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\ihead{Reductions for the subdivision algebra}
\ohead{\today}
\begin{document}

\title{$t$-unique reductions for M\'{e}sz\'aros's subdivision algebra}
\author{Darij Grinberg}
\date{version 8.0, April 20, 2019 (post-publication update of pre-editorial version,
with corrections)}
\maketitle

\begin{abstract}
\textbf{Abstract.} Fix a commutative ring $\mathbf{k}$, two elements $\beta
\in\mathbf{k}$ and $\alpha\in\mathbf{k}$ and a positive integer $n$. Let
$\mathcal{X}$ be the polynomial ring over $\mathbf{k}$ in the $n\left(
n-1\right)  /2$ indeterminates $x_{i,j}$ for all $1\leq i<j\leq n$. Consider
the ideal $\mathcal{J}$ of $\mathcal{X}$ generated by all polynomials of the
form $x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  -\alpha$ for
$1\leq i<j<k\leq n$. The quotient algebra $\mathcal{X}/\mathcal{J}$ (at least
for a certain choice of $\mathbf{k}$, $\beta$ and $\alpha$) has been
introduced by Karola M\'{e}sz\'{a}ros in \cite{Meszar09} as a commutative
analogue of Anatol Kirillov's quasi-classical Yang-Baxter algebra. A monomial
in $\mathcal{X}$ is said to be \textit{pathless} if it has no divisors of the
form $x_{i,j}x_{j,k}$ with $1\leq i<j<k\leq n$. The residue classes of these
pathless monomials span the $\mathbf{k}$-module $\mathcal{X}/\mathcal{J}$, but
(in general) are $\mathbf{k}$-linearly dependent. More combinatorially:
Reducing a given $p\in\mathcal{X}$ modulo the ideal $\mathcal{J}$ by applying
replacements of the form $x_{i,j}x_{j,k}\mapsto x_{i,k}\left(  x_{i,j}%
+x_{j,k}+\beta\right)  +\alpha$ always eventually leads to a $\mathbf{k}%
$-linear combination of pathless monomials, but the result may depend on the
choices made in the process.

More recently, the study of Grothendieck polynomials has led Laura Escobar and
Karola M\'{e}sz\'{a}ros \cite[\S 5]{EscMes15} to defining a $\mathbf{k}%
$-algebra homomorphism $D$ from $\mathcal{X}$ into the polynomial ring
$\mathbf{k}\left[  t_{1},t_{2},\ldots,t_{n-1}\right]  $ that sends each
$x_{i,j}$ to $t_{i}$. We show the following fact (generalizing a conjecture of
M\'{e}sz\'{a}ros): If $p\in\mathcal{X}$, and if $q\in\mathcal{X}$ is a
$\mathbf{k}$-linear combination of pathless monomials satisfying $p\equiv
q\operatorname{mod}\mathcal{J}$, then $D\left(  q\right)  $ does not depend on
$q$ (as long as $\beta$, $\alpha$ and $p$ are fixed). Thus, the above way of
reducing a $p\in\mathcal{X}$ modulo $\mathcal{J}$ may lead to different
results, but all of them become identical once $D$ is applied.

We also find an actual basis of the $\mathbf{k}$-module $\mathcal{X}%
/\mathcal{J}$, using what we call \textit{forkless monomials}.

\end{abstract}
\tableofcontents

\section*{Introduction}

The main result of this paper is probably best illustrated by an example:

\begin{example}
\label{exam.intro}Let us play a solitaire game. Fix a positive integer $n$ and
two numbers $\beta\in\mathbb{Q}$ and $\alpha\in\mathbb{Q}$, and let
$\mathcal{X}$ be the ring $\mathbb{Q}\left[  x_{i,j}\ \mid\ 1\leq i<j\leq
n\right]  $ of polynomials with rational coefficients in the $n\left(
n-1\right)  /2$ indeterminates $x_{i,j}$ with $1\leq i<j\leq n$. (For example,
if $n=4$, then $\mathcal{X}=\mathbb{Q}\left[  x_{1,2},x_{1,3},x_{1,4}%
,x_{2,3},x_{2,4},x_{3,4}\right]  $.)

Start with any polynomial $p\in\mathcal{X}$. The allowed move is the
following: Pick a monomial $\mathfrak{m}$ that appears (with nonzero
coefficient) in $p$ and that is divisible by $x_{i,j}x_{j,k}$ for some $1\leq
i<j<k\leq n$. For example, $x_{1,2}x_{1,3}x_{2,4}$ is such a monomial (if it
appears in $p$ and if $n\geq4$), because it is divisible by $x_{i,j}x_{j,k}$
for $\left(  i,j,k\right)  =\left(  1,2,4\right)  $. Choose one triple
$\left(  i,j,k\right)  $ with $1\leq i<j<k\leq n$ and $x_{i,j}x_{j,k}%
\mid\mathfrak{m}$ (sometimes, there are several choices). Now, replace this
monomial $\mathfrak{m}$ by $\dfrac{x_{i,k}\left(  x_{i,j}+x_{j,k}%
+\beta\right)  +\alpha}{x_{i,j}x_{j,k}}\mathfrak{m}$ in $p$.

Thus, each move modifies the polynomial, replacing a monomial by a sum of four
monomials (or fewer, if $\beta$ or $\alpha$ is $0$). The game ends when no
more moves are possible (i.e., no monomial $\mathfrak{m}$ appearing in your
polynomial is divisible by $x_{i,j}x_{j,k}$ for any $1\leq i<j<k\leq n$).

It is easy to see that this game (a thinly veiled reduction procedure modulo
an ideal of $\mathcal{X}$) always ends after finitely many moves. Here is one
instance of this game being played, for $n=4$ and $\beta=1$ and $\alpha=0$ and
starting with the polynomial $p=x_{1,2}x_{2,3}x_{3,4}$:%
\begin{align}
&  x_{1,2}x_{2,3}x_{3,4}\nonumber\\
&  \mapsto x_{1,3}\left(  x_{1,2}+x_{2,3}+1\right)  x_{3,4}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we chose }\mathfrak{m}%
=x_{1,2}x_{2,3}x_{3,4}\text{ and }\left(  i,j,k\right)  =\left(  1,2,3\right)
\right) \nonumber\\
&  =x_{1,2}x_{1,3}x_{3,4}+x_{1,3}x_{2,3}x_{3,4}+x_{1,3}x_{3,4}\nonumber\\
&  \mapsto x_{1,2}x_{1,4}\left(  x_{1,3}+x_{3,4}+1\right)  +x_{1,3}%
x_{2,3}x_{3,4}+x_{1,3}x_{3,4}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we chose }\mathfrak{m}%
=x_{1,2}x_{1,3}x_{3,4}\text{ and }\left(  i,j,k\right)  =\left(  1,3,4\right)
\right) \nonumber\\
&  =x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,2}x_{1,4}+x_{1,3}%
x_{2,3}x_{3,4}+x_{1,3}x_{3,4}\nonumber\\
&  \mapsto x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,2}x_{1,4}%
+x_{1,3}x_{2,4}\left(  x_{2,3}+x_{3,4}+1\right)  +x_{1,3}x_{3,4}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we chose }\mathfrak{m}%
=x_{1,3}x_{2,3}x_{3,4}\text{ and }\left(  i,j,k\right)  =\left(  2,3,4\right)
\right) \nonumber\\
&  =x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,2}x_{1,4}+x_{1,3}%
x_{2,3}x_{2,4}+x_{1,3}x_{2,4}x_{3,4}+x_{1,3}x_{2,4}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +x_{1,3}x_{3,4}\nonumber\\
&  \mapsto x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,2}x_{1,4}%
+x_{1,3}x_{2,3}x_{2,4}+x_{1,3}x_{2,4}x_{3,4}+x_{1,3}x_{2,4}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +x_{1,4}\left(  x_{1,3}+x_{3,4}+1\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we chose }\mathfrak{m}%
=x_{1,3}x_{3,4}\text{ and }\left(  i,j,k\right)  =\left(  1,3,4\right)
\right) \nonumber\\
&  =x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,2}x_{1,4}+x_{1,3}%
x_{2,3}x_{2,4}+x_{1,3}x_{2,4}x_{3,4}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +x_{1,3}x_{2,4}+x_{1,3}x_{1,4}+x_{1,4}x_{3,4}%
+x_{1,4}\nonumber\\
&  \mapsto x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,2}x_{1,4}%
+x_{1,3}x_{2,3}x_{2,4}+x_{2,4}x_{1,4}\left(  x_{1,3}+x_{3,4}+1\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +x_{1,3}x_{2,4}+x_{1,3}x_{1,4}+x_{1,4}x_{3,4}%
+x_{1,4}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we chose }\mathfrak{m}%
=x_{1,3}x_{2,4}x_{3,4}\text{ and }\left(  i,j,k\right)  =\left(  1,3,4\right)
\right) \nonumber\\
&  =x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,2}x_{1,4}+x_{1,3}%
x_{2,3}x_{2,4}+x_{1,3}x_{1,4}x_{2,4}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +x_{1,4}x_{2,4}x_{3,4}+x_{1,4}x_{2,4}+x_{1,3}%
x_{2,4}+x_{1,3}x_{1,4}+x_{1,4}x_{3,4}+x_{1,4}. \label{eq.exam.intro.res1}%
\end{align}
The game ends at this polynomial, since there are no more moves to be done.

A standard question about games like this is: Is the state obtained at the end
of the game (i.e., in our case, the polynomial after the game has ended)
independent of the choices made during the game? In our case, the answer is
\textquotedblleft no\textquotedblright\ (in general, for $n\geq4$). Indeed,
the reader can easily verify that the above game could have led to a different
result if we had made different choices.

However, something else turns out to be independent of the choices. Namely,
let us transform the polynomial at the end of the game further by applying the
substitution $x_{i,j}\mapsto t_{i}$ (where $t_{1},t_{2},\ldots,t_{n-1}$ are
new indeterminates). For example, doing this to the polynomial
(\ref{eq.exam.intro.res1}) results in%
\begin{align*}
&  t_{1}t_{1}t_{1}+t_{1}t_{1}t_{3}+t_{1}t_{1}+t_{1}t_{2}t_{2}+t_{1}t_{1}%
t_{2}+t_{1}t_{2}t_{3}+t_{1}t_{2}+t_{1}t_{2}+t_{1}t_{1}+t_{1}t_{3}+t_{1}\\
&  =t_{1}\left(  2t_{1}+2t_{2}+t_{3}+t_{1}^{2}+t_{2}^{2}+t_{1}t_{2}+t_{1}%
t_{3}+t_{2}t_{3}+1\right)  .
\end{align*}
According to a conjecture of M\'{e}sz\'{a}ros, the result of this substitution
is indeed independent of the choices made during the game (as long as $p$ is fixed).
\end{example}

Why would one play a game like this? The reduction rule $\mathfrak{m}%
\mapsto\dfrac{x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  }{x_{i,j}x_{j,k}%
}\mathfrak{m}$ (this is a particular case of our above rule, when $\alpha$ is
set to $0$) has appeared in Karola M\'{e}sz\'{a}ros's study \cite{Meszar09} of
the abelianization of Anatol Kirillov's quasi-classical Yang-Baxter algebra
(see, e.g., \cite{Kirill16} for a recent survey of the latter and its many
variants); it has a long prehistory (some of which is surveyed in Section
\ref{subsect.arnold} below), starting with Vladimir Arnold's 1971 work
\cite{Arnold71} on the braid arrangement. To define this
abelianization\footnote{The notations used in this Introduction are meant to
be provisional. In the rest of this paper, we shall work with different
notations (and in a more general setting), which will be introduced in Section
\ref{sect.states}.}, we let $\beta$ be an indeterminate (unlike in Example
\ref{exam.intro}, where it was an element of $\mathbb{Q}$). Furthermore, fix a
positive integer $n$. The abelianization of the ($n$-th) quasi-classical
Yang-Baxter algebra is the commutative $\mathbb{Q}\left[  \beta\right]
$-algebra $\mathcal{S}\left(  A_{n}\right)  $ with%
\begin{align*}
\text{generators}\ \ \ \ \ \ \ \ \ \  &  x_{i,j}\ \ \ \ \ \ \ \ \ \ \text{for
all }1\leq i<j\leq n\ \ \ \ \ \ \ \ \ \ \text{and}\\
\text{relations}\ \ \ \ \ \ \ \ \ \  &  x_{i,j}x_{j,k}=x_{i,k}\left(
x_{i,j}+x_{j,k}+\beta\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }1\leq
i<j<k\leq n.
\end{align*}
A natural question is to find an explicit basis of $\mathcal{S}\left(
A_{n}\right)  $ (as a $\mathbb{Q}$-vector space, or, if possible, as a
$\mathbb{Q}\left[  \beta\right]  $-module). One might try constructing such a
basis using a reduction algorithm (or \textquotedblleft straightening
law\textquotedblright) that takes any element of $\mathcal{S}\left(
A_{n}\right)  $ (written as any polynomial in the generators $x_{i,j}$) and
rewrites it in a \textquotedblleft normal form\textquotedblright. The most
obvious way one could try to construct such a reduction algorithm is by
repeatedly rewriting products of the form $x_{i,j}x_{j,k}$ (with $1\leq
i<j<k\leq n$) as $x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  $, until this
is no longer possible. This is precisely the game that we played in Example
\ref{exam.intro} (with the only difference that $\beta$ is now an
indeterminate, not a number). Unfortunately, the result of the game turns out
to depend on the choices made while playing it; consequently, the
\textquotedblleft normal form\textquotedblright\ it constructs is not
literally a normal form, and instead of a basis of $\mathcal{S}\left(
A_{n}\right)  $ we only obtain a spanning set.\footnote{Surprisingly, a
similar reduction algorithm \textbf{does} work for the (non-abelianized)
quasi-classical Yang-Baxter algebra itself. This is one of M\'{e}sz\'{a}ros's
results (\cite[Theorem 30]{Meszar09}).}

Nevertheless, the result of the game is not meaningless. The idea to
substitute $t_{i}$ for $x_{i,j}$ (in the result, not in the original
polynomial!) seems to have appeared in work of Postnikov, Stanley and
M\'{e}sz\'{a}ros; some concrete formulas (for specific values of the initial
polynomial and specific values of $\beta$) appear in \cite[Exercise
A22]{Stanle15} (resulting in Catalan and Narayana numbers). Recent work on
Grothendieck polynomials by Anatol Kirillov (\cite[\S 4]{Kirill13},
\cite{Kirill16}) and by Laura Escobar and Karola M\'{e}sz\'{a}ros
\cite[\S 5]{EscMes15} has again brought up the notion of substituting $t_{i}$
for $x_{i,j}$ in the polynomial obtained at the end of the game. This has led
M\'{e}sz\'{a}ros to the conjecture that, after this substitution, the
resulting polynomial no longer depends on the choices made during the game.
She has proven this conjecture for a certain class of polynomials (those
corresponding to \textquotedblleft noncrossing trees\textquotedblright).

