\right)  \wedge\cdots\wedge f_{n}\left(  \xi\right)  \right)  =f_{1}\left(
\xi\right)  \wedge f_{2}\left(  \xi\right)  \wedge\cdots\wedge f_{n}\left(
\xi\right)  .
\end{align*}
Hence,%
\begin{align*}
&  f_{1}\left(  \xi\right)  \wedge f_{2}\left(  \xi\right)  \wedge\cdots\wedge
f_{n}\left(  \xi\right) \\
&  =\alpha^{\prime}\left(  f_{1}\left(  \xi\right)  \wedge f_{2}\left(
\xi\right)  \wedge\cdots\wedge f_{n}\left(  \xi\right)  \right) \\
&  =\operatorname*{Res}\left(  \dfrac{f_{1}}{P},\dfrac{f_{2}}{P},\ldots
,\dfrac{f_{n}}{P}\right)  \xi^{n-1}\wedge\xi^{n-2}\wedge\cdots\wedge\xi^{0}%
\end{align*}
(by the definition of $\alpha^{\prime}$). Hence, (4.5) is proven.

\item \textbf{page 839, \S 5.1:} Replace \textquotedblleft isomorphism of
$S$-algebras $S\left[  \xi\right]  \rightarrow S\left[  X_{i}\right]
$\textquotedblright\ by \textquotedblleft homomorphism of $S$-algebras
$S\left[  \xi\right]  \rightarrow A\left[  X_{1},X_{2},\ldots,X_{n}\right]
$\textquotedblright. (It is true that this homomorphism restricts to an
isomorphism of $S$-algebras $S\left[  \xi\right]  \rightarrow S\left[
X_{i}\right]  $, but this is neither needed nor easy to prove at this point.)

\item \textbf{page 839, \S 5.1:} It is worth mentioning here that you are
considering $\bigwedge\nolimits_{S}^{n}S\left[  \xi\right]  $ to be equipped
with its natural structure (not its symmetric structure) throughout \S 5.

\item \textbf{page 839, \S 5.2:} Replace \textquotedblleft when $F_{i}%
=T^{n-i}$ for $i=1,2,\ldots,n$\textquotedblright\ by \textquotedblleft when
$\left(  F_{i}=T^{n-i}\text{ for }i=1,2,\ldots,n\right)  $\textquotedblright.
(The parentheses are meant to clarify the logical structure of this sentence:

\begin{statement}
\textquotedblleft The function $R$ is equal to $1$ when $\left(  F_{i}%
=T^{n-i}\text{ for }i=1,2,\ldots,n\right)  $\textquotedblright,
\end{statement}

not

\begin{statement}
\textquotedblleft$\left(  \text{The function }R\text{ is equal to }1\text{
when }F_{i}=T^{n-i}\right)  $ for $i=1,2,\ldots,n$\textquotedblright.
\end{statement}

\item \textbf{page 839, \S 5.2:} You write: \textquotedblleft It follows
immediately that\textquotedblright. I don't find this obvious enough to
deserve the word \textquotedblleft immediately\textquotedblright. The argument
you are tacitly making here is essentially the argument you have done in
\S 4.2 in order to prove (4.5); it is not in any way made unnecessary by the
slight change of viewpoint done in \S 5.

\item \textbf{page 839, \S 5.2:} \textquotedblleft prove that the generator
$\xi^{n-1}\wedge\cdots\wedge\xi^{0}$ has no $S$-torsion\textquotedblright%
\ $\rightarrow$ \textquotedblleft prove that the generator $X^{n-1}%
\wedge\cdots\wedge X^{0}$ has no $S$-torsion\textquotedblright. (For the
generator $\xi^{n-1}\wedge\cdots\wedge\xi^{0}$, this is obvious, but that is
not the generator you need here.)

\item \textbf{page 839, \S 5.2:} \textquotedblleft Under the composition of
the map (4.4) with the alternator\textquotedblright\ $\rightarrow$
\textquotedblleft Under the alternator\textquotedblright. (The map (4.4) is
not needed here.)

\item \textbf{page 839, \S 5.2:} \textquotedblleft$\bigwedge\nolimits^{n}%
A\left[  X\right]  $\textquotedblright\ $\rightarrow$ \textquotedblleft%
$\bigwedge\nolimits_{A}^{n}A\left[  X\right]  $\textquotedblright.

\item \textbf{page 839, \S 5.2:} \textquotedblleft the generator $\xi
^{n-1}\wedge\cdots\wedge\xi^{0}$ is mapped\textquotedblright\ $\rightarrow$
\textquotedblleft the generator $X^{n-1}\wedge\cdots\wedge X^{0}$ is
mapped\textquotedblright.

\item \textbf{page 839, \S 5.2:} After \textquotedblleft the generator itself
has no $S$-torsion\textquotedblright, add \textquotedblleft(since the
alternator map $\operatorname*{alt}$ is $S$-linear)\textquotedblright.

\item \textbf{page 839, \S 5.3:} \textquotedblleft Therefore the target is a
free $S$-module of rank $1$\textquotedblright\ $\rightarrow$ \textquotedblleft
Therefore the target is generated (as an $S$-module) by the regular polynomial
$\Delta$. Hence, it is a free $S$-module of rank $1$\textquotedblright.

\item \textbf{page 839, \S 5.3:} \textquotedblleft It follows that both maps
are isomorphisms.\textquotedblright\ $\rightarrow$ \textquotedblleft Hence the
composite map in (5.2) is a surjective $S$-linear map between two free
$S$-modules of rank $1$, and thus is an isomorphism. It follows that both maps
in (5.2) are isomorphisms (since they are surjective).\textquotedblright

\item \textbf{page 839, \S 5.3:} I would replace \textquotedblleft times
$\Delta$ or, equivalently, an anti-symmetric\textquotedblright\ by
\textquotedblleft times $\Delta$. Equivalently, an
anti-symmetric\textquotedblright.

\item \textbf{page 840, \S 6.1:} After \textquotedblleft see [2, VI 6.5], [24,
2.1], [6], or [25]\textquotedblright, I would also add a reference to
\cite[\S 1.3]{LakTho12} (where $A_{r}$ is denoted by $S_{r}$).

\item \textbf{page 840, \S 6.1:} In the sentence that defines $\partial
^{1,\ldots,r}$, replace \textquotedblleft if $h_{i}=n-i$ for $i=1,\ldots
,r$\textquotedblright\ by \textquotedblleft if $h_{i}=n-i$ for all
$i=1,\ldots,r$\textquotedblright\ (to prevent misunderstanding).

\item \textbf{page 840, \S 6.1:} \textquotedblleft we write $\partial
=\partial^{1,\ldots,n}$\textquotedblright\ $\rightarrow$ \textquotedblleft we
write $\partial=\partial^{1,\ldots,n-1}$\textquotedblright.

\item \textbf{page 840, \S 6.1:} Replace \textquotedblleft$A_{r}\left[
T\right]  \left[  [T^{-1}\right]  \rightarrow A\left[  T\right]  \left[
\left[  T^{-1}\right]  \right]  $\textquotedblright\ by \textquotedblleft%
$A_{r}\left[  T\right]  \left[  \left[  T^{-1}\right]  \right]  \rightarrow
A\left[  T\right]  \left[  \left[  T^{-1}\right]  \right]  $\textquotedblright.

\item \textbf{page 841, proof of Lemma 6.1:} After \textquotedblleft Now,
$p\left(  T\right)  $ is an $A$-linear combination of monomials $T^{i}%
$\textquotedblright, add \textquotedblleft with $i\leq n$, and in this
combination the monomial $T^{n}$ has coefficient $1$\textquotedblright.

\item \textbf{page 841, proof of Lemma 6.1:} \textquotedblleft we see,
that\textquotedblright\ $\rightarrow$ \textquotedblleft we see
that\textquotedblright.

\item \textbf{page 841, Proposition 6.2:} This proposition is correct, but it
is insufficient for what you want to use it for (namely, proving Proposition
6.3). In order to make it stronger, I suggest removing the words
\textquotedblleft of degree $t$\textquotedblright\ (so $q$ can have any
degree). This necessitates a minor tweak in the proof (see below).

\item \textbf{page 842, (6.6):} On the right hand side of (6.6), add a comma
before \textquotedblleft$\dfrac{g_{t}}{q}$\textquotedblright.

\item \textbf{page 842, proof of Proposition 6.2:} The last paragraph of this
proof is no longer correct now that I have generalized it. So let me suggest
an alternative to this last paragraph:

\textquotedblleft Thus, we know that the left hand side of (6.6) vanishes if
$q$ divides some $g_{j}$, and is $A$-linear in each $g_{j}$. Hence, the left
hand side of (6.6) does not change if we add a multiple of $q$ to some $g_{j}%
$. The same holds for the right hand side (for the same reason). Thus, we can
replace each polynomial $g_{j}$ by its remainder modulo $\left(  q\right)  $.
Hence, we can WLOG assume that all polynomials $g_{j}$ have degree $<\deg q$.
Assume this. Since both sides of (6.6) are $A$-linear in each $g_{j}$, we can
furthermore assume that each $g_{j}$ is a single monomial: that is, there is a
$t$-tuple $\left(  m_{1},m_{2},\ldots,m_{t}\right)  \in\left\{  0,1,\ldots
,\deg q-1\right\}  ^{t}$ such that each $j$ satisfies $g_{j}=T^{m_{j}}$.
Assume this, too. Furthermore, both sides of (6.6) vanish if two of the
$g_{j}$ are equal; thus, we can WLOG assume that $m_{1},m_{2},\ldots,m_{t}$
are distinct. Assume this. Finally, both sides of (6.6) are anti-symmetric in
the $g_{j}$; hence, we can WLOG assume that $m_{1}\geq m_{2}\geq\cdots\geq
m_{t}$ (since otherwise, we can just permute $m_{1},m_{2},\ldots,m_{t}$ so
that this holds). Assume this. Combining $m_{1}\geq m_{2}\geq\cdots\geq m_{t}$
with the fact that $m_{1},m_{2},\ldots,m_{t}$ are distinct, we obtain
$m_{1}>m_{2}>\cdots>m_{t}$. Therefore, $m_{1}+1\geq m_{2}+2\geq\cdots\geq
m_{t}+t$. Hence, each $j\in\left\{  1,2,\ldots,t\right\}  $ satisfies%
\begin{align*}
m_{j}+j  &  \leq m_{1}+1\leq\deg q\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\left(  m_{1},m_{2},\ldots,m_{t}\right)  \in\left\{
0,1,\ldots,\deg q-1\right\}  ^{t}\\
\text{and thus }m_{1}\in\left\{  0,1,\ldots,\deg q-1\right\}  \text{, so that
}m_{1}\leq\deg q-1
\end{array}
\right)
\end{align*}
and thus%
\[
m_{j}\leq\deg q-j.
\]
Thus, for each $j\in\left\{  1,2,\ldots,t\right\}  $, the polynomial
$g_{j}=T^{m_{j}}$ is monic of degree at most $\deg q-j$ (since $m_{j}\leq\deg
q-j$), and therefore can be written in the form%
\begin{equation}
g_{j}=c_{j}T^{\deg q-j}+\left(  \text{lower order terms}\right)  ,
\label{p842.pf.p6.2.4}%
\end{equation}
where \textquotedblleft$\left(  \text{lower order terms}\right)
$\textquotedblright\ means an $A$-linear combination of monomials $T^{u}$ with
$u<\deg q-j$, and where $c_{j}\in A$ is either $0$ or $1$ (depending on
whether $m_{j}<\deg q-j$ or $m_{j}=\deg q-j$). Consider these $c_{j}$.

Thus, for each $j\in\left\{  1,2,\ldots,t\right\}  $, the Laurent series
$\dfrac{g_{j}}{q}\in A\left[  T\right]  \left[  \left[  T^{-1}\right]
\right]  $ has the form%
\begin{align}
\dfrac{g_{j}}{q}  &  =\dfrac{c_{j}T^{\deg q-j}+\left(  \text{lower order
terms}\right)  }{q}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{p842.pf.p6.2.4}%
)}\right) \nonumber\\
&  =c_{j}T^{-j}+\left(  \text{lower order terms}\right)  ,
\label{p842.pf.p6.2.tria1}%
\end{align}
where \textquotedblleft$\left(  \text{lower order terms}\right)
$\textquotedblright\ means an $A$-linear combination of monomials $T^{u}$ with
$u<-j$. (Here, we have used the fact that $q$ is a monic polynomials of degree
$\deg q$, and therefore division by $q$ lowers the leading term of any Laurent
series by the degree $q$.)

From (\ref{p842.pf.p6.2.tria1}), we see that the matrix used in defining the
residue $\operatorname*{Res}\left(  \dfrac{g_{1}}{q},\ldots,\dfrac{g_{t}}%
{q}\right)  $ is upper-triangular, with diagonal entries $c_{1},c_{2}%
,\ldots,c_{t}$. Hence, its determinant is given by%
\begin{equation}
\operatorname*{Res}\left(  \dfrac{g_{1}}{q},\ldots,\dfrac{g_{t}}{q}\right)
=c_{1}c_{2}\cdots c_{t}. \label{p842.pf.p6.2.rhs}%
\end{equation}


Also, $p_{r}q$ is a monic polynomial of degree $r+\deg q$ (since $p_{r}$ and
$q$ are monic polynomials of degrees $r$ and $\deg q$, respectively). Now, for
each $j\in\left\{  1,2,\ldots,t\right\}  $, the Laurent series $\dfrac{g_{j}%
}{p_{r}q}\in A\left[  T\right]  \left[  \left[  T^{-1}\right]  \right]  $ has
the form%
\begin{align}
\dfrac{g_{j}}{p_{r}q}  &  =\dfrac{c_{j}T^{\deg q-j}+\left(  \text{lower order
terms}\right)  }{p_{r}q}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{p842.pf.p6.2.4})}\right) \nonumber\\
&  =c_{j}T^{-r-j}+\left(  \text{lower order terms}\right)  ,
\label{p842.pf.p6.2.tria2}%
\end{align}
where \textquotedblleft$\left(  \text{lower order terms}\right)
$\textquotedblright\ means an $A$-linear combination of monomials $T^{u}$ with
$u<-r-j$. (Here, we have used the fact that $p_{r}q$ is a monic polynomial of
degree $r+\deg q$, and therefore division by $p_{r}q$ lowers the leading term
any Laurent series by the degree $r+\deg q$.)

Furthermore, for each $i\in\left\{  1,2,\ldots,r\right\}  $, the Laurent
series $\dfrac{1}{p_{i}}\in A\left[  T\right]  \left[  \left[  T^{-1}\right]
\right]  $ has the form%
\begin{equation}
\dfrac{1}{p_{i}}=T^{-i}+\left(  \text{lower order terms}\right)  ,
\label{p842.pf.p6.2.tria3}%
\end{equation}
where \textquotedblleft$\left(  \text{lower order terms}\right)
$\textquotedblright\ means an $A$-linear combination of monomials $T^{u}$ with
$u<-i$. (This is simply because $p_{i}$ is a monic polynomial of degree $i$.)

From (\ref{p842.pf.p6.2.tria3}) and (\ref{p842.pf.p6.2.tria2}), we see that
the matrix used in defining the residue $\operatorname*{Res}\left(  \dfrac
{1}{p_{1}},\ldots,\dfrac{1}{p_{r}},\dfrac{g_{1}}{p_{r}q},\ldots,\dfrac{g_{t}%
}{p_{r}q}\right)  $ is upper-triangular, with diagonal entries
$\underbrace{1,1,\ldots,1}_{r\text{ times}},c_{1},c_{2},\ldots,c_{t}$. Hence,
its determinant is given by%
\[
\operatorname*{Res}\left(  \dfrac{1}{p_{1}},\ldots,\dfrac{1}{p_{r}}%
,\dfrac{g_{1}}{p_{r}q},\ldots,\dfrac{g_{t}}{p_{r}q}\right)
=\underbrace{1\cdot1\cdots1}_{r\text{ times}}c_{1}c_{2}\cdots c_{t}=c_{1}%
c_{2}\cdots c_{t}.
\]
Comparing this with (\ref{p842.pf.p6.2.rhs}), we obtain
\[
\operatorname*{Res}\left(  \dfrac{1}{p_{1}},\ldots,\dfrac{1}{p_{r}}%
,\dfrac{g_{1}}{p_{r}q},\ldots,\dfrac{g_{t}}{p_{r}q}\right)
=\operatorname*{Res}\left(  \dfrac{g_{1}}{q},\ldots,\dfrac{g_{t}}{q}\right)
.
\]
This proves (6.6), and thus completes the proof of Proposition
6.2.\textquotedblright

\item \textbf{page 842, proof of Proposition 6.3:} Replace \textquotedblleft
the corresponding composition $\partial^{2,\ldots,r}$\textquotedblright\ by
\textquotedblleft the corresponding composition $\partial^{2,\ldots
,r}:=\partial^{2}\circ\cdots\circ\partial^{r}$\textquotedblright, since
strictly speaking you have not defined $\partial^{2,\ldots,r}$ yet (not that
it isn't very obvious).

\item \textbf{page 842, (6.8):} On the right hand side of the first line of
(6.8), replace \textquotedblleft$f\left(  \xi_{1}\right)  $\textquotedblright%
\ by \textquotedblleft$f_{1}\left(  \xi_{1}\right)  $\textquotedblright.

\item \textbf{page 842, proof of Proposition 6.3:} \textquotedblleft follows
from equation (6.7) applied with $r:=1$ and $q:=q_{1}$\textquotedblright%
\ $\rightarrow$ \textquotedblleft follows from equation (6.3) applied with
$A:=A_{1}$, $r:=1$, $a_{1}:=\xi_{1}$, $q:=q_{1}$, $t:=r-1$ and $g_{j}%
:=f_{j-1}$ (since $p_{1}=T-\xi_{1}$ and $p=\left(  T-\xi_{1}\right)  \cdot
q_{1}=p_{1}q_{1}$)\textquotedblright.

Note that this relies on the generalized version of Proposition 6.2 suggested
above, since $q_{1}$ has degree $n-1$, not $r-1$ (in general).

\item \textbf{page 843, \S 7.1:} You write: \textquotedblleft It is easy to
check that the $S$-algebra $A\left[  X_{1},\ldots,X_{n}\right]  $ satisfies
the universal properties of the splitting algebra of the \textit{generic}
polynomial $P=\left(  T-X_{1}\right)  \cdots\left(  T-X_{n}\right)
=T^{n}-c_{1}T^{n-1}+\cdots+\left(  -1\right)  ^{n}c_{n}$ over $S$ with
$X_{1},\ldots,X_{n}$ as universal roots\textquotedblright.

Let me spell out what this means and actually check that it is true.

First of all, the universal property of the splitting algebra of a polynomial
has been stated in \cite{LakTho12} (more precisely, \cite[\S 1.2]{LakTho12}
defines factorization algebras through their universal property, and
\cite[\S 1.3]{LakTho12} defines splitting algebras as a particular case of
factorization algebras). Applying this property to the $S$-algebra $A\left[
X_{1},\ldots,X_{n}\right]  $, we see that the claim that \textquotedblleft the
$S$-algebra $A\left[  X_{1},\ldots,X_{n}\right]  $ satisfies the universal
properties of the splitting algebra of the \textit{generic} polynomial
$P=\left(  T-X_{1}\right)  \cdots\left(  T-X_{n}\right)  =T^{n}-c_{1}%
T^{n-1}+\cdots+\left(  -1\right)  ^{n}c_{n}$ over $S$ with $X_{1},\ldots
,X_{n}$ as universal roots\textquotedblright\ boils down to the following statement:

\begin{statement}
\textit{Statement 1:} Let $B$ be any $S$-algebra. Let $p=\varphi_{1}%
\varphi_{2}\cdots\varphi_{n}$ be any factorization of $p$ over $B\left[
T\right]  $ into monic linear polynomials $\varphi_{1},\varphi_{2}%
,\ldots,\varphi_{n}\in B\left[  T\right]  $. Then, there is a unique
$S$-algebra homomorphism $\gamma:A\left[  X_{1},\ldots,X_{n}\right]
\rightarrow B$ such that the induced map\footnote{Here and in the following,
we are using the following notation: If $P$ and $Q$ are two rings, and if
$\alpha:P\rightarrow Q$ is any ring homomorphism, then $\alpha\left[
T\right]  $ shall denote the ring homomorphism from $P\left[  T\right]  $ to
$Q\left[  T\right]  $ that is defined by%
\[
\left(  \alpha\left[  T\right]  \right)  \left(  \sum_{i=0}^{m}p_{i}%
T^{i}\right)  =\sum_{i=0}^{m}\alpha\left(  p_{i}\right)  T^{i}%
\ \ \ \ \ \ \ \ \ \ \text{for all }m\in\mathbb{N}\text{ and }p_{0}%
,p_{1},\ldots,p_{m}\in P.
\]
(Thus, roughly speaking, $\alpha\left[  T\right]  $ is the map that transforms
a polynomial $p\in P\left[  T\right]  $ by applying $\alpha$ to each
coefficient of this polynomial.) This map $\alpha\left[  T\right]  $ is said
to be \textit{induced} by $\alpha$.
\par
If both rings $P$ and $Q$ are $W$-algebras for some commutative ring $W$, and
if $\alpha:P\rightarrow Q$ is a $W$-algebra homomorphism, then the induced map
$\alpha\left[  T\right]  $ is a $W\left[  T\right]  $-algebra homomorphism.}
$\gamma\left[  T\right]  :\left(  A\left[  X_{1},\ldots,X_{n}\right]  \right)
\left[  T\right]  \rightarrow B\left[  T\right]  $ maps $T-X_{i}$ to
$\varphi_{i}$ for all $i\in\left\{  1,2,\ldots,n\right\}  $.
\end{statement}

Thus, it suffices to prove Statement 1.