The main purpose of this paper is to establish M\'{e}sz\'{a}ros's conjecture
in the general case. We shall, in fact, work in greater generality than all
previously published sources. First, instead of the relation $x_{i,j}%
x_{j,k}=x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  $, we shall consider the
\textquotedblleft deformed\textquotedblright\ relation $x_{i,j}x_{j,k}%
=x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  +\alpha$; the idea of this
deformation again goes back to the work of Anatol Kirillov (see, e.g.,
\cite[Definition 5.1 (1)]{Kirill16} for a noncommutative variant of the
quotient ring $\mathcal{X}/\mathcal{J}$, which he calls the \textquotedblleft
associative quasi-classical Yang--Baxter algebra of weight $\left(
\alpha,\beta\right)  $\textquotedblright). Instead of requiring $\beta$ to be
either a rational number (as in Example \ref{exam.intro}) or an indeterminate
over $\mathbb{Q}$ (as in the definition of $\mathcal{S}\left(  A_{n}\right)
$), we shall let $\beta$ be any element of the ground ring, which in turn will
be an arbitrary commutative ring $\mathbf{k}$. Rather than working in an
algebra like $\mathcal{S}\left(  A_{n}\right)  $, we shall work in the
polynomial ring $\mathcal{X}=\mathbf{k}\left[  x_{i,j}\ \mid\ 1\leq i<j\leq
n\right]  $, and study the ideal $\mathcal{J}$ generated by all elements of
the form $x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  -\alpha$
for $1\leq i<j<k\leq n$. Instead of focussing on the reduction algorithm, we
shall generally study polynomials in $\mathcal{X}$ that are congruent to each
other modulo the ideal $\mathcal{J}$. A monomial in $\mathcal{X}$ will be
called \textquotedblleft pathless\textquotedblright\ if it is not divisible by
any monomial of the form $x_{i,j}x_{j,k}$ with $i<j<k$. A polynomial in
$\mathcal{X}$ will be called \textquotedblleft pathless\textquotedblright\ if
all monomials appearing in it are pathless. Thus, \textquotedblleft
pathless\textquotedblright\ polynomials are precisely the polynomials
$p\in\mathcal{X}$ for which the game in Example \ref{exam.intro} would end
immediately if started at $p$.

Our main result (Theorem~\ref{thm.t-red.unique}) will show that if
$p\in\mathcal{X}$ is a polynomial, and if $q\in\mathcal{X}$ is a pathless
polynomial congruent to $p$ modulo $\mathcal{J}$, then the image of $q$ under
the substitution $x_{i,j}\mapsto t_{i}$ does not depend on $q$ (but only on
$\alpha$, $\beta$ and $p$). This, in particular, yields M\'{e}sz\'{a}ros's
conjecture; but it is a stronger result, because it does not require that $q$
is obtained from $p$ by playing the game from Example \ref{exam.intro} (all we
ask for is that $q$ be pathless and congruent to $p$ modulo $\mathcal{J}$),
and of course because of the more general setting.

After the proof of Theorem~\ref{thm.t-red.unique}, we shall rewrite the
definition of $\mathcal{J}$ (and of $\mathcal{X}$) in a more symmetric form
(Section \ref{sect.symmetry}). Then, we shall also answer the (easier)
question of finding a basis for the quotient ring $\mathcal{X}/\mathcal{J}$
(Proposition \ref{prop.forkless.basis}). This basis will be obtained using an
explicit Gr\"{o}bner basis of the ideal $\mathcal{J}$.

We shall close with further considerations, open questions and connections to
previous research.

A recent preprint by M\'{e}sz\'{a}ros and St. Dizier \cite{MesDiz17} proves a
fact \cite[Theorem A]{MesDiz17} which, translated into our language, confirms
the conjecture stated in Example \ref{exam.intro} at least in the case when
$\alpha=0$ and the game is started with a monomial $p$. This might provide a
different route to some of our results. (The arguments in \cite{MesDiz17} are
of combinatorial nature, involving flows on graphs, and so is the language
used in \cite{MesDiz17}; in particular, monomials are encoded by graphs.)

\subsection{Acknowledgments}

The \texttt{SageMath} computer algebra system \cite{SageMath} was of great
service during the development of the results below. Conversations with Nick
Early have led me to the ideas in Subsection \ref{subsect.arnold}, and Victor
Reiner has helped me concretize them. This paper has furthermore profited from
enlightening comments by Ricky Liu, Karola M\'{e}sz\'{a}ros, Nicholas
Proudfoot, Travis Scrimshaw, Richard Stanley, two anonymous referees and an
anonymous editor.

\subsection{Remark on alternative versions}

\begin{vershort}
This paper also has a detailed version \cite{verlong}, which includes some
proofs that have been omitted from the present version (mostly straightforward
computations and basic properties of Gr\"{o}bner bases).
\end{vershort}

\begin{verlong}
You are reading the detailed version of this paper. For the standard version
(which is shorter by virtue of omitting some straightforward or well-known
proofs), see \cite{vershort}.
\end{verlong}

In a previous version
(\href{https://arxiv.org/abs/1704.00839v2}{arXiv:1704.00839v2}) of this paper,
a weaker version of the main result was proven (which corresponds to the case
$\alpha=0$ in our notations). The proof used a somewhat different construction
(involving formal power series instead of Laurent series, and a different map
$A$).

\section{\label{sect.states}Definitions and results}

Let us now start from scratch, and set the stage for the main result.

\begin{definition}
Let $\mathbb{N}=\left\{  0,1,2,\ldots\right\}  $.

Let $\left[  m\right]  $ be the set $\left\{  1,2,\ldots,m\right\}  $ for each
$m\in\mathbb{N}$.

Let $\mathbf{k}$ be a commutative ring. (We fix $\mathbf{k}$ throughout this paper.)

Fix two elements $\beta$ and $\alpha$ of $\mathbf{k}$.

The word \textquotedblleft monomial\textquotedblright\ shall always mean an
element of a free abelian monoid (written multiplicatively). For example, the
monomials in two indeterminates $x$ and $y$ are the elements of the form
$x^{i}y^{j}$ with $\left(  i,j\right)  \in\mathbb{N}^{2}$. Thus, monomials do
not include coefficients (and are not bound to a specific base ring).
\end{definition}

\begin{definition}
Fix a positive integer $n$. Let $\mathcal{X}$ be the polynomial ring
\[
\mathbf{k}\left[  x_{i,j}\ \mid\ \left(  i,j\right)  \in\left[  n\right]
^{2}\text{ satisfying }i<j\right]  .
\]
This is a polynomial ring in $n\left(  n-1\right)  /2$ indeterminates
$x_{i,j}$ over $\mathbf{k}$.

We shall use the notation $\mathfrak{M}$ for the set of all monomials in these
indeterminates $x_{i,j}$. Notice that $\mathfrak{M}$ is an abelian monoid
under multiplication.
\end{definition}

\begin{definition}
A monomial $\mathfrak{m}\in\mathfrak{M}$ is said to be \textit{pathless} if
there exists no triple $\left(  i,j,k\right)  \in\left[  n\right]  ^{3}$
satisfying $i<j<k$ and $x_{i,j}x_{j,k}\mid\mathfrak{m}$ (as monomials).

A polynomial $p\in\mathcal{X}$ is said to be \textit{pathless} if it is a
$\mathbf{k}$-linear combination of pathless monomials.
\end{definition}

\begin{definition}
Let $\mathcal{J}$ be the ideal of $\mathcal{X}$ generated by all elements of
the form $x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  -\alpha$
for $\left(  i,j,k\right)  \in\left[  n\right]  ^{3}$ satisfying $i<j<k$.
\end{definition}

The following fact is easy to check:

\begin{proposition}
\label{prop.path-red.span}Let $p\in\mathcal{X}$. Then, there exists a pathless
polynomial $q\in\mathcal{X}$ such that $p\equiv q\operatorname{mod}%
\mathcal{J}$.
\end{proposition}

In general, this $q$ is not unique.\footnote{For instance, if $\mathbf{k}%
=\mathbb{Z}$, $\beta=1$, $\alpha=0$ and $n=4$, then
\begin{align*}
q_{1}  &  =x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}%
+x_{1,3}x_{1,4}+x_{1,3}x_{1,4}x_{2,4}+x_{1,3}x_{2,3}x_{2,4}\\
&  \ \ \ \ \ \ \ \ \ \ +x_{1,3}x_{2,4}+x_{1,4}+x_{1,4}x_{2,4}+x_{1,4}%
x_{2,4}x_{3,4}+x_{1,4}x_{3,4}%
\end{align*}
and%
\begin{align*}
q_{2}  &  =x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}%
+x_{1,3}x_{1,4}+x_{1,3}x_{1,4}x_{2,3}\\
&  \ \ \ \ \ \ \ \ \ \ +x_{1,4}+x_{1,4}x_{2,3}+x_{1,4}x_{2,3}x_{2,4}%
+x_{1,4}x_{2,4}+x_{1,4}x_{2,4}x_{3,4}+x_{1,4}x_{3,4}%
\end{align*}
are two pathless polynomials $q\in\mathcal{X}$ satisfying $x_{1,2}%
x_{2,3}x_{3,4}\equiv q\operatorname{mod}\mathcal{J}$, but they are not
identical.}

\begin{vershort}
We shall roughly outline a proof of Proposition \ref{prop.path-red.span} now;
a detailed writeup of this proof can be found in the detailed version
\cite{verlong} of this paper.
\end{vershort}

\begin{verlong}
We shall roughly outline a proof of Proposition \ref{prop.path-red.span} now;
a more detailed writeup of this proof can be found at the end of this section.
\end{verlong}

\begin{proof}
[Proof of Proposition \ref{prop.path-red.span} (sketched).]The \textit{weight}
of a monomial $\prod_{\substack{\left(  i,j\right)  \in\left[  n\right]
^{2};\\i<j}}x_{i,j}^{a_{i,j}}\in\mathfrak{M}$ shall mean the nonnegative
integer $\sum_{\substack{\left(  i,j\right)  \in\left[  n\right]  ^{2}%
;\\i<j}}a_{i,j}\left(  n-j+i\right)  $. If we have a monomial $\mathfrak{m}%
\in\mathfrak{M}$ that is not pathless, then we can find a triple $\left(
i,j,k\right)  \in\left[  n\right]  ^{3}$ satisfying $i<j<k$ and $x_{i,j}%
x_{j,k}\mid\mathfrak{m}$; then, we can replace $\mathfrak{m}$ by a polynomial
$\widetilde{\mathfrak{m}}=\mathfrak{m}\cdot\dfrac{x_{i,k}\left(
x_{i,j}+x_{j,k}+\beta\right)  +\alpha}{x_{i,j}x_{j,k}}$, which is congruent to
$\mathfrak{m}$ modulo $\mathcal{J}$ but has the property that all monomials
appearing in it have a smaller weight than $\mathfrak{m}$. This gives rise to
a recursive algorithm\footnote{or \textquotedblleft straightening
law\textquotedblright, as algorithms of this kind are commonly called in
algebraic combinatorics} for reducing a polynomial modulo the ideal
$\mathcal{J}$. The procedure will necessarily terminate (although its result
might depend on the order of operation); the polynomial resulting at its end
will be pathless.
\end{proof}

The ideal $\mathcal{J}$ is relevant to the so-called \textit{subdivision
algebra of root polytopes} (denoted by $\mathcal{S}\left(  \beta\right)  $ in
\cite[\S 5]{EscMes15} and $\mathcal{S}\left(  A_{n}\right)  $ in
\cite[\S 1]{Meszar09}). Namely, this latter algebra is defined as the quotient
$\mathcal{X}/\mathcal{J}$ for a certain choice of $\mathbf{k}$, $\beta$ and
$\alpha$ (namely, for the choice where $\mathbf{k}$ is a univariate polynomial
ring over $\mathbb{Q}$, where $\beta$ is the indeterminate in $\mathbf{k}$,
and where $\alpha=0$). This algebra was first introduced by M\'{e}sz\'{a}ros
in \cite{Meszar09} as the abelianization of Anatol Kirillov's quasi-classical
Yang-Baxter algebra.

In \cite[\S 5 and Appendix A]{EscMes15}, Escobar and M\'{e}sz\'{a}ros
(motivated by computations of Grothendieck polynomials) consider the result of
substituting $t_{i}$ for each variable $x_{i,j}$ in a polynomial
$f\in\mathcal{X}$. In our language, this leads to the following definition:

\begin{definition}
Let $\mathcal{T}^{\prime}$ be the polynomial ring $\mathbf{k}\left[
t_{1},t_{2},\ldots,t_{n-1}\right]  $. We define a $\mathbf{k}$-algebra
homomorphism $D:\mathcal{X}\rightarrow\mathcal{T}^{\prime}$ by
\[
D\left(  x_{i,j}\right)  =t_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(
i,j\right)  \in\left[  n\right]  ^{2}\text{ satisfying }i<j.
\]

\end{definition}

The goal of this paper is to prove the following fact, which (in a less
general setting) was conjectured by Karola M\'{e}sz\'{a}ros in a 2015 talk at MIT:

\begin{theorem}
\label{thm.t-red.unique}Let $p\in\mathcal{X}$. Consider any pathless
polynomial $q\in\mathcal{X}$ such that $p\equiv q\operatorname{mod}%
\mathcal{J}$. Then, $D\left(  q\right)  $ does not depend on the choice of $q$
(but merely on the choice of $\alpha$, $\beta$ and $p$).
\end{theorem}

It is not generally true that $D\left(  q\right)  =D\left(  p\right)  $; thus,
Theorem~\ref{thm.t-red.unique} does not follow from a simple \textquotedblleft
invariant\textquotedblright.

\begin{verlong}
We shall prove Theorem~\ref{thm.t-red.unique} in the next section. But first,
let us pay a debt and explain the proof of Proposition
\ref{prop.path-red.span} in full detail.