We shall prove it by showing a slightly nicer version of it first:

\begin{statement}
\textit{Statement 2:} Let $B$ be any $S$-algebra. Let $u_{1},u_{2}%
,\ldots,u_{n}\in B$ be any elements such that%
\begin{equation}
P=\left(  T-u_{1}\right)  \left(  T-u_{2}\right)  \cdots\left(  T-u_{n}%
\right)  \ \ \ \ \ \ \ \ \ \ \text{in }B\left[  T\right]  .
\label{p843.St2.eq.ass}%
\end{equation}
Then, there is a unique $S$-algebra homomorphism $\gamma:A\left[  X_{1}%
,\ldots,X_{n}\right]  \rightarrow B$ such that
\begin{equation}
\left(  \gamma\left(  X_{i}\right)  =u_{i}\ \ \ \ \ \ \ \ \ \ \text{for all
}i\in\left\{  1,2,\ldots,n\right\}  \right)  . \label{p843.St2.eq.1}%
\end{equation}

\end{statement}

[\textit{Proof of Statement 2:} First of all, let us recall the universal
property of the polynomial ring $A\left[  X_{1},\ldots,X_{n}\right]  $. This
property shows that there is a unique $A$-algebra homomorphism $\gamma
:A\left[  X_{1},\ldots,X_{n}\right]  \rightarrow B$ satisfying
(\ref{p843.St2.eq.1}). Denote this $\gamma$ by $\eta$. Thus, $\eta:A\left[
X_{1},\ldots,X_{n}\right]  \rightarrow B$ is an $A$-algebra homomorphism and
satisfies%
\begin{equation}
\left(  \eta\left(  X_{i}\right)  =u_{i}\ \ \ \ \ \ \ \ \ \ \text{for all
}i\in\left\{  1,2,\ldots,n\right\}  \right)  . \label{p843.St2.pf.1}%
\end{equation}


Next, we shall show that $\eta$ is actually an $S$-algebra homomorphism. Let
us consider the $A\left[  T\right]  $-algebra homomorphism%
\[
\eta\left[  T\right]  :\left(  A\left[  X_{1},\ldots,X_{n}\right]  \right)
\left[  T\right]  \rightarrow B\left[  T\right]
\]
induced by $\eta$. (This homomorphism $\eta\left[  T\right]  $ simply applies
$\eta$ to each coefficient of the polynomial it acts upon.)

\begin{vershort}
For each $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{align*}
\left(  \eta\left[  T\right]  \right)  \left(  T-X_{i}\right)   &
=T-\underbrace{\eta\left(  X_{i}\right)  }_{\substack{=u_{i}\\\text{(by
(\ref{p843.St2.pf.1}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition
of }\eta\left[  T\right]  \right) \\
&  =T-u_{i}.
\end{align*}

\end{vershort}

\begin{verlong}
For each $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{align*}
\left(  \eta\left[  T\right]  \right)  \left(  T-X_{i}\right)   &
=\underbrace{\left(  \eta\left[  T\right]  \right)  \left(  T\right)
}_{\substack{=T\\\text{(by the definition}\\\text{of }\eta\left[  T\right]
\text{)}}}-\underbrace{\left(  \eta\left[  T\right]  \right)  \left(
X_{i}\right)  }_{\substack{=\eta\left(  X_{i}\right)  \\\text{(by the
definition}\\\text{of }\eta\left[  T\right]  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta\left[  T\right]  \text{ is a
ring homomorphism}\right) \\
&  =T-\underbrace{\eta\left(  X_{i}\right)  }_{\substack{=u_{i}\\\text{(by
(\ref{p843.St2.pf.1}))}}}=T-u_{i}.
\end{align*}

\end{verlong}

\noindent Multiplying these equalities for all $i\in\left\{  1,2,\ldots
,n\right\}  $, we obtain%
\begin{align}
&  \left(  \eta\left[  T\right]  \right)  \left(  T-X_{1}\right)  \cdot\left(
\eta\left[  T\right]  \right)  \left(  T-X_{2}\right)  \cdot\cdots\cdot\left(
\eta\left[  T\right]  \right)  \left(  T-X_{n}\right) \nonumber\\
&  =\left(  T-u_{1}\right)  \left(  T-u_{2}\right)  \cdots\left(
T-u_{n}\right)  . \label{p843.St2.pf.2}%
\end{align}


On the other hand, $B$ is an $S$-algebra. Thus, the map%
\[
\iota:S\rightarrow B,\ \ \ \ \ \ \ \ \ \ s\mapsto s\cdot1_{B}%
\]
is an $S$-algebra homomorphism, and therefore an $A$-algebra homomorphism. It
thus induces an $A\left[  T\right]  $-algebra homomorphism $\iota\left[
T\right]  :S\left[  T\right]  \rightarrow B\left[  T\right]  $. Note that the
\textquotedblleft$P$\textquotedblright\ on the left hand side of the equality
(\ref{p843.St2.eq.ass}) actually stands not for the polynomial $P\in S\left[
T\right]  $ itself, but rather for its image $\left(  \iota\left[  T\right]
\right)  \left(  P\right)  $ under this homomorphism; thus,
(\ref{p843.St2.eq.ass}) rewrites as
\[
\left(  \iota\left[  T\right]  \right)  \left(  P\right)  =\left(
T-u_{1}\right)  \left(  T-u_{2}\right)  \cdots\left(  T-u_{n}\right)  .
\]
Comparing this with (\ref{p843.St2.pf.2}), we find%
\begin{align}
&  \left(  \eta\left[  T\right]  \right)  \left(  T-X_{1}\right)  \cdot\left(
\eta\left[  T\right]  \right)  \left(  T-X_{2}\right)  \cdot\cdots\cdot\left(
\eta\left[  T\right]  \right)  \left(  T-X_{n}\right) \nonumber\\
&  =\left(  \iota\left[  T\right]  \right)  \left(  P\right)  .
\label{p843.St2.pf.3}%
\end{align}


From $P=T^{n}-c_{1}T^{n-1}+\cdots+\left(  -1\right)  ^{n}c_{n}$, we obtain%
\[
\left(  \eta\left[  T\right]  \right)  \left(  P\right)  =T^{n}-\eta\left(
c_{1}\right)  T^{n-1}+\cdots+\left(  -1\right)  ^{n}\eta\left(  c_{n}\right)
\]
(by the definition of $\eta\left[  T\right]  $). Hence,%
\begin{align*}
&  T^{n}-\eta\left(  c_{1}\right)  T^{n-1}+\cdots+\left(  -1\right)  ^{n}%
\eta\left(  c_{n}\right) \\
&  =\left(  \eta\left[  T\right]  \right)  \left(  P\right)  =\left(
\eta\left[  T\right]  \right)  \left(  \left(  T-X_{1}\right)  \left(
T-X_{2}\right)  \cdots\left(  T-X_{n}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }P=\left(  T-X_{1}\right)  \left(
T-X_{2}\right)  \cdots\left(  T-X_{n}\right)  \right) \\
&  =\left(  \eta\left[  T\right]  \right)  \left(  T-X_{1}\right)
\cdot\left(  \eta\left[  T\right]  \right)  \left(  T-X_{2}\right)
\cdot\cdots\cdot\left(  \eta\left[  T\right]  \right)  \left(  T-X_{n}\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta\left[  T\right]  \text{ is a
ring homomorphism}\right) \\
&  =\left(  \iota\left[  T\right]  \right)  \left(  P\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{p843.St2.pf.3})}\right) \\
&  =\left(  \iota\left[  T\right]  \right)  \left(  T^{n}-c_{1}T^{n-1}%
+\cdots+\left(  -1\right)  ^{n}c_{n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }P=T^{n}-c_{1}T^{n-1}%
+\cdots+\left(  -1\right)  ^{n}c_{n}\right) \\
&  =T^{n}-\iota\left(  c_{1}\right)  T^{n-1}+\cdots+\left(  -1\right)
^{n}\iota\left(  c_{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\iota\left[  T\right]  \right)  .
\end{align*}
This is an equality between two polynomials in $B\left[  T\right]  $.
Comparing the coefficients on both sides of this equality, we find that%
\[
\left(  -1\right)  ^{i}\eta\left(  c_{i}\right)  =\left(  -1\right)  ^{i}%
\iota\left(  c_{i}\right)  \ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{
1,2,\ldots,n\right\}  .
\]
In other words,%
\begin{equation}
\eta\left(  c_{i}\right)  =\iota\left(  c_{i}\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  1,2,\ldots,n\right\}  .
\label{p843.St2.pf.etai}%
\end{equation}
Now, $\iota$ is an $S$-algebra homomorphism and thus an $A$-algebra
homomorphism (since $A$ is a subring of $S$). Hence, $\eta$ and $\iota$ are
two $A$-algebra homomorphisms. These two homomorphisms $\eta$ and $\iota$ are
equal on the $n$ elements $c_{1},c_{2},\ldots,c_{n}$ (by
(\ref{p843.St2.pf.etai})); thus, they are equal on a generating set of the
$A$-algebra $S$ (since the $n$ elements $c_{1},c_{2},\ldots,c_{n}$ form a
generating set of the $A$-algebra $S$\ \ \ \ \footnote{by the Fundamental
Theorem on Symmetric Polynomials}). Therefore, these two homomorphisms must be
identical\footnote{Here we are using the following fact: If two $A$-algebra
homomorphisms have the same domain and the same codomain, and are equal on a
generating set of their domain, then they must be identical.}. In other words,
$\eta=\iota$. Hence, $\eta$ is an $S$-algebra homomorphism (since $\iota$ is
an $S$-algebra homomorphism). Therefore, $\eta$ is an an $S$-algebra
homomorphism $\gamma:A\left[  X_{1},\ldots,X_{n}\right]  \rightarrow B$
satisfying (\ref{p843.St2.eq.1}) (since $\eta\left(  X_{i}\right)  =u_{i}$ for
all $i\in\left\{  1,2,\ldots,n\right\}  $). Thus, there exists \textbf{at
least one} $S$-algebra homomorphism $\gamma:A\left[  X_{1},\ldots
,X_{n}\right]  \rightarrow B$ satisfying (\ref{p843.St2.eq.1}) (namely,
$\gamma=\eta$).

On the other hand, it is easy to see that there exists \textbf{at most one}
$S$-algebra homomorphism $\gamma:A\left[  X_{1},\ldots,X_{n}\right]
\rightarrow B$ satisfying (\ref{p843.St2.eq.1})\footnote{\textit{Proof.} The
universal property of the polynomial ring $A\left[  X_{1},\ldots,X_{n}\right]
$ shows that there exists exactly one $A$-algebra homomorphism $\gamma
:A\left[  X_{1},\ldots,X_{n}\right]  \rightarrow B$ satisfying
(\ref{p843.St2.eq.1}). Hence, a fortiori, there exists at most one such
$A$-algebra homomorphism. Thus, there exists at most one $S$-algebra
homomorphism $\gamma:A\left[  X_{1},\ldots,X_{n}\right]  \rightarrow B$
satisfying (\ref{p843.St2.eq.1}) (because any $S$-algebra homomorphism
$\gamma:A\left[  X_{1},\ldots,X_{n}\right]  \rightarrow B$ is automatically an
$A$-algebra homomorphism $\gamma:A\left[  X_{1},\ldots,X_{n}\right]
\rightarrow B$). Qed.}.

\begin{vershort}
Combining the previous two sentences, we conclude that there is
\textbf{exactly} one $S$-algebra homomorphism $\gamma:A\left[  X_{1}%
,\ldots,X_{n}\right]  \rightarrow B$ satisfying (\ref{p843.St2.eq.1}). This
proves Statement 2.]
\end{vershort}

\begin{verlong}
Now we have proven the following two observations:

\begin{itemize}
\item There exists \textbf{at least one} $S$-algebra homomorphism
$\gamma:A\left[  X_{1},\ldots,X_{n}\right]  \rightarrow B$ satisfying
(\ref{p843.St2.eq.1}).

\item There exists \textbf{at most one} $S$-algebra homomorphism
$\gamma:A\left[  X_{1},\ldots,X_{n}\right]  \rightarrow B$ satisfying
(\ref{p843.St2.eq.1}).
\end{itemize}

Combining these two observations, we conclude that there is \textbf{exactly}
one $S$-algebra homomorphism $\gamma:A\left[  X_{1},\ldots,X_{n}\right]
\rightarrow B$ satisfying (\ref{p843.St2.eq.1}). In other words, there is a
unique $S$-algebra homomorphism $\gamma:A\left[  X_{1},\ldots,X_{n}\right]
\rightarrow B$ satisfying (\ref{p843.St2.eq.1}). This proves Statement 2.]
\end{verlong}

[\textit{Proof of Statement 1:} For each $i\in\left\{  1,2,\ldots,n\right\}
$, we can write the polynomial $\varphi_{i}\in B\left[  T\right]  $ in the
form $\varphi_{i}=T-u_{i}$ for some $u_{i}\in B$ (because $\varphi_{i}$ is a
monic linear polynomial over $B$, and since every monic linear polynomial over
$B$ can be written in this form). Consider this $u_{i}$. Thus, $u_{1}%
,u_{2},\ldots,u_{n}$ are $n$ elements of $B$. Now, in $B\left[  T\right]  $,
we have%
\[
p=\varphi_{1}\varphi_{2}\cdots\varphi_{n}=\left(  T-u_{1}\right)  \left(
T-u_{2}\right)  \cdots\left(  T-u_{n}\right)
\]
(since $\varphi_{i}=T-u_{i}$ for each $i\in\left\{  1,2,\ldots,n\right\}  $).
Hence, Statement 2 yields that there is a unique $S$-algebra homomorphism
$\gamma:A\left[  X_{1},\ldots,X_{n}\right]  \rightarrow B$ such that
\[
\left(  \gamma\left(  X_{i}\right)  =u_{i}\ \ \ \ \ \ \ \ \ \ \text{for all
}i\in\left\{  1,2,\ldots,n\right\}  \right)  .
\]


\begin{vershort}
Now, for any $S$-algebra homomorphism $\gamma:A\left[  X_{1},\ldots
,X_{n}\right]  \rightarrow B$, we have the following chain of equivalences:%
\begin{align}
&  \ \left(  \text{the induced map }\gamma\left[  T\right]  :\left(  A\left[
X_{1},\ldots,X_{n}\right]  \right)  \left[  T\right]  \rightarrow B\left[
T\right]  \right. \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \left.  \text{maps }T-X_{i}\text{ to }\varphi
_{i}\text{ for all }i\in\left\{  1,2,\ldots,n\right\}  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \underbrace{\left(  \gamma\left[  T\right]
\right)  \left(  T-X_{i}\right)  }_{\substack{=T-\gamma\left(  X_{i}\right)
\\\text{(by the definition of }\gamma\left[  T\right]  \text{)}}%
}=\underbrace{\varphi_{i}}_{=T-u_{i}}\text{ for all }i\in\left\{
1,2,\ldots,n\right\}  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \underbrace{T-\gamma\left(  X_{i}\right)
=T-u_{i}}_{\Longleftrightarrow\ \left(  \gamma\left(  X_{i}\right)
=u_{i}\right)  }\text{ for all }i\in\left\{  1,2,\ldots,n\right\}  \right)
\nonumber\\
&  \Longleftrightarrow\ \left(  \gamma\left(  X_{i}\right)  =u_{i}\text{ for
all }i\in\left\{  1,2,\ldots,n\right\}  \right)  .
\label{p843.St1.pf.short.equivalence}%
\end{align}


Now, recall that there is a unique $S$-algebra homomorphism $\gamma:A\left[
X_{1},\ldots,X_{n}\right]  \rightarrow B$ such that
\[
\left(  \gamma\left(  X_{i}\right)  =u_{i}\ \ \ \ \ \ \ \ \ \ \text{for all
}i\in\left\{  1,2,\ldots,n\right\}  \right)  .
\]
In view of the equivalence (\ref{p843.St1.pf.short.equivalence}), we can
restate this as follows: There is a unique $S$-algebra homomorphism
$\gamma:A\left[  X_{1},\ldots,X_{n}\right]  \rightarrow B$ such that the
induced map $\gamma\left[  T\right]  :\left(  A\left[  X_{1},\ldots
,X_{n}\right]  \right)  \left[  T\right]  \rightarrow B\left[  T\right]  $
maps $T-X_{i}$ to $\varphi_{i}$ for all $i\in\left\{  1,2,\ldots,n\right\}  $.
This proves Statement 1.]
\end{vershort}

\begin{verlong}
If $\gamma:A\left[  X_{1},\ldots,X_{n}\right]  \rightarrow B$ is any
$S$-algebra homomorphism, then the induced map $\gamma\left[  T\right]
:\left(  A\left[  X_{1},\ldots,X_{n}\right]  \right)  \left[  T\right]
\rightarrow B\left[  T\right]  $ satisfies%
\begin{align}
\left(  \gamma\left[  T\right]  \right)  \left(  T-X_{i}\right)   &
=\underbrace{\left(  \gamma\left[  T\right]  \right)  \left(  T\right)
}_{\substack{=T\\\text{(by the definition}\\\text{of }\gamma\left[  T\right]
\text{)}}}-\underbrace{\left(  \gamma\left[  T\right]  \right)  \left(
X_{i}\right)  }_{\substack{=\gamma\left(  X_{i}\right)  \\\text{(by the
definition}\\\text{of }\gamma\left[  T\right]  \text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\gamma\left[  T\right]  \text{ is
a ring homomorphism}\right) \nonumber\\
&  =T-\gamma\left(  X_{i}\right)  . \label{p843.St1.pf.4}%
\end{align}


Now, for any $S$-algebra homomorphism $\gamma:A\left[  X_{1},\ldots
,X_{n}\right]  \rightarrow B$, we have the following chain of equivalences:%
\begin{align}
&  \ \left(  \text{the induced map }\gamma\left[  T\right]  :\left(  A\left[
X_{1},\ldots,X_{n}\right]  \right)  \left[  T\right]  \rightarrow B\left[
T\right]  \right. \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \left.  \text{maps }T-X_{i}\text{ to }\varphi
_{i}\text{ for all }i\in\left\{  1,2,\ldots,n\right\}  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \underbrace{\left(  \gamma\left[  T\right]
\right)  \left(  T-X_{i}\right)  }_{\substack{=T-\gamma\left(  X_{i}\right)
\\\text{(by (\ref{p843.St1.pf.4}))}}}=\underbrace{\varphi_{i}}_{=T-u_{i}%
}\text{ for all }i\in\left\{  1,2,\ldots,n\right\}  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \underbrace{T-\gamma\left(  X_{i}\right)
=T-u_{i}}_{\Longleftrightarrow\ \left(  \gamma\left(  X_{i}\right)
=u_{i}\right)  }\text{ for all }i\in\left\{  1,2,\ldots,n\right\}  \right)
\nonumber\\
&  \Longleftrightarrow\ \left(  \gamma\left(  X_{i}\right)  =u_{i}\text{ for
all }i\in\left\{  1,2,\ldots,n\right\}  \right)  .
\label{p843.St1.pf.equivalence}%
\end{align}


Now, recall that there is a unique $S$-algebra homomorphism $\gamma:A\left[
X_{1},\ldots,X_{n}\right]  \rightarrow B$ such that
\[
\left(  \gamma\left(  X_{i}\right)  =u_{i}\ \ \ \ \ \ \ \ \ \ \text{for all
}i\in\left\{  1,2,\ldots,n\right\}  \right)  .
\]
In view of the equivalence (\ref{p843.St1.pf.equivalence}), we can restate
this as follows: There is a unique $S$-algebra homomorphism $\gamma:A\left[
X_{1},\ldots,X_{n}\right]  \rightarrow B$ such that the induced map
$\gamma\left[  T\right]  :\left(  A\left[  X_{1},\ldots,X_{n}\right]  \right)
\left[  T\right]  \rightarrow B\left[  T\right]  $ maps $T-X_{i}$ to
$\varphi_{i}$ for all $i\in\left\{  1,2,\ldots,n\right\}  $. This proves
Statement 1.]
\end{verlong}

\item \textbf{page 843, Proposition 7.1:} After \textquotedblleft is an
isomorphism\textquotedblright, add \textquotedblleft sending $X^{n-1}%
\wedge\cdots\wedge X^{0}$ to $1$\textquotedblright\ (since you end up using
this later, in \S 7.2).

\item \textbf{page 843, proof of Proposition 7.1:} After \textquotedblleft So
the $S$-module $\bigwedge\nolimits_{A}^{n}A\left[  X\right]  $ is generated by
$X^{h_{1}}\wedge\cdots\wedge X^{h_{n}}$ for $0\leq h_{i}\leq n-i$%
\textquotedblright, add \textquotedblleft(since this $S$-module is a quotient
of $\bigotimes\nolimits_{A}^{n}A\left[  X\right]  =A\left[  X_{1},\ldots
,X_{n}\right]  $)\textquotedblright.

\item \textbf{page 843, \S 7.2:} Replace \textquotedblleft$f_{1}\left(
X_{1}\right)  \wedge\cdots\wedge f_{n}\left(  X_{n}\right)  $%
\textquotedblright\ by \textquotedblleft$f_{1}\left(  X\right)  \wedge
\cdots\wedge f_{n}\left(  X\right)  $\textquotedblright.
\end{itemize}

\section{Appendix: Some alternative proofs}

\subsection{\label{sect.pf.2.1}An alternative proof of Proposition 2.1}

\begin{fineprint}
\begin{proof}
[Alternative proof of Proposition 2.1.]Let $\mathbb{N}=\left\{  0,1,2,\ldots
\right\}  $, and let%
\begin{align*}
\mathcal{K}_{h}  &  =\left\{  \left(  j_{1},j_{2},\ldots,j_{n}\right)
\in\mathbb{N}^{n}\ \mid\ \left(  j_{i}\geq h_{i}\text{ for all }i\in\left\{
1,2,\ldots,n\right\}  \right)  \right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{and }j_{1}+j_{2}+\cdots+j_{n}=h_{1}%
+h_{2}+\cdots+h_{n}+h\right\}  .
\end{align*}
If $\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}$, then we say
that $\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ is \textit{interlacing} if
each $k\in\left\{  2,3,\ldots,n\right\}  $ satisfies $h_{k-1}>j_{k}$. Thus%
\begin{align*}
&  \left\{  \left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}%
\ \mid\ \left(  j_{1},j_{2},\ldots,j_{n}\right)  \text{ is interlacing}%
\right\} \\
&  =\left\{  \left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}%
\ \mid\ h_{k-1}>j_{k}\text{ for all }k\in\left\{  2,3,\ldots,n\right\}
\right\} \\
&  =\left\{  \left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathbb{N}^{n}%
\ \mid\ \left(  h_{k-1}>j_{k}\text{ for all }k\in\left\{  2,3,\ldots
,n\right\}  \right)  \right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{and }\left(  j_{i}\geq h_{i}\text{ for
all }i\in\left\{  1,2,\ldots,n\right\}  \right)  \right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{and }j_{1}+j_{2}+\cdots+j_{n}=h_{1}%
+h_{2}+\cdots+h_{n}+h\right\}  \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\mathcal{K}_{h}\right) \\
&  =\left\{  \left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathbb{N}^{n}%
\ \mid\ j_{1}+j_{2}+\cdots+j_{n}=h_{1}+h_{2}+\cdots+h_{n}+h\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{and }\underbrace{\left(  h_{k-1}%
>j_{k}\text{ for all }k\in\left\{  2,3,\ldots,n\right\}  \right)  \text{ and
}\left(  j_{i}\geq h_{i}\text{ for all }i\in\left\{  1,2,\ldots,n\right\}
\right)  }_{\Longleftrightarrow\ \left(  j_{1}\geq h_{1}>j_{2}\geq
h_{2}>\cdots>j_{n}\geq h_{n}\right)  }\right\} \\
&  =\left\{  \left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathbb{N}^{n}%
\ \mid\ j_{1}+j_{2}+\cdots+j_{n}=h_{1}+h_{2}+\cdots+h_{n}+h\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{and }j_{1}\geq h_{1}>j_{2}\geq
h_{2}>\cdots>j_{n}\geq h_{n}\right\} \\
&  =\mathcal{J}_{h}%
\end{align*}
(by the definition of $\mathcal{J}_{h}$). Hence,%
\begin{equation}
\sum_{\substack{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}%
_{h}\\\text{is interlacing}}}=\sum_{\left(  j_{1},j_{2},\ldots,j_{n}\right)
\in\mathcal{J}_{h}} \label{pf.p2.1.a-i}%
\end{equation}
(an equality of summation signs).

Also, the map%
\begin{align*}
\mathcal{I}_{h}  &  \rightarrow\mathcal{K}_{h},\\
\left(  i_{1},i_{2},\ldots,i_{n}\right)   &  \mapsto\left(  h_{1}+i_{1}%
,h_{2}+i_{2},\ldots,h_{n}+i_{n}\right)
\end{align*}
is well-defined and a bijection (this follows easily from the definitions of
$\mathcal{I}_{h}$ and $\mathcal{K}_{h}$). Hence, we can substitute $\left(
j_{1},j_{2},\ldots,j_{n}\right)  $ for $\left(  h_{1}+i_{1},h_{2}+i_{2}%
,\ldots,h_{n}+i_{n}\right)  $ in the sum \newline$\sum_{\left(  i_{1}%
,i_{2},\ldots,i_{n}\right)  \in\mathcal{I}_{h}}X^{h_{1}+i_{1}}\wedge
X^{h_{2}+i_{2}}\wedge\cdots\wedge X^{h_{n}+i_{n}}$. We thus obtain
\begin{align}
&  \sum_{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}_{h}}%
X^{h_{1}+i_{1}}\wedge X^{h_{2}+i_{2}}\wedge\cdots\wedge X^{h_{n}+i_{n}%
}\nonumber\\
&  =\sum_{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}}%
X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}. \label{pf.p2.1.a-all}%
\end{align}


If $\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}$, then we say
that $\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ is \textit{non-interlacing}
if $\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ is not interlacing.

Thus, if $\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ is non-interlacing, then
there exists some $k\in\left\{  2,3,\ldots,n\right\}  $ that satisfies
$h_{k-1}\leq j_{k}$ (because otherwise, $\left(  j_{1},j_{2},\ldots
,j_{n}\right)  $ would be interlacing). The \textbf{largest} such $k$ will be
called the \textit{violation} of $\left(  j_{1},j_{2},\ldots,j_{n}\right)  $.

If an $n$-tuple $\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}$
is non-interlacing and has violation $k$, then:

\begin{itemize}
\item we say that $\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ is
\textit{degenerate} if $j_{k}=j_{k-1}$;

\item we say that $\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ is
\textit{non-degenerate} if $j_{k}\neq j_{k-1}$.
\end{itemize}

Clearly, any non-interlacing $n$-tuple $\left(  j_{1},j_{2},\ldots
,j_{n}\right)  \in\mathcal{K}_{h}$ is either degenerate or non-degenerate (but
not both).

If $\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}$ is
non-interlacing and degenerate, then%
\[
X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}=0
\]
(because if we let $k$ denote the violation of $\left(  j_{1},j_{2}%
,\ldots,j_{n}\right)  $, then the degenerateness of $\left(  j_{1}%
,j_{2},\ldots,j_{n}\right)  $ yields $j_{k}=j_{k-1}$, and therefore there are
two equal elements among \newline$X^{j_{1}},X^{j_{2}},\ldots,X^{j_{n}}$).
Thus,%
\begin{equation}
\sum_{\substack{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}%
_{h}\\\text{is non-interlacing}\\\text{and degenerate}}}\underbrace{X^{j_{1}%
}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}}_{=0}=0. \label{pf.p2.1.a-ni-d}%
\end{equation}


If $\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}$ is
non-interlacing and non-degenerate, and if $k$ is the violation of $\left(
j_{1},j_{2},\ldots,j_{n}\right)  $, then we have $j_{k}\geq h_{k-1}$ (since
the definition of \textquotedblleft violation\textquotedblright\ yields
$h_{k-1}\leq j_{k}$) and $j_{k-1}\geq h_{k}$ (since $j_{k-1}\geq h_{k-1}\geq
h_{k}$). Hence, in this case, the $n$-tuple $\left(  j_{1},j_{2}%
,\ldots,j_{k-2},j_{k},j_{k-1},j_{k+1},j_{k+2},\ldots,j_{n}\right)  $ (obtained
from $\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ by swapping the $\left(
k-1\right)  $-st and $k$-th entries) still belongs to $\mathcal{K}_{h}$.
Moreover, this $n$-tuple \newline$\left(  j_{1},j_{2},\ldots,j_{k-2}%
,j_{k},j_{k-1},j_{k+1},j_{k+2},\ldots,j_{n}\right)  $ is non-interlacing
(since $k\in\left\{  2,3,\ldots,n\right\}  $ satisfies $h_{k-1}\leq j_{k-1}$)
and has violation $k$ (because this $k$ is still the largest such $k$; indeed,
no $i>k$ satisfies $h_{i-1}\leq j_{i}$), and thus is non-degenerate (since
$j_{k}\neq j_{k-1}$ yields $j_{k-1}\neq j_{k}$).

Thus, we can define a map
\begin{align*}
\Phi &  :\left\{  \text{non-degenerate non-interlacing }\left(  j_{1}%
,j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}\right\} \\
&  \rightarrow\left\{  \text{non-degenerate non-interlacing }\left(
j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}\right\}
\end{align*}
as follows: If $\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}$
is non-interlacing and non-degenerate, and if $k$ is the violation of $\left(
j_{1},j_{2},\ldots,j_{n}\right)  $, then we set%
\[
\Phi\left(  j_{1},j_{2},\ldots,j_{n}\right)  =\left(  j_{1},j_{2}%
,\ldots,j_{k-2},j_{k},j_{k-1},j_{k+1},j_{k+2},\ldots,j_{n}\right)
\]
(that is, $\Phi$ swaps the $\left(  k-1\right)  $-st and $k$-th entries of
$\left(  j_{1},j_{2},\ldots,j_{n}\right)  $, while leaving all other entries
unchanged). The previous paragraph shows that this map $\Phi$ is well-defined
(i.e., if $\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}$ is
non-interlacing and non-degenerate, then so is $\Phi\left(  j_{1},j_{2}%
,\ldots,j_{n}\right)  $) and preserves the violation (i.e., if $\left(
j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}$ is non-interlacing and
non-degenerate, then the violation of $\left(  j_{1},j_{2},\ldots
,j_{n}\right)  $ is also the violation of $\Phi\left(  j_{1},j_{2}%
,\ldots,j_{n}\right)  $). Thus $\Phi$ is an involution (that is, $\Phi
\circ\Phi=\operatorname*{id}$), because applying $\Phi$ to $\Phi\left(
j_{1},j_{2},\ldots,j_{n}\right)  $ will swap the same two entries that were
swapped in the definition of $\Phi\left(  j_{1},j_{2},\ldots,j_{n}\right)  $
and therefore recover the original $n$-tuple $\left(  j_{1},j_{2},\ldots
,j_{n}\right)  $. Moreover, if $\left(  j_{1},j_{2},\ldots,j_{n}\right)
\in\mathcal{K}_{h}$ is non-interlacing and non-degenerate, then%
\[
\Phi\left(  j_{1},j_{2},\ldots,j_{n}\right)  \neq\left(  j_{1},j_{2}%
,\ldots,j_{n}\right)  .
\]
\footnote{\textit{Proof.} Let $\left(  j_{1},j_{2},\ldots,j_{n}\right)
\in\mathcal{K}_{h}$ be non-interlacing and non-degenerate. Let $k$ be the
violation of $\left(  j_{1},j_{2},\ldots,j_{n}\right)  $. Thus, $j_{k}\neq
j_{k-1}$ (since $\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ is
non-degenerate). Thus, the $\left(  k-1\right)  $-st entry of $\Phi\left(
j_{1},j_{2},\ldots,j_{n}\right)  $ is distinct from the $\left(  k-1\right)
$-st entry of $\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ (since the former
entry is $j_{k}$, while the latter entry is $j_{k-1}$). Thus, $\Phi\left(
j_{1},j_{2},\ldots,j_{n}\right)  \neq\left(  j_{1},j_{2},\ldots,j_{n}\right)
$.} In other words, the involution $\Phi$ has no fixed points. Finally, if
$\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}$ is
non-interlacing and non-degenerate, and if $\left(  j_{1}^{\prime}%
,j_{2}^{\prime},\ldots,j_{n}^{\prime}\right)  =\Phi\left(  j_{1},j_{2}%
,\ldots,j_{n}\right)  $, then%
\[
X^{j_{1}^{\prime}}\wedge X^{j_{2}^{\prime}}\wedge\cdots\wedge X^{j_{n}%
^{\prime}}=-X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}%
\]
(because the $n$-tuple $\left(  X^{j_{1}^{\prime}},X^{j_{2}^{\prime}}%
,\ldots,X^{j_{n}^{\prime}}\right)  $ is obtained from the $n$-tuple $\left(
X^{j_{1}},X^{j_{2}},\ldots,X^{j_{n}}\right)  $ by swapping the $\left(
k-1\right)  $-st and $k$-th entries). Thus, the fixed-point-free involution
$\Phi$ pairs up the addends of the sum%
\[
\sum_{\substack{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}%
_{h}\\\text{is non-interlacing}\\\text{and non-degenerate}}}X^{j_{1}}\wedge
X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}%
\]
into pairs of mutually cancelling addends. Consequently, this sum is $0$. In
other words, we have%
\begin{equation}
\sum_{\substack{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}%
_{h}\\\text{is non-interlacing}\\\text{and non-degenerate}}}X^{j_{1}}\wedge
X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}=0. \label{pf.p2.1.a-ni-nd}%
\end{equation}


Now, every non-interlacing $\left(  j_{1},j_{2},\ldots,j_{n}\right)
\in\mathcal{K}_{h}$ is either degenerate or non-degenerate (but not both).
Hence,%
\begin{align}
&  \sum_{\substack{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}%
_{h}\\\text{is non-interlacing}}}X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge
X^{j_{n}}\nonumber\\
&  =\underbrace{\sum_{\substack{\left(  j_{1},j_{2},\ldots,j_{n}\right)
\in\mathcal{K}_{h}\\\text{is non-interlacing}\\\text{and degenerate}}%
}X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}}%
_{\substack{=0\\\text{(by (\ref{pf.p2.1.a-ni-d}))}}}+\underbrace{\sum
_{\substack{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}%
_{h}\\\text{is non-interlacing}\\\text{and non-degenerate}}}X^{j_{1}}\wedge
X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}}_{\substack{=0\\\text{(by
(\ref{pf.p2.1.a-ni-nd}))}}}\nonumber\\
&  =0. \label{pf.p2.1.a-ni}%
\end{align}


But each $\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}$ is
either interlacing or non-interlacing (but not both). Hence,%
\begin{align*}
&  \sum_{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}}X^{j_{1}%
}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}\\
&  =\sum_{\substack{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in
\mathcal{K}_{h}\\\text{is interlacing}}}X^{j_{1}}\wedge X^{j_{2}}\wedge
\cdots\wedge X^{j_{n}}+\underbrace{\sum_{\substack{\left(  j_{1},j_{2}%
,\ldots,j_{n}\right)  \in\mathcal{K}_{h}\\\text{is non-interlacing}}}X^{j_{1}%
}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}}_{\substack{=0\\\text{(by
(\ref{pf.p2.1.a-ni}))}}}\\
&  =\sum_{\substack{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in
\mathcal{K}_{h}\\\text{is interlacing}}}X^{j_{1}}\wedge X^{j_{2}}\wedge
\cdots\wedge X^{j_{n}}=\sum_{\left(  j_{1},j_{2},\ldots,j_{n}\right)
\in\mathcal{J}_{h}}X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}%
\end{align*}
(by (\ref{pf.p2.1.a-i})). Hence, the equality (2.1) becomes%
\begin{align*}
s_{h}\left(  X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{n}}\right)   &
=\sum_{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}_{h}}%
X^{h_{1}+i_{1}}\wedge X^{h_{2}+i_{2}}\wedge\cdots\wedge X^{h_{n}+i_{n}}\\
&  =\sum_{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{K}_{h}}%
X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.p2.1.a-all})}\right) \\
&  =\sum_{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{J}_{h}}%
X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}.
\end{align*}
This proves Proposition 2.1.
\end{proof}

\begin{noncompile}
\textit{Old proof of Proposition 2.1.} If $\left(  i_{1},i_{2},\ldots
,i_{n}\right)  \in\mathcal{I}_{h}$, then we say that $\left(  i_{1}%
,i_{2},\ldots,i_{n}\right)  $ is \textit{interlacing} if each $k\in\left\{
2,3,\ldots,n\right\}  $ satisfies $h_{k-1}>h_{k}+i_{k}$. In other words,
$\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ is interlacing if and only if%
\[
h_{1}+i_{1}\geq h_{1}>h_{2}+i_{2}\geq h_{2}>\cdots>h_{n}+i_{n}\geq h_{n}.
\]
Thus, there is a bijection%
\begin{align*}
\left\{  \text{interlacing }\left(  i_{1},i_{2},\ldots,i_{n}\right)
\in\mathcal{I}_{h}\right\}   &  \rightarrow\mathcal{J}_{h},\\
\left(  i_{1},i_{2},\ldots,i_{n}\right)   &  \mapsto\left(  h_{1}+i_{1}%
,h_{2}+i_{2},\ldots,h_{n}+i_{n}\right)  .
\end{align*}
This shows that%
\begin{align}
&  \sum_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}%
_{h}\\\text{is interlacing}}}X^{h_{1}+i_{1}}\wedge X^{h_{2}+i_{2}}\wedge
\cdots\wedge X^{h_{n}+i_{n}}\nonumber\\
&  =\sum_{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{J}_{h}}%
X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}. \label{pf.p2.1.1}%
\end{align}


If $\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}_{h}$, then we say
that $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ is \textit{non-interlacing}
if $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ is not interlacing.

Thus, if $\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}_{h}$ is
non-interlacing, then there exists some $k\in\left\{  2,3,\ldots,n\right\}  $
that satisfies $h_{k-1}\leq h_{k}+i_{k}$. The \textbf{largest} such $k$ will
be called the \textit{violation} of $\left(  i_{1},i_{2},\ldots,i_{n}\right)
$. Equivalently, the violation of $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $
is the largest $k\in\left\{  2,3,\ldots,n\right\}  $ that satisfies $i_{k}\geq
h_{k-1}-h_{k}$.

If an $n$-tuple $\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}_{h}$
is non-interlacing and has violation $k$, then:

\begin{itemize}
\item we say that $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ is
\textit{degenerate} if $h_{k}+i_{k}=h_{k-1}+i_{k-1}$;

\item we say that $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ is
\textit{non-degenerate} if $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ is not degenerate.
\end{itemize}

If $\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}_{h}$ is
non-interlacing and degenerate, then%
\[
X^{h_{1}+i_{1}}\wedge X^{h_{2}+i_{2}}\wedge\cdots\wedge X^{h_{n}+i_{n}}=0
\]
(because if we let $k$ denote the violation of $\left(  i_{1},i_{2}%
,\ldots,i_{n}\right)  $, then the degenerateness of $\left(  i_{1}%
,i_{2},\ldots,i_{n}\right)  $ yields $h_{k}+i_{k}=h_{k-1}+i_{k-1}$, and
therefore there are two equal elements among $X^{h_{1}+i_{1}},X^{h_{2}+i_{2}%
},\ldots,X^{h_{n}+i_{n}}$). Thus,%
\begin{equation}
\sum_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}%
_{h}\\\text{is non-interlacing}\\\text{and degenerate}}}\underbrace{X^{h_{1}%
+i_{1}}\wedge X^{h_{2}+i_{2}}\wedge\cdots\wedge X^{h_{n}+i_{n}}}_{=0}=0.
\label{pf.p2.1.2}%
\end{equation}


Now, we define a map
\begin{align*}
\Phi &  :\left\{  \text{non-degenerate non-interlacing }\left(  i_{1}%
,i_{2},\ldots,i_{n}\right)  \in\mathcal{I}_{h}\right\} \\
&  \rightarrow\left\{  \text{non-degenerate non-interlacing }\left(
i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}_{h}\right\}
\end{align*}
as follows: If $\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}_{h}$
is non-interlacing and non-degenerate, and if $k$ is the violation of $\left(
i_{1},i_{2},\ldots,i_{n}\right)  $, then we set%
\begin{align*}
&  \Phi\left(  i_{1},i_{2},\ldots,i_{n}\right) \\
&  =\left(  i_{1},i_{2},\ldots,i_{k-2},h_{k}+i_{k}-h_{k-1},h_{k-1}%
+i_{k-1}-h_{k},i_{k+1},i_{k+2},\ldots,i_{n}\right)
\end{align*}
(that is, $\Phi$ replaces the $\left(  k-1\right)  $-st and $k$-th entries of
$\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ by $h_{k}+i_{k}-h_{k-1}$ and
$h_{k-1}+i_{k-1}-h_{k}$, while leaving all other entries unchanged). It is
easy to see that this map $\Phi$ is well-defined (i.e., if $\left(
i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}_{h}$ is non-interlacing and
non-degenerate, then so is $\Phi\left(  i_{1},i_{2},\ldots,i_{n}\right)  $)
and preserves the violation (i.e., if $\left(  i_{1},i_{2},\ldots
,i_{n}\right)  \in\mathcal{I}_{h}$ is non-interlacing and non-degenerate, then
the violation of $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ is also the
violation of $\Phi\left(  i_{1},i_{2},\ldots,i_{n}\right)  $). This makes it
very easy to see that $\Phi$ is an involution (that is, $\Phi\circ
\Phi=\operatorname*{id}$). Moreover, if $\left(  i_{1},i_{2},\ldots
,i_{n}\right)  \in\mathcal{I}_{h}$ is non-interlacing and non-degenerate, then%
\[
\Phi\left(  i_{1},i_{2},\ldots,i_{n}\right)  \neq\left(  i_{1},i_{2}%
,\ldots,i_{n}\right)  .
\]
\footnote{\textit{Proof.} Let $\left(  i_{1},i_{2},\ldots,i_{n}\right)
\in\mathcal{I}_{h}$ be non-interlacing and non-degenerate. Let $k$ be the
violation of $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $. If we had
$h_{k}+i_{k}=h_{k-1}+i_{k-1}$, then $\left(  i_{1},i_{2},\ldots,i_{n}\right)
$ would be degenerate. Thus, we don't have $h_{k}+i_{k}=h_{k-1}+i_{k-1}$.
Hence, we have $h_{k}+i_{k}\neq h_{k-1}+i_{k-1}$, so that $h_{k}+i_{k}%
-h_{k-1}\neq i_{k-1}$. Thus, the $\left(  k-1\right)  $-th entry of
$\Phi\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ is distinct from the $\left(
k-1\right)  $-th entry of $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ (since
the former entry is $h_{k}+i_{k}-h_{k-1}$, while the latter entry is $i_{k-1}%
$). Thus, $\Phi\left(  i_{1},i_{2},\ldots,i_{n}\right)  \neq\left(
i_{1},i_{2},\ldots,i_{n}\right)  $.} In other words, the involution $\Phi$ has
no fixed points. Finally, if $\left(  i_{1},i_{2},\ldots,i_{n}\right)
\in\mathcal{I}_{h}$ is non-interlacing and non-degenerate, and if $\left(
j_{1},j_{2},\ldots,j_{n}\right)  =\Phi\left(  i_{1},i_{2},\ldots,i_{n}\right)
$, then%
\[
X^{h_{1}+j_{1}}\wedge X^{h_{2}+j_{2}}\wedge\cdots\wedge X^{h_{n}+j_{n}%
}=-X^{h_{1}+i_{1}}\wedge X^{h_{2}+i_{2}}\wedge\cdots\wedge X^{h_{n}+i_{n}}%
\]
(because the $n$-tuple $\left(  X^{h_{1}+j_{1}},X^{h_{2}+j_{2}},\ldots
,X^{h_{n}+j_{n}}\right)  $ is obtained from the $n$-tuple $\left(
X^{h_{1}+i_{1}},X^{h_{2}+i_{2}},\ldots,X^{h_{n}+i_{n}}\right)  $ by swapping
the $\left(  k-1\right)  $-st and $k$-th entries). Thus, the fixed-point-free
involution $\Phi$ pairs up the addends of the sum%
\[
\sum_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}%
_{h}\\\text{is non-interlacing}\\\text{and non-degenerate}}}X^{h_{1}+i_{1}%
}\wedge X^{h_{2}+i_{2}}\wedge\cdots\wedge X^{h_{n}+i_{n}}%
\]
into pairs of mutually cancelling addends. Consequently, this sum is $0$. In
other words, we have%
\begin{equation}
\sum_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}%
_{h}\\\text{is non-interlacing}\\\text{and non-degenerate}}}X^{h_{1}+i_{1}%
}\wedge X^{h_{2}+i_{2}}\wedge\cdots\wedge X^{h_{n}+i_{n}}=0. \label{pf.p2.1.3}%
\end{equation}


Now, every non-interlacing $\left(  i_{1},i_{2},\ldots,i_{n}\right)
\in\mathcal{I}_{h}$ is either degenerate or non-degenerate (but not both).
Hence,%
\begin{align}
&  \sum_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}%
_{h}\\\text{is non-interlacing}}}X^{h_{1}+i_{1}}\wedge X^{h_{2}+i_{2}}%
\wedge\cdots\wedge X^{h_{n}+i_{n}}\nonumber\\
&  =\underbrace{\sum_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)
\in\mathcal{I}_{h}\\\text{is non-interlacing}\\\text{and degenerate}}%
}X^{h_{1}+i_{1}}\wedge X^{h_{2}+i_{2}}\wedge\cdots\wedge X^{h_{n}+i_{n}}%
}_{\substack{=0\\\text{(by (\ref{pf.p2.1.2}))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum_{\substack{\left(  i_{1},i_{2}%
,\ldots,i_{n}\right)  \in\mathcal{I}_{h}\\\text{is non-interlacing}\\\text{and
non-degenerate}}}X^{h_{1}+i_{1}}\wedge X^{h_{2}+i_{2}}\wedge\cdots\wedge
X^{h_{n}+i_{n}}}_{\substack{=0\\\text{(by (\ref{pf.p2.1.3}))}}}\nonumber\\
&  =0. \label{pf.p2.1.4}%
\end{align}


But each $\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}_{h}$ is
either interlacing or non-interlacing (but not both). Hence,%
\begin{align*}
&  \sum_{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}_{h}}%
X^{h_{1}+i_{1}}\wedge X^{h_{2}+i_{2}}\wedge\cdots\wedge X^{h_{n}+i_{n}}\\
&  =\sum_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in
\mathcal{I}_{h}\\\text{is interlacing}}}X^{h_{1}+i_{1}}\wedge X^{h_{2}+i_{2}%
}\wedge\cdots\wedge X^{h_{n}+i_{n}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum_{\substack{\left(  i_{1},i_{2}%
,\ldots,i_{n}\right)  \in\mathcal{I}_{h}\\\text{is non-interlacing}}%
}X^{h_{1}+i_{1}}\wedge X^{h_{2}+i_{2}}\wedge\cdots\wedge X^{h_{n}+i_{n}}%
}_{\substack{=0\\\text{(by (\ref{pf.p2.1.4}))}}}\\
&  =\sum_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in
\mathcal{I}_{h}\\\text{is interlacing}}}X^{h_{1}+i_{1}}\wedge X^{h_{2}+i_{2}%
}\wedge\cdots\wedge X^{h_{n}+i_{n}}\\
&  =\sum_{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{J}_{h}}%
X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}%
\end{align*}
(by (\ref{pf.p2.1.1})). Hence, the equality (2.1) becomes%
\begin{align*}
&  s_{h}\left(  X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{n}}\right)
\\
&  =\sum_{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathcal{I}_{h}}%
X^{h_{1}+i_{1}}\wedge X^{h_{2}+i_{2}}\wedge\cdots\wedge X^{h_{n}+i_{n}}\\
&  =\sum_{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{J}_{h}}%
X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}.
\end{align*}
This proves Proposition 2.1. $\blacksquare$
\end{noncompile}
\end{fineprint}

\subsection{\label{sect.pf.2.2}An alternative proof of Corollary 2.2}

The following proof of Corollary 2.2 is not substantially different from the
one in your paper, but it is a lot more explicit and requires less
combinatorial skill to understand.

\begin{fineprint}
I will break the proof up into several lemmas. First, some definitions are needed:

\begin{definition}
Let $\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{m}\right)  \in\mathbb{N}^{m}$ be
an $m$-tuple of nonnegative integers. We say that $\mathbf{j}$ is
\textit{nonincreasing} if $j_{1}\geq j_{2}\geq\cdots\geq j_{m}$.
\end{definition}

\begin{definition}
Fix $n\in\mathbb{N}$. We let $\mathcal{NI}$ denote the set of all
nonincreasing $n$-tuples of nonnegative integers.
\end{definition}

Thus, $\mathcal{NI}\subseteq\mathbb{N}^{n}$ (since each element of
$\mathcal{NI}$ is an $n$-tuple of nonnegative integers, thus belongs to
$\mathbb{N}^{n}$).

\begin{definition}
Let $\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{m}\right)  \in\mathbb{N}^{m}$ be
an $m$-tuple of nonnegative integers. We define $\left\vert \mathbf{j}%
\right\vert $ to be the nonnegative integer $j_{1}+j_{2}+\cdots+j_{m}$.
\end{definition}

\begin{definition}
Let $\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathbb{N}^{n}$ be
an $n$-tuple of nonnegative integers. We define $X^{\mathbf{j}}$ to be the
$n$-vector $X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}\in
\bigwedge\nolimits_{A}^{n}A\left[  X\right]  $.
\end{definition}

\begin{definition}
Let $\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathbb{N}^{n}$
and $\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n}\right)  \in\mathbb{N}^{n}$ be
two $n$-tuples of nonnegative integers. We write $\mathbf{j}\oslash\mathbf{h}$
if and only if we have $j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots>j_{n}\geq
h_{n}$ (that is, if and only if we have $j_{i}\geq h_{i}$ for each
$i\in\left\{  1,2,\ldots,n\right\}  $ and $h_{i}>j_{i+1}$ for each
$i\in\left\{  1,2,\ldots,n+1\right\}  $).
\end{definition}

We can now rewrite Proposition 2.1 as follows:

\begin{lemma}
\label{lem.2.2a}Let $\mathbf{h}\in\mathbb{N}^{n}$ be nonincreasing. Let
$h\in\mathbb{Z}$. Then,%
\[
s_{h}X^{\mathbf{h}}=\sum_{\substack{\mathbf{j}\in\mathcal{NI};\\\mathbf{j}%
\oslash\mathbf{h};\\\left\vert \mathbf{j}\right\vert =\left\vert
\mathbf{h}\right\vert +h}}X^{\mathbf{j}}.
\]
(Here, we are following the convention that $s_{h}=0$ when $h<0$.)
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.2.2a}.]Write the $n$-tuple $\mathbf{h}\in
\mathbb{N}^{n}$ in the form $\mathbf{h}=\left(  h_{1},h_{2},\ldots
,h_{n}\right)  $. Hence, $\left\vert \mathbf{h}\right\vert =h_{1}+h_{2}%
+\cdots+h_{n}$ (by the definition of $\left\vert \mathbf{h}\right\vert $) and
$X^{\mathbf{h}}=X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{n}}$ (by the
definition of $X^{\mathbf{h}}$). Also, $\left(  h_{1},h_{2},\ldots
,h_{n}\right)  =\mathbf{h}$ is nonincreasing; in other words, $h_{1}\geq
h_{2}\geq\cdots\geq h_{n}$.