We begin with a few definitions:

\begin{definition}
Let $\mathcal{X}_{\operatorname*{pathless}}$ denote the $\mathbf{k}$-submodule
of $\mathcal{X}$ spanned by all pathless monomials $\mathfrak{m}%
\in\mathfrak{M}$. Thus, $\mathcal{X}_{\operatorname*{pathless}}$ is the set of
all pathless polynomials $f\in\mathcal{X}$.
\end{definition}

\begin{definition}
\label{def.span-detail.weight}Let $\mathfrak{m}\in\mathfrak{M}$ be a monomial.
The \textit{weight} of $\mathfrak{m}$ is defined to be $\sum
_{\substack{\left(  i,j\right)  \in\left[  n\right]  ^{2};\\i<j}%
}a_{i,j}\left(  n-j+i\right)  $, where the monomial $\mathfrak{m}$ has been
written in the form $\mathfrak{m}=\prod_{\substack{\left(  i,j\right)
\in\left[  n\right]  ^{2};\\i<j}}x_{i,j}^{a_{i,j}}$ (with $a_{i,j}%
\in\mathbb{N}$). This weight is an integer, and will be denoted by
$\operatorname*{weight}\mathfrak{m}$.
\end{definition}

Observe the following properties of weights:

\begin{lemma}
\label{lem.span-detail.weight.triv}\textbf{(a)} For any $\left(  i,j\right)
\in\left[  n\right]  ^{2}$ satisfying $i<j$, we have $\operatorname*{weight}%
\left(  x_{i,j}\right)  =n-j+i$.

\textbf{(b)} If $\mathfrak{p}\in\mathfrak{M}$ and $\mathfrak{q}\in
\mathfrak{M}$ are two monomials, then $\operatorname*{weight}\left(
\mathfrak{pq}\right)  =\operatorname*{weight}\mathfrak{p}%
+\operatorname*{weight}\mathfrak{q}$.

\textbf{(c)} If $\mathfrak{m}\in\mathfrak{M}$ is a monomial, then
$\operatorname*{weight}\mathfrak{m}\in\mathbb{N}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.span-detail.weight.triv}.]Parts \textbf{(a)} and
\textbf{(b)} of Lemma \ref{lem.span-detail.weight.triv} are left to the reader.

Let us observe that every $\left(  i,j\right)  \in\left[  n\right]  ^{2}$
satisfies%
\begin{equation}
n-\underbrace{j}_{\substack{\leq n\\\text{(since }j\in\left[  n\right]
\text{)}}}+\underbrace{i}_{\substack{>0\\\text{(since }i\in\left[  n\right]
\text{)}}}\geq n-n+0=0. \label{pf.lem.span-detail.weight.triv.1}%
\end{equation}


\textbf{(c)} Let $\mathfrak{m}\in\mathfrak{M}$ be a monomial. Write the
monomial $\mathfrak{m}$ in the form $\mathfrak{m}=\prod_{\substack{\left(
i,j\right)  \in\left[  n\right]  ^{2};\\i<j}}x_{i,j}^{a_{i,j}}$ (with
$a_{i,j}\in\mathbb{N}$). Then, the definition of $\operatorname*{weight}%
\mathfrak{m}$ yields%
\[
\operatorname*{weight}\mathfrak{m}=\sum_{\substack{\left(  i,j\right)
\in\left[  n\right]  ^{2};\\i<j}}\underbrace{a_{i,j}}_{\geq0}%
\underbrace{\left(  n-j+i\right)  }_{\substack{\geq0\\\text{(by
(\ref{pf.lem.span-detail.weight.triv.1}))}}}\geq\sum_{\substack{\left(
i,j\right)  \in\left[  n\right]  ^{2};\\i<j}}0\cdot0=0.
\]
Hence, $\operatorname*{weight}\mathfrak{m}\in\mathbb{N}$. This proves Lemma
\ref{lem.span-detail.weight.triv} \textbf{(c)}.
\end{proof}

\begin{lemma}
\label{lem.span-detail.1}Let $\mathfrak{m}\in\mathfrak{M}$ be a monomial.
Then, $\mathfrak{m}\in\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.span-detail.1}.]We shall prove Lemma
\ref{lem.span-detail.1} by strong induction over $\operatorname*{weight}%
\mathfrak{m}$.\ \ \ \ \footnote{This is legitimate, since Lemma
\ref{lem.span-detail.weight.triv} \textbf{(c)} shows that
$\operatorname*{weight}\mathfrak{m}\in\mathbb{N}$ in the situation of Lemma
\ref{lem.span-detail.1}.} Thus, we fix any $N\in\mathbb{N}$, and we assume (as
the induction hypothesis) that Lemma \ref{lem.span-detail.1} holds in the case
when $\operatorname*{weight}\mathfrak{m}<N$. We then must show that Lemma
\ref{lem.span-detail.1} holds in the case when $\operatorname*{weight}%
\mathfrak{m}=N$.

We have assumed that Lemma \ref{lem.span-detail.1} holds in the case when
$\operatorname*{weight}\mathfrak{m}<N$. In other words,%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{if }\mathfrak{m}\in\mathfrak{M}\text{ is a monomial such that
}\operatorname*{weight}\mathfrak{m}<N\text{,}\\
\text{then }\mathfrak{m}\in\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}%
\end{array}
\right)  . \label{pf.lem.span-detail.1.indhyp}%
\end{equation}


Now, fix a monomial $\mathfrak{m}\in\mathfrak{M}$ such that
$\operatorname*{weight}\mathfrak{m}=N$. We shall show that $\mathfrak{m}%
\in\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}$.

If $\mathfrak{m}$ is pathless, then this is obvious\footnote{\textit{Proof.}
Assume that $\mathfrak{m}$ is pathless. We must then show that $\mathfrak{m}%
\in\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}$.
\par
But recall that the $\mathbf{k}$-module $\mathcal{X}_{\operatorname*{pathless}%
}$ is spanned by the pathless monomials. Thus, $\mathfrak{m}\in\mathcal{X}%
_{\operatorname*{pathless}}$ (since $\mathfrak{m}$ is a pathless monomial).
Hence, $\mathfrak{m}\in\mathcal{X}_{\operatorname*{pathless}}\subseteq
\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}$, qed.}. Hence, for the
rest of this proof, we WLOG assume that $\mathfrak{m}$ is not pathless. In
other words, there exists a triple $\left(  i,j,k\right)  \in\left[  n\right]
^{3}$ satisfying $i<j<k$ and $x_{i,j}x_{j,k}\mid\mathfrak{m}$ (as monomials).
Consider such a triple $\left(  i,j,k\right)  $.

\begin{noncompile}
From $i\in\left[  n\right]  $, we obtain $i\geq1$. From $j\in\left[  n\right]
$, we obtain $j\leq n$. Hence, $i<j\leq n$, so that $n>i\geq1$. Thus, $n-1>0$.
\end{noncompile}

We have $x_{i,j}x_{j,k}\mid\mathfrak{m}$ (as monomials). In other words, there
exists a monomial $\mathfrak{n}\in\mathfrak{M}$ such that $\mathfrak{m}%
=x_{i,j}x_{j,k}\mathfrak{n}$. Consider this $\mathfrak{n}$.

We have $x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)
-\alpha\in\mathcal{J}$ (since $x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}%
+x_{j,k}+\beta\right)  -\alpha$ is one of the designated generators of the
ideal $\mathcal{J}$). Thus, \newline$x_{i,j}x_{j,k}\equiv x_{i,k}\left(
x_{i,j}+x_{j,k}+\beta\right)  +\alpha\operatorname{mod}\mathcal{J}$. Now,%
\begin{align*}
\mathfrak{m}  &  =\underbrace{x_{i,j}x_{j,k}}_{\equiv x_{i,k}\left(
x_{i,j}+x_{j,k}+\beta\right)  +\alpha\operatorname{mod}\mathcal{J}%
}\mathfrak{n}\equiv\left(  x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)
+\alpha\right)  \mathfrak{n}\\
&  =x_{i,k}x_{i,j}\mathfrak{n}+x_{i,k}x_{j,k}\mathfrak{n}+\beta x_{i,k}%
\mathfrak{n}+\alpha\mathfrak{n}\operatorname{mod}\mathcal{J}.
\end{align*}
In other words,%
\begin{equation}
\mathfrak{m}\in x_{i,k}x_{i,j}\mathfrak{n}+x_{i,k}x_{j,k}\mathfrak{n}+\beta
x_{i,k}\mathfrak{n}+\alpha\mathfrak{n}+\mathcal{J}.
\label{pf.lem.span-detail.1.3}%
\end{equation}
We shall now analyze the four monomials $x_{i,k}x_{i,j}\mathfrak{n}$,
$x_{i,k}x_{j,k}\mathfrak{n}$, $x_{i,k}\mathfrak{n}$ and $\mathfrak{n}$ on the
right hand side of (\ref{pf.lem.span-detail.1.3}):

\begin{itemize}
\item We have $\operatorname*{weight}\left(  x_{i,k}x_{i,j}\mathfrak{n}%
\right)  <N$\ \ \ \ \footnote{\textit{Proof.} We have $\mathfrak{m}%
=x_{i,j}x_{j,k}\mathfrak{n}=x_{j,k}x_{i,j}\mathfrak{n}$ and thus%
\begin{align*}
\operatorname*{weight}\underbrace{\mathfrak{m}}_{=x_{j,k}x_{i,j}\mathfrak{n}}
&  =\operatorname*{weight}\left(  x_{j,k}x_{i,j}\mathfrak{n}\right)
=\underbrace{\operatorname*{weight}\left(  x_{j,k}\right)  }%
_{\substack{=n-k+j\\\text{(by Lemma \ref{lem.span-detail.weight.triv}
\textbf{(a),}}\\\text{applied to }j\text{ and }k\text{ instead of }i\text{ and
}j\text{)}}}+\operatorname*{weight}\left(  x_{i,j}\mathfrak{n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.span-detail.weight.triv} \textbf{(b)}, applied to }\mathfrak{p}%
=x_{j,k}\text{ and }\mathfrak{q}=x_{i,j}\mathfrak{n}\right) \\
&  =n-k+j+\operatorname*{weight}\left(  x_{i,j}\mathfrak{n}\right)  .
\end{align*}
But Lemma \ref{lem.span-detail.weight.triv} \textbf{(b)} (applied to
$\mathfrak{p}=x_{i,k}$ and $\mathfrak{q}=x_{i,j}\mathfrak{n}$) shows that%
\begin{align*}
\operatorname*{weight}\left(  x_{i,k}x_{i,j}\mathfrak{n}\right)   &
=\underbrace{\operatorname*{weight}\left(  x_{i,k}\right)  }%
_{\substack{=n-k+i\\\text{(by Lemma \ref{lem.span-detail.weight.triv}
\textbf{(a),}}\\\text{applied to }k\text{ instead of }j\text{)}}%
}+\operatorname*{weight}\left(  x_{i,j}\mathfrak{n}\right)
=n-k+\underbrace{i}_{<j}+\operatorname*{weight}\left(  x_{i,j}\mathfrak{n}%
\right) \\
&  <n-k+j+\operatorname*{weight}\left(  x_{i,j}\mathfrak{n}\right)
=\operatorname*{weight}\mathfrak{m}=N,
\end{align*}
qed.}. Hence, (\ref{pf.lem.span-detail.1.indhyp}) (applied to $x_{i,k}%
x_{i,j}\mathfrak{n}$ instead of $\mathfrak{m}$) shows that
\[
x_{i,k}x_{i,j}\mathfrak{n}\in\mathcal{X}_{\operatorname*{pathless}%
}+\mathcal{J}.
\]


\item We have $\operatorname*{weight}\left(  x_{i,k}x_{j,k}\mathfrak{n}%
\right)  <N$\ \ \ \ \footnote{\textit{Proof.} We have%
\begin{align*}
\operatorname*{weight}\underbrace{\mathfrak{m}}_{=x_{i,j}x_{j,k}\mathfrak{n}}
&  =\operatorname*{weight}\left(  x_{i,j}x_{j,k}\mathfrak{n}\right)
=\underbrace{\operatorname*{weight}\left(  x_{i,j}\right)  }%
_{\substack{=n-j+i\\\text{(by Lemma \ref{lem.span-detail.weight.triv}
\textbf{(a)})}}}+\operatorname*{weight}\left(  x_{j,k}\mathfrak{n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.span-detail.weight.triv} \textbf{(b)}, applied to }\mathfrak{p}%
=x_{i,j}\text{ and }\mathfrak{q}=x_{j,k}\mathfrak{n}\right) \\
&  =n-j+i+\operatorname*{weight}\left(  x_{j,k}\mathfrak{n}\right)  .
\end{align*}
But Lemma \ref{lem.span-detail.weight.triv} \textbf{(b)} (applied to
$\mathfrak{p}=x_{i,k}$ and $\mathfrak{q}=x_{j,k}\mathfrak{n}$) shows that%
\begin{align*}
\operatorname*{weight}\left(  x_{i,k}x_{j,k}\mathfrak{n}\right)   &
=\underbrace{\operatorname*{weight}\left(  x_{i,k}\right)  }%
_{\substack{=n-k+i\\\text{(by Lemma \ref{lem.span-detail.weight.triv}
\textbf{(a),}}\\\text{applied to }k\text{ instead of }j\text{)}}%
}+\operatorname*{weight}\left(  x_{j,k}\mathfrak{n}\right)  =n-\underbrace{k}%
_{\substack{>j\\\text{(since }j<k\text{)}}}+i+\operatorname*{weight}\left(
x_{j,k}\mathfrak{n}\right) \\
&  <n-j+i+\operatorname*{weight}\left(  x_{j,k}\mathfrak{n}\right)
=\operatorname*{weight}\mathfrak{m}=N,
\end{align*}
qed.}. Hence, (\ref{pf.lem.span-detail.1.indhyp}) (applied to $x_{i,k}%
x_{j,k}\mathfrak{n}$ instead of $\mathfrak{m}$) shows that
\[
x_{i,k}x_{j,k}\mathfrak{n}\in\mathcal{X}_{\operatorname*{pathless}%
}+\mathcal{J}.
\]