\begin{vershort}
It is easy to see that Lemma \ref{lem.2.2a} holds when $h<0$%
\ \ \ \ \footnote{\textit{Proof.} Assume that $h<0$. Thus, $s_{h}=0$, so that
$s_{h}X^{\mathbf{h}}=0$.
\par
On the other hand, we claim that there exists no $\mathbf{j}\in\mathcal{NI}$
satisfying $\mathbf{j}\oslash\mathbf{h}$ and $\left\vert \mathbf{j}\right\vert
=\left\vert \mathbf{h}\right\vert +h$.
\par
Indeed, let $\mathbf{j}\in\mathcal{NI}$ be such that $\mathbf{j}%
\oslash\mathbf{h}$ and $\left\vert \mathbf{j}\right\vert =\left\vert
\mathbf{h}\right\vert +h$. Then, $\mathbf{j}$ is a nonincreasing $n$-tuple of
nonnegative integers (since $\mathbf{j}\in\mathcal{NI}$). Write this $n$-tuple
$\mathbf{j}$ in the form $\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{n}\right)
$.
\par
Recall that $\mathbf{j}\oslash\mathbf{h}$; in other words, $j_{1}\geq
h_{1}>j_{2}\geq h_{2}>\cdots>j_{n}\geq h_{n}$ (since this is how
\textquotedblleft$\mathbf{j}\oslash\mathbf{h}$\textquotedblright\ is defined).
Hence, $j_{i}\geq h_{i}$ for each $i\in\left\{  1,2,\ldots,n\right\}  $.
Adding up these $n$ inequalities, we obtain $j_{1}+j_{2}+\cdots+j_{n}\geq
h_{1}+h_{2}+\cdots+h_{n}$. The definition of $\left\vert \mathbf{j}\right\vert
$ yields $\left\vert \mathbf{j}\right\vert =j_{1}+j_{2}+\cdots+j_{n}\geq
h_{1}+h_{2}+\cdots+h_{n}=\left\vert \mathbf{h}\right\vert $. This contradicts
$\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}\right\vert
+\underbrace{h}_{<0}<\left\vert \mathbf{h}\right\vert $.
\par
Now, forget that we fixed $\mathbf{j}$. We thus have found a contradiction for
each $\mathbf{j}\in\mathcal{NI}$ satisfying $\mathbf{j}\oslash\mathbf{h}$ and
$\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}\right\vert +h$.
Hence, no such $\mathbf{j}$ exists. Thus, the sum $\sum_{\substack{\mathbf{j}%
\in\mathcal{NI};\\\mathbf{j}\oslash\mathbf{h};\\\left\vert \mathbf{j}%
\right\vert =\left\vert \mathbf{h}\right\vert +h}}X^{\mathbf{j}}$ is empty,
and thus equals $0$. Comparing this with $s_{h}X^{\mathbf{h}}=0$, we find
$s_{h}X^{\mathbf{h}}=\sum_{\substack{\mathbf{j}\in\mathcal{NI};\\\mathbf{j}%
\oslash\mathbf{h};\\\left\vert \mathbf{j}\right\vert =\left\vert
\mathbf{h}\right\vert +h}}X^{\mathbf{j}}$. Thus, Lemma \ref{lem.2.2a} is
proven (under the assumption that $h<0$).}. Hence, for the rest of this proof,
we WLOG assume that $h\geq0$.
\end{vershort}

\begin{verlong}
It is easy to see that Lemma \ref{lem.2.2a} holds when $h<0$%
\ \ \ \ \footnote{\textit{Proof.} Assume that $h<0$. Thus, $s_{h}=0$, so that
$s_{h}X^{\mathbf{h}}=0X^{\mathbf{h}}=0$.
\par
On the other hand, we claim that there exists no $\mathbf{j}\in\mathcal{NI}$
satisfying $\mathbf{j}\oslash\mathbf{h}$ and $\left\vert \mathbf{j}\right\vert
=\left\vert \mathbf{h}\right\vert +h$.
\par
Indeed, let $\mathbf{j}\in\mathcal{NI}$ be such that $\mathbf{j}%
\oslash\mathbf{h}$ and $\left\vert \mathbf{j}\right\vert =\left\vert
\mathbf{h}\right\vert +h$. Then, $\mathbf{j}$ is a nonincreasing $n$-tuple of
nonnegative integers (since $\mathbf{j}\in\mathcal{NI}$). Write this $n$-tuple
$\mathbf{j}$ in the form $\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{n}\right)
$. Hence, $\left\vert \mathbf{j}\right\vert =j_{1}+j_{2}+\cdots+j_{n}$ (by the
definition of $\left\vert \mathbf{j}\right\vert $).
\par
Also, recall that $\mathbf{j}\oslash\mathbf{h}$; in other words, $j_{1}\geq
h_{1}>j_{2}\geq h_{2}>\cdots>j_{n}\geq h_{n}$ (since this is how
\textquotedblleft$\mathbf{j}\oslash\mathbf{h}$\textquotedblright\ is defined).
Hence, $j_{i}\geq h_{i}$ for each $i\in\left\{  1,2,\ldots,n\right\}  $.
Adding up these $n$ inequalities, we obtain $j_{1}+j_{2}+\cdots+j_{n}\geq
h_{1}+h_{2}+\cdots+h_{n}$. Hence, $\left\vert \mathbf{j}\right\vert
=j_{1}+j_{2}+\cdots+j_{n}\geq h_{1}+h_{2}+\cdots+h_{n}=\left\vert
\mathbf{h}\right\vert $. This contradicts $\left\vert \mathbf{j}\right\vert
=\left\vert \mathbf{h}\right\vert +\underbrace{h}_{<0}<\left\vert
\mathbf{h}\right\vert $.
\par
Now, forget that we fixed $\mathbf{j}$. We thus have found a contradiction for
each $\mathbf{j}\in\mathcal{NI}$ satisfying $\mathbf{j}\oslash\mathbf{h}$ and
$\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}\right\vert +h$.
Hence, no such $\mathbf{j}$ exists. Thus, the sum $\sum_{\substack{\mathbf{j}%
\in\mathcal{NI};\\\mathbf{j}\oslash\mathbf{h};\\\left\vert \mathbf{j}%
\right\vert =\left\vert \mathbf{h}\right\vert +h}}X^{\mathbf{j}}$ is empty,
and thus equals $0$. In other words, $\sum_{\substack{\mathbf{j}%
\in\mathcal{NI};\\\mathbf{j}\oslash\mathbf{h};\\\left\vert \mathbf{j}%
\right\vert =\left\vert \mathbf{h}\right\vert +h}}X^{\mathbf{j}}=0$. Comparing
this with $s_{h}X^{\mathbf{h}}=0$, we find $s_{h}X^{\mathbf{h}}=\sum
_{\substack{\mathbf{j}\in\mathcal{NI};\\\mathbf{j}\oslash\mathbf{h}%
;\\\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}\right\vert
+h}}X^{\mathbf{j}}$. Thus, Lemma \ref{lem.2.2a} is proven (under the
assumption that $h<0$).}. Hence, for the rest of this proof, we WLOG assume
that $h\geq0$.
\end{verlong}

Therefore, Proposition 2.1 yields%
\begin{align}
&  s_{h}\left(  X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{n}}\right)
\nonumber\\
&  =\sum_{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{J}_{h}}%
X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}, \label{pf.l2.2a.1}%
\end{align}
where $\mathcal{J}_{h}$ is the set of all $n$-tuples $\left(  j_{1}%
,j_{2},\ldots,j_{n}\right)  \in\mathbb{N}^{n}$ satisfying $j_{1}+j_{2}%
+\cdots+j_{n}=h_{1}+h_{2}+\cdots+h_{n}+h$ and $j_{1}\geq h_{1}>j_{2}\geq
h_{2}>\cdots>j_{n}\geq h_{n}$. Consider this set $\mathcal{J}_{h}$.

\begin{vershort}
For any $n$-tuple $\mathbf{j}\in\mathbb{N}^{n}$, we have the logical
equivalence%
\begin{equation}
\left(  \mathbf{j}\in\mathcal{J}_{h}\right)  \ \Longleftrightarrow\ \left(
\mathbf{j}\in\mathcal{NI}\text{ and }\mathbf{j}\oslash\mathbf{h}\text{ and
}\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}\right\vert +h\right)
\label{pf.l2.2a.short.equivalence}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.l2.2a.short.equivalence}):} Let
$\mathbf{j}\in\mathbb{N}^{n}$ be an $n$-tuple. Write $\mathbf{j}$ in the form
$\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{n}\right)  $. Then, we have
$\left\vert \mathbf{j}\right\vert =j_{1}+j_{2}+\cdots+j_{n}$ (by the
definition of $\left\vert \mathbf{j}\right\vert $). Moreover, $\mathbf{j}%
=\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathbb{N}^{n}$; hence, we have
the chain of equivalences%
\begin{align*}
\left(  \mathbf{j}\in\mathcal{NI}\right)  \  &  \Longleftrightarrow\ \left(
\mathbf{j}\text{ is nonincreasing}\right)  \ \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\mathcal{NI}\right) \\
&  \Longleftrightarrow\ \left(  j_{1}\geq j_{2}\geq\cdots\geq j_{n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of \textquotedblleft
nonincreasing\textquotedblright}\right)
\end{align*}
and the equivalence%
\[
\left(  \mathbf{j}\oslash\mathbf{h}\right)  \ \Longleftrightarrow\ \left(
j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots>j_{n}\geq h_{n}\right)
\]
(by the definition of \textquotedblleft$\mathbf{j}\oslash\mathbf{h}%
$\textquotedblright), since $\mathbf{j}=\left(  j_{1},j_{2},\ldots
,j_{n}\right)  $ and $\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n}\right)  $.
\par
On the other hand, we know that $\left(  j_{1},j_{2},\ldots,j_{n}\right)
\in\mathbb{N}^{n}$. Hence, we have the equivalence%
\begin{align}
&  \ \left(  \left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{J}%
_{h}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots
>j_{n}\geq h_{n}\text{ and }j_{1}+j_{2}+\cdots+j_{n}=h_{1}+h_{2}+\cdots
+h_{n}+h\right)  \label{pf.l2.2a.short.equivalence.pf.3}%
\end{align}
(by the definition of $\mathcal{J}_{h}$).
\par
Now, we have the following chain of equivalences:%
\begin{align*}
&  \ \left(  \mathbf{j}\in\mathcal{NI}\text{ and }\mathbf{j}\oslash
\mathbf{h}\text{ and }\left\vert \mathbf{j}\right\vert =\left\vert
\mathbf{h}\right\vert +h\right) \\
&  \Longleftrightarrow\ \underbrace{\left(  \mathbf{j}\in\mathcal{NI}\right)
}_{\Longleftrightarrow\ \left(  j_{1}\geq j_{2}\geq\cdots\geq j_{n}\right)
}\wedge\underbrace{\left(  \mathbf{j}\oslash\mathbf{h}\right)  }%
_{\Longleftrightarrow\ \left(  j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots
>j_{n}\geq h_{n}\right)  }\wedge\left(  \underbrace{\left\vert \mathbf{j}%
\right\vert }_{=j_{1}+j_{2}+\cdots+j_{n}}=\underbrace{\left\vert
\mathbf{h}\right\vert }_{=h_{1}+h_{2}+\cdots+h_{n}}+h\right) \\
&  \Longleftrightarrow\ \underbrace{\left(  j_{1}\geq j_{2}\geq\cdots\geq
j_{n}\right)  \wedge\left(  j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots>j_{n}\geq
h_{n}\right)  }_{\substack{\Longleftrightarrow\ \left(  j_{1}\geq h_{1}%
>j_{2}\geq h_{2}>\cdots>j_{n}\geq h_{n}\right)  \\\text{(since the chain of
inequalities }\left(  j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots>j_{n}\geq
h_{n}\right)  \\\text{clearly implies }\left(  j_{1}\geq j_{2}\geq\cdots\geq
j_{n}\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \wedge\left(  j_{1}+j_{2}+\cdots+j_{n}=h_{1}%
+h_{2}+\cdots+h_{n}+h\right) \\
&  \Longleftrightarrow\ \left(  j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots
>j_{n}\geq h_{n}\right)  \wedge\left(  j_{1}+j_{2}+\cdots+j_{n}=h_{1}%
+h_{2}+\cdots+h_{n}+h\right) \\
&  \Longleftrightarrow\ \left(  j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots
>j_{n}\geq h_{n}\text{ and }j_{1}+j_{2}+\cdots+j_{n}=h_{1}+h_{2}+\cdots
+h_{n}+h\right) \\
&  \Longleftrightarrow\ \left(  \left(  j_{1},j_{2},\ldots,j_{n}\right)
\in\mathcal{J}_{h}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.l2.2a.short.equivalence.pf.3})}\right) \\
&  \Longleftrightarrow\ \left(  \mathbf{j}\in\mathcal{J}_{h}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  j_{1},j_{2},\ldots
,j_{n}\right)  =\mathbf{j}\right)  .
\end{align*}
This proves (\ref{pf.l2.2a.short.equivalence}).}.

Now, recall that $\mathcal{J}_{h}\subseteq\mathbb{N}^{n}$. Hence, we have the
following equality of summation signs:%
\begin{align*}
\sum_{\mathbf{j}\in\mathcal{J}_{h}}  &  =\sum_{\substack{\mathbf{j}%
\in\mathbb{N}^{n};\\\mathbf{j}\in\mathcal{J}_{h}}}=\sum_{\substack{\mathbf{j}%
\in\mathbb{N}^{n};\\\mathbf{j}\in\mathcal{NI};\\\mathbf{j}\oslash
\mathbf{h};\\\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}%
\right\vert +h}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because for any }\mathbf{j}\in\mathbb{N}^{n}\text{, we have the}\\
\text{logical equivalence (\ref{pf.l2.2a.short.equivalence})}%
\end{array}
\right) \\
&  =\sum_{\substack{\mathbf{j}\in\mathcal{NI};\\\mathbf{j}\oslash
\mathbf{h};\\\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}%
\right\vert +h}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathcal{NI}%
\subseteq\mathbb{N}^{n}\right)  .
\end{align*}
Hence,%
\[
\sum_{\mathbf{j}\in\mathcal{J}_{h}}X^{\mathbf{j}}=\sum_{\substack{\mathbf{j}%
\in\mathcal{NI};\\\mathbf{j}\oslash\mathbf{h};\\\left\vert \mathbf{j}%
\right\vert =\left\vert \mathbf{h}\right\vert +h}}X^{\mathbf{j}},
\]
so that%
\begin{align*}
\sum_{\substack{\mathbf{j}\in\mathcal{NI};\\\mathbf{j}\oslash\mathbf{h}%
;\\\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}\right\vert
+h}}X^{\mathbf{j}}  &  =\sum_{\mathbf{j}\in\mathcal{J}_{h}}X^{\mathbf{j}}%
=\sum_{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{J}_{h}%
}\underbrace{X^{\left(  j_{1},j_{2},\ldots,j_{n}\right)  }}%
_{\substack{=X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}\\\text{(by
the definition of }X^{\left(  j_{1},j_{2},\ldots,j_{n}\right)  }\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}\mathbf{j}\text{ as }\left(  j_{1},j_{2},\ldots,j_{n}\right)  \right) \\
&  =\sum_{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{J}_{h}}%
X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}.
\end{align*}

\end{vershort}

\begin{verlong}
For any $n$-tuple $\mathbf{j}\in\mathbb{N}^{n}$, we have the logical
equivalence%
\begin{equation}
\left(  \mathbf{j}\in\mathcal{J}_{h}\right)  \ \Longleftrightarrow\ \left(
\mathbf{j}\in\mathcal{NI}\text{ and }\mathbf{j}\oslash\mathbf{h}\text{ and
}\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}\right\vert +h\right)
\label{pf.l2.2a.equivalence}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.l2.2a.equivalence}):} Let $\mathbf{j}%
\in\mathbb{N}^{n}$ be an $n$-tuple. Write $\mathbf{j}$ in the form
$\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{n}\right)  $. Then, we have
$\left\vert \mathbf{j}\right\vert =j_{1}+j_{2}+\cdots+j_{n}$ (by the
definition of $\left\vert \mathbf{j}\right\vert $). Moreover, we have the
equivalence $\left(  \mathbf{j}\in\mathcal{NI}\right)  \ \Longleftrightarrow
\ \left(  \mathbf{j}\text{ is nonincreasing}\right)  $ (by the definition of
$\mathcal{NI}$), since $\mathbf{j}$ is an $n$-tuple of nonnegative integers.
Furthermore, we have the equivalence%
\[
\left(  \mathbf{j}\text{ is nonincreasing}\right)  \ \Longleftrightarrow
\ \left(  j_{1}\geq j_{2}\geq\cdots\geq j_{n}\right)
\]
(by the definition of \textquotedblleft nonincreasing\textquotedblright),
since $\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{n}\right)  $. Furthermore, we
have the equivalence
\[
\left(  \mathbf{j}\oslash\mathbf{h}\right)  \ \Longleftrightarrow\ \left(
j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots>j_{n}\geq h_{n}\right)
\]
(by the definition of \textquotedblleft$\mathbf{j}\oslash\mathbf{h}%
$\textquotedblright), since $\mathbf{j}=\left(  j_{1},j_{2},\ldots
,j_{n}\right)  $ and $\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n}\right)  $.
\par
On the other hand, we know that $\left(  j_{1},j_{2},\ldots,j_{n}\right)
\in\mathbb{N}^{n}$. Hence, we have the equivalence%
\begin{align}
&  \ \left(  \left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{J}%
_{h}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots
>j_{n}\geq h_{n}\text{ and }j_{1}+j_{2}+\cdots+j_{n}=h_{1}+h_{2}+\cdots
+h_{n}+h\right)  \label{pf.l2.2a.equivalence.pf.3}%
\end{align}
(by the definition of $\mathcal{J}_{h}$).
\par
Now, we have the following chain of equivalences:%
\begin{align*}
&  \ \left(  \mathbf{j}\in\mathcal{NI}\text{ and }\mathbf{j}\oslash
\mathbf{h}\text{ and }\left\vert \mathbf{j}\right\vert =\left\vert
\mathbf{h}\right\vert +h\right) \\
&  \Longleftrightarrow\ \underbrace{\left(  \mathbf{j}\in\mathcal{NI}\right)
}_{\substack{\Longleftrightarrow\ \left(  \mathbf{j}\text{ is nonincreasing}%
\right)  \\\Longleftrightarrow\ \left(  j_{1}\geq j_{2}\geq\cdots\geq
j_{n}\right)  }}\wedge\underbrace{\left(  \mathbf{j}\oslash\mathbf{h}\right)
}_{\Longleftrightarrow\ \left(  j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots
>j_{n}\geq h_{n}\right)  }\wedge\left(  \underbrace{\left\vert \mathbf{j}%
\right\vert }_{=j_{1}+j_{2}+\cdots+j_{n}}=\underbrace{\left\vert
\mathbf{h}\right\vert }_{=h_{1}+h_{2}+\cdots+h_{n}}+h\right) \\
&  \Longleftrightarrow\ \underbrace{\left(  j_{1}\geq j_{2}\geq\cdots\geq
j_{n}\right)  \wedge\left(  j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots>j_{n}\geq
h_{n}\right)  }_{\substack{\Longleftrightarrow\ \left(  j_{1}\geq h_{1}%
>j_{2}\geq h_{2}>\cdots>j_{n}\geq h_{n}\right)  \\\text{(since the chain of
inequalities }\left(  j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots>j_{n}\geq
h_{n}\right)  \\\text{clearly implies }\left(  j_{1}\geq j_{2}\geq\cdots\geq
j_{n}\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \wedge\left(  j_{1}+j_{2}+\cdots+j_{n}=h_{1}%
+h_{2}+\cdots+h_{n}+h\right) \\
&  \Longleftrightarrow\ \left(  j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots
>j_{n}\geq h_{n}\right)  \wedge\left(  j_{1}+j_{2}+\cdots+j_{n}=h_{1}%
+h_{2}+\cdots+h_{n}+h\right) \\
&  \Longleftrightarrow\ \left(  j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots
>j_{n}\geq h_{n}\text{ and }j_{1}+j_{2}+\cdots+j_{n}=h_{1}+h_{2}+\cdots
+h_{n}+h\right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\left(  j_{1},j_{2},\ldots
,j_{n}\right)  }_{=\mathbf{j}}\in\mathcal{J}_{h}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.l2.2a.equivalence.pf.3})}\right)
\\
&  \Longleftrightarrow\ \left(  \mathbf{j}\in\mathcal{J}_{h}\right)  .
\end{align*}
This proves (\ref{pf.l2.2a.equivalence}).}.

Now, recall that $\mathcal{J}_{h}\subseteq\mathbb{N}^{n}$ (since
$\mathcal{J}_{h}$ was defined as a set of certain $n$-tuples $\left(
j_{1},j_{2},\ldots,j_{n}\right)  \in\mathbb{N}^{n}$). Hence, we have the
following equality of summation signs:%
\begin{align*}
\sum_{\mathbf{j}\in\mathcal{J}_{h}}  &  =\sum_{\substack{\mathbf{j}%
\in\mathbb{N}^{n};\\\mathbf{j}\in\mathcal{J}_{h}}}=\sum_{\substack{\mathbf{j}%
\in\mathbb{N}^{n};\\\mathbf{j}\in\mathcal{NI};\\\mathbf{j}\oslash
\mathbf{h};\\\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}%
\right\vert +h}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because for any }\mathbf{j}\in\mathbb{N}^{n}\text{, we have the}\\
\text{logical equivalence (\ref{pf.l2.2a.equivalence})}%
\end{array}
\right) \\
&  =\sum_{\substack{\mathbf{j}\in\mathcal{NI};\\\mathbf{j}\oslash
\mathbf{h};\\\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}%
\right\vert +h}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathcal{NI}%
\subseteq\mathbb{N}^{n}\right)  .
\end{align*}
Hence,%
\[
\sum_{\mathbf{j}\in\mathcal{J}_{h}}X^{\mathbf{j}}=\sum_{\substack{\mathbf{j}%
\in\mathcal{NI};\\\mathbf{j}\oslash\mathbf{h};\\\left\vert \mathbf{j}%
\right\vert =\left\vert \mathbf{h}\right\vert +h}}X^{\mathbf{j}},
\]
so that%
\begin{align*}
\sum_{\substack{\mathbf{j}\in\mathcal{NI};\\\mathbf{j}\oslash\mathbf{h}%
;\\\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}\right\vert
+h}}X^{\mathbf{j}}  &  =\sum_{\mathbf{j}\in\mathcal{J}_{h}}X^{\mathbf{j}}%
=\sum_{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{J}_{h}%
}\underbrace{X^{\left(  j_{1},j_{2},\ldots,j_{n}\right)  }}%
_{\substack{=X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}\\\text{(by
the definition of }X^{\left(  j_{1},j_{2},\ldots,j_{n}\right)  }\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\left(  j_{1},j_{2},\ldots,j_{n}\right)
\text{ for }\mathbf{j}\\
\text{in the sum (since every element of }\mathcal{J}_{h}\\
\text{is an }n\text{-tuple }\left(  j_{1},j_{2},\ldots,j_{n}\right)
\in\mathbb{N}^{n}\text{)}%
\end{array}
\right) \\
&  =\sum_{\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in\mathcal{J}_{h}}%
X^{j_{1}}\wedge X^{j_{2}}\wedge\cdots\wedge X^{j_{n}}.
\end{align*}

\end{verlong}

\noindent Comparing this with (\ref{pf.l2.2a.1}), we obtain%
\begin{equation}
s_{h}\left(  X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{n}}\right)
=\sum_{\substack{\mathbf{j}\in\mathcal{NI};\\\mathbf{j}\oslash\mathbf{h}%
;\\\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}\right\vert
+h}}X^{\mathbf{j}}.\nonumber
\end{equation}
In view of $X^{\mathbf{h}}=X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge
X^{h_{n}}$, this rewrites as
\[
s_{h}X^{\mathbf{h}}=\sum_{\substack{\mathbf{j}\in\mathcal{NI};\\\mathbf{j}%
\oslash\mathbf{h};\\\left\vert \mathbf{j}\right\vert =\left\vert
\mathbf{h}\right\vert +h}}X^{\mathbf{j}}.
\]
This proves Lemma \ref{lem.2.2a}.
\end{proof}

Next, we introduce another notation:

\begin{definition}
Let $\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{m}\right)  \in\mathbb{N}^{m}$ be
an $m$-tuple of nonnegative integers. Let $i\in\left\{  1,2,\ldots,m\right\}
$. Then, \textbf{$h$}$^{\sim i}$ is defined to be the $\left(  m-1\right)
$-tuple $\left(  h_{1},h_{2},\ldots,h_{i-1},h_{i+1},\ldots,h_{m}\right)
\in\mathbb{N}^{m-1}$ of nonnegative integers. (This is obtained from
\textbf{$h$} by removing the $i$-th entry.)
\end{definition}

\begin{lemma}
\label{lem.2.2b}Let $\mathbf{j}\in\mathbb{N}^{n}$ and $\mathbf{h}\in
\mathbb{N}^{n+1}$ be two nonincreasing tuples such that $\left\vert
\mathbf{j}\right\vert >\left\vert \mathbf{h}\right\vert -n$. Then,%
\[
\sum_{\substack{i\in\left\{  1,2,\ldots,n+1\right\}  ;\\\mathbf{j}%
\oslash\mathbf{h}^{\sim i}}}\left(  -1\right)  ^{i}=0.
\]

\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.2.2b}.]Write the $n$-tuple $\mathbf{j}\in
\mathbb{N}^{n}$ in the form $\mathbf{j}=\left(  j_{1},j_{2},\ldots
,j_{n}\right)  $. We extend the $n$-tuple $\left(  j_{1},j_{2},\ldots
,j_{n}\right)  \in\mathbb{N}^{n}\subseteq\mathbb{Z}^{n}$ to an $\left(
n+1\right)  $-tuple $\left(  j_{1},j_{2},\ldots,j_{n+1}\right)  \in
\mathbb{Z}^{n+1}$ by setting $j_{n+1}=-1$.

Write the $\left(  n+1\right)  $-tuple $\mathbf{h}\in\mathbb{N}^{n+1}$ in the
form $\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n+1}\right)  $. Thus,
$h_{1}\geq h_{2}\geq\cdots\geq h_{n+1}$ (since $\mathbf{h}$ is nonincreasing).

Define the two sets%
\begin{align}
J  &  =\left\{  i\in\left\{  1,2,\ldots,n+1\right\}  \ \mid\ \mathbf{j}%
\oslash\mathbf{h}^{\sim i}\text{ and }j_{i}\geq h_{i}\right\}
\ \ \ \ \ \ \ \ \ \ \text{and}\label{pf.l2.2b.short.J=}\\
H  &  =\left\{  i\in\left\{  1,2,\ldots,n+1\right\}  \ \mid\ \mathbf{j}%
\oslash\mathbf{h}^{\sim i}\text{ and }j_{i}<h_{i}\right\}  .
\label{pf.l2.2b.short.H=}%
\end{align}


Each $i\in\left\{  1,2,\ldots,n+1\right\}  $ satisfies either $j_{i}\geq
h_{i}$ or $j_{i}<h_{i}$ (but not both at the same time). Hence, we can split
the sum $\sum_{\substack{i\in\left\{  1,2,\ldots,n+1\right\}  ;\\\mathbf{j}%
\oslash\mathbf{h}^{\sim i}}}\left(  -1\right)  ^{i}$ as follows:%
\begin{align}
\sum_{\substack{i\in\left\{  1,2,\ldots,n+1\right\}  ;\\\mathbf{j}%
\oslash\mathbf{h}^{\sim i}}}\left(  -1\right)  ^{i}  &  =\underbrace{\sum
_{\substack{i\in\left\{  1,2,\ldots,n+1\right\}  ;\\\mathbf{j}\oslash
\mathbf{h}^{\sim i};\\j_{i}\geq h_{i}}}}_{\substack{=\sum_{i\in J}\\\text{(by
(\ref{pf.l2.2b.short.J=}))}}}\left(  -1\right)  ^{i}+\underbrace{\sum
_{\substack{i\in\left\{  1,2,\ldots,n+1\right\}  ;\\\mathbf{j}\oslash
\mathbf{h}^{\sim i};\\j_{i}<h_{i}}}}_{\substack{=\sum_{i\in H}\\\text{(by
(\ref{pf.l2.2b.short.H=}))}}}\left(  -1\right)  ^{i}\nonumber\\
&  =\sum_{i\in J}\left(  -1\right)  ^{i}+\sum_{i\in H}\left(  -1\right)  ^{i}.
\label{pf.l2.2b.short.sum-decomposed}%
\end{align}


Now, we claim that
\begin{equation}
u+1\in H\ \ \ \ \ \ \ \ \ \ \text{for each }u\in J\text{.}
\label{pf.l2.2b.short.u+1}%
\end{equation}


[\textit{Proof of (\ref{pf.l2.2b.short.u+1}):} Let $u\in J$. We must prove
that $u+1\in H$.