\item We have $\operatorname*{weight}\left(  x_{i,k}\mathfrak{n}\right)
<N$\ \ \ \ \footnote{\textit{Proof.} We have%
\begin{align*}
\operatorname*{weight}\underbrace{\mathfrak{m}}_{=x_{i,j}x_{j,k}\mathfrak{n}}
&  =\operatorname*{weight}\left(  x_{i,j}x_{j,k}\mathfrak{n}\right)
=\underbrace{\operatorname*{weight}\left(  x_{i,j}\right)  }%
_{\substack{=n-j+i\\\text{(by Lemma \ref{lem.span-detail.weight.triv}
\textbf{(a)})}}}+\underbrace{\operatorname*{weight}\left(  x_{j,k}%
\mathfrak{n}\right)  }_{\substack{=\operatorname*{weight}\left(
x_{j,k}\right)  +\operatorname*{weight}\mathfrak{n}\\\text{(by Lemma
\ref{lem.span-detail.weight.triv} \textbf{(b)}, applied to }\mathfrak{p}%
=x_{j,k}\text{ and }\mathfrak{q}=\mathfrak{n}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.span-detail.weight.triv} \textbf{(b)}, applied to }\mathfrak{p}%
=x_{i,j}\text{ and }\mathfrak{q}=x_{j,k}\mathfrak{n}\right) \\
&  =n-j+i+\underbrace{\operatorname*{weight}\left(  x_{j,k}\right)
}_{\substack{=n-k+j\\\text{(by Lemma \ref{lem.span-detail.weight.triv}
\textbf{(a),}}\\\text{applied to }j\text{ and }k\text{ instead of }i\text{ and
}j\text{)}}}+\operatorname*{weight}\mathfrak{n}\\
&  =\underbrace{n-j+i+n-k+j}_{=n+n-k+i}+\operatorname*{weight}\mathfrak{n}%
=\underbrace{n}_{>0}+n-k+i+\operatorname*{weight}\mathfrak{n}\\
&  >n-k+i+\operatorname*{weight}\mathfrak{n}.
\end{align*}
But Lemma \ref{lem.span-detail.weight.triv} \textbf{(b)} (applied to
$\mathfrak{p}=x_{i,k}$ and $\mathfrak{q}=\mathfrak{n}$) yields%
\begin{align*}
\operatorname*{weight}\left(  x_{i,k}\mathfrak{n}\right)   &
=\underbrace{\operatorname*{weight}\left(  x_{i,k}\right)  }%
_{\substack{=n-k+i\\\text{(by Lemma \ref{lem.span-detail.weight.triv}
\textbf{(a),}}\\\text{applied to }k\text{ instead of }j\text{)}}%
}+\operatorname*{weight}\mathfrak{n}=n-k+i+\operatorname*{weight}%
\mathfrak{n}\\
&  <\operatorname*{weight}\mathfrak{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\operatorname*{weight}\mathfrak{m}>n-k+i+\operatorname*{weight}%
\mathfrak{n}\right) \\
&  =N,
\end{align*}
qed.}. Hence, (\ref{pf.lem.span-detail.1.indhyp}) (applied to $x_{i,k}%
\mathfrak{n}$ instead of $\mathfrak{m}$) shows that
\[
x_{i,k}\mathfrak{n}\in\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}.
\]


\item We have $\operatorname*{weight}\mathfrak{n}<N$%
\ \ \ \ \footnote{\textit{Proof.} Lemma \ref{lem.span-detail.weight.triv}
\textbf{(c)} (applied to $\mathfrak{m}=x_{j,k}$) yields
$\operatorname*{weight}\left(  x_{j,k}\right)  \in\mathbb{N}$. Thus,
$\operatorname*{weight}\left(  x_{j,k}\right)  \geq0$. Also, Lemma
\ref{lem.span-detail.weight.triv} \textbf{(a)} yields $\operatorname*{weight}%
\left(  x_{i,j}\right)  =n-\underbrace{j}_{\leq n}+\underbrace{i}%
_{>0}>n-n+0=0$. Now,%
\begin{align*}
\operatorname*{weight}\underbrace{\mathfrak{m}}_{=x_{i,j}x_{j,k}\mathfrak{n}}
&  =\operatorname*{weight}\left(  x_{i,j}x_{j,k}\mathfrak{n}\right)
=\underbrace{\operatorname*{weight}\left(  x_{i,j}\right)  }_{>0}%
+\underbrace{\operatorname*{weight}\left(  x_{j,k}\mathfrak{n}\right)
}_{\substack{=\operatorname*{weight}\left(  x_{j,k}\right)
+\operatorname*{weight}\mathfrak{n}\\\text{(by Lemma
\ref{lem.span-detail.weight.triv} \textbf{(b)}, applied to }\mathfrak{p}%
=x_{j,k}\text{ and }\mathfrak{q}=\mathfrak{n}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.span-detail.weight.triv} \textbf{(b)}, applied to }\mathfrak{p}%
=x_{i,j}\text{ and }\mathfrak{q}=x_{j,k}\mathfrak{n}\right) \\
&  >0+\underbrace{\operatorname*{weight}\left(  x_{j,k}\right)  }_{\geq
0}+\operatorname*{weight}\mathfrak{n}\geq0+0+\operatorname*{weight}%
\mathfrak{n}=\operatorname*{weight}\mathfrak{n}.
\end{align*}
Hence, $\operatorname*{weight}\mathfrak{n}<\operatorname*{weight}%
\mathfrak{m}=N$, qed.}. Hence, (\ref{pf.lem.span-detail.1.indhyp}) (applied to
$\mathfrak{n}$ instead of $\mathfrak{m}$) shows that
\[
\mathfrak{n}\in\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}.
\]

\end{itemize}

Now, (\ref{pf.lem.span-detail.1.3}) becomes%
\begin{align*}
\mathfrak{m}  &  \in\underbrace{x_{i,k}x_{i,j}\mathfrak{n}}_{\in
\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}}+\underbrace{x_{i,k}%
x_{j,k}\mathfrak{n}}_{\in\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}%
}+\beta\underbrace{x_{i,k}\mathfrak{n}}_{\in\mathcal{X}%
_{\operatorname*{pathless}}+\mathcal{J}}+\alpha\underbrace{\mathfrak{n}}%
_{\in\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}}%
+\underbrace{\mathcal{J}}_{\subseteq\mathcal{X}_{\operatorname*{pathless}%
}+\mathcal{J}}\\
&  \subseteq\left(  \mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}\right)
+\left(  \mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}\right)
+\beta\left(  \mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\alpha\left(  \mathcal{X}_{\operatorname*{pathless}%
}+\mathcal{J}\right)  +\left(  \mathcal{X}_{\operatorname*{pathless}%
}+\mathcal{J}\right) \\
&  \subseteq\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}%
\end{align*}
(since $\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}$ is a $\mathbf{k}$-module).

Now, let us forget that we fixed $\mathfrak{m}$. We thus have shown that if
$\mathfrak{m}\in\mathfrak{M}$ is a monomial such that $\operatorname*{weight}%
\mathfrak{m}=N$, then $\mathfrak{m}\in\mathcal{X}_{\operatorname*{pathless}%
}+\mathcal{J}$. In other words, Lemma \ref{lem.span-detail.1} holds in the
case when $\operatorname*{weight}\mathfrak{m}=N$. This completes the induction
step. Thus, Lemma \ref{lem.span-detail.1} is proven.
\end{proof}

Now, we can prove Proposition \ref{prop.path-red.span}:

\begin{proof}
[Proof of Proposition \ref{prop.path-red.span}.]Lemma \ref{lem.span-detail.1}
shows that $\mathfrak{m}\in\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}$
for each $\mathfrak{m}\in\mathfrak{M}$. In other words, $\mathfrak{M}%
\subseteq\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}$ (where we
consider $\mathfrak{M}$ as being embedded into the polynomial ring
$\mathcal{X}$).

For any subset $\mathcal{W}$ of $\mathcal{X}$, we let $\operatorname*{span}%
\mathcal{W}$ denote the $\mathbf{k}$-submodule of $\mathcal{X}$ spanned by
$\mathcal{W}$. The set $\mathfrak{M}$ spans the $\mathbf{k}$-module
$\mathcal{X}$ (since each polynomial $f\in\mathcal{X}$ is a $\mathbf{k}%
$-linear combination of monomials $\mathfrak{m}\in\mathfrak{M}$). In other
words, $\mathcal{X}=\operatorname*{span}\mathfrak{M}$. But from $\mathfrak{M}%
\subseteq\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}$, we obtain
$\operatorname*{span}\mathfrak{M}\subseteq\operatorname*{span}\left(
\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}\right)  =\mathcal{X}%
_{\operatorname*{pathless}}+\mathcal{J}$ (since $\mathcal{X}%
_{\operatorname*{pathless}}+\mathcal{J}$ is a $\mathbf{k}$-submodule of
$\mathcal{X}$). Hence, $\mathcal{X}=\operatorname*{span}\mathfrak{M}%
\subseteq\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}$.

\begin{noncompile}
In other words, $\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}$ contains
$\mathfrak{M}$ as a subset.

But the set $\mathfrak{M}$ is a spanning set of the $\mathbf{k}$-module
$\mathcal{X}$ (since each polynomial $f\in\mathcal{X}$ is a $\mathbf{k}%
$-linear combination of monomials $\mathfrak{m}\in\mathfrak{M}$). Therefore,
$\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}$ contains a spanning set
of $\mathcal{X}$ as a subset (namely, the spanning set $\mathfrak{M}$).
Therefore, $\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}$ must contain
the whole $\mathcal{X}$ as a subset (since $\mathcal{X}%
_{\operatorname*{pathless}}+\mathcal{J}$ is a $\mathbf{k}$-submodule of
$\mathcal{X}$). In other words, $\mathcal{X}\subseteq\mathcal{X}%
_{\operatorname*{pathless}}+\mathcal{J}$.
\end{noncompile}

Now, $p\in\mathcal{X}\subseteq\mathcal{X}_{\operatorname*{pathless}%
}+\mathcal{J}$. In other words, there exist $u\in\mathcal{X}%
_{\operatorname*{pathless}}$ and $v\in\mathcal{J}$ such that $p=u+v$. Consider
these $u$ and $v$.

We have $p=u+\underbrace{v}_{\in\mathcal{J}}\in u+\mathcal{J}$. In other
words, $p\equiv u\operatorname{mod}\mathcal{J}$. But $\mathcal{X}%
_{\operatorname*{pathless}}$ is the set of all pathless polynomials in
$\mathcal{X}$. Thus, $u$ is a pathless polynomial (since $u\in\mathcal{X}%
_{\operatorname*{pathless}}$). Hence, there exists a pathless polynomial
$q\in\mathcal{X}$ such that $p\equiv q\operatorname{mod}\mathcal{J}$ (namely,
$q=u$). This proves Proposition \ref{prop.path-red.span}.
\end{proof}
\end{verlong}

\section{\label{sect.proof}The proof}

\subsection{Preliminaries}

The proof of Theorem~\ref{thm.t-red.unique} will occupy most of this paper. It
proceeds in several steps. First, we shall define four $\mathbf{k}$-algebras
$\mathcal{Q}$, $\mathcal{T}^{\prime}\left[  \left[  w\right]  \right]  $,
$\mathcal{T}$ and $\mathcal{T}\left[  \left[  w\right]  \right]  $ (with
$\mathcal{T}^{\prime}$ being a subalgebra of $\mathcal{T}$) and three
$\mathbf{k}$-linear maps $A$, $B$ and $E$ (with $A$ and $E$ being $\mathbf{k}%
$-algebra homomorphisms) forming a diagram%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%\calX\ar[r]^{A} \ar[dr]_{D} & \calQ\ar[r]^-{B} & \calT\left[\left
%[w\right]\right] \\
%& \calT^\prime\ar[r]_-{E} & \calT^\prime\left[\left[w\right]\right] \arinj[u]
%}}}%
%BeginExpansion
\xymatrix{
\calX\ar[r]^{A} \ar[dr]_{D} & \calQ\ar[r]^-{B} & \calT\left[\left
[w\right]\right] \\
& \calT^\prime\ar[r]_-{E} & \calT^\prime\left[\left[w\right]\right] \arinj[u]
}%
%EndExpansion
\]
(where the vertical arrow is a canonical injection) that is \textbf{not}
commutative. We shall eventually show that:

\begin{itemize}
\item (Proposition \ref{prop.A.killsJ} below) the homomorphism $A$ annihilates
the ideal $\mathcal{J}$,

\item (Proposition \ref{prop.E.inj} below) the homomorphism $E$ is injective, and

\item (Corollary \ref{cor.pathless.D} below) each pathless polynomial $q$
satisfies $\left(  E\circ D\right)  \left(  q\right)  =\left(  B\circ
A\right)  \left(  q\right)  $ (the equation makes sense since $\mathcal{T}%
^{\prime}\left[  \left[  w\right]  \right]  \subseteq\mathcal{T}\left[
\left[  w\right]  \right]  $).
\end{itemize}

These three facts will allow us to prove Theorem~\ref{thm.t-red.unique}.
Indeed, the first and the third will imply that each pathless polynomial in
$\mathcal{J}$ is annihilated by $E\circ D$; because of the second, this will
show that it is also annihilated by $D$; and from here,
Theorem~\ref{thm.t-red.unique} will easily follow.

\subsection{The algebra $\mathcal{Q}$ of Laurent series}

Let us begin by defining the notion of (formal) Laurent series in $n$
indeterminates $r_{1},r_{2},\ldots,r_{n}$. This is somewhat slippery terrain,
and it is easy to accidentally get a non-working definition (e.g., a notion of
\textquotedblleft Laurent series\textquotedblright\ not closed under
multiplication, or not allowing multiplication at all), but there are also
several different working definitions (see, e.g., \cite{MonKau13} for a
systematic treatment revealing many degrees of freedom). The definition we
shall give here has been tailored to make our constructions work.

We begin by defining a $\mathbf{k}$-module $\mathcal{Q}^{\pm}$ of
\textquotedblleft two-sided infinite formal power series over $\mathbf{k}%
$\textquotedblright; this is not going to be a ring:

\begin{definition}
\label{def.Qbig}Consider $n$ distinct symbols $r_{1},r_{2},\ldots,r_{n}$. Let
$\mathfrak{R}$ denote the free abelian group on these $n$ symbols, written
multiplicatively. (That is, $\mathfrak{R}$ is the free $\mathbb{Z}$-module on
$n$ generators $r_{1},r_{2},\ldots,r_{n}$, but with the addition renamed as
multiplication.) The elements of $\mathfrak{R}$ thus have the form
$r_{1}^{a_{1}}r_{2}^{a_{2}}\cdots r_{n}^{a_{n}}$ for $\left(  a_{1}%
,a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}$; we shall refer to such
elements as \textit{Laurent monomials} in the symbols $r_{1},r_{2}%
,\ldots,r_{n}$.