We have $u\in J$. In view of (\ref{pf.l2.2b.short.J=}), this means that $u$ is
an element of $\left\{  1,2,\ldots,n+1\right\}  $ that satisfies
$\mathbf{j}\oslash\mathbf{h}^{\sim u}$ and $j_{u}\geq h_{u}$.

Hence, $j_{u}\geq h_{u}\geq0$ (since $\left(  h_{1},h_{2},\ldots
,h_{n+1}\right)  =\mathbf{h}\in\mathbb{N}^{n+1}$). If we had $u=n+1$, then we
would have $j_{u}=j_{n+1}=-1<0$, which would contradict $j_{u}\geq0$. Hence,
$u\neq n+1$. Combining this with $u\in\left\{  1,2,\ldots,n+1\right\}  $, we
obtain $u\in\left\{  1,2,\ldots,n\right\}  $. Hence, $u+1\in\left\{
1,2,\ldots,n+1\right\}  $. Thus, the $n$-tuple $\mathbf{h}^{\sim\left(
u+1\right)  }$ is well-defined.

Recall that $\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n+1}\right)  $. Hence,
the definitions of $\mathbf{h}^{\sim u}$ and $\mathbf{h}^{\sim\left(
u+1\right)  }$ yield%
\begin{align*}
\mathbf{h}^{\sim u}  &  =\left(  h_{1},h_{2},\ldots,h_{u-1},h_{u+1}%
,\ldots,h_{n+1}\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
\mathbf{h}^{\sim\left(  u+1\right)  }  &  =\left(  h_{1},h_{2},\ldots
,h_{u},h_{u+2},\ldots,h_{n+1}\right)  .
\end{align*}


We have $\mathbf{j}\oslash\mathbf{h}^{\sim u}$. In view of $\mathbf{j}=\left(
j_{1},j_{2},\ldots,j_{n}\right)  $ and $\mathbf{h}^{\sim u}=\left(
h_{1},h_{2},\ldots,h_{u-1},h_{u+1},\ldots,h_{n+1}\right)  $, this rewrites as%
\[
j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots>j_{u-1}\geq h_{u-1}>j_{u}\geq
h_{u+1}>j_{u+1}\geq h_{u+2}>\cdots>j_{n}\geq h_{n+1}%
\]
(by the definition of the notation \textquotedblleft$\mathbf{j}\oslash
\mathbf{h}^{\sim u}$\textquotedblright). We can split this chain of
inequalities into three pieces as follows:%
\begin{align}
j_{1}  &  \geq h_{1}>j_{2}\geq h_{2}>\cdots>j_{u-1}\geq h_{u-1}>j_{u}%
;\label{pf.l2.2b.short.u+1.pf.chain-a}\\
j_{u}  &  \geq h_{u+1}>j_{u+1};\nonumber\\
j_{u+1}  &  \geq h_{u+2}>j_{u+2}\geq h_{u+3}>\cdots>j_{n}\geq h_{n+1}.
\label{pf.l2.2b.short.u+1.pf.chain-c}%
\end{align}


From $h_{1}\geq h_{2}\geq\cdots\geq h_{n+1}$, we obtain $h_{u}\geq h_{u+1}$,
so that $h_{u}\geq h_{u+1}>j_{u+1}$. Hence,%
\begin{equation}
j_{u}\geq h_{u}>j_{u+1}. \label{pf.l2.2b.short.u+1.pf.4b}%
\end{equation}


We can splice the three chains of inequalities
(\ref{pf.l2.2b.short.u+1.pf.chain-a}), (\ref{pf.l2.2b.short.u+1.pf.4b}) and
(\ref{pf.l2.2b.short.u+1.pf.chain-c}) together into one long chain:%
\[
j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots>j_{u}\geq h_{u}>j_{u+1}\geq
h_{u+2}>j_{u+2}\geq h_{u+3}>\cdots>j_{n}\geq h_{n+1}.
\]
This rewrites as $\mathbf{j}\oslash\mathbf{h}^{\sim\left(  u+1\right)  }$ (by
the definition of the notation \textquotedblleft$\mathbf{j}\oslash
\mathbf{h}^{\sim\left(  u+1\right)  }$\textquotedblright, since $\mathbf{j}%
=\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ and $\mathbf{h}^{\sim\left(
u+1\right)  }=\left(  h_{1},h_{2},\ldots,h_{u},h_{u+2},\ldots,h_{n+1}\right)
$).

Also, $j_{u+1}<h_{u+1}$ (since $h_{u+1}>j_{u+1}$). Now, we know that $u+1$ is
an element of $\left\{  1,2,\ldots,n+1\right\}  $ and satisfies $\mathbf{j}%
\oslash\mathbf{h}^{\sim\left(  u+1\right)  }$ and $j_{u+1}<h_{u+1}$. In view
of (\ref{pf.l2.2b.short.H=}), this rewrites as $u+1\in H$. This proves
(\ref{pf.l2.2b.short.u+1}).]

Next, we claim that%
\begin{equation}
u-1\in J\ \ \ \ \ \ \ \ \ \ \text{for each }u\in H\text{.}
\label{pf.l2.2b.short.u-1}%
\end{equation}


[\textit{Proof of (\ref{pf.l2.2b.short.u-1}):} Let $u\in H$. We must prove
that $u-1\in J$.

We have $u\in H$. In view of (\ref{pf.l2.2b.short.H=}), this means that $u$ is
an element of $\left\{  1,2,\ldots,n+1\right\}  $ and satisfies $\mathbf{j}%
\oslash\mathbf{h}^{\sim u}$ and $j_{u}<h_{u}$.

It is not hard to see that $u\neq1$\ \ \ \ \footnote{\textit{Proof.} Assume
the contrary. Thus, $u=1$. Hence, $\mathbf{h}^{\sim u}=\mathbf{h}^{\sim
1}=\left(  h_{2},h_{3},\ldots,h_{n+1}\right)  $ (by the definition of
$\mathbf{h}^{\sim1}$, because $\mathbf{h}=\left(  h_{1},h_{2},\ldots
,h_{n+1}\right)  $). Also, $j_{u}<h_{u}$; this rewrites as $j_{1}<h_{1}$
(since $u=1$). In other words, $h_{1}>j_{1}$.
\par
But recall that $\mathbf{j}\oslash\mathbf{h}^{\sim u}$. In view of
$\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ and $\mathbf{h}^{\sim
u}=\left(  h_{2},h_{3},\ldots,h_{n+1}\right)  $, this rewrites as%
\[
j_{1}\geq h_{2}>j_{2}\geq h_{3}>\cdots>j_{n}\geq h_{n+1}%
\]
(by the definition of the notation \textquotedblleft$\mathbf{j}\oslash
\mathbf{h}^{\sim u}$\textquotedblright). Thus, in particular, we have
$h_{i}>j_{i}$ for each $i\in\left\{  2,3,\ldots,n\right\}  $. This inequality
also holds for $i=1$ (since $h_{1}>j_{1}$), and thus holds for all
$i\in\left\{  1,2,\ldots,n\right\}  $. Hence, for each $i\in\left\{
1,2,\ldots,n\right\}  $, we have $h_{i}\geq j_{i}+1$ (because $h_{i}>j_{i}$,
but both $h_{i}$ and $j_{i}$ are integers). Hence,
\[
\sum_{i=1}^{n}h_{i}\geq\sum_{i=1}^{n}\left(  j_{i}+1\right)  =\underbrace{\sum
_{i=1}^{n}j_{i}}_{=j_{1}+j_{2}+\cdots+j_{n}}+\underbrace{\sum_{i=1}^{n}1}%
_{=n}=\left(  j_{1}+j_{2}+\cdots+j_{n}\right)  +n.
\]
But $\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ and thus
$\left\vert \mathbf{j}\right\vert =j_{1}+j_{2}+\cdots+j_{n}$ (by the
definition of $\left\vert \mathbf{j}\right\vert $). Hence,%
\begin{align*}
\sum_{i=1}^{n}h_{i}  &  \geq\underbrace{\left(  j_{1}+j_{2}+\cdots
+j_{n}\right)  }_{=\left\vert \mathbf{j}\right\vert >\left\vert \mathbf{h}%
\right\vert -n}+n>\left\vert \mathbf{h}\right\vert -n+n=\left\vert
\mathbf{h}\right\vert =h_{1}+h_{2}+\cdots+h_{n+1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\left\vert
\mathbf{h}\right\vert \text{, since }\mathbf{h}=\left(  h_{1},h_{2}%
,\ldots,h_{n+1}\right)  \right) \\
&  =\underbrace{\left(  h_{1}+h_{2}+\cdots+h_{n}\right)  }_{=\sum_{i=1}%
^{n}h_{i}}+\underbrace{h_{n+1}}_{\substack{\geq0\\\text{(since }\left(
h_{1},h_{2},\ldots,h_{n+1}\right)  =\mathbf{h}\in\mathbb{N}^{n+1}\text{)}%
}}\geq\sum_{i=1}^{n}h_{i}.
\end{align*}
This is absurd. This contradiction shows that our assumption was false. Qed.}.
Combining this with $u\in\left\{  1,2,\ldots,n+1\right\}  $, we obtain
$u\in\left\{  2,3,\ldots,n+1\right\}  $, so that $u-1\in\left\{
1,2,\ldots,n+1\right\}  $. Hence, the $n$-tuple $\mathbf{h}^{\sim\left(
u-1\right)  }$ is well-defined.

Recall that $\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n+1}\right)  $. Hence,
the definitions of $\mathbf{h}^{\sim u}$ and $\mathbf{h}^{\sim\left(
u-1\right)  }$ yield%
\begin{align*}
\mathbf{h}^{\sim u}  &  =\left(  h_{1},h_{2},\ldots,h_{u-1},h_{u+1}%
,\ldots,h_{n+1}\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
\mathbf{h}^{\sim\left(  u-1\right)  }  &  =\left(  h_{1},h_{2},\ldots
,h_{u-2},h_{u},\ldots,h_{n+1}\right)  .
\end{align*}


We have $\mathbf{j}\oslash\mathbf{h}^{\sim u}$. In view of $\mathbf{j}=\left(
j_{1},j_{2},\ldots,j_{n}\right)  $ and $\mathbf{h}^{\sim u}=\left(
h_{1},h_{2},\ldots,h_{u-1},h_{u+1},\ldots,h_{n+1}\right)  $, this rewrites as%
\[
j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots>j_{u-1}\geq h_{u-1}>j_{u}\geq
h_{u+1}>j_{u+1}\geq h_{u+2}>\cdots>j_{n}\geq h_{n+1}%
\]
(by the definition of the notation \textquotedblleft$\mathbf{j}\oslash
\mathbf{h}^{\sim u}$\textquotedblright). We can split this chain of
inequalities into three pieces as follows:\footnote{We are using $u\in\left\{
2,3,\ldots,n+1\right\}  $ here.}%
\begin{align}
j_{1}  &  \geq h_{1}>j_{2}\geq h_{2}>\cdots>j_{u-2}\geq h_{u-2}>j_{u-1}%
;\label{pf.l2.2b.short.u-1.pf.chain-a}\\
j_{u-1}  &  \geq h_{u-1}>j_{u};\nonumber\\
j_{u}  &  \geq h_{u+1}>j_{u+1}\geq h_{u+2}>\cdots>j_{n}\geq h_{n+1}.
\label{pf.l2.2b.short.u-1.pf.chain-c}%
\end{align}


From $h_{1}\geq h_{2}\geq\cdots\geq h_{n+1}$, we obtain $h_{u-1}\geq h_{u}$,
so that $j_{u-1}\geq h_{u-1}\geq h_{u}$. Hence,%
\begin{equation}
j_{u-1}\geq h_{u}>j_{u}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }j_{u}%
<h_{u}\right)  . \label{pf.l2.2b.short.u-1.pf.4b}%
\end{equation}


We can splice the three chains of inequalities
(\ref{pf.l2.2b.short.u-1.pf.chain-a}), (\ref{pf.l2.2b.short.u-1.pf.4b}) and
(\ref{pf.l2.2b.short.u-1.pf.chain-c}) together into one long chain:%
\[
j_{1}\geq h_{1}>j_{2}\geq h_{2}>\cdots>j_{u-1}\geq h_{u}>j_{u}\geq
h_{u+1}>j_{u+1}\geq h_{u+2}>\cdots>j_{n}\geq h_{n+1}.
\]
This rewrites as $\mathbf{j}\oslash\mathbf{h}^{\sim\left(  u-1\right)  }$ (by
the definition of the notation \textquotedblleft$\mathbf{j}\oslash
\mathbf{h}^{\sim\left(  u-1\right)  }$\textquotedblright, since $\mathbf{j}%
=\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ and $\mathbf{h}^{\sim\left(
u-1\right)  }=\left(  h_{1},h_{2},\ldots,h_{u-2},h_{u},\ldots,h_{n+1}\right)
$).

Also, $j_{u-1}\geq h_{u-1}$ (as we have seen). Now, we know that $u-1$ is an
element of $\left\{  1,2,\ldots,n+1\right\}  $ and satisfies $\mathbf{j}%
\oslash\mathbf{h}^{\sim\left(  u-1\right)  }$ and $j_{u-1}\geq h_{u-1}$. In
view of (\ref{pf.l2.2b.short.J=}), this rewrites as $u-1\in J$. This proves
(\ref{pf.l2.2b.short.u-1}).]

Now, define a map%
\[
\alpha:J\rightarrow H,\ \ \ \ \ \ \ \ \ \ u\mapsto u+1.
\]
(This map is well-defined, due to (\ref{pf.l2.2b.short.u+1}).)

Also, define a map%
\[
\beta:H\rightarrow J,\ \ \ \ \ \ \ \ \ \ u\mapsto u-1.
\]
(This map is well-defined, due to (\ref{pf.l2.2b.short.u-1}).)

Clearly, the maps $\alpha$ and $\beta$ are mutually inverse. Thus, the map
$\alpha$ is invertible, i.e., is a bijection. Now,
(\ref{pf.l2.2b.short.sum-decomposed}) becomes%
\begin{align*}
\sum_{\substack{i\in\left\{  1,2,\ldots,n+1\right\}  ;\\\mathbf{j}%
\oslash\mathbf{h}^{\sim i}}}\left(  -1\right)  ^{i}  &  =\sum_{i\in J}\left(
-1\right)  ^{i}+\underbrace{\sum_{i\in H}\left(  -1\right)  ^{i}%
}_{\substack{=\sum_{i\in J}\left(  -1\right)  ^{\alpha\left(  i\right)
}\\\text{(here, we have}\\\text{substituted }\alpha\left(  i\right)  \text{
for }i\\\text{in the sum, since the}\\\text{map }\alpha:J\rightarrow H\text{
is a bijection)}}}=\sum_{i\in J}\left(  -1\right)  ^{i}+\sum_{i\in
J}\underbrace{\left(  -1\right)  ^{\alpha\left(  i\right)  }}%
_{\substack{=\left(  -1\right)  ^{i+1}\\\text{(since }\alpha\left(  i\right)
=i+1\\\text{(by the definition of }\alpha\text{))}}}\\
&  =\sum_{i\in J}\left(  -1\right)  ^{i}+\sum_{i\in J}\underbrace{\left(
-1\right)  ^{i+1}}_{=-\left(  -1\right)  ^{i}}=\sum_{i\in J}\left(  -1\right)
^{i}-\sum_{i\in J}\left(  -1\right)  ^{i}=0.
\end{align*}
This proves Lemma \ref{lem.2.2b}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.2.2b}.]Write the $n$-tuple $\mathbf{j}\in
\mathbb{N}^{n}$ in the form $\mathbf{j}=\left(  j_{1},j_{2},\ldots
,j_{n}\right)  $. We extend the $n$-tuple $\left(  j_{1},j_{2},\ldots
,j_{n}\right)  \in\mathbb{N}^{n}\subseteq\mathbb{Z}^{n}$ to an $\left(
n+1\right)  $-tuple $\left(  j_{1},j_{2},\ldots,j_{n+1}\right)  \in
\mathbb{Z}^{n+1}$ by setting $j_{n+1}=-1$.

Write the $\left(  n+1\right)  $-tuple $\mathbf{h}\in\mathbb{N}^{n+1}$ in the
form $\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n+1}\right)  $. Thus,
$h_{1}\geq h_{2}\geq\cdots\geq h_{n+1}$ (since $\mathbf{h}$ is nonincreasing).

Define the two sets%
\begin{equation}
J=\left\{  i\in\left\{  1,2,\ldots,n+1\right\}  \ \mid\ \mathbf{j}%
\oslash\mathbf{h}^{\sim i}\text{ and }j_{i}\geq h_{i}\right\}
\label{pf.l2.2b.J=}%
\end{equation}
and%
\begin{equation}
H=\left\{  i\in\left\{  1,2,\ldots,n+1\right\}  \ \mid\ \mathbf{j}%
\oslash\mathbf{h}^{\sim i}\text{ and }j_{i}<h_{i}\right\}  .
\label{pf.l2.2b.H=}%
\end{equation}


Each $i\in\left\{  1,2,\ldots,n+1\right\}  $ satisfies either $j_{i}\geq
h_{i}$ or $j_{i}<h_{i}$ (but not both at the same time). Hence, we can split
the sum $\sum_{\substack{i\in\left\{  1,2,\ldots,n+1\right\}  ;\\\mathbf{j}%
\oslash\mathbf{h}^{\sim i}}}\left(  -1\right)  ^{i}$ as follows:%
\begin{align}
&  \sum_{\substack{i\in\left\{  1,2,\ldots,n+1\right\}  ;\\\mathbf{j}%
\oslash\mathbf{h}^{\sim i}}}\left(  -1\right)  ^{i}\nonumber\\
&  =\underbrace{\sum_{\substack{i\in\left\{  1,2,\ldots,n+1\right\}
;\\\mathbf{j}\oslash\mathbf{h}^{\sim i};\\j_{i}\geq h_{i}}}}_{\substack{=\sum
_{i\in J}\\\text{(since }J=\left\{  i\in\left\{  1,2,\ldots,n+1\right\}
\ \mid\ \mathbf{j}\oslash\mathbf{h}^{\sim i}\text{ and }j_{i}\geq
h_{i}\right\}  \text{)}}}\left(  -1\right)  ^{i}+\underbrace{\sum
_{\substack{i\in\left\{  1,2,\ldots,n+1\right\}  ;\\\mathbf{j}\oslash
\mathbf{h}^{\sim i};\\j_{i}<h_{i}}}}_{\substack{=\sum_{i\in H}\\\text{(since
}H=\left\{  i\in\left\{  1,2,\ldots,n+1\right\}  \ \mid\ \mathbf{j}%
\oslash\mathbf{h}^{\sim i}\text{ and }j_{i}<h_{i}\right\}  \text{)}}}\left(
-1\right)  ^{i}\nonumber\\
&  =\sum_{i\in J}\left(  -1\right)  ^{i}+\sum_{i\in H}\left(  -1\right)  ^{i}.
\label{pf.l2.2b.sum-decomposed}%
\end{align}


Now, we claim that
\begin{equation}
u+1\in H\ \ \ \ \ \ \ \ \ \ \text{for each }u\in J\text{.}
\label{pf.l2.2b.u+1}%
\end{equation}


[\textit{Proof of (\ref{pf.l2.2b.u+1}):} Let $u\in J$. We must prove that
$u+1\in H$.

We have $u\in J=\left\{  i\in\left\{  1,2,\ldots,n+1\right\}  \ \mid
\ \mathbf{j}\oslash\mathbf{h}^{\sim i}\text{ and }j_{i}\geq h_{i}\right\}  $.
In other words, $u$ is an $i\in\left\{  1,2,\ldots,n+1\right\}  $ satisfying
$\mathbf{j}\oslash\mathbf{h}^{\sim i}$ and $j_{i}\geq h_{i}$. In other words,
$u$ is an element of $\left\{  1,2,\ldots,n+1\right\}  $ and satisfies
$\mathbf{j}\oslash\mathbf{h}^{\sim u}$ and $j_{u}\geq h_{u}$.

Hence, $j_{u}\geq h_{u}\geq0$ (since $\left(  h_{1},h_{2},\ldots
,h_{n+1}\right)  =\mathbf{h}\in\mathbb{N}^{n+1}$). If we had $u=n+1$, then we
would have $j_{u}=j_{n+1}=-1<0$, which would contradict $j_{u}\geq0$. Thus, we
cannot have $u=n+1$. Hence, $u\neq n+1$. Combining this with $u\in\left\{
1,2,\ldots,n+1\right\}  $, we obtain $u\in\left\{  1,2,\ldots,n+1\right\}
\setminus\left\{  n+1\right\}  =\left\{  1,2,\ldots,n\right\}  $. Hence,
$u+1\in\left\{  2,3,\ldots,n+1\right\}  \subseteq\left\{  1,2,\ldots
,n+1\right\}  $. Thus, the $n$-tuple $\mathbf{h}^{\sim\left(  u+1\right)  }$
is well-defined.

Write the $n$-tuple $\mathbf{h}^{\sim u}$ as
\[
\mathbf{h}^{\sim u}=\left(  h_{1}^{\prime},h_{2}^{\prime},\ldots,h_{n}%
^{\prime}\right)  .
\]
Thus,%
\begin{equation}
\left(  h_{1}^{\prime},h_{2}^{\prime},\ldots,h_{n}^{\prime}\right)
=\mathbf{h}^{\sim u}=\left(  h_{1},h_{2},\ldots,h_{u-1},h_{u+1},\ldots
,h_{n+1}\right)  \label{pf.l2.2b.u+1.pf.1}%
\end{equation}
(by the definition of $\mathbf{h}^{\sim u}$, since $\mathbf{h}=\left(
h_{1},h_{2},\ldots,h_{n+1}\right)  $). Hence,%
\begin{align}
\left(  h_{1}^{\prime},h_{2}^{\prime},\ldots,h_{u-1}^{\prime}\right)   &
=\left(  h_{1},h_{2},\ldots,h_{u-1}\right)  ;\label{pf.l2.2b.u+1.pf.1a}\\
h_{u}^{\prime}  &  =h_{u+1};\label{pf.l2.2b.u+1.pf.1b}\\
\left(  h_{u+1}^{\prime},h_{u+2}^{\prime},\ldots,h_{n}^{\prime}\right)   &
=\left(  h_{u+2},h_{u+3},\ldots,h_{n+1}\right)  \label{pf.l2.2b.u+1.pf.1c}%
\end{align}
(since $u\in\left\{  1,2,\ldots,n\right\}  $).

Write the $n$-tuple $\mathbf{h}^{\sim\left(  u+1\right)  }$ as
\[
\mathbf{h}^{\sim\left(  u+1\right)  }=\left(  h_{1}^{\prime\prime}%
,h_{2}^{\prime\prime},\ldots,h_{n}^{\prime\prime}\right)  .
\]
Thus,%
\begin{equation}
\left(  h_{1}^{\prime\prime},h_{2}^{\prime\prime},\ldots,h_{n}^{\prime\prime
}\right)  =\mathbf{h}^{\sim\left(  u+1\right)  }=\left(  h_{1},h_{2}%
,\ldots,h_{u},h_{u+2},\ldots,h_{n+1}\right)  \label{pf.l2.2b.u+1.pf.2}%
\end{equation}
(by the definition of $\mathbf{h}^{\sim\left(  u+1\right)  }$, since
$\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n+1}\right)  $). Hence,%
\begin{align}
\left(  h_{1}^{\prime\prime},h_{2}^{\prime\prime},\ldots,h_{u-1}^{\prime
\prime}\right)   &  =\left(  h_{1},h_{2},\ldots,h_{u-1}\right)
;\label{pf.l2.2b.u+1.pf.2a}\\
h_{u}^{\prime\prime}  &  =h_{u};\label{pf.l2.2b.u+1.pf.2b}\\
\left(  h_{u+1}^{\prime\prime},h_{u+2}^{\prime\prime},\ldots,h_{n}%
^{\prime\prime}\right)   &  =\left(  h_{u+2},h_{u+3},\ldots,h_{n+1}\right)
\label{pf.l2.2b.u+1.pf.2c}%
\end{align}
(since $u\in\left\{  1,2,\ldots,n\right\}  $).