Informally, we let $\mathcal{Q}^{\pm}$ denote the $\mathbf{k}$-module of all
\textquotedblleft infinite $\mathbf{k}$-linear combinations\textquotedblright%
\ of Laurent monomials. Formally speaking, we define $\mathcal{Q}^{\pm}$ as
the direct product $\prod_{\mathfrak{r}\in\mathfrak{R}}\mathbf{k}$ of copies
of $\mathbf{k}$ indexed by Laurent monomials. We want to write each element
$\left(  \lambda_{\mathfrak{r}}\right)  _{\mathfrak{r}\in\mathfrak{R}}\in
\prod_{\mathfrak{r}\in\mathfrak{R}}\mathbf{k}$ of this direct product as the
formal $\mathbf{k}$-linear combination $\sum_{\mathfrak{r}\in\mathfrak{R}%
}\lambda_{\mathfrak{r}}\mathfrak{r}$; in order for this to work, we make
several further conventions: First, we identify each Laurent monomial
$\mathfrak{s}\in\mathfrak{R}$ with the element $\left(  \delta_{\mathfrak{s}%
,\mathfrak{r}}\right)  _{\mathfrak{r}\in\mathfrak{R}}$ of $\mathcal{Q}^{\pm}$
(where $\delta_{\mathfrak{s},\mathfrak{r}}$ is the Kronecker delta). Second,
we equip the $\mathbf{k}$-module $\mathcal{Q}^{\pm}$ with a topology: namely,
the product topology, defined by recalling that it is a direct product
$\prod_{\mathfrak{r}\in\mathfrak{R}}\mathbf{k}$ of copies of $\mathbf{k}$
(each of which is equipped with the discrete topology). Having made these
conventions, we can easily verify that each element $\left(  \lambda
_{\mathfrak{r}}\right)  _{\mathfrak{r}\in\mathfrak{R}}$ of $\prod
_{\mathfrak{r}\in\mathfrak{R}}\mathbf{k}=\mathcal{Q}^{\pm}$ is indeed
identical with the infinite sum $\sum_{\mathfrak{r}\in\mathfrak{R}}%
\lambda_{\mathfrak{r}}\mathfrak{r}$ (which makes sense because of the topology
on $\mathcal{Q}^{\pm}$). As usual, if $f=\left(  \lambda_{\mathfrak{r}%
}\right)  _{\mathfrak{r}\in\mathfrak{R}}$ is an element of $\mathcal{Q}^{\pm}%
$, then $\lambda_{\mathfrak{r}}$ (for a given $\mathfrak{r}\in\mathfrak{R}$)
will be called the \textit{coefficient} of $\mathfrak{r}$ in $f$ and denoted
by $\left[  \mathfrak{r}\right]  f$.
\end{definition}

As we know, the Laurent monomials in $\mathfrak{R}$ have the form
$r_{1}^{a_{1}}r_{2}^{a_{2}}\cdots r_{n}^{a_{n}}$ for $\left(  a_{1}%
,a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}$; thus, sums of the form
$\sum_{\mathfrak{r}\in\mathfrak{R}}\lambda_{\mathfrak{r}}\mathfrak{r}$ can
also be rewritten in the form $\sum_{\left(  a_{1},a_{2},\ldots,a_{n}\right)
\in\mathbb{Z}^{n}}\lambda_{a_{1},a_{2},\ldots,a_{n}}r_{1}^{a_{1}}r_{2}^{a_{2}%
}\cdots r_{n}^{a_{n}}$; this is the usual way in which elements of
$\mathcal{Q}^{\pm}$ are written.

For example, for $n=1$, an element of $\mathcal{Q}^{\pm}$ will have the form
$\sum_{a\in\mathbb{Z}}\lambda_{a}r_{1}^{a}$ for some family $\left(
\lambda_{a}\right)  _{a\in\mathbb{Z}}$ of elements of $\mathbf{k}$. Already in
this simple situation, we see that $\mathcal{Q}^{\pm}$ is not a ring (or, at
least, the usual recipe for multiplying power series does not work in
$\mathcal{Q}^{\pm}$): Multiplying $\sum_{a\in\mathbb{Z}}r_{1}^{a}$ with itself
would result in
\[
\left(  \sum_{a\in\mathbb{Z}}r_{1}^{a}\right)  \left(  \sum_{a\in\mathbb{Z}%
}r_{1}^{a}\right)  =\sum_{\left(  a,b\right)  \in\mathbb{Z}^{2}}r_{1}^{a+b},
\]
which is not a convergent sum in any reasonable topology (it contains each
Laurent monomial infinitely many times). We shall define Laurent series as a
subring of $\mathcal{Q}^{\pm}$:

\begin{definition}
\label{def.laurent}\textbf{(a)} If $d$ is an integer and $\mathfrak{r}%
\in\mathfrak{R}$ is a Laurent monomial, then we say that $\mathfrak{r}$
\textit{lives above }$d$ if and only if $\mathfrak{r}=r_{1}^{a_{1}}%
r_{2}^{a_{2}}\cdots r_{n}^{a_{n}}$ for some $\left(  a_{1},a_{2},\ldots
,a_{n}\right)  \in\left\{  d,d+1,d+2,\ldots\right\}  ^{n}$.

\textbf{(b)} If $d$ is an integer and $f$ is an element of $\mathcal{Q}^{\pm}%
$, then we say that $f$ is \textit{supported above }$d$ if and only if every
$\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}\setminus\left\{
d,d+1,d+2,\ldots\right\}  ^{n}$ satisfies $\left[  r_{1}^{a_{1}}r_{2}^{a_{2}%
}\cdots r_{n}^{a_{n}}\right]  f=0$. In other words, $f$ is supported above $d$
if and only if $f$ is an infinite $\mathbf{k}$-linear combination of Laurent
monomials that live above $d$.

\textbf{(c)} An element $f\in\mathcal{Q}^{\pm}$ is said to be a
\textit{Laurent series} if and only if there exists some $d\in\mathbb{Z}$ such
that $f$ is supported above $d$.

\textbf{(d)} We let $\mathbf{k}\left(  \left(  r_{1},r_{2},\ldots
,r_{n}\right)  \right)  $ denote the $\mathbf{k}$-submodule of $\mathcal{Q}%
^{\pm}$ consisting of all Laurent series.

\textbf{(e)} A multiplication can be defined on $\mathbf{k}\left(  \left(
r_{1},r_{2},\ldots,r_{n}\right)  \right)  $ by extending the multiplication in
the group $\mathfrak{R}$ (in such a way that the resulting map is bilinear and
continuous). Explicitly, this means that if $f=\sum_{\mathfrak{r}%
\in\mathfrak{R}}\lambda_{\mathfrak{r}}\mathfrak{r}$ and $g=\sum_{\mathfrak{r}%
\in\mathfrak{R}}\mu_{\mathfrak{r}}\mathfrak{r}$ are two Laurent series, then
their product $fg$ is defined as the Laurent series%
\[
\left(  \sum_{\mathfrak{r}\in\mathfrak{R}}\lambda_{\mathfrak{r}}%
\mathfrak{r}\right)  \left(  \sum_{\mathfrak{r}\in\mathfrak{R}}\mu
_{\mathfrak{r}}\mathfrak{r}\right)  =\sum_{\mathfrak{u}\in\mathfrak{R}}%
\sum_{\mathfrak{v}\in\mathfrak{R}}\lambda_{\mathfrak{u}}\mu_{\mathfrak{v}%
}\mathfrak{uv}=\sum_{\mathfrak{r}\in\mathfrak{R}}\left(  \sum
_{\substack{\left(  \mathfrak{u},\mathfrak{v}\right)  \in\mathfrak{R}%
^{2};\\\mathfrak{uv}=\mathfrak{r}}}\lambda_{\mathfrak{u}}\mu_{\mathfrak{v}%
}\right)  \mathfrak{r}.
\]
The inner sum $\sum_{\substack{\left(  \mathfrak{u},\mathfrak{v}\right)
\in\mathfrak{R}^{2};\\\mathfrak{uv}=\mathfrak{r}}}\lambda_{\mathfrak{u}}%
\mu_{\mathfrak{v}}$ here is well-defined, because all but finitely many of its
addends are zero. (In fact, if $f$ is supported above $d$, and if $g$ is
supported above $e$, then (for each given $\mathfrak{r}\in\mathfrak{R}$) there
are only finitely many pairs $\left(  \mathfrak{u},\mathfrak{v}\right)
\in\mathfrak{R}^{2}$ such that $\mathfrak{uv}=\mathfrak{r}$ and $\mathfrak{u}$
lives above $d$ and $\mathfrak{v}$ lives above $e$; but these are the only
pairs that can contribute nonzero addends to the sum $\sum_{\substack{\left(
\mathfrak{u},\mathfrak{v}\right)  \in\mathfrak{R}^{2};\\\mathfrak{uv}%
=\mathfrak{r}}}\lambda_{\mathfrak{u}}\mu_{\mathfrak{v}}$.)

Thus, $\mathbf{k}\left(  \left(  r_{1},r_{2},\ldots,r_{n}\right)  \right)  $
becomes a $\mathbf{k}$-algebra with unity $1=r_{1}^{0}r_{2}^{0}\cdots
r_{n}^{0}$. We denote this $\mathbf{k}$-algebra by $\mathcal{Q}$. Note that
$\mathcal{Q}$ is a topological $\mathbf{k}$-algebra; its topology is inherited
from $\mathcal{Q}^{\pm}$.

\textbf{(f)} An element $f\in\mathcal{Q}^{\pm}$ is said to be a \textit{formal
power series} if and only if $f$ is supported above $0$.

\textbf{(g)} We let $\mathbf{k}\left[  \left[  r_{1},r_{2},\ldots
,r_{n}\right]  \right]  $ denote the $\mathbf{k}$-submodule of $\mathcal{Q}%
^{\pm}$ consisting of all formal power series. Thus, $\mathbf{k}\left[
\left[  r_{1},r_{2},\ldots,r_{n}\right]  \right]  \subseteq\mathcal{Q}%
\subseteq\mathcal{Q}^{\pm}$.
\end{definition}

We now define certain Laurent monomials $q_{1},q_{2},\ldots,q_{n}$ that we
shall often use:

\begin{definition}
\label{def.Q'}For each $i\in\left[  n\right]  $, we define a Laurent monomial
$q_{i}$ in the indeterminates $r_{1},r_{2},\ldots,r_{n}$ by $q_{i}%
=r_{i}r_{i+1}\cdots r_{n}$. Notice that this $q_{i}$ is an actual monomial,
not only a Laurent monomial.

Notice that each Laurent monomial in $\mathfrak{R}$ belongs to $\mathcal{Q}$.
Each of the elements $q_{1},q_{2},\ldots,q_{n}$ of $\mathcal{Q}$ is a Laurent
monomial, and thus has an inverse (in $\mathfrak{R}$ and thus also in
$\mathcal{Q}$). Hence, it makes sense to speak of quotients such as
$q_{i}/q_{j}$ for $1\leq i\leq j\leq n$. Explicitly, $q_{i}/q_{j}=r_{i}%
r_{i+1}\cdots r_{j-1}$ whenever $1\leq i\leq j\leq n$. Thus, for any
$i\in\left[  n\right]  $ and $j\in\left[  n\right]  $ satisfying $i<j$, the
difference $1-q_{i}/q_{j}=1-r_{i}r_{i+1}\cdots r_{j-1}$ is a formal power
series in $\mathbf{k}\left[  \left[  r_{1},r_{2},\ldots,r_{n}\right]  \right]
$ having constant term $1$; it is therefore invertible in $\mathbf{k}\left[
\left[  r_{1},r_{2},\ldots,r_{n}\right]  \right]  $.
\end{definition}

It is easy to see that%
\begin{equation}
q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}=r_{1}^{a_{1}}r_{2}^{a_{1}%
+a_{2}}r_{3}^{a_{1}+a_{2}+a_{3}}\cdots r_{n}^{a_{1}+a_{2}+\cdots+a_{n}}
\label{eq.Q'.reindex}%
\end{equation}
for all $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}$. Also,%
\begin{equation}
r_{1}^{b_{1}}r_{2}^{b_{2}}\cdots r_{n}^{b_{n}}=q_{1}^{b_{1}}q_{2}^{b_{2}%
-b_{1}}q_{3}^{b_{3}-b_{2}}\cdots q_{n}^{b_{n}-b_{n-1}} \label{eq.Q'.reindex2}%
\end{equation}
for all $\left(  b_{1},b_{2},\ldots,b_{n}\right)  \in\mathbb{Z}^{n}$. Thus,
each Laurent monomial $\mathfrak{r}\in\mathfrak{R}$ can be written uniquely in
the form $q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ with $\left(
a_{1},a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}$. Thus,\newline$\left(
q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}\right)  _{\left(  a_{1}%
,a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}}$ is a topological
basis\footnote{The notion of a \textquotedblleft topological
basis\textquotedblright\ that we are using here has nothing to do with the
concept of a basis of a topology (also known as \textquotedblleft
base\textquotedblright). Instead, it is merely an analogue of the concept of a
basis of a $\mathbf{k}$-module. It is defined as follows:
\par
A \textit{topological basis} of a topological $\mathbf{k}$-module
$\mathcal{M}$ means a family $\left(  m_{s}\right)  _{s\in\mathfrak{S}}%
\in\mathcal{M}^{\mathfrak{S}}$ with the following two properties:
\par
\begin{itemize}
\item For each family $\left(  \lambda_{s}\right)  _{s\in\mathfrak{S}}%
\in\mathbf{k}^{\mathfrak{S}}$, the sum $\sum_{s\in\mathfrak{S}}\lambda
_{s}m_{s}$ converges with respect to the topology on $\mathcal{M}$. (Such a
sum is called an \textit{infinite} $\mathbf{k}$\textit{-linear combination} of
the family $\left(  m_{s}\right)  _{s\in\mathfrak{S}}$.)
\par
\item Each element of $\mathcal{M}$ can be uniquely represented in the form
$\sum_{s\in\mathfrak{S}}\lambda_{s}m_{s}$ for some family $\left(  \lambda
_{s}\right)  _{s\in\mathfrak{S}}\in\mathbf{k}^{\mathfrak{S}}$.
\end{itemize}
\par
For example, $\left(  r_{1}^{b_{1}}r_{2}^{b_{2}}\cdots r_{n}^{b_{n}}\right)
_{\left(  b_{1},b_{2},\ldots,b_{n}\right)  \in\mathbb{N}^{n}}$ is a
topological basis of the topological $\mathbf{k}$-module $\mathbf{k}\left[
\left[  r_{1},r_{2},\ldots,r_{n}\right]  \right]  $, because each power series
can be uniquely represented as an infinite $\mathbf{k}$-linear combination of
all the monomials.} of the $\mathbf{k}$-module $\mathcal{Q}^{\pm}$.