Comparing (\ref{pf.l2.2b.u+1.pf.1a}) with (\ref{pf.l2.2b.u+1.pf.2a}), we
obtain%
\[
\left(  h_{1}^{\prime},h_{2}^{\prime},\ldots,h_{u-1}^{\prime}\right)  =\left(
h_{1}^{\prime\prime},h_{2}^{\prime\prime},\ldots,h_{u-1}^{\prime\prime
}\right)  .
\]
In other words,%
\begin{equation}
h_{i}^{\prime}=h_{i}^{\prime\prime}\ \ \ \ \ \ \ \ \ \ \text{for each }%
i\in\left\{  1,2,\ldots,u-1\right\}  . \label{pf.l2.2b.u+1.pf.1a=2a}%
\end{equation}
Comparing (\ref{pf.l2.2b.u+1.pf.1c}) with (\ref{pf.l2.2b.u+1.pf.2c}), we
obtain%
\[
\left(  h_{u+1}^{\prime},h_{u+2}^{\prime},\ldots,h_{n}^{\prime}\right)
=\left(  h_{u+1}^{\prime\prime},h_{u+2}^{\prime\prime},\ldots,h_{n}%
^{\prime\prime}\right)  .
\]
In other words,%
\begin{equation}
h_{i}^{\prime}=h_{i}^{\prime\prime}\ \ \ \ \ \ \ \ \ \ \text{for each }%
i\in\left\{  u+1,u+2,\ldots,n\right\}  . \label{pf.l2.2b.u+1.pf.1c=2c}%
\end{equation}


We have $\mathbf{j}\oslash\mathbf{h}^{\sim u}$. In view of $\mathbf{j}=\left(
j_{1},j_{2},\ldots,j_{n}\right)  $ and $\mathbf{h}^{\sim u}=\left(
h_{1}^{\prime},h_{2}^{\prime},\ldots,h_{n}^{\prime}\right)  $, this rewrites
as%
\[
\left(  j_{1},j_{2},\ldots,j_{n}\right)  \oslash\left(  h_{1}^{\prime}%
,h_{2}^{\prime},\ldots,h_{n}^{\prime}\right)  .
\]
In other words,%
\[
j_{1}\geq h_{1}^{\prime}>j_{2}\geq h_{2}^{\prime}>\cdots>j_{n}\geq
h_{n}^{\prime}%
\]
(by the definition of the notation \textquotedblleft$\left(  j_{1}%
,j_{2},\ldots,j_{n}\right)  \oslash\left(  h_{1}^{\prime},h_{2}^{\prime
},\ldots,h_{n}^{\prime}\right)  $\textquotedblright). We can break this chain
of inequalities into three sub-chains: Namely,%
\begin{align}
j_{1}  &  \geq h_{1}^{\prime}>j_{2}\geq h_{2}^{\prime}>\cdots>j_{u-1}\geq
h_{u-1}^{\prime}>j_{u};\label{pf.l2.2b.u+1.pf.3a}\\
j_{u}  &  \geq h_{u}^{\prime}>j_{u+1};\label{pf.l2.2b.u+1.pf.3b}\\
j_{u+1}  &  \geq h_{u+1}^{\prime}>j_{u+2}\geq h_{u+2}^{\prime}>\cdots
>j_{n}\geq h_{n}^{\prime}. \label{pf.l2.2b.u+1.pf.3c}%
\end{align}
(Here, of course, the chain of inequalities (\ref{pf.l2.2b.u+1.pf.3a}) is
regarded as vacuously true when $u=1$, and the chain of inequalities
(\ref{pf.l2.2b.u+1.pf.3c}) is regarded as vacuously true when $u=n$. We
already know that $u\in\left\{  1,2,\ldots,n\right\}  $, so that
(\ref{pf.l2.2b.u+1.pf.3b}) always makes sense.)

Each of the $h_{i}^{\prime}$ appearing in the chain of inequalities
(\ref{pf.l2.2b.u+1.pf.3a}) equals the corresponding $h_{i}^{\prime\prime}$ (by
(\ref{pf.l2.2b.u+1.pf.1a=2a})). Hence, we can rewrite
(\ref{pf.l2.2b.u+1.pf.3a}) as follows:%
\begin{equation}
j_{1}\geq h_{1}^{\prime\prime}>j_{2}\geq h_{2}^{\prime\prime}>\cdots
>j_{u-1}\geq h_{u-1}^{\prime\prime}>j_{u}. \label{pf.l2.2b.u+1.pf.4a}%
\end{equation}


From $h_{1}\geq h_{2}\geq\cdots\geq h_{n+1}$, we obtain $h_{u}\geq h_{u+1}$.
From (\ref{pf.l2.2b.u+1.pf.1b}), we obtain $h_{u+1}=h_{u}^{\prime}>j_{u+1}$
(by (\ref{pf.l2.2b.u+1.pf.3b})). Furthermore, (\ref{pf.l2.2b.u+1.pf.2b})
yields $h_{u}^{\prime\prime}=h_{u}\geq h_{u+1}>j_{u+1}$.

We have $j_{u}\geq h_{u}=h_{u}^{\prime\prime}$ (by (\ref{pf.l2.2b.u+1.pf.2b}%
)). Hence,%
\begin{equation}
j_{u}\geq h_{u}^{\prime\prime}>j_{u+1}. \label{pf.l2.2b.u+1.pf.4b}%
\end{equation}


Each of the $h_{i}^{\prime}$ appearing in the chain of inequalities
(\ref{pf.l2.2b.u+1.pf.3c}) equals the corresponding $h_{i}^{\prime\prime}$ (by
(\ref{pf.l2.2b.u+1.pf.1c=2c})). Hence, we can rewrite
(\ref{pf.l2.2b.u+1.pf.3c}) as follows:%
\begin{equation}
j_{u+1}\geq h_{u+1}^{\prime\prime}>j_{u+2}\geq h_{u+2}^{\prime\prime}%
>\cdots>j_{n}\geq h_{n}^{\prime\prime}. \label{pf.l2.2b.u+1.pf.4c}%
\end{equation}


We can splice the three chains of inequalities (\ref{pf.l2.2b.u+1.pf.4a}),
(\ref{pf.l2.2b.u+1.pf.4b}) and (\ref{pf.l2.2b.u+1.pf.4c}) together into one
long chain:%
\[
j_{1}\geq h_{1}^{\prime\prime}>j_{2}\geq h_{2}^{\prime\prime}>\cdots>j_{n}\geq
h_{n}^{\prime\prime}.
\]
In other words,%
\[
\left(  j_{1},j_{2},\ldots,j_{n}\right)  \oslash\left(  h_{1}^{\prime\prime
},h_{2}^{\prime\prime},\ldots,h_{n}^{\prime\prime}\right)  .
\]
(by the definition of the notation \textquotedblleft$\left(  j_{1}%
,j_{2},\ldots,j_{n}\right)  \oslash\left(  h_{1}^{\prime\prime},h_{2}%
^{\prime\prime},\ldots,h_{n}^{\prime\prime}\right)  $\textquotedblright). In
view of $\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ and
$\mathbf{h}^{\sim\left(  u+1\right)  }=\left(  h_{1}^{\prime\prime}%
,h_{2}^{\prime\prime},\ldots,h_{n}^{\prime\prime}\right)  $, this rewrites as
$\mathbf{j}\oslash\mathbf{h}^{\sim\left(  u+1\right)  }$.

Also, $j_{u+1}<h_{u+1}$ (since $h_{u+1}>j_{u+1}$). Now, we know that $u+1$ is
an element of $\left\{  1,2,\ldots,n+1\right\}  $ and satisfies $\mathbf{j}%
\oslash\mathbf{h}^{\sim\left(  u+1\right)  }$ and $j_{u+1}<h_{u+1}$. In other
words, $u+1$ is an $i\in\left\{  1,2,\ldots,n+1\right\}  $ satisfying
$\mathbf{j}\oslash\mathbf{h}^{\sim i}$ and $j_{i}<h_{i}$. In other words,
$u+1\in\left\{  i\in\left\{  1,2,\ldots,n+1\right\}  \ \mid\ \mathbf{j}%
\oslash\mathbf{h}^{\sim i}\text{ and }j_{i}<h_{i}\right\}  $. In view of
(\ref{pf.l2.2b.H=}), this rewrites as $u+1\in H$. This proves
(\ref{pf.l2.2b.u+1}).]

Next, we claim that%
\begin{equation}
u-1\in J\ \ \ \ \ \ \ \ \ \ \text{for each }u\in H\text{.}
\label{pf.l2.2b.u-1}%
\end{equation}


[\textit{Proof of (\ref{pf.l2.2b.u-1}):} Let $u\in H$. We must prove that
$u-1\in J$.

We have $u\in H=\left\{  i\in\left\{  1,2,\ldots,n+1\right\}  \ \mid
\ \mathbf{j}\oslash\mathbf{h}^{\sim i}\text{ and }j_{i}<h_{i}\right\}  $. In
other words, $u$ is an $i\in\left\{  1,2,\ldots,n+1\right\}  $ satisfying
$\mathbf{j}\oslash\mathbf{h}^{\sim i}$ and $j_{i}<h_{i}$. In other words, $u$
is an element of $\left\{  1,2,\ldots,n+1\right\}  $ and satisfies
$\mathbf{j}\oslash\mathbf{h}^{\sim u}$ and $j_{u}<h_{u}$.

It is not hard to see that $u\neq1$\ \ \ \ \footnote{\textit{Proof.} Assume
the contrary. Thus, $u=1$. Hence, $\mathbf{h}^{\sim u}=\mathbf{h}^{\sim
1}=\left(  h_{2},h_{3},\ldots,h_{n+1}\right)  $ (by the definition of
$\mathbf{h}^{\sim1}$, because $\mathbf{h}=\left(  h_{1},h_{2},\ldots
,h_{n+1}\right)  $). Also, $j_{u}<h_{u}$; this rewrites as $j_{1}<h_{1}$
(since $u=1$). In other words, $h_{1}>j_{1}$.
\par
But recall that $\mathbf{j}\oslash\mathbf{h}^{\sim u}$. In view of
$\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ and $\mathbf{h}^{\sim
u}=\left(  h_{2},h_{3},\ldots,h_{n+1}\right)  $, this rewrites as $\left(
j_{1},j_{2},\ldots,j_{n}\right)  \oslash\left(  h_{2},h_{3},\ldots
,h_{n+1}\right)  $. In other words,%
\[
j_{1}\geq h_{2}>j_{2}\geq h_{3}>\cdots>j_{n}\geq h_{n+1}%
\]
(by the definition of the notation \textquotedblleft$\left(  j_{1}%
,j_{2},\ldots,j_{n}\right)  \oslash\left(  h_{2},h_{3},\ldots,h_{n+1}\right)
$\textquotedblright). Thus, in particular, we have $h_{i}>j_{i}$ for each
$i\in\left\{  2,3,\ldots,n\right\}  $. This inequality also holds for $i=1$
(since $h_{1}>j_{1}$), and thus holds for all $i\in\left\{  1,2,\ldots
,n\right\}  $. In other words, we have $h_{i}>j_{i}$ for all $i\in\left\{
1,2,\ldots,n\right\}  $. Hence, for each $i\in\left\{  1,2,\ldots,n\right\}
$, we have $h_{i}\geq j_{i}+1$ (because $h_{i}>j_{i}$, but both $h_{i}$ and
$j_{i}$ are integers). Hence,
\[
\sum_{i=1}^{n}h_{i}\geq\sum_{i=1}^{n}\left(  j_{i}+1\right)  =\underbrace{\sum
_{i=1}^{n}j_{i}}_{=j_{1}+j_{2}+\cdots+j_{n}}+\underbrace{\sum_{i=1}^{n}1}%
_{=n}=\left(  j_{1}+j_{2}+\cdots+j_{n}\right)  +n.
\]
But $\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ and thus
$\left\vert \mathbf{j}\right\vert =j_{1}+j_{2}+\cdots+j_{n}$ (by the
definition of $\left\vert \mathbf{j}\right\vert $). Hence,%
\begin{align*}
\sum_{i=1}^{n}h_{i}  &  \geq\underbrace{\left(  j_{1}+j_{2}+\cdots
+j_{n}\right)  }_{=\left\vert \mathbf{j}\right\vert >\left\vert \mathbf{h}%
\right\vert -n}+n>\left\vert \mathbf{h}\right\vert -n+n=\left\vert
\mathbf{h}\right\vert =h_{1}+h_{2}+\cdots+h_{n+1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\left\vert
\mathbf{h}\right\vert \text{, since }\mathbf{h}=\left(  h_{1},h_{2}%
,\ldots,h_{n+1}\right)  \right) \\
&  =\underbrace{\left(  h_{1}+h_{2}+\cdots+h_{n}\right)  }_{=\sum_{i=1}%
^{n}h_{i}}+\underbrace{h_{n+1}}_{\substack{\geq0\\\text{(since }\left(
h_{1},h_{2},\ldots,h_{n+1}\right)  =\mathbf{h}\in\mathbb{N}^{n+1}\text{)}%
}}\geq\sum_{i=1}^{n}h_{i}.
\end{align*}
This is absurd (since a number cannot be greater than itself). This
contradiction shows that our assumption was false. Qed.}. Combining this with
$u\in\left\{  1,2,\ldots,n+1\right\}  $, we obtain $u\in\left\{
1,2,\ldots,n+1\right\}  \setminus\left\{  1\right\}  =\left\{  2,3,\ldots
,n+1\right\}  $, so that $u-1\in\left\{  1,2,\ldots,n\right\}  \subseteq
\left\{  1,2,\ldots,n+1\right\}  $. Hence, the $n$-tuple $\mathbf{h}%
^{\sim\left(  u-1\right)  }$ is well-defined.

Write the $n$-tuple $\mathbf{h}^{\sim u}$ as
\[
\mathbf{h}^{\sim u}=\left(  h_{1}^{\prime},h_{2}^{\prime},\ldots,h_{n}%
^{\prime}\right)  .
\]
Thus,%
\begin{equation}
\left(  h_{1}^{\prime},h_{2}^{\prime},\ldots,h_{n}^{\prime}\right)
=\mathbf{h}^{\sim u}=\left(  h_{1},h_{2},\ldots,h_{u-1},h_{u+1},\ldots
,h_{n+1}\right)  \label{pf.l2.2b.u-1.pf.1}%
\end{equation}
(by the definition of $\mathbf{h}^{\sim u}$, since $\mathbf{h}=\left(
h_{1},h_{2},\ldots,h_{n+1}\right)  $). Hence,%
\begin{align}
\left(  h_{1}^{\prime},h_{2}^{\prime},\ldots,h_{u-2}^{\prime}\right)   &
=\left(  h_{1},h_{2},\ldots,h_{u-2}\right)  ;\label{pf.l2.2b.u-1.pf.1a}\\
h_{u-1}^{\prime}  &  =h_{u-1};\label{pf.l2.2b.u-1.pf.1b}\\
\left(  h_{u}^{\prime},h_{u+1}^{\prime},\ldots,h_{n}^{\prime}\right)   &
=\left(  h_{u+1},h_{u+2},\ldots,h_{n+1}\right)  \label{pf.l2.2b.u-1.pf.1c}%
\end{align}
(since $u-1\in\left\{  1,2,\ldots,n\right\}  $).

Write the $n$-tuple $\mathbf{h}^{\sim\left(  u-1\right)  }$ as
\[
\mathbf{h}^{\sim\left(  u-1\right)  }=\left(  h_{1}^{\prime\prime}%
,h_{2}^{\prime\prime},\ldots,h_{n}^{\prime\prime}\right)  .
\]
Thus,%
\begin{equation}
\left(  h_{1}^{\prime\prime},h_{2}^{\prime\prime},\ldots,h_{n}^{\prime\prime
}\right)  =\mathbf{h}^{\sim\left(  u-1\right)  }=\left(  h_{1},h_{2}%
,\ldots,h_{u-2},h_{u},\ldots,h_{n+1}\right)  \label{pf.l2.2b.u-1.pf.2}%
\end{equation}
(by the definition of $\mathbf{h}^{\sim\left(  u-1\right)  }$, since
$\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n+1}\right)  $). Hence,%
\begin{align}
\left(  h_{1}^{\prime\prime},h_{2}^{\prime\prime},\ldots,h_{u-2}^{\prime
\prime}\right)   &  =\left(  h_{1},h_{2},\ldots,h_{u-2}\right)
;\label{pf.l2.2b.u-1.pf.2a}\\
h_{u-1}^{\prime\prime}  &  =h_{u};\label{pf.l2.2b.u-1.pf.2b}\\
\left(  h_{u}^{\prime\prime},h_{u+1}^{\prime\prime},\ldots,h_{n}^{\prime
\prime}\right)   &  =\left(  h_{u+1},h_{u+2},\ldots,h_{n+1}\right)
\label{pf.l2.2b.u-1.pf.2c}%
\end{align}
(since $u-1\in\left\{  1,2,\ldots,n\right\}  $).

Comparing (\ref{pf.l2.2b.u-1.pf.1a}) with (\ref{pf.l2.2b.u-1.pf.2a}), we
obtain%
\[
\left(  h_{1}^{\prime},h_{2}^{\prime},\ldots,h_{u-2}^{\prime}\right)  =\left(
h_{1}^{\prime\prime},h_{2}^{\prime\prime},\ldots,h_{u-2}^{\prime\prime
}\right)  .
\]
In other words,%
\begin{equation}
h_{i}^{\prime}=h_{i}^{\prime\prime}\ \ \ \ \ \ \ \ \ \ \text{for each }%
i\in\left\{  1,2,\ldots,u-2\right\}  . \label{pf.l2.2b.u-1.pf.1a=2a}%
\end{equation}
Comparing (\ref{pf.l2.2b.u-1.pf.1c}) with (\ref{pf.l2.2b.u-1.pf.2c}), we
obtain%
\[
\left(  h_{u}^{\prime},h_{u+1}^{\prime},\ldots,h_{n}^{\prime}\right)  =\left(
h_{u}^{\prime\prime},h_{u+1}^{\prime\prime},\ldots,h_{n}^{\prime\prime
}\right)  .
\]
In other words,%
\begin{equation}
h_{i}^{\prime}=h_{i}^{\prime\prime}\ \ \ \ \ \ \ \ \ \ \text{for each }%
i\in\left\{  u,u+1,\ldots,n\right\}  . \label{pf.l2.2b.u-1.pf.1c=2c}%
\end{equation}


We have $\mathbf{j}\oslash\mathbf{h}^{\sim u}$. In view of $\mathbf{j}=\left(
j_{1},j_{2},\ldots,j_{n}\right)  $ and $\mathbf{h}^{\sim u}=\left(
h_{1}^{\prime},h_{2}^{\prime},\ldots,h_{n}^{\prime}\right)  $, this rewrites
as%
\[
\left(  j_{1},j_{2},\ldots,j_{n}\right)  \oslash\left(  h_{1}^{\prime}%
,h_{2}^{\prime},\ldots,h_{n}^{\prime}\right)  .
\]
In other words,%
\[
j_{1}\geq h_{1}^{\prime}>j_{2}\geq h_{2}^{\prime}>\cdots>j_{n}\geq
h_{n}^{\prime}%
\]
(by the definition of the notation \textquotedblleft$\left(  j_{1}%
,j_{2},\ldots,j_{n}\right)  \oslash\left(  h_{1}^{\prime},h_{2}^{\prime
},\ldots,h_{n}^{\prime}\right)  $\textquotedblright). We can break this chain
of inequalities into three sub-chains: Namely,%
\begin{align}
j_{1}  &  \geq h_{1}^{\prime}>j_{2}\geq h_{2}^{\prime}>\cdots>j_{u-2}\geq
h_{u-2}^{\prime}>j_{u-1};\label{pf.l2.2b.u-1.pf.3a}\\
j_{u-1}  &  \geq h_{u-1}^{\prime}>j_{u};\label{pf.l2.2b.u-1.pf.3b}\\
j_{u}  &  \geq h_{u}^{\prime}>j_{u+1}\geq h_{u+1}^{\prime}>\cdots>j_{n}\geq
h_{n}^{\prime}. \label{pf.l2.2b.u-1.pf.3c}%
\end{align}
(Here, of course, the chain of inequalities (\ref{pf.l2.2b.u-1.pf.3a}) is
regarded as vacuously true when $u=2$, and the chain of inequalities
(\ref{pf.l2.2b.u-1.pf.3c}) is regarded as vacuously true when $u=n+1$. We
already know that $u-1\in\left\{  1,2,\ldots,n\right\}  $, so that
(\ref{pf.l2.2b.u-1.pf.3b}) always makes sense.)

Each of the $h_{i}^{\prime}$ appearing in the chain of inequalities
(\ref{pf.l2.2b.u-1.pf.3a}) equals the corresponding $h_{i}^{\prime\prime}$ (by
(\ref{pf.l2.2b.u-1.pf.1a=2a})). Hence, we can rewrite
(\ref{pf.l2.2b.u-1.pf.3a}) as follows:%
\begin{equation}
j_{1}\geq h_{1}^{\prime\prime}>j_{2}\geq h_{2}^{\prime\prime}>\cdots
>j_{u-2}\geq h_{u-2}^{\prime\prime}>j_{u-1}. \label{pf.l2.2b.u-1.pf.4a}%
\end{equation}


From $h_{1}\geq h_{2}\geq\cdots\geq h_{n+1}$, we obtain $h_{u-1}\geq h_{u}$.
From (\ref{pf.l2.2b.u-1.pf.1b}), we obtain $h_{u-1}^{\prime}=h_{u-1}$. Now,
(\ref{pf.l2.2b.u-1.pf.3b}) yields $j_{u-1}\geq h_{u-1}^{\prime}=h_{u-1}\geq
h_{u}$. Thus, $j_{u-1}\geq h_{u}>j_{u}$ (since $j_{u}<h_{u}$). In view of
(\ref{pf.l2.2b.u-1.pf.2b}), this rewrites as%
\begin{equation}
j_{u-1}\geq h_{u-1}^{\prime\prime}>j_{u}. \label{pf.l2.2b.u-1.pf.4b}%
\end{equation}


Each of the $h_{i}^{\prime}$ appearing in the chain of inequalities
(\ref{pf.l2.2b.u-1.pf.3c}) equals the corresponding $h_{i}^{\prime\prime}$ (by
(\ref{pf.l2.2b.u-1.pf.1c=2c})). Hence, we can rewrite
(\ref{pf.l2.2b.u-1.pf.3c}) as follows:%
\begin{equation}
j_{u}\geq h_{u}^{\prime\prime}>j_{u+1}\geq h_{u+1}^{\prime\prime}>\cdots
>j_{n}\geq h_{n}^{\prime\prime}. \label{pf.l2.2b.u-1.pf.4c}%
\end{equation}


We can splice the three chains of inequalities (\ref{pf.l2.2b.u-1.pf.4a}),
(\ref{pf.l2.2b.u-1.pf.4b}) and (\ref{pf.l2.2b.u-1.pf.4c}) together into one
long chain:%
\[
j_{1}\geq h_{1}^{\prime\prime}>j_{2}\geq h_{2}^{\prime\prime}>\cdots>j_{n}\geq
h_{n}^{\prime\prime}.
\]
In other words,%
\[
\left(  j_{1},j_{2},\ldots,j_{n}\right)  \oslash\left(  h_{1}^{\prime\prime
},h_{2}^{\prime\prime},\ldots,h_{n}^{\prime\prime}\right)  .
\]
(by the definition of the notation \textquotedblleft$\left(  j_{1}%
,j_{2},\ldots,j_{n}\right)  \oslash\left(  h_{1}^{\prime\prime},h_{2}%
^{\prime\prime},\ldots,h_{n}^{\prime\prime}\right)  $\textquotedblright). In
view of $\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ and
$\mathbf{h}^{\sim\left(  u-1\right)  }=\left(  h_{1}^{\prime\prime}%
,h_{2}^{\prime\prime},\ldots,h_{n}^{\prime\prime}\right)  $, this rewrites as
$\mathbf{j}\oslash\mathbf{h}^{\sim\left(  u-1\right)  }$.

Also, $j_{u-1}\geq h_{u-1}$ (as we have seen). Now, we know that $u-1$ is an
element of $\left\{  1,2,\ldots,n+1\right\}  $ and satisfies $\mathbf{j}%
\oslash\mathbf{h}^{\sim\left(  u-1\right)  }$ and $j_{u-1}\geq h_{u-1}$. In
other words, $u-1$ is an $i\in\left\{  1,2,\ldots,n+1\right\}  $ satisfying
$\mathbf{j}\oslash\mathbf{h}^{\sim i}$ and $j_{i}\geq h_{i}$. In other words,
$u-1\in\left\{  i\in\left\{  1,2,\ldots,n+1\right\}  \ \mid\ \mathbf{j}%
\oslash\mathbf{h}^{\sim i}\text{ and }j_{i}\geq h_{i}\right\}  $. In view of
(\ref{pf.l2.2b.J=}), this rewrites as $u-1\in J$. This proves
(\ref{pf.l2.2b.u-1}).]

Now, define a map%
\[
\alpha:J\rightarrow H,\ \ \ \ \ \ \ \ \ \ u\mapsto u+1.
\]
(This map is well-defined, due to (\ref{pf.l2.2b.u+1}).)

Also, define a map%
\[
\beta:H\rightarrow J,\ \ \ \ \ \ \ \ \ \ u\mapsto u-1.
\]
(This map is well-defined, due to (\ref{pf.l2.2b.u-1}).)

We have $\alpha\circ\beta=\operatorname*{id}$\ \ \ \ \footnote{\textit{Proof.}
For each $u\in H$, we have%
\begin{align*}
\left(  \alpha\circ\beta\right)  \left(  u\right)   &  =\alpha\left(
\beta\left(  u\right)  \right)  =\underbrace{\beta\left(  u\right)
}_{\substack{=u-1\\\text{(by the definition of }\beta\text{)}}%
}+1\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\alpha\right) \\
&  =\left(  u-1\right)  +1=u=\operatorname*{id}\left(  u\right)  .
\end{align*}
In other words, $\alpha\circ\beta=\operatorname*{id}$.} and $\beta\circ
\alpha=\operatorname*{id}$\ \ \ \ \footnote{\textit{Proof.} For each $u\in J$,
we have%
\begin{align*}
\left(  \beta\circ\alpha\right)  \left(  u\right)   &  =\beta\left(
\alpha\left(  u\right)  \right)  =\underbrace{\alpha\left(  u\right)
}_{\substack{=u+1\\\text{(by the definition of }\alpha\text{)}}%
}-1\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\beta\right) \\
&  =\left(  u+1\right)  -1=u=\operatorname*{id}\left(  u\right)  .
\end{align*}
In other words, $\beta\circ\alpha=\operatorname*{id}$.}. Hence, the maps
$\alpha$ and $\beta$ are mutually inverse. Thus, the map $\alpha$ is
invertible, i.e., is a bijection. Now, (\ref{pf.l2.2b.sum-decomposed}) becomes%
\begin{align*}
\sum_{\substack{i\in\left\{  1,2,\ldots,n+1\right\}  ;\\\mathbf{j}%
\oslash\mathbf{h}^{\sim i}}}\left(  -1\right)  ^{i}  &  =\sum_{i\in J}\left(
-1\right)  ^{i}+\underbrace{\sum_{i\in H}\left(  -1\right)  ^{i}%
}_{\substack{=\sum_{i\in J}\left(  -1\right)  ^{\alpha\left(  i\right)
}\\\text{(here, we have substituted }\alpha\left(  i\right)  \text{ for
}i\\\text{in the sum, since the map }\alpha:J\rightarrow H\text{ is a
bijection)}}}\\
&  =\sum_{i\in J}\left(  -1\right)  ^{i}+\sum_{i\in J}\underbrace{\left(
-1\right)  ^{\alpha\left(  i\right)  }}_{\substack{=\left(  -1\right)
^{i+1}\\\text{(since }\alpha\left(  i\right)  =i+1\\\text{(by the definition
of }\alpha\text{))}}}=\sum_{i\in J}\left(  -1\right)  ^{i}+\sum_{i\in
J}\underbrace{\left(  -1\right)  ^{i+1}}_{=-\left(  -1\right)  ^{i}}\\
&  =\sum_{i\in J}\left(  -1\right)  ^{i}-\sum_{i\in J}\left(  -1\right)
^{i}=0.
\end{align*}
This proves Lemma \ref{lem.2.2b}.
\end{proof}
\end{verlong}

\begin{lemma}
\label{lem.2.2c}Let $\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n+1}\right)
\in\mathbb{N}^{n+1}$ be a nonincreasing tuple. Let $q\in\mathbb{Z}$ be such
that $q<n$. Then,%
\[
\sum_{i=1}^{n+1}\left(  -1\right)  ^{i}s_{h_{i}-q}X^{\mathbf{h}^{\sim i}}=0.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.2.2c}.]The $\left(  n+1\right)  $-tuple $\left(
h_{1},h_{2},\ldots,h_{n+1}\right)  $ is nonincreasing. In other words,
$h_{1}\geq h_{2}\geq\cdots\geq h_{n+1}$.

Let $i\in\left\{  1,2,\ldots,n+1\right\}  $. Then, from $h_{1}\geq h_{2}%
\geq\cdots\geq h_{n+1}$, we obtain $h_{1}\geq h_{2}\geq\cdots\geq h_{i-1}\geq
h_{i+1}\geq\cdots\geq h_{n+1}$.

\begin{vershort}
In other words, the $n$-tuple $\mathbf{h}^{\sim i}$ is nonincreasing (since
\newline$\mathbf{h}^{\sim i}=\left(  h_{1},h_{2},\ldots,h_{i-1},h_{i+1}%
,\ldots,h_{n+1}\right)  $). Moreover, it is easy to see that $\left\vert
\mathbf{h}^{\sim i}\right\vert +h_{i}=\left\vert \mathbf{h}\right\vert $.
\end{vershort}

\begin{verlong}
We have $\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n+1}\right)  $. Hence, the
definition of $\left\vert \mathbf{h}\right\vert $ yields
\begin{align}
\left\vert \mathbf{h}\right\vert  &  =h_{1}+h_{2}+\cdots+h_{n+1}=\sum
_{j\in\left\{  1,2,\ldots,n+1\right\}  }h_{j}\nonumber\\
&  =h_{i}+\sum_{\substack{j\in\left\{  1,2,\ldots,n+1\right\}  ;\\j\neq
i}}h_{j} \label{pf.l2.2c.1}%
\end{align}
(here, we have split off the addend for $j=i$ from the sum).

But the definition of $\mathbf{h}^{\sim i}$ yields $\mathbf{h}^{\sim
i}=\left(  h_{1},h_{2},\ldots,h_{i-1},h_{i+1},\ldots,h_{n+1}\right)  $ (since
$\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n+1}\right)  $). Hence, the
definition of $\left\vert \mathbf{h}^{\sim i}\right\vert $ yields%
\[
\left\vert \mathbf{h}^{\sim i}\right\vert =h_{1}+h_{2}+\cdots+h_{i-1}%
+h_{i+1}+\cdots+h_{n+1}=\sum_{\substack{j\in\left\{  1,2,\ldots,n+1\right\}
;\\j\neq i}}h_{j}.
\]
Hence,
\[
\left\vert \mathbf{h}^{\sim i}\right\vert +h_{i}=\left(  \sum_{\substack{j\in
\left\{  1,2,\ldots,n+1\right\}  ;\\j\neq i}}h_{j}\right)  +h_{i}=h_{i}%
+\sum_{\substack{j\in\left\{  1,2,\ldots,n+1\right\}  ;\\j\neq i}%
}h_{j}=\left\vert \mathbf{h}\right\vert
\]
(by (\ref{pf.l2.2c.1})).

Recall that $h_{1}\geq h_{2}\geq\cdots\geq h_{i-1}\geq h_{i+1}\geq\cdots\geq
h_{n+1}$. In other words, the $n$-tuple $\mathbf{h}^{\sim i}$ is nonincreasing
(by the definition of \textquotedblleft nonincreasing\textquotedblright, since
\newline$\mathbf{h}^{\sim i}=\left(  h_{1},h_{2},\ldots,h_{i-1},h_{i+1}%
,\ldots,h_{n+1}\right)  $).
\end{verlong}

Thus, Lemma \ref{lem.2.2a} (applied to $\mathbf{h}^{\sim i}$ and $h_{i}-q$
instead of $\mathbf{h}$ and $h$) yields%
\begin{equation}
s_{h_{i}-q}X^{\mathbf{h}^{\sim i}}=\sum_{\substack{\mathbf{j}\in
\mathcal{NI};\\\mathbf{j}\oslash\mathbf{h}^{\sim i};\\\left\vert
\mathbf{j}\right\vert =\left\vert \mathbf{h}^{\sim i}\right\vert +h_{i}%
-q}}X^{\mathbf{j}}=\sum_{\substack{\mathbf{j}\in\mathcal{NI};\\\mathbf{j}%
\oslash\mathbf{h}^{\sim i};\\\left\vert \mathbf{j}\right\vert =\left\vert
\mathbf{h}\right\vert -q}}X^{\mathbf{j}} \label{pf.l2.2c.3}%
\end{equation}
(since $\left\vert \mathbf{h}^{\sim i}\right\vert +h_{i}=\left\vert
\mathbf{h}\right\vert $).

Now, forget that we fixed $i$. We thus have proven the equality
(\ref{pf.l2.2c.3}) for each $i\in\left\{  1,2,\ldots,n+1\right\}  $.

Recall that $q<n$; thus, $\left\vert \mathbf{h}\right\vert -q>\left\vert
\mathbf{h}\right\vert -n$. Now,%
\begin{align*}
&  \underbrace{\sum_{i=1}^{n+1}}_{=\sum_{i\in\left\{  1,2,\ldots,n+1\right\}
}}\left(  -1\right)  ^{i}\underbrace{s_{h_{i}-q}X^{\mathbf{h}^{\sim i}}%
}_{\substack{=\sum_{\substack{\mathbf{j}\in\mathcal{NI};\\\mathbf{j}%
\oslash\mathbf{h}^{\sim i};\\\left\vert \mathbf{j}\right\vert =\left\vert
\mathbf{h}\right\vert -q}}X^{\mathbf{j}}\\\text{(by (\ref{pf.l2.2c.3}))}}}\\
&  =\sum_{i\in\left\{  1,2,\ldots,n+1\right\}  }\left(  -1\right)  ^{i}%
\sum_{\substack{\mathbf{j}\in\mathcal{NI};\\\mathbf{j}\oslash\mathbf{h}^{\sim
i};\\\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}\right\vert
-q}}X^{\mathbf{j}}=\underbrace{\sum_{i\in\left\{  1,2,\ldots,n+1\right\}
}\sum_{\substack{\mathbf{j}\in\mathcal{NI};\\\mathbf{j}\oslash\mathbf{h}^{\sim
i};\\\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}\right\vert -q}%
}}_{=\sum_{\substack{\mathbf{j}\in\mathcal{NI};\\\left\vert \mathbf{j}%
\right\vert =\left\vert \mathbf{h}\right\vert -q}}\sum_{\substack{i\in\left\{
1,2,\ldots,n+1\right\}  ;\\\mathbf{j}\oslash\mathbf{h}^{\sim i}}}}\left(
-1\right)  ^{i}X^{\mathbf{j}}\\
&  =\sum_{\substack{\mathbf{j}\in\mathcal{NI};\\\left\vert \mathbf{j}%
\right\vert =\left\vert \mathbf{h}\right\vert -q}}\sum_{\substack{i\in\left\{
1,2,\ldots,n+1\right\}  ;\\\mathbf{j}\oslash\mathbf{h}^{\sim i}}}\left(
-1\right)  ^{i}X^{\mathbf{j}}=\sum_{\substack{\mathbf{j}\in\mathcal{NI}%
;\\\left\vert \mathbf{j}\right\vert =\left\vert \mathbf{h}\right\vert
-q}}\underbrace{\left(  \sum_{\substack{i\in\left\{  1,2,\ldots,n+1\right\}
;\\\mathbf{j}\oslash\mathbf{h}^{\sim i}}}\left(  -1\right)  ^{i}\right)
}_{\substack{=0\\\text{(by Lemma \ref{lem.2.2b}}\\\text{(since }\left\vert
\mathbf{j}\right\vert =\left\vert \mathbf{h}\right\vert -q>\left\vert
\mathbf{h}\right\vert -n\text{))}}}X^{\mathbf{j}}\\
&  =\sum_{\substack{\mathbf{j}\in\mathcal{NI};\\\left\vert \mathbf{j}%
\right\vert =\left\vert \mathbf{h}\right\vert -q}}0X^{\mathbf{j}}=0.
\end{align*}
This proves Lemma \ref{lem.2.2c}.
\end{proof}

\begin{lemma}
\label{lem.2.2d}Let $\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n+1}\right)
\in\mathbb{N}^{n+1}$ be any $\left(  n+1\right)  $-tuple. Let $q\in\mathbb{Z}$
be such that $q<n$. Then,%
\[
\sum_{i=1}^{n+1}\left(  -1\right)  ^{i}s_{h_{i}-q}X^{\mathbf{h}^{\sim i}}=0.
\]

\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.2.2d}.]If we swap two adjacent entries of the
$\left(  n+1\right)  $-tuple $\mathbf{h}$ (say, the $r$-th and the $\left(
r+1\right)  $-st entry, where $r$ is some element of $\left\{  1,2,\ldots
,n\right\}  $), then the sum $\sum_{i=1}^{n+1}\left(  -1\right)  ^{i}%
s_{h_{i}-q}X^{\mathbf{h}^{\sim i}}$ flips its sign (indeed, all addends of
this sum except for the $r$-th and $\left(  r+1\right)  $-st one flip their
sign, whereas the $r$-th and the $\left(  r+1\right)  $-th addends trade
places and also flip their signs). Therefore, the sum $\sum_{i=1}^{n+1}\left(
-1\right)  ^{i}s_{h_{i}-q}X^{\mathbf{h}^{\sim i}}$ is anti-symmetric in the
$h_{i}$ (since any permutation of the entries of the $\left(  n+1\right)
$-tuple $\mathbf{h}$ can be achieved by repeatedly swapping adjacent entries).
Hence, in proving that this sum equals $0$, we can WLOG assume that $h_{1}\geq
h_{2}\geq\cdots\geq h_{n+1}$. Assume this. In other words, the $\left(
n+1\right)  $-tuple $\mathbf{h}$ is nonincreasing (since $\mathbf{h}=\left(
h_{1},h_{2},\ldots,h_{n+1}\right)  $). Hence, Lemma \ref{lem.2.2c} yields
$\sum_{i=1}^{n+1}\left(  -1\right)  ^{i}s_{h_{i}-q}X^{\mathbf{h}^{\sim i}}=0$.
This proves Lemma \ref{lem.2.2d}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.2.2d}.]We make the following two observations:

\begin{itemize}
\item If we swap two adjacent entries of the $\left(  n+1\right)  $-tuple
$\mathbf{h}$ (say, the $r$-th and the $\left(  r+1\right)  $-st entry, where
$r$ is some element of $\left\{  1,2,\ldots,n\right\}  $), then the sum
$\sum_{i=1}^{n+1}\left(  -1\right)  ^{i}s_{h_{i}-q}X^{\mathbf{h}^{\sim i}}$
flips its sign (indeed, all addends of this sum except for the $r$-th and
$\left(  r+1\right)  $-st one flip their sign, whereas the $r$-th and the
$\left(  r+1\right)  $-th addends trade places and also flip their signs).

\item If two adjacent entries of the $\left(  n+1\right)  $-tuple $\mathbf{h}$
(say, the $r$-th and the $\left(  r+1\right)  $-st entry, where $r$ is some
element of $\left\{  1,2,\ldots,n\right\}  $) are equal, then the sum
$\sum_{i=1}^{n+1}\left(  -1\right)  ^{i}s_{h_{i}-q}X^{\mathbf{h}^{\sim i}}$
vanishes (because all addends of this sum except for the $r$-th and $\left(
r+1\right)  $-st one vanish, whereas the $r$-th and the $\left(  r+1\right)
$-th addends cancel one another).
\end{itemize}

\noindent These two observations show that the sum $\sum_{i=1}^{n+1}\left(
-1\right)  ^{i}s_{h_{i}-q}X^{\mathbf{h}^{\sim i}}$ is alternating in the
$h_{i}$. Hence, in proving that this sum equals $0$, we can WLOG assume that
$h_{1}>h_{2}>\cdots>h_{n+1}$. Assume this. Thus, $h_{1}\geq h_{2}\geq
\cdots\geq h_{n+1}$. In other words, the $\left(  n+1\right)  $-tuple
$\mathbf{h}$ is nonincreasing (by the definition of \textquotedblleft
nonincreasing\textquotedblright, since $\mathbf{h}=\left(  h_{1},h_{2}%
,\ldots,h_{n+1}\right)  $). Hence, Lemma \ref{lem.2.2c} yields $\sum
_{i=1}^{n+1}\left(  -1\right)  ^{i}s_{h_{i}-q}X^{\mathbf{h}^{\sim i}}=0$. This
proves Lemma \ref{lem.2.2d}.
\end{proof}
\end{verlong}

\begin{proof}
[Alternative proof of Corollary 2.2.]Let $m\in\left\{  1,2,\ldots,n\right\}  $
and $\left(  h_{1},h_{2},\ldots,h_{m}\right)  \in\mathbb{N}^{m}$. Let us
extend the $m$-tuple $\left(  h_{1},h_{2},\ldots,h_{m}\right)  \in
\mathbb{N}^{m}$ to an $\left(  n+1\right)  $-tuple $\left(  h_{1},h_{2}%
,\ldots,h_{n+1}\right)  \in\mathbb{Z}^{n+1}$ by setting
\begin{equation}
\left(  h_{i}=n+1-i\ \ \ \ \ \ \ \ \ \ \text{for each }i>m\right)
.\label{p834.pf.c2.2.hi=n-i}%
\end{equation}


\begin{vershort}
Thus,%
\[
\left(  h_{1},h_{2},\ldots,h_{n+1}\right)  =\left(  h_{1},h_{2},\ldots
,h_{m},n-m,n-m-1,\ldots,0\right)  \in\mathbb{N}^{n+1}.
\]
Denote this $\left(  n+1\right)  $-tuple $\left(  h_{1},h_{2},\ldots
,h_{n+1}\right)  \in\mathbb{N}^{n+1}$ by $\mathbf{h}$. Thus, $\mathbf{h}%
=\left(  h_{1},h_{2},\ldots,h_{n+1}\right)  $. From (\ref{p834.pf.c2.2.hi=n-i}%
), we also obtain%
\begin{align*}
\left(  h_{m+1},h_{m+2},\ldots,h_{n+1}\right)   &  =\left(  n-m,n-m-1,\ldots
,0\right)  \ \ \ \ \ \ \ \ \ \ \text{and thus}\\
\left(  h_{m+2},h_{m+3},\ldots,h_{n+1}\right)   &  =\left(  n-m-1,n-m-2,\ldots
,0\right)  .
\end{align*}


We have $m\leq n\leq n+1$. Furthermore, $m-n\leq0$ (since $m\leq n$), so that
$0\in\left\{  m-n,m-n+1,\ldots,0\right\}  $.

Also, $n-m<n$ (since $m>0$). Hence, Lemma \ref{lem.2.2d} (applied to $q=n-m$)
yields
\[
\sum_{i=1}^{n+1}\left(  -1\right)  ^{i}s_{h_{i}-\left(  n-m\right)
}X^{\mathbf{h}^{\sim i}}=0.
\]
Hence,%
\[
0=\sum_{i=1}^{n+1}\left(  -1\right)  ^{i}s_{h_{i}-\left(  n-m\right)
}X^{\mathbf{h}^{\sim i}}=\sum_{i=1}^{m}\left(  -1\right)  ^{i}s_{h_{i}-\left(
n-m\right)  }X^{\mathbf{h}^{\sim i}}+\sum_{i=m+1}^{n+1}\left(  -1\right)
^{i}s_{h_{i}-\left(  n-m\right)  }X^{\mathbf{h}^{\sim i}}%
\]
(here, we have split the sum at $i=m+1$, since $0\leq m\leq n+1$). Thus,%
\begin{equation}
\sum_{i=m+1}^{n+1}\left(  -1\right)  ^{i}s_{h_{i}-\left(  n-m\right)
}X^{\mathbf{h}^{\sim i}}=-\sum_{i=1}^{m}\left(  -1\right)  ^{i}s_{h_{i}%
-\left(  n-m\right)  }X^{\mathbf{h}^{\sim i}}.\label{p834.pf.c2.2.short.3}%
\end{equation}


But every $i\in\left\{  m+1,m+2,\ldots,n+1\right\}  $ satisfies $i>m$ and thus
$h_{i}=n+1-i$ (by (\ref{p834.pf.c2.2.hi=n-i})) and therefore%
\begin{equation}
\underbrace{h_{i}}_{=n+1-i}-\left(  n-m\right)  =\left(  n+1-i\right)
-\left(  n-m\right)  =m+1-i.\label{p834.pf.c2.2.short.4}%
\end{equation}


Hence,%
\begin{align*}
&  \sum_{i=m+1}^{n+1}\left(  -1\right)  ^{i}\underbrace{s_{h_{i}-\left(
n-m\right)  }}_{\substack{=s_{m+1-i}\\\text{(by (\ref{p834.pf.c2.2.short.4}%
))}}}X^{\mathbf{h}^{\sim i}}\\
&  =\sum_{i=m+1}^{n+1}\left(  -1\right)  ^{i}s_{m+1-i}X^{\mathbf{h}^{\sim i}%
}=\sum_{i=m-n}^{0}\left(  -1\right)  ^{m+1-i}s_{i}X^{\mathbf{h}^{\sim\left(
m+1-i\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }m+1-i\text{
for }i\text{ in the sum}\right)  \\
&  =\underbrace{\left(  -1\right)  ^{m+1-0}}_{=\left(  -1\right)  ^{m+1}%
}\underbrace{s_{0}}_{=1}X^{\mathbf{h}^{\sim\left(  m+1-0\right)  }}%
+\sum_{i=m-n}^{-1}\left(  -1\right)  ^{m+1-i}\underbrace{s_{i}}%
_{\substack{=0\\\text{(since }i\leq-1<0\text{)}}}X^{\mathbf{h}^{\sim\left(
m+1-i\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }i=0\text{ from the sum,}\\
\text{since }0\in\left\{  m-n,m-n+1,\ldots,0\right\}
\end{array}
\right)  \\
&  =\left(  -1\right)  ^{m+1}X^{\mathbf{h}^{\sim\left(  m+1-0\right)  }%
}+\underbrace{\sum_{i=m-n}^{-1}\left(  -1\right)  ^{m+1-i}0X^{\mathbf{h}%
^{\sim\left(  m+1-i\right)  }}}_{=0}=\left(  -1\right)  ^{m+1}X^{\mathbf{h}%
^{\sim\left(  m+1-0\right)  }}.
\end{align*}
Comparing this equality with (\ref{p834.pf.c2.2.short.3}), we find%
\[
\left(  -1\right)  ^{m+1}X^{\mathbf{h}^{\sim\left(  m+1-0\right)  }}%
=-\sum_{i=1}^{m}\left(  -1\right)  ^{i}s_{h_{i}-\left(  n-m\right)
}X^{\mathbf{h}^{\sim i}}.
\]
Multiplying both sides of this equality with $\left(  -1\right)  ^{m+1}$, we
find%
\begin{align}
X^{\mathbf{h}^{\sim\left(  m+1-0\right)  }} &  =\underbrace{-\left(
-1\right)  ^{m+1}}_{=\left(  -1\right)  ^{m}}\sum_{i=1}^{m}\left(  -1\right)
^{i}\underbrace{s_{h_{i}-\left(  n-m\right)  }}_{=s_{h_{i}-n+m}}%
X^{\mathbf{h}^{\sim i}}=\left(  -1\right)  ^{m}\sum_{i=1}^{m}\left(
-1\right)  ^{i}s_{h_{i}-n+m}X^{\mathbf{h}^{\sim i}}\nonumber\\
&  =\sum_{i=1}^{m}\left(  -1\right)  ^{m+i}s_{h_{i}-n+m}X^{\mathbf{h}^{\sim
i}}.\label{p834.pf.c2.2.short.7}%
\end{align}
But%
\[
\mathbf{h}^{\sim\left(  m+1-0\right)  }=\mathbf{h}^{\sim\left(  m+1\right)
}=\left(  h_{1},h_{2},\ldots,h_{m},h_{m+2},h_{m+3},\ldots,h_{n+1}\right)
\]
(since $\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n+1}\right)  $). Hence, the
definition of $X^{\mathbf{h}^{\sim\left(  m+1-0\right)  }}$ yields%
\begin{align*}
X^{\mathbf{h}^{\sim\left(  m+1-0\right)  }} &  =X^{h_{1}}\wedge X^{h_{2}%
}\wedge\cdots\wedge X^{h_{m}}\wedge\underbrace{X^{h_{m+2}}\wedge X^{h_{m+3}%
}\wedge\cdots\wedge X^{h_{n+1}}}_{\substack{=X^{n-m-1}\wedge X^{n-m-2}%
\wedge\cdots\wedge X^{0}\\\text{(since }\left(  h_{m+2},h_{m+3},\ldots
,h_{n+1}\right)  =\left(  n-m-1,n-m-2,\ldots,0\right)  \text{)}}}\\
&  =X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{m}}\wedge X^{n-m-1}%
\wedge X^{n-m-2}\wedge\cdots\wedge X^{0}.
\end{align*}
Comparing this with (\ref{p834.pf.c2.2.short.7}), we obtain%
\begin{align}
&  X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{m}}\wedge X^{n-m-1}\wedge
X^{n-m-2}\wedge\cdots\wedge X^{0}\nonumber\\
&  =\sum_{i=1}^{m}\left(  -1\right)  ^{m+i}s_{h_{i}-n+m}X^{\mathbf{h}^{\sim
i}}.\label{p834.pf.c2.2.short.9}%
\end{align}


Also, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\[
\mathbf{h}^{\sim i}=\left(  h_{1},h_{2},\ldots,h_{i-1},h_{i+1},\ldots
,h_{n+1}\right)
\]
(since $\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n+1}\right)  $) and therefore%
\begin{align*}
X^{\mathbf{h}^{\sim i}} &  =X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge
X^{h_{i-1}}\wedge X^{h_{i+1}}\wedge\cdots\wedge X^{h_{n+1}}\\
&  =X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{i-1}}\wedge X^{h_{i+1}%
}\wedge\cdots\wedge X^{h_{m}}\wedge\underbrace{X^{h_{m+1}}\wedge X^{h_{m+2}%
}\wedge\cdots\wedge X^{h_{n+1}}}_{\substack{=X^{n-m}\wedge X^{n-m-1}%
\wedge\cdots\wedge X^{0}\\\text{(since }\left(  h_{m+1},h_{m+2},\ldots
,h_{n+1}\right)  =\left(  n-m,n-m-1,\ldots,0\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\leq m\right)  \\
&  =X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{i-1}}\wedge X^{h_{i+1}%
}\wedge\cdots\wedge X^{h_{m}}\wedge X^{n-m}\wedge X^{n-m-1}\wedge\cdots\wedge
X^{0}.
\end{align*}
Hence, (\ref{p834.pf.c2.2.short.9}) rewrites as%
\begin{align*}
&  X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{m}}\wedge X^{n-m-1}\wedge
X^{n-m-2}\wedge\cdots\wedge X^{0}\\
&  =\sum_{i=1}^{m}\left(  -1\right)  ^{m+i}s_{h_{i}-n+m}\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  X^{h_{1}}\wedge X^{h_{2}}\wedge
\cdots\wedge X^{h_{i-1}}\wedge X^{h_{i+1}}\wedge\cdots\wedge X^{h_{m}}\wedge
X^{n-m}\wedge X^{n-m-1}\wedge\cdots\wedge X^{0}\right)  .
\end{align*}
This proves Corollary 2.2.
\end{vershort}

\begin{verlong}
\noindent Thus,%
\begin{align}
&  \left(  h_{1},h_{2},\ldots,h_{n+1}\right)  \nonumber\\
&  =\left(  h_{1},h_{2},\ldots,h_{m},\left(  n+1\right)  -\left(  m+1\right)
,\left(  n+1\right)  -\left(  m+2\right)  ,\ldots,\left(  n+1\right)  -\left(
n+1\right)  \right)  \nonumber\\
&  =\left(  h_{1},h_{2},\ldots,h_{m},n-m,n-m-1,\ldots,0\right)
.\label{p834.pf.c2.2.1}%
\end{align}
Thus, $\left(  h_{1},h_{2},\ldots,h_{n+1}\right)  \in\mathbb{N}^{n+1}$ (since
$\left(  h_{1},h_{2},\ldots,h_{m}\right)  \in\mathbb{N}^{m}$ and since all of
the new entries $n-m,n-m-1,\ldots,0$ belong to $\mathbb{N}$). Denote this
$\left(  n+1\right)  $-tuple $\left(  h_{1},h_{2},\ldots,h_{n+1}\right)
\in\mathbb{N}^{n+1}$ by $\mathbf{h}$. Thus, $\mathbf{h}=\left(  h_{1}%
,h_{2},\ldots,h_{n+1}\right)  $.