\subsection{The algebra homomorphism $A:\mathcal{X}\rightarrow\mathcal{Q}$}

\begin{definition}
\label{def.A}Define a $\mathbf{k}$-algebra homomorphism $A:\mathcal{X}%
\rightarrow\mathcal{Q}$ by%
\[
A\left(  x_{i,j}\right)  =-\dfrac{q_{i}+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}%
}\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  i,j\right)  \in\left[  n\right]
^{2}\text{ satisfying }i<j.
\]
Notice that this is well-defined, since all denominators appearing here are
invertible (indeed, $q_{j}$ is an invertible Laurent monomial in
$\mathfrak{R}$, and $1-q_{i}/q_{j}$ is an invertible formal power series in
$\mathbf{k}\left[  \left[  r_{1},r_{2},\ldots,r_{n}\right]  \right]  $).
\end{definition}

\begin{proposition}
\label{prop.A.killsJ}We have $A\left(  \mathcal{J}\right)  =0$.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.A.killsJ}.]The ideal $\mathcal{J}$ of
$\mathcal{X}$ is generated by all elements of the form $x_{i,j}x_{j,k}%
-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  -\alpha$ for all triples
$\left(  i,j,k\right)  \in\left[  n\right]  ^{3}$ satisfying $i<j<k$. Thus, it
suffices to show that $A\left(  x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}%
+x_{j,k}+\beta\right)  -\alpha\right)  =0$ for all such triples. But this is a
straightforward computation (see \cite{verlong} for the details).
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.A.killsJ}.]The ideal $\mathcal{J}$ of
$\mathcal{X}$ is generated by all elements of the form $x_{i,j}x_{j,k}%
-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  -\alpha$ for $\left(
i,j,k\right)  \in\left[  n\right]  ^{3}$ satisfying $i<j<k$. Thus, it suffices
to show that $A\left(  x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}%
+\beta\right)  -\alpha\right)  =0$ for all triples $\left(  i,j,k\right)
\in\left[  n\right]  ^{3}$ satisfying $i<j<k$. So let us fix such a triple.

The definition of $A$ yields $A\left(  x_{i,j}\right)  =-\dfrac{q_{i}%
+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}}$. Similarly, $A\left(  x_{j,k}\right)
=-\dfrac{q_{j}+\beta+\alpha/q_{k}}{1-q_{j}/q_{k}}$ and $A\left(
x_{i,k}\right)  =-\dfrac{q_{i}+\beta+\alpha/q_{k}}{1-q_{i}/q_{k}}$.

But $A$ is a $\mathbf{k}$-algebra homomorphism. Thus,%
\begin{align*}
&  A\left(  x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)
-\alpha\right) \\
&  =\underbrace{A\left(  x_{i,j}\right)  }_{=-\dfrac{q_{i}+\beta+\alpha/q_{j}%
}{1-q_{i}/q_{j}}}\underbrace{A\left(  x_{j,k}\right)  }_{=-\dfrac{q_{j}%
+\beta+\alpha/q_{k}}{1-q_{j}/q_{k}}}\\
&  \ \ \ \ \ \ \ \ \ \ -\underbrace{A\left(  x_{i,k}\right)  }_{=-\dfrac
{q_{i}+\beta+\alpha/q_{k}}{1-q_{i}/q_{k}}}\left(  \underbrace{A\left(
x_{i,j}\right)  }_{=-\dfrac{q_{i}+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}}%
}+\underbrace{A\left(  x_{j,k}\right)  }_{=-\dfrac{q_{j}+\beta+\alpha/q_{k}%
}{1-q_{j}/q_{k}}}+\beta\right)  -\alpha\\
&  =\left(  -\dfrac{q_{i}+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}}\right)  \left(
-\dfrac{q_{j}+\beta+\alpha/q_{k}}{1-q_{j}/q_{k}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  -\dfrac{q_{i}+\beta+\alpha/q_{k}}%
{1-q_{i}/q_{k}}\right)  \left(  \left(  -\dfrac{q_{i}+\beta+\alpha/q_{j}%
}{1-q_{i}/q_{j}}\right)  +\left(  -\dfrac{q_{j}+\beta+\alpha/q_{k}}%
{1-q_{j}/q_{k}}\right)  +\beta\right)  -\alpha\\
&  =0,
\end{align*}
as an unpleasant but straightforward computation reveals. This proves
Proposition \ref{prop.A.killsJ}.
\end{proof}
\end{verlong}

\subsection{The algebras $\mathcal{T}$ and $\mathcal{T}\left[  \left[
w\right]  \right]  $ of power series}

\begin{definition}
\textbf{(a)} Let $\mathcal{T}$ be the topological $\mathbf{k}$-algebra
$\mathbf{k}\left[  \left[  t_{1},t_{2},\ldots,t_{n}\right]  \right]  $. This
is the ring of formal power series in the $n$ indeterminates $t_{1}%
,t_{2},\ldots,t_{n}$ over $\mathbf{k}$.

The topology on $\mathcal{T}$ shall be the usual one (i.e., the one defined
similarly to the one on $\mathcal{Q}^{\pm}$).

\textbf{(b)} We shall regard the canonical injections%
\[
\mathcal{T}^{\prime}=\mathbf{k}\left[  t_{1},t_{2},\ldots,t_{n-1}\right]
\hookrightarrow\mathbf{k}\left[  t_{1},t_{2},\ldots,t_{n}\right]
\hookrightarrow\mathbf{k}\left[  \left[  t_{1},t_{2},\ldots,t_{n}\right]
\right]  =\mathcal{T}%
\]
as inclusions. Thus, $\mathcal{T}^{\prime}$ becomes a $\mathbf{k}$-subalgebra
of $\mathcal{T}$. Hence, $D:\mathcal{X}\rightarrow\mathcal{T}^{\prime}$
becomes a $\mathbf{k}$-algebra homomorphism $\mathcal{X}\rightarrow
\mathcal{T}$.

\textbf{(c)} We consider the $\mathbf{k}$-algebras $\mathcal{T}\left[  \left[
w\right]  \right]  $ and $\mathcal{T}^{\prime}\left[  \left[  w\right]
\right]  $. These are the $\mathbf{k}$-algebras of formal power series in a
(new) indeterminate $w$ over $\mathcal{T}$ and over $\mathcal{T}^{\prime}$,
respectively. We endow the $\mathbf{k}$-algebra $\mathcal{T}\left[  \left[
w\right]  \right]  $ with a topology defined as the product topology, where
$\mathcal{T}\left[  \left[  w\right]  \right]  $ is identified with a direct
product of infinitely many copies of $\mathcal{T}$ (each of which is equipped
with the topology we previously defined).
\end{definition}

\subsection{The continuous $\mathbf{k}$-linear map $B:\mathcal{Q}%
\rightarrow\mathcal{T}\left[  \left[  w\right]  \right]  $}

We have $\mathcal{T}=\mathbf{k}\left[  \left[  t_{1},t_{2},\ldots
,t_{n}\right]  \right]  $. Thus, $\mathcal{T}\left[  \left[  w\right]
\right]  $ can be regarded as the ring of formal power series in the $n+1$
indeterminates $t_{1},t_{2},\ldots,t_{n},w$ over $\mathbf{k}$. (Strictly
speaking, this should say that there is a canonical topological $\mathbf{k}%
$-algebra isomorphism from $\mathcal{T}\left[  \left[  w\right]  \right]  $ to
the latter ring). Let us now show a simple lemma:

\begin{lemma}
\label{lem.B.wd1}Let $\mathfrak{m}$ be a monomial in the indeterminates
$t_{1},t_{2},\ldots,t_{n},w$ (with nonnegative exponents). Then, there exist
only finitely many $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}%
^{n}$ satisfying
\begin{equation}
\left(  \prod_{\substack{i\in\left[  n\right]  ;\\a_{i}>0}}t_{i}^{a_{i}%
}\right)  \left(  \prod_{\substack{i\in\left[  n\right]  ;\\a_{i}<0}%
}w^{-a_{i}}\right)  =\mathfrak{m}. \label{eq.lem.B.wd1.eq}%
\end{equation}

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.B.wd1}.]Write $\mathfrak{m}$ in the form
$\mathfrak{m}=\left(  \prod_{i\in\left[  n\right]  }t_{i}^{b_{i}}\right)
w^{c}$ for some nonnegative integers $b_{1},b_{2},\ldots,b_{n},c$. Let $S$ be
the finite set $\left\{  -c,-c+1,\ldots,0\right\}  \cup\left\{  b_{1}%
,b_{2},\ldots,b_{n}\right\}  $. Hence, $S^{n}$ is also a finite set.

We want to prove that there exist only finitely many $\left(  a_{1}%
,a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}$ satisfying
(\ref{eq.lem.B.wd1.eq}). We shall show that each such $\left(  a_{1}%
,a_{2},\ldots,a_{n}\right)  $ belongs to the set $S^{n}$.

Indeed, let $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}$
satisfy (\ref{eq.lem.B.wd1.eq}). We must show that $\left(  a_{1},a_{2}%
,\ldots,a_{n}\right)  $ belongs to $S^{n}$.

Let $j\in\left[  n\right]  $. We want to prove that $a_{j}\in S$.

We know that (\ref{eq.lem.B.wd1.eq}) holds. Thus%
\[
\left(  \prod_{\substack{i\in\left[  n\right]  ;\\a_{i}>0}}t_{i}^{a_{i}%
}\right)  \left(  \prod_{\substack{i\in\left[  n\right]  ;\\a_{i}<0}%
}w^{-a_{i}}\right)  =\mathfrak{m}=\left(  \prod_{i\in\left[  n\right]  }%
t_{i}^{b_{i}}\right)  w^{c}.
\]


\begin{verlong}
Thus,%
\[
\left(  \prod_{i\in\left[  n\right]  }t_{i}^{b_{i}}\right)  w^{c}=\left(
\prod_{\substack{i\in\left[  n\right]  ;\\a_{i}>0}}t_{i}^{a_{i}}\right)
\left(  \prod_{\substack{i\in\left[  n\right]  ;\\a_{i}<0}}w^{-a_{i}}\right)
=\left(  \prod_{\substack{i\in\left[  n\right]  ;\\a_{i}>0}}t_{i}^{a_{i}%
}\right)  w^{\sum_{\substack{i\in\left[  n\right]  ;\\a_{i}<0}}\left(
-a_{i}\right)  }.
\]

\end{verlong}

\noindent This is an equality between two monomials in the indeterminates
$t_{1},t_{2},\ldots,t_{n},w$. Comparing exponents on both sides of this
equality, we find that%
\begin{equation}
b_{i}=%
\begin{cases}
a_{i}, & \text{if }a_{i}>0;\\
0, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left[  n\right]
\label{pf.lem.B.wd1.bi=}%
\end{equation}
and%
\begin{equation}
c=\sum_{\substack{i\in\left[  n\right]  ;\\a_{i}<0}}\left(  -a_{i}\right)  .
\label{pf.lem.B.wd1.c=}%
\end{equation}


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

\textit{Case 1:} We have $a_{j}<0$.

\textit{Case 2:} We have $a_{j}=0$.

\textit{Case 3:} We have $a_{j}>0$.

\begin{vershort}
Let us first consider Case 1. In this case, we have $a_{j}<0$. Thus, $-a_{j}$
is one of the addends in the sum $\sum_{\substack{i\in\left[  n\right]
;\\a_{i}<0}}\left(  -a_{i}\right)  $. Since this sum is greater or equal to
each of its addends (because its addends are positive), we thus obtain
$\sum_{\substack{i\in\left[  n\right]  ;\\a_{i}<0}}\left(  -a_{i}\right)
\geq-a_{j}$. Hence, (\ref{pf.lem.B.wd1.c=}) becomes $c=\sum_{\substack{i\in
\left[  n\right]  ;\\a_{i}<0}}\left(  -a_{i}\right)  \geq-a_{j}$. In other
words, $a_{j}\geq-c$. Combining this with $a_{j}<0$, we find
\[
a_{j}\in\left\{  -c,-c+1,\ldots,-1\right\}  \subseteq\left\{  -c,-c+1,\ldots
,0\right\}  \cup\left\{  b_{1},b_{2},\ldots,b_{n}\right\}  =S.
\]
Thus, $a_{j}\in S$ is proven in Case 1.
\end{vershort}

\begin{verlong}
Let us first consider Case 1. In this case, we have $a_{j}<0$. Thus, we can
split off the addend for $i=j$ from the sum $\sum_{\substack{i\in\left[
n\right]  ;\\a_{i}<0}}\left(  -a_{i}\right)  $. We thus find%
\[
\sum_{\substack{i\in\left[  n\right]  ;\\a_{i}<0}}\left(  -a_{i}\right)
=\left(  -a_{j}\right)  +\sum_{\substack{i\in\left[  n\right]  ;\\a_{i}%
<0;\\i\neq j}}\underbrace{\left(  -a_{i}\right)  }_{\substack{>0\\\text{(since
}a_{i}<0\text{)}}}\geq\left(  -a_{j}\right)  +\underbrace{\sum_{\substack{i\in
\left[  n\right]  ;\\a_{i}<0;\\i\neq j}}0}_{=0}=-a_{j}.
\]
Hence, (\ref{pf.lem.B.wd1.c=}) becomes $c=\sum_{\substack{i\in\left[
n\right]  ;\\a_{i}<0}}\left(  -a_{i}\right)  \geq-a_{j}$. In other words,
$a_{j}\geq-c$. Combining this with $a_{j}<0$, we find
\begin{align*}
a_{j}  &  \in\left\{  -c,-c+1,\ldots,-1\right\}  \subseteq\left\{
-c,-c+1,\ldots,0\right\} \\
&  \subseteq\left\{  -c,-c+1,\ldots,0\right\}  \cup\left\{  b_{1},b_{2}%
,\ldots,b_{n}\right\}  =S.
\end{align*}
Thus, $a_{j}\in S$ is proven in Case 1.
\end{verlong}

Let us now consider Case 2. In this case, we have $a_{j}=0$. Hence,
\[
a_{j}\in\left\{  -c,-c+1,\ldots,0\right\}  \subseteq\left\{  -c,-c+1,\ldots
,0\right\}  \cup\left\{  b_{1},b_{2},\ldots,b_{n}\right\}  =S.
\]
Thus, $a_{j}\in S$ is proven in Case 2.

Let us finally consider Case 3. In this case, we have $a_{j}>0$. Applying
(\ref{pf.lem.B.wd1.bi=}) to $i=j$, we find%
\[
b_{j}=%
\begin{cases}
a_{j}, & \text{if }a_{j}>0;\\
0, & \text{otherwise}%
\end{cases}
=a_{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a_{j}>0\right)  ,
\]
so that%
\[
a_{j}=b_{j}\in\left\{  b_{1},b_{2},\ldots,b_{n}\right\}  \subseteq\left\{
-c,-c+1,\ldots,0\right\}  \cup\left\{  b_{1},b_{2},\ldots,b_{n}\right\}  =S.
\]
Thus, $a_{j}\in S$ is proven in Case 3.

We have now proven $a_{j}\in S$ in all three Cases 1, 2 and 3. Hence,
$a_{j}\in S$ always holds.

Forget that we have fixed $j$. We thus have shown that $a_{j}\in S$ for each
$j\in\left[  n\right]  $. In other words, $\left(  a_{1},a_{2},\ldots
,a_{n}\right)  $ belongs to $S^{n}$.

Now, forget that we have fixed $\left(  a_{1},a_{2},\ldots,a_{n}\right)  $. We
thus have shown that each $\left(  a_{1},a_{2},\ldots,a_{n}\right)
\in\mathbb{Z}^{n}$ satisfying (\ref{eq.lem.B.wd1.eq}) belongs to $S^{n}$.
Since $S^{n}$ is a finite set, this shows that there exist only finitely many
$\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}$ satisfying
(\ref{eq.lem.B.wd1.eq}). This proves Lemma \ref{lem.B.wd1}.
\end{proof}

\begin{definition}
We define a continuous $\mathbf{k}$-linear map $B:\mathcal{Q}^{\pm}%
\rightarrow\mathcal{T}\left[  \left[  w\right]  \right]  $ by setting%
\begin{align*}
B\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}\right)   &  =\left(
\prod_{\substack{i\in\left[  n\right]  ;\\a_{i}>0}}t_{i}^{a_{i}}\right)
\left(  \prod_{\substack{i\in\left[  n\right]  ;\\a_{i}<0}}w^{-a_{i}}\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \text{for each }\left(  a_{1},a_{2},\ldots
,a_{n}\right)  \in\mathbb{Z}^{n}.
\end{align*}
This is well-defined, since $\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\right)  _{\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in
\mathbb{Z}^{n}}$ is a topological basis of $\mathcal{Q}^{\pm}$, and because of
Lemma \ref{lem.B.wd1} (which guarantees convergence when the map $B$ is
applied to an infinite $\mathbf{k}$-linear combination of Laurent monomials).

The $\mathbf{k}$-linear map $B:\mathcal{Q}^{\pm}\rightarrow\mathcal{T}\left[
\left[  w\right]  \right]  $ can be restricted to the $\mathbf{k}$-submodule
$\mathcal{Q}$ of $\mathcal{Q}^{\pm}$. We denote this restriction by $B$ as
well. In the following, we shall only be concerned with this restriction.
\end{definition}

Of course, $B$ is (in general) not a $\mathbf{k}$-algebra homomorphism.

\subsection{The $\mathbf{k}$-algebra monomorphism $E:\mathcal{T}^{\prime
}\rightarrow\mathcal{T}^{\prime}\left[  \left[  w\right]  \right]  $}

\begin{definition}
We define a $\mathbf{k}$-algebra homomorphism $E:\mathcal{T}^{\prime
}\rightarrow\mathcal{T}^{\prime}\left[  \left[  w\right]  \right]  $ by%
\[
E\left(  t_{i}\right)  =-\dfrac{t_{i}+\beta+\alpha w}{1-t_{i}w}%
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left[  n-1\right]  .
\]
This is well-defined (by the universal property of the polynomial ring
$\mathcal{T}^{\prime}=\mathbf{k}\left[  t_{1},t_{2},\ldots,t_{n-1}\right]  $),
because for each $i\in\left[  n-1\right]  $, the power series $1-t_{i}w$ is
invertible in $\mathcal{T}^{\prime}\left[  \left[  w\right]  \right]  $
(indeed, its constant term is $1$).
\end{definition}

\begin{proposition}
\label{prop.E.inj}The homomorphism $E$ is injective.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.E.inj}.]Let $F:\mathcal{T}^{\prime}\left[
\left[  w\right]  \right]  \rightarrow\mathcal{T}^{\prime}$ be the
$\mathcal{T}^{\prime}$-algebra homomorphism that sends each formal power
series $f\in\mathcal{T}^{\prime}\left[  \left[  w\right]  \right]  $ (regarded
as a formal power series in the single indeterminate $w$ over $\mathcal{T}%
^{\prime}$) to its constant term $f\left(  0\right)  \in\mathcal{T}^{\prime}$.
Thus, $F$ is a $\mathbf{k}$-algebra homomorphism, and it sends $w$ to $0$
while sending each element of $\mathcal{T}^{\prime}$ to itself.

Let $G:\mathcal{T}^{\prime}\rightarrow\mathcal{T}^{\prime}$ be the
$\mathbf{k}$-algebra homomorphism that sends $t_{i}$ to $-t_{i}-\beta$ for
each $i\in\left[  n-1\right]  $. (This is well-defined by the universal
property of the polynomial ring $\mathcal{T}^{\prime}=\mathbf{k}\left[
t_{1},t_{2},\ldots,t_{n-1}\right]  $.)

Notice that the map $G\circ F\circ E:\mathcal{T}^{\prime}\rightarrow
\mathcal{T}^{\prime}$ is a $\mathbf{k}$-algebra homomorphism (since it is the
composition of the three $\mathbf{k}$-algebra homomorphisms $E,F,G$).

For each $i\in\left[  n-1\right]  $, we have%
\begin{align*}
\left(  F\circ E\right)  \left(  t_{i}\right)   &  =F\left(
\underbrace{E\left(  t_{i}\right)  }_{=-\dfrac{t_{i}+\beta+\alpha w}{1-t_{i}%
w}}\right)  =F\left(  -\dfrac{t_{i}+\beta+\alpha w}{1-t_{i}w}\right)
=-\dfrac{t_{i}+\beta+\alpha F\left(  w\right)  }{1-t_{i}F\left(  w\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }F\text{ is a }\mathcal{T}^{\prime
}\text{-algebra homomorphism}\right) \\
&  =-\dfrac{t_{i}+\beta+\alpha0}{1-t_{i}0}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }F\left(  w\right)  =0\right) \\
&  =-\left(  t_{i}+\beta\right)
\end{align*}
and thus%
\begin{align*}
\left(  G\circ F\circ E\right)  \left(  t_{i}\right)   &  =G\left(
\underbrace{\left(  F\circ E\right)  \left(  t_{i}\right)  }_{=-\left(
t_{i}+\beta\right)  }\right)  =G\left(  -\left(  t_{i}+\beta\right)  \right)
=-\left(  \underbrace{G\left(  t_{i}\right)  }_{=-t_{i}-\beta}+\beta\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }G\text{ is a }\mathbf{k}%
\text{-algebra homomorphism}\right) \\
&  =-\left(  -t_{i}-\beta+\beta\right)  =t_{i}=\operatorname*{id}\left(
t_{i}\right)  .
\end{align*}
Hence, the two $\mathbf{k}$-algebra homomorphisms $G\circ F\circ
E:\mathcal{T}^{\prime}\rightarrow\mathcal{T}^{\prime}$ and $\operatorname*{id}%
:\mathcal{T}^{\prime}\rightarrow\mathcal{T}^{\prime}$ agree on the generating
set $\left\{  t_{1},t_{2},\ldots,t_{n-1}\right\}  $ of the $\mathbf{k}%
$-algebra $\mathcal{T}^{\prime}$. Thus, these two homomorphisms must be
identical. In other words, $G\circ F\circ E=\operatorname*{id}$. Hence, the
map $E$ has a left inverse, and thus is injective. This proves Proposition
\ref{prop.E.inj}.
\end{proof}

Thus, we have defined the following spaces and maps between them:%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%\calX\ar[r]^{A} \ar[dr]_{D} & \calQ\ar[r]^-{B} & \calT\left[\left
%[w\right]\right] \\
%& \calT^\prime\ar[r]_-{E} & \calT^\prime\left[\left[w\right]\right] \arinj[u]
%}}}%
%BeginExpansion
\xymatrix{
\calX\ar[r]^{A} \ar[dr]_{D} & \calQ\ar[r]^-{B} & \calT\left[\left
[w\right]\right] \\
& \calT^\prime\ar[r]_-{E} & \calT^\prime\left[\left[w\right]\right] \arinj[u]
}%
%EndExpansion
\]
(but this is \textbf{not} a commutative diagram). It is worth reminding
ourselves that $A$, $D$ and $E$ are $\mathbf{k}$-algebra homomorphisms, but
$B$ (in general) is not.

\subsection{Pathless monomials and subsets $S$ of $\left[  n-1\right]  $}

Next, we want to study the action of the compositions $B\circ A$ and $E\circ
D$ on pathless monomials. We first introduce some more notations:

\begin{definition}
\label{def.Ssubset}Let $S$ be a subset of $\left[  n-1\right]  $.

\textbf{(a)} Let $\mathfrak{P}_{S}$ be the set of all pairs $\left(
i,j\right)  \in S\times\left(  \left[  n\right]  \setminus S\right)  $
satisfying $i<j$.

\textbf{(b)} A monomial $\mathfrak{m}\in\mathfrak{M}$ is said to be
$S$\textit{-friendly} if it is a product of some of the indeterminates
$x_{i,j}$ with $\left(  i,j\right)  \in\mathfrak{P}_{S}$. In other words, a
monomial $\mathfrak{m}\in\mathfrak{M}$ is $S$-friendly if and only if every
indeterminate $x_{i,j}$ that appears in $\mathfrak{m}$ satisfies $i\in S$ and
$j\notin S$.

We let $\mathfrak{M}_{S}$ denote the set of all $S$-friendly monomials.

\textbf{(c)} We let $\mathcal{X}_{S}$ denote the polynomial ring
$\mathbf{k}\left[  x_{i,j}\ \mid\ \left(  i,j\right)  \in\mathfrak{P}%
_{S}\right]  $. This is clearly a subring of $\mathcal{X}$. The $\mathbf{k}%
$-module $\mathcal{X}_{S}$ has a basis consisting of all $S$-friendly
monomials $\mathfrak{m}\in\mathfrak{M}$.

\textbf{(d)} An $n$-tuple $\left(  a_{1},a_{2},\ldots,a_{n}\right)
\in\mathbb{Z}^{n}$ is said to be $S$\textit{-adequate} if and only if it
satisfies $\left(  a_{i}\geq0\text{ for all }i\in S\right)  $ and $\left(
a_{i}\leq0\text{ for all }i\in\left[  n\right]  \setminus S\right)  $. We let
$\mathcal{Q}_{S}$ denote the subset of $\mathcal{Q}$ consisting of all
infinite $\mathbf{k}$-linear combinations of the Laurent monomials
$q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ for $S$-adequate $n$-tuples
$\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}$ (as long as these
combinations belong to $\mathcal{Q}$). It is easy to see that $\mathcal{Q}%
_{S}$ is a topological $\mathbf{k}$-subalgebra of $\mathcal{Q}$ (since the
entrywise sum of two $S$-adequate $n$-tuples is $S$-adequate again).

(At this point, it is helpful to recall once again that the $q_{1}%
,q_{2},\ldots,q_{n}$ are not indeterminates, but rather monomials defined by
$q_{i}=r_{i}r_{i+1}\cdots r_{n}$. But their products $q_{1}^{a_{1}}%
q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ are Laurent monomials. Explicitly, they can
be rewritten as products of the $r_{1},r_{2},\ldots,r_{n}$ using
(\ref{eq.Q'.reindex}). Thus, it is easy to see that the elements of
$\mathcal{Q}_{S}$ are the infinite $\mathbf{k}$-linear combinations of the
Laurent monomials $r_{1}^{b_{1}}r_{2}^{b_{2}}\cdots r_{n}^{b_{n}}$ for all
$\left(  b_{1},b_{2},\ldots,b_{n}\right)  \in\mathbb{Z}^{n}$ satisfying
$\left(  b_{i}\geq b_{i-1}\text{ for all }i\in S\right)  $ and $\left(
b_{i}\leq b_{i-1}\text{ for all }i\in\left[  n\right]  \setminus S\right)  $,
where we set $b_{0}=0$, as long as these combinations belong to $\mathcal{Q}$.
But we won't need this characterization.)

\textbf{(e)} We let $\mathcal{T}_{S}$ denote the topological $\mathbf{k}%
$-algebra $\mathbf{k}\left[  \left[  t_{i}\ \mid\ i\in S\right]  \right]  $.
This is a topological subalgebra of $\mathcal{T}$. Hence, the ring
$\mathcal{T}_{S}\left[  \left[  w\right]  \right]  $ (that is, the ring of
formal power series in the (single) variable $w$ over $\mathcal{T}_{S}$) is a
topological $\mathbf{k}$-subalgebra of the similarly-defined ring
$\mathcal{T}\left[  \left[  w\right]  \right]  $.

\begin{vershort}
\textbf{(f)} We define a $\mathbf{k}$-algebra homomorphism $A_{S}%
:\mathcal{X}_{S}\rightarrow\mathcal{Q}_{S}$ by%
\[
A_{S}\left(  x_{i,j}\right)  =-\dfrac{q_{i}+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}%
}\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  i,j\right)  \in\mathfrak{P}_{S}.
\]
This is well-defined, because for each $\left(  i,j\right)  \in\mathfrak{P}%
_{S}$, the power series $-\dfrac{q_{i}+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}}$
does indeed belong to $\mathcal{Q}_{S}$ (indeed, a look at the monomials
reveals that both series $-\left(  q_{i}+\beta+\alpha/q_{j}\right)  $ and
$\dfrac{1}{1-q_{i}/q_{j}}=\sum_{k\geq0}\left(  q_{i}/q_{j}\right)  ^{k}$
belong to $\mathcal{Q}_{S}$, and therefore so does their product, which is
$-\dfrac{q_{i}+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}}$).
\end{vershort}

\begin{verlong}
\textbf{(f)} We define a $\mathbf{k}$-algebra homomorphism $A_{S}%
:\mathcal{X}_{S}\rightarrow\mathcal{Q}_{S}$ by%
\[
A_{S}\left(  x_{i,j}\right)  =-\dfrac{q_{i}+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}%
}\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  i,j\right)  \in\mathfrak{P}_{S}.
\]
Notice that this is well-defined\footnotemark.
\end{verlong}

\textbf{(g)} We define a continuous $\mathbf{k}$-linear map $B_{S}%
:\mathcal{Q}_{S}\rightarrow\mathcal{T}_{S}\left[  \left[  w\right]  \right]  $
by setting%
\begin{align*}
B_{S}\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}\right)   &
=\left(  \prod_{i\in S}t_{i}^{a_{i}}\right)  \left(  \prod_{i\in\left[
n\right]  \setminus S}w^{-a_{i}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \text{for each }S\text{-adequate }\left(  a_{1}%
,a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}.
\end{align*}
This is well-defined, as we will see below (in Proposition
\ref{prop.pathless.BSwd} \textbf{(b)}).