From (\ref{p834.pf.c2.2.1}), we obtain
\begin{align*}
\left(  h_{m+1},h_{m+2},\ldots,h_{n+1}\right)   &  =\left(  n-m,n-m-1,\ldots
,0\right)  \ \ \ \ \ \ \ \ \ \ \text{and thus}\\
\left(  h_{m+2},h_{m+3},\ldots,h_{n+1}\right)   &  =\left(  n-m-1,n-m-2,\ldots
,0\right)  .
\end{align*}


We have $m\leq n\leq n+1$. Furthermore, $m-n\leq0$ (since $m\leq n$), so that
$0\in\left\{  m-n,m-n+1,\ldots,0\right\}  $.

Also, we have $m>0$ and thus $n-\underbrace{m}_{>0}<n$. Hence, Lemma
\ref{lem.2.2d} (applied to $q=n-m$) yields
\[
\sum_{i=1}^{n+1}\left(  -1\right)  ^{i}s_{h_{i}-\left(  n-m\right)
}X^{\mathbf{h}^{\sim i}}=0.
\]
Hence,%
\[
0=\sum_{i=1}^{n+1}\left(  -1\right)  ^{i}s_{h_{i}-\left(  n-m\right)
}X^{\mathbf{h}^{\sim i}}=\sum_{i=1}^{m}\left(  -1\right)  ^{i}s_{h_{i}-\left(
n-m\right)  }X^{\mathbf{h}^{\sim i}}+\sum_{i=m+1}^{n+1}\left(  -1\right)
^{i}s_{h_{i}-\left(  n-m\right)  }X^{\mathbf{h}^{\sim i}}%
\]
(here, we have split the sum at $i=m+1$, since $0\leq m\leq n+1$). Thus,%
\begin{equation}
\sum_{i=m+1}^{n+1}\left(  -1\right)  ^{i}s_{h_{i}-\left(  n-m\right)
}X^{\mathbf{h}^{\sim i}}=-\sum_{i=1}^{m}\left(  -1\right)  ^{i}s_{h_{i}%
-\left(  n-m\right)  }X^{\mathbf{h}^{\sim i}}.\label{p834.pf.c2.2.3}%
\end{equation}


But every $i\in\left\{  m+1,m+2,\ldots,n+1\right\}  $ satisfies $i\geq m+1>m$
and thus $h_{i}=n+1-i$ (by (\ref{p834.pf.c2.2.hi=n-i})) and therefore%
\begin{equation}
\underbrace{h_{i}}_{=n+1-i}-\left(  n-m\right)  =\left(  n+1-i\right)
-\left(  n-m\right)  =m+1-i.\label{p834.pf.c2.2.4}%
\end{equation}
Hence,%
\begin{align*}
&  \sum_{i=m+1}^{n+1}\left(  -1\right)  ^{i}\underbrace{s_{h_{i}-\left(
n-m\right)  }}_{\substack{=s_{m+1-i}\\\text{(by (\ref{p834.pf.c2.2.4}))}%
}}X^{\mathbf{h}^{\sim i}}\\
&  =\sum_{i=m+1}^{n+1}\left(  -1\right)  ^{i}s_{m+1-i}X^{\mathbf{h}^{\sim i}%
}=\underbrace{\sum_{i=\left(  m+1\right)  -\left(  n+1\right)  }^{0}}%
_{=\sum_{i=m-n}^{0}}\left(  -1\right)  ^{m+1-i}\underbrace{s_{m+1-\left(
m+1-i\right)  }}_{\substack{=s_{i}\\\text{(since }m+1-\left(  m+1-i\right)
=i\text{)}}}X^{\mathbf{h}^{\sim\left(  m+1-i\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }m+1-i\text{
for }i\text{ in the sum}\right)  \\
&  =\sum_{i=m-n}^{0}\left(  -1\right)  ^{m+1-i}s_{i}X^{\mathbf{h}^{\sim\left(
m+1-i\right)  }}\\
&  =\underbrace{\left(  -1\right)  ^{m+1-0}}_{=\left(  -1\right)  ^{m+1}%
}\underbrace{s_{0}}_{=1}X^{\mathbf{h}^{\sim\left(  m+1-0\right)  }}%
+\sum_{i=m-n}^{-1}\left(  -1\right)  ^{m+1-i}\underbrace{s_{i}}%
_{\substack{=0\\\text{(since }i\leq-1<0\text{)}}}X^{\mathbf{h}^{\sim\left(
m+1-i\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }i=0\text{ from the sum,}\\
\text{since }0\in\left\{  m-n,m-n+1,\ldots,0\right\}
\end{array}
\right)  \\
&  =\left(  -1\right)  ^{m+1}X^{\mathbf{h}^{\sim\left(  m+1-0\right)  }%
}+\underbrace{\sum_{i=m-n}^{-1}\left(  -1\right)  ^{m+1-i}0X^{\mathbf{h}%
^{\sim\left(  m+1-i\right)  }}}_{=0}=\left(  -1\right)  ^{m+1}X^{\mathbf{h}%
^{\sim\left(  m+1-0\right)  }}.
\end{align*}
Comparing this equality with (\ref{p834.pf.c2.2.3}), we find%
\[
\left(  -1\right)  ^{m+1}X^{\mathbf{h}^{\sim\left(  m+1-0\right)  }}%
=-\sum_{i=1}^{m}\left(  -1\right)  ^{i}s_{h_{i}-\left(  n-m\right)
}X^{\mathbf{h}^{\sim i}}.
\]
Multiplying both sides of this equality with $\left(  -1\right)  ^{m+1}$, we
find%
\begin{align*}
&  \left(  -1\right)  ^{m+1}\left(  -1\right)  ^{m+1}X^{\mathbf{h}%
^{\sim\left(  m+1-0\right)  }}\\
&  =\underbrace{-\left(  -1\right)  ^{m+1}}_{=\left(  -1\right)  ^{m}}%
\sum_{i=1}^{m}\left(  -1\right)  ^{i}s_{h_{i}-\left(  n-m\right)
}X^{\mathbf{h}^{\sim i}}\\
&  =\left(  -1\right)  ^{m}\sum_{i=1}^{m}\left(  -1\right)  ^{i}%
s_{h_{i}-\left(  n-m\right)  }X^{\mathbf{h}^{\sim i}}=\sum_{i=1}%
^{m}\underbrace{\left(  -1\right)  ^{m}\left(  -1\right)  ^{i}}_{=\left(
-1\right)  ^{m+i}}\underbrace{s_{h_{i}-\left(  n-m\right)  }}%
_{\substack{=s_{h_{i}-n+m}\\\text{(since }h_{i}-\left(  n-m\right)
=h_{i}-n+m\text{)}}}X^{\mathbf{h}^{\sim i}}\\
&  =\sum_{i=1}^{m}\left(  -1\right)  ^{m+i}s_{h_{i}-n+m}X^{\mathbf{h}^{\sim
i}}.
\end{align*}
Comparing this with%
\[
\underbrace{\left(  -1\right)  ^{m+1}\left(  -1\right)  ^{m+1}}%
_{\substack{=\left(  \left(  -1\right)  ^{m+1}\right)  ^{2}=\left(  -1\right)
^{\left(  m+1\right)  \cdot2}=1\\\text{(since }\left(  m+1\right)
\cdot2\text{ is even)}}}X^{\mathbf{h}^{\sim\left(  m+1-0\right)  }%
}=X^{\mathbf{h}^{\sim\left(  m+1-0\right)  }},
\]
we obtain%
\begin{equation}
X^{\mathbf{h}^{\sim\left(  m+1-0\right)  }}=\sum_{i=1}^{m}\left(  -1\right)
^{m+i}s_{h_{i}-n+m}X^{\mathbf{h}^{\sim i}}.\label{p834.pf.c2.2.7}%
\end{equation}
But%
\[
\mathbf{h}^{\sim\left(  m+1-0\right)  }=\mathbf{h}^{\sim\left(  m+1\right)
}=\left(  h_{1},h_{2},\ldots,h_{m},h_{m+2},h_{m+3},\ldots,h_{n+1}\right)
\]
(by the definition of $\mathbf{h}^{\sim\left(  m+1\right)  }$, since
$\mathbf{h}=\left(  h_{1},h_{2},\ldots,h_{n+1}\right)  $). Hence, the
definition of $X^{\mathbf{h}^{\sim\left(  m+1-0\right)  }}$ yields%
\begin{align*}
X^{\mathbf{h}^{\sim\left(  m+1-0\right)  }} &  =X^{h_{1}}\wedge X^{h_{2}%
}\wedge\cdots\wedge X^{h_{m}}\wedge\underbrace{X^{h_{m+2}}\wedge X^{h_{m+3}%
}\wedge\cdots\wedge X^{h_{n+1}}}_{\substack{=X^{n-m-1}\wedge X^{n-m-2}%
\wedge\cdots\wedge X^{0}\\\text{(since }\left(  h_{m+2},h_{m+3},\ldots
,h_{n+1}\right)  =\left(  n-m-1,n-m-2,\ldots,0\right)  \text{)}}}\\
&  =X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{m}}\wedge X^{n-m-1}%
\wedge X^{n-m-2}\wedge\cdots\wedge X^{0}.
\end{align*}
Comparing this with (\ref{p834.pf.c2.2.7}), we obtain%
\begin{align}
&  X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{m}}\wedge X^{n-m-1}\wedge
X^{n-m-2}\wedge\cdots\wedge X^{0}\nonumber\\
&  =\sum_{i=1}^{m}\left(  -1\right)  ^{m+i}s_{h_{i}-n+m}X^{\mathbf{h}^{\sim
i}}.\label{p834.pf.c2.2.9}%
\end{align}


Also, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\[
\mathbf{h}^{\sim i}=\left(  h_{1},h_{2},\ldots,h_{i-1},h_{i+1},\ldots
,h_{n+1}\right)
\]
(by the definition of $\mathbf{h}^{\sim i}$, since $\mathbf{h}=\left(
h_{1},h_{2},\ldots,h_{n+1}\right)  $) and therefore%
\begin{align*}
&  X^{\mathbf{h}^{\sim i}}\\
&  =X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{i-1}}\wedge X^{h_{i+1}%
}\wedge\cdots\wedge X^{h_{n+1}}\\
&  =X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{i-1}}\wedge X^{h_{i+1}%
}\wedge\cdots\wedge X^{h_{m}}\wedge\underbrace{X^{h_{m+1}}\wedge X^{h_{m+2}%
}\wedge\cdots\wedge X^{h_{n+1}}}_{\substack{=X^{n-m}\wedge X^{n-m-1}%
\wedge\cdots\wedge X^{0}\\\text{(since }\left(  h_{m+1},h_{m+2},\ldots
,h_{n+1}\right)  =\left(  n-m,n-m-1,\ldots,0\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\leq m\right)  \\
&  =X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{i-1}}\wedge X^{h_{i+1}%
}\wedge\cdots\wedge X^{h_{m}}\wedge X^{n-m}\wedge X^{n-m-1}\wedge\cdots\wedge
X^{0}.
\end{align*}
Hence, (\ref{p834.pf.c2.2.9}) rewrites as%
\begin{align*}
&  X^{h_{1}}\wedge X^{h_{2}}\wedge\cdots\wedge X^{h_{m}}\wedge X^{n-m-1}\wedge
X^{n-m-2}\wedge\cdots\wedge X^{0}\\
&  =\sum_{i=1}^{m}\left(  -1\right)  ^{m+i}s_{h_{i}-n+m}\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  X^{h_{1}}\wedge X^{h_{2}}\wedge
\cdots\wedge X^{h_{i-1}}\wedge X^{h_{i+1}}\wedge\cdots\wedge X^{h_{m}}\wedge
X^{n-m}\wedge X^{n-m-1}\wedge\cdots\wedge X^{0}\right)  .
\end{align*}
This proves Corollary 2.2.
\end{verlong}
\end{proof}
\end{fineprint}

\subsection{\label{sect.pf.2.3}An alternative proof of Lemma 2.3}

\begin{fineprint}
\begin{proof}
[Alternative proof of Lemma 2.3.]We must prove that each $f\in S$ satisfying
\newline$f\cdot\left(  X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge X^{0}\right)
=0$ satisfies $f=0$. So let $f\in S$ be such that \newline$f\cdot\left(
X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge X^{0}\right)  =0$. We must prove that
$f=0$.

Consider the alternator map $\operatorname*{alt}:\bigwedge\nolimits_{A}%
^{n}A\left[  X\right]  \rightarrow A\left[  X_{1},X_{2},\ldots,X_{n}\right]
$. This map $\operatorname*{alt}$ is $S$-linear (by Proposition 1.3). Hence,%
\begin{align}
f\cdot\operatorname*{alt}\left(  X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge
X^{0}\right)   &  =\operatorname*{alt}\left(  \underbrace{f\cdot\left(
X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge X^{0}\right)  }_{=0}\right)
\nonumber\\
&  =\operatorname*{alt}0=0. \label{pf.l2.3.alt.0}%
\end{align}
But the definition of $\operatorname*{alt}$ yields\footnote{Here, we are using
the notation $\left(  -1\right)  ^{\sigma}$ for the sign of a permutation
$\sigma$.}
\begin{align}
&  \operatorname*{alt}\left(  X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge
X^{0}\right) \nonumber\\
&  =\sum_{\sigma\in\mathfrak{S}_{n}}\left(  -1\right)  ^{\sigma}%
X_{\sigma\left(  1\right)  }^{n-1}X_{\sigma\left(  2\right)  }^{n-2}\cdots
X_{\sigma\left(  n\right)  }^{n-n}. \label{pf.l2.3.alt.1}%
\end{align}
Now, we claim that this element $\operatorname*{alt}\left(  X^{n-1}\wedge
X^{n-2}\wedge\cdots\wedge X^{0}\right)  $ of $A\left[  X_{1},X_{2}%
,\ldots,X_{n}\right]  $ is regular (i.e., not a zero-divisor). Here are two
ways to prove this:

[\textit{First proof of the fact that }$\operatorname*{alt}\left(
X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge X^{0}\right)  $ \textit{is regular:}
Equip the set $\mathbb{N}^{n}$ with the lexicographic order; this is the total
order in which
\begin{align*}
&  \left(  h_{1},h_{2},\ldots,h_{n}\right)  >\left(  k_{1},k_{2},\ldots
,k_{n}\right)  \ \ \ \ \ \ \ \ \ \ \text{if and only if}\\
&  \text{the first non-zero term in the sequence }h_{1}-k_{1},h_{2}%
-k_{2},\ldots,h_{n}-k_{n}\text{ is positive.}%
\end{align*}
Now, (\ref{pf.l2.3.alt.1}) yields
\begin{align}
&  \operatorname*{alt}\left(  X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge
X^{0}\right) \nonumber\\
&  =\sum_{\sigma\in\mathfrak{S}_{n}}\left(  -1\right)  ^{\sigma}%
X_{\sigma\left(  1\right)  }^{n-1}X_{\sigma\left(  2\right)  }^{n-2}\cdots
X_{\sigma\left(  n\right)  }^{n-n}\nonumber\\
&  =X_{1}^{n-1}X_{2}^{n-2}\cdots X_{n}^{n-n}+\left(  \text{lower order
terms}\right)  , \label{pf.l2.3.alt.reg.1st.1}%
\end{align}
where \textquotedblleft$\left(  \text{lower order terms}\right)
$\textquotedblright\ means an $A$-linear combination of monomials
$X_{1}^{v_{1}}X_{2}^{v_{2}}\cdots X_{n}^{v_{n}}$ with $\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  <\left(  n-1,n-2,\ldots,n-n\right)  $. (Of course,
the \textquotedblleft$<$\textquotedblright\ sign here refers to the
lexicographic order on $\mathbb{N}^{n}$.)

On the other hand, let $g\in A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ be
nonzero. Then, $g$ has at least one nonzero coefficient. Hence, we can find
some nonzero $c\in A$ and some $\left(  g_{1},g_{2},\ldots,g_{n}\right)
\in\mathbb{N}^{n}$ such that%
\begin{equation}
g=cX_{1}^{g_{1}}X_{2}^{g_{2}}\cdots X_{n}^{g_{n}}+\left(  \text{lower order
terms}\right)  , \label{pf.l2.3.alt.reg.1st.2}%
\end{equation}
where \textquotedblleft$\left(  \text{lower order terms}\right)
$\textquotedblright\ means an $A$-linear combination of monomials
$X_{1}^{u_{1}}X_{2}^{u_{2}}\cdots X_{n}^{u_{n}}$ with $\left(  u_{1}%
,u_{2},\ldots,u_{n}\right)  <\left(  g_{1},g_{2},\ldots,g_{n}\right)  $.
Consider this $c$ and this $\left(  g_{1},g_{2},\ldots,g_{n}\right)  $.

But it is easy to see that the lexicographic order on $\mathbb{N}^{n}$
respects entrywise addition of $n$-tuples in $\mathbb{N}^{n}$ (which, of
course, corresponds to multiplication of monomials in $A\left[  X_{1}%
,X_{2},\ldots,X_{n}\right]  $). To be more precise: If four $n$-tuples
\[
\left(  u_{1},u_{2},\ldots,u_{n}\right)  ,\ \left(  p_{1},p_{2},\ldots
,p_{n}\right)  ,\ \left(  v_{1},v_{2},\ldots,v_{n}\right)  \text{ and }\left(
q_{1},q_{2},\ldots,q_{n}\right)
\]
in $\mathbb{N}^{n}$ satisfy%
\begin{align}
\left(  u_{1},u_{2},\ldots,u_{n}\right)   &  \leq\left(  p_{1},p_{2}%
,\ldots,p_{n}\right)  \text{ and}\label{pf.l2.3.alt.reg.1st.3}\\
\left(  v_{1},v_{2},\ldots,v_{n}\right)   &  \leq\left(  q_{1},q_{2}%
,\ldots,q_{n}\right)  , \label{pf.l2.3.alt.reg.1st.4}%
\end{align}
then we have%
\[
\left(  u_{1},u_{2},\ldots,u_{n}\right)  +\left(  v_{1},v_{2},\ldots
,v_{n}\right)  \leq\left(  p_{1},p_{2},\ldots,p_{n}\right)  +\left(
q_{1},q_{2},\ldots,q_{n}\right)
\]
(where the addition of $n$-tuples is entrywise), and this inequality becomes
an equality only when both (\ref{pf.l2.3.alt.reg.1st.3}) and
(\ref{pf.l2.3.alt.reg.1st.4}) become equalities.

Hence, if we multiply the equalities (\ref{pf.l2.3.alt.reg.1st.2}) and
(\ref{pf.l2.3.alt.reg.1st.1}), then we find%
\begin{align*}
&  g\cdot\operatorname*{alt}\left(  X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge
X^{0}\right) \\
&  =cX_{1}^{g_{1}+\left(  n-1\right)  }X_{2}^{g_{2}+\left(  n-2\right)
}\cdots X_{n}^{g_{n}+\left(  n-n\right)  }+\left(  \text{lower order
terms}\right)  ,
\end{align*}
where \textquotedblleft$\left(  \text{lower order terms}\right)
$\textquotedblright\ means an $A$-linear combination of monomials
$X_{1}^{u_{1}}X_{2}^{u_{2}}\cdots X_{n}^{u_{n}}$ with $\left(  u_{1}%
,u_{2},\ldots,u_{n}\right)  <\left(  g_{1}+\left(  n-1\right)  ,g_{2}+\left(
n-2\right)  ,\ldots,g_{n}+\left(  n-2\right)  \right)  $. Thus, the
coefficient of the monomial $X_{1}^{g_{1}+\left(  n-1\right)  }X_{2}%
^{g_{2}+\left(  n-2\right)  }\cdots X_{n}^{g_{n}+\left(  n-n\right)  }$ in the
polynomial \newline$g\cdot\operatorname*{alt}\left(  X^{n-1}\wedge
X^{n-2}\wedge\cdots\wedge X^{0}\right)  $ is $c$, which is nonzero. Hence, the
polynomial \newline$g\cdot\operatorname*{alt}\left(  X^{n-1}\wedge
X^{n-2}\wedge\cdots\wedge X^{0}\right)  $ has at least one nonzero
coefficient. Thus, this polynomial $g\cdot\operatorname*{alt}\left(
X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge X^{0}\right)  $ is nonzero.

Now, forget that we fixed $g$. We thus have proven that $g\cdot
\operatorname*{alt}\left(  X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge
X^{0}\right)  $ is nonzero whenever $g\in A\left[  X_{1},X_{2},\ldots
,X_{n}\right]  $ is nonzero. In other words, the element $\operatorname*{alt}%
\left(  X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge X^{0}\right)  $ of $A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $ is regular. Qed.]

[\textit{Second proof of the fact that }$\operatorname*{alt}\left(
X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge X^{0}\right)  $ \textit{is regular:}
From (\ref{pf.l2.3.alt.1}), we obtain%
\begin{align}
&  \operatorname*{alt}\left(  X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge
X^{0}\right) \nonumber\\
&  =\sum_{\sigma\in\mathfrak{S}_{n}}\left(  -1\right)  ^{\sigma}%
X_{\sigma\left(  1\right)  }^{n-1}X_{\sigma\left(  2\right)  }^{n-2}\cdots
X_{\sigma\left(  n\right)  }^{n-n}\nonumber\\
&  =\det\left(  \left(  X_{j}^{n-i}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of a
determinant}\right) \nonumber\\
&  =\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)
\label{pf.l2.3.alt.reg.2nd.1}%
\end{align}
(by the well-known formula for the Vandermonde determinant). But the
polynomial $\prod_{1\leq i<j\leq n}\left(  X_{i}-X_{j}\right)  $ is a regular
element of $A\left[  X_{1},X_{2},\ldots,X_{n}\right]  $ (this is well-known;
see, e.g., \cite[Corollary 4.4]{regpol}). In view of
(\ref{pf.l2.3.alt.reg.2nd.1}), this rewrites as follows: The polynomial
$\operatorname*{alt}\left(  X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge
X^{0}\right)  $ is a regular element of $A\left[  X_{1},X_{2},\ldots
,X_{n}\right]  $. Qed.]

Either way, we have now shown that the element $\operatorname*{alt}\left(
X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge X^{0}\right)  $ of $A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $ is regular. Hence, from
(\ref{pf.l2.3.alt.0}), we obtain $f=0$ (since $f\in S\subseteq A\left[
X_{1},X_{2},\ldots,X_{n}\right]  $).

Now, forget that we fixed $f$. We thus have shown that each $f\in S$
satisfying $f\cdot\left(  X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge
X^{0}\right)  =0$ satisfies $f=0$. In other words, the annihilator of
$X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge X^{0}$ in $S$ is zero. In other
words, the annihilator of $X^{n-1}\wedge X^{n-2}\wedge\cdots\wedge X^{0}$ in
$A\left[  X_{1},X_{2},\ldots,X_{n}\right]  ^{\operatorname*{sym}}$ is zero
(since $S=A\left[  X_{1},X_{2},\ldots,X_{n}\right]  ^{\operatorname*{sym}}$).
This proves Lemma 2.3.
\end{proof}
\end{fineprint}

\begin{thebibliography}{99999999}                                                                                         %


\bibitem[Grinbe19]{regpol}Darij Grinberg, \textit{Regular elements of a ring,
monic polynomials and \textquotedblleft lcm-coprimality\textquotedblright}, 16
July 2019.\newline\url{http://www.cip.ifi.lmu.de/~grinberg/algebra/regpol.pdf}

\bibitem[LakTho12]{LakTho12}Dan Laksov, Anders Thorup, \textit{Splitting
Algebras and Schubert Calculus}, Indiana University Mathematics Journal, Vol.
61, No. 3 (2012).
\end{thebibliography}


\end{document}