\textbf{(h)} We let $\mathcal{T}_{S}^{\prime}$ denote the $\mathbf{k}$-algebra
$\mathbf{k}\left[  t_{i}\ \mid\ i\in S\right]  $. This is a $\mathbf{k}%
$-subalgebra of $\mathcal{T}^{\prime}$. Hence, the ring $\mathcal{T}%
_{S}^{\prime}\left[  \left[  w\right]  \right]  $ (that is, the ring of formal
power series in the (single) variable $w$ over $\mathcal{T}_{S}^{\prime}$) is
a $\mathbf{k}$-subalgebra of the similarly-defined ring $\mathcal{T}^{\prime
}\left[  \left[  w\right]  \right]  $.

\textbf{(i)} We define a $\mathbf{k}$-algebra homomorphism $D_{S}%
:\mathcal{X}_{S}\rightarrow\mathcal{T}_{S}^{\prime}$ by%
\[
D_{S}\left(  x_{i,j}\right)  =t_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }\left(
i,j\right)  \in\mathfrak{P}_{S}.
\]
This is well-defined, since each $\left(  i,j\right)  \in\mathfrak{P}_{S}$
satisfies $i\in S$.

\textbf{(j)} We define a $\mathbf{k}$-algebra homomorphism $E_{S}%
:\mathcal{T}_{S}^{\prime}\rightarrow\mathcal{T}_{S}^{\prime}\left[  \left[
w\right]  \right]  $ by%
\[
E_{S}\left(  t_{i}\right)  =-\dfrac{t_{i}+\beta+\alpha w}{1-t_{i}%
w}\ \ \ \ \ \ \ \ \ \ \text{for each }i\in S.
\]
This is well-defined (by the universal property of the polynomial ring
$\mathcal{T}_{S}^{\prime}$), because for each $i\in S$, the power series
$1-t_{i}w$ is invertible in $\mathcal{T}_{S}^{\prime}\left[  \left[  w\right]
\right]  $ (indeed, its constant term is $1$).
\end{definition}

\begin{verlong}
\footnotetext{\textit{Proof.} Let $\left(  i,j\right)  \in\mathfrak{P}_{S}$.
The Laurent series $q_{i}$ belongs to $\mathcal{Q}_{S}$, because the $n$-tuple
$\left(  0,0,\ldots,0,1,0,0,\ldots0\right)  $ (where the $1$ stands in the
$i$-th position) is $S$-adequate. The Laurent series $\alpha/q_{j}$ belongs to
$\mathcal{Q}_{S}$, because the $n$-tuple $\left(  0,0,\ldots,0,-1,0,0,\ldots
0\right)  $ (where the $-1$ stands in the $j$-th position) is $S$-adequate.
The Laurent series $\beta$ (being a constant) is also in $\mathcal{Q}_{S}$.
Hence, $q_{i}+\beta+\alpha/q_{j}$ (being a sum of three elements of
$\mathcal{Q}_{S}$) is also in $\mathcal{Q}_{S}$.
\par
For each $k\geq0$, the Laurent monomial $\left(  q_{i}/q_{j}\right)  ^{k}$
belongs to $\mathcal{Q}_{S}$, because the $n$-tuple $\left(  0,0,\ldots
,0,k,0,0,\ldots,0,-k,0,0,\ldots,0\right)  $ (where the $k$ stands in the
$i$-th position, and the $-k$ stands in the $j$-th position) is $S$-adequate.
Hence, $\sum_{k\geq0}\left(  q_{i}/q_{j}\right)  ^{k}\in\mathcal{Q}_{S}$. In
view of $\sum_{k\geq0}\left(  q_{i}/q_{j}\right)  ^{k}=\dfrac{1}{1-q_{i}%
/q_{j}}$, this rewrites as $\dfrac{1}{1-q_{i}/q_{j}}\in\mathcal{Q}_{S}$. So we
know that both Laurent series $q_{i}+\beta+\alpha/q_{j}$ and $\dfrac
{1}{1-q_{i}/q_{j}}$ belong to $\mathcal{Q}_{S}$. Therefore, so does their
product $\left(  q_{i}+\beta+\alpha/q_{j}\right)  \cdot\dfrac{1}{1-q_{i}%
/q_{j}}=\dfrac{q_{i}+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}}$. Hence,
$-\dfrac{q_{i}+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}}$ also belongs to
$\mathcal{Q}_{S}$. This proves that $A_{S}$ is well-defined (via the universal
property of the polynomial ring $\mathcal{X}_{S}$).}
\end{verlong}

\begin{proposition}
\label{prop.pathless.BSwd}Let $S$ be a subset of $\left[  n-1\right]  $.

\textbf{(a)} We have%
\begin{equation}
B\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}\right)  =\left(
\prod_{i\in S}t_{i}^{a_{i}}\right)  \left(  \prod_{i\in\left[  n\right]
\setminus S}w^{-a_{i}}\right)  \label{eq.prop.pathless.BSwd.a.eq}%
\end{equation}
for each $S$-adequate $n$-tuple $\left(  a_{1},a_{2},\ldots,a_{n}\right)
\in\mathbb{Z}^{n}$.

\textbf{(b)} The map $B_{S}$ (defined in Definition \ref{def.Ssubset}
\textbf{(g)}) is well-defined.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.pathless.BSwd}.]\textbf{(a)} Let $\left(
a_{1},a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}$ be an $S$-adequate
$n$-tuple. We must prove (\ref{eq.prop.pathless.BSwd.a.eq}).

\begin{vershort}
The $n$-tuple $\left(  a_{1},a_{2},\ldots,a_{n}\right)  $ is $S$-adequate.
Thus, $\left(  a_{i}\geq0\text{ for all }i\in S\right)  $ and \newline$\left(
a_{i}\leq0\text{ for all }i\in\left[  n\right]  \setminus S\right)  $. Hence,
each $i\in\left[  n\right]  $ satisfying $a_{i}>0$ must belong to $S$. Also,
each $i\in\left[  n\right]  $ satisfying $a_{i}<0$ must belong to $\left[
n\right]  \setminus S$ (since $a_{i}\geq0$ for all $i\in S$).
\end{vershort}

\begin{verlong}
The $n$-tuple $\left(  a_{1},a_{2},\ldots,a_{n}\right)  $ is $S$-adequate.
Thus, $\left(  a_{i}\geq0\text{ for all }i\in S\right)  $ and \newline$\left(
a_{i}\leq0\text{ for all }i\in\left[  n\right]  \setminus S\right)  $. In
particular, $\left(  a_{i}\leq0\text{ for all }i\in\left[  n\right]  \setminus
S\right)  $. Hence, each $i\in\left[  n\right]  $ satisfying $a_{i}>0$ must
belong to $S$ (because otherwise, $i$ would belong to $\left[  n\right]
\setminus S$, and therefore would have to satisfy $a_{i}\leq0$, which would
contradict $a_{i}>0$).

On the other hand, $\left(  a_{i}\geq0\text{ for all }i\in S\right)  $. Hence,
each $i\in\left[  n\right]  $ satisfying $a_{i}<0$ must belong to $\left[
n\right]  \setminus S$ (because otherwise, $i$ would belong to $S$, and
therefore would have to satisfy $a_{i}\geq0$, which would contradict $a_{i}<0$).
\end{verlong}

\begin{vershort}
Now, the definition of the map $B$ yields
\[
B\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}\right)
=\underbrace{\left(  \prod_{\substack{i\in\left[  n\right]  ;\\a_{i}>0}%
}t_{i}^{a_{i}}\right)  }_{\substack{=\prod_{\substack{i\in S;\\a_{i}>0}%
}t_{i}^{a_{i}}\\\text{(since each }i\in\left[  n\right]  \\\text{satisfying
}a_{i}>0\\\text{must belong to }S\text{)}}}\underbrace{\left(  \prod
_{\substack{i\in\left[  n\right]  ;\\a_{i}<0}}w^{-a_{i}}\right)
}_{\substack{=\prod_{\substack{i\in\left[  n\right]  \setminus S;\\a_{i}%
<0}}w^{-a_{i}}\\\text{(since each }i\in\left[  n\right]  \\\text{satisfying
}a_{i}<0\\\text{must belong to }\left[  n\right]  \setminus S\text{)}%
}}=\left(  \prod_{\substack{i\in S;\\a_{i}>0}}t_{i}^{a_{i}}\right)  \left(
\prod_{\substack{i\in\left[  n\right]  \setminus S;\\a_{i}<0}}w^{-a_{i}%
}\right)  .
\]
Comparing this with%
\begin{align*}
&  \underbrace{\left(  \prod_{i\in S}t_{i}^{a_{i}}\right)  }%
_{\substack{=\left(  \prod_{\substack{i\in S;\\a_{i}=0}}t_{i}^{a_{i}}\right)
\left(  \prod_{\substack{i\in S;\\a_{i}>0}}t_{i}^{a_{i}}\right)
\\\text{(since each }i\in S\text{ satisfies either }a_{i}=0\text{ or }%
a_{i}>0\\\text{(since }a_{i}\geq0\text{ for all }i\in S\text{))}%
}}\underbrace{\left(  \prod_{i\in\left[  n\right]  \setminus S}w^{-a_{i}%
}\right)  }_{\substack{=\left(  \prod_{\substack{i\in\left[  n\right]
\setminus S;\\a_{i}=0}}w^{-a_{i}}\right)  \left(  \prod_{\substack{i\in\left[
n\right]  \setminus S;\\a_{i}<0}}w^{-a_{i}}\right)  \\\text{(since each }%
i\in\left[  n\right]  \setminus S\text{ satisfies either }a_{i}=0\text{ or
}a_{i}<0\\\text{(since }a_{i}\leq0\text{ for all }i\in\left[  n\right]
\setminus S\text{))}}}\\
&  =\left(  \prod_{\substack{i\in S;\\a_{i}=0}}\underbrace{t_{i}^{a_{i}}%
}_{\substack{=1\\\text{(since }a_{i}=0\text{)}}}\right)  \left(
\prod_{\substack{i\in S;\\a_{i}>0}}t_{i}^{a_{i}}\right)  \left(
\prod_{\substack{i\in\left[  n\right]  \setminus S;\\a_{i}=0}%
}\underbrace{w^{-a_{i}}}_{\substack{=1\\\text{(since }a_{i}=0\text{)}%
}}\right)  \left(  \prod_{\substack{i\in\left[  n\right]  \setminus
S;\\a_{i}<0}}w^{-a_{i}}\right) \\
&  =\left(  \prod_{\substack{i\in S;\\a_{i}>0}}t_{i}^{a_{i}}\right)  \left(
\prod_{\substack{i\in\left[  n\right]  \setminus S;\\a_{i}<0}}w^{-a_{i}%
}\right)  ,
\end{align*}
we obtain $B\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}\right)
=\left(  \prod_{i\in S}t_{i}^{a_{i}}\right)  \left(  \prod_{i\in\left[
n\right]  \setminus S}w^{-a_{i}}\right)  $. This proves Proposition
\ref{prop.pathless.BSwd} \textbf{(a)}.
\end{vershort}

\begin{verlong}
Now, the definition of the map $B$ yields
\[
B\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}\right)
=\underbrace{\left(  \prod_{\substack{i\in\left[  n\right]  ;\\a_{i}>0}%
}t_{i}^{a_{i}}\right)  }_{\substack{=\prod_{\substack{i\in S;\\a_{i}>0}%
}t_{i}^{a_{i}}\\\text{(since each }i\in\left[  n\right]  \\\text{satisfying
}a_{i}>0\\\text{must belong to }S\text{)}}}\underbrace{\left(  \prod
_{\substack{i\in\left[  n\right]  ;\\a_{i}<0}}w^{-a_{i}}\right)
}_{\substack{=\prod_{\substack{i\in\left[  n\right]  \setminus S;\\a_{i}%
<0}}w^{-a_{i}}\\\text{(since each }i\in\left[  n\right]  \\\text{satisfying
}a_{i}<0\\\text{must belong to }\left[  n\right]  \setminus S\text{)}%
}}=\left(  \prod_{\substack{i\in S;\\a_{i}>0}}t_{i}^{a_{i}}\right)  \left(
\prod_{\substack{i\in\left[  n\right]  \setminus S;\\a_{i}<0}}w^{-a_{i}%
}\right)  .
\]
Comparing this with%
\begin{align*}
&  \underbrace{\left(  \prod_{i\in S}t_{i}^{a_{i}}\right)  }%
_{\substack{=\left(  \prod_{\substack{i\in S;\\a_{i}=0}}t_{i}^{a_{i}}\right)
\left(  \prod_{\substack{i\in S;\\a_{i}>0}}t_{i}^{a_{i}}\right)
\\\text{(since each }i\in S\text{ satisfies either }a_{i}=0\text{ or }%
a_{i}>0\\\text{(since }a_{i}\geq0\text{ for all }i\in S\text{))}%
}}\underbrace{\left(  \prod_{i\in\left[  n\right]  \setminus S}w^{-a_{i}%
}\right)  }_{\substack{=\left(  \prod_{\substack{i\in\left[  n\right]
\setminus S;\\a_{i}=0}}w^{-a_{i}}\right)  \left(  \prod_{\substack{i\in\left[
n\right]  \setminus S;\\a_{i}<0}}w^{-a_{i}}\right)  \\\text{(since each }%
i\in\left[  n\right]  \setminus S\text{ satisfies either }a_{i}=0\text{ or
}a_{i}<0\\\text{(since }a_{i}\leq0\text{ for all }i\in\left[  n\right]
\setminus S\text{))}}}\\
&  =\left(  \prod_{\substack{i\in S;\\a_{i}=0}}\underbrace{t_{i}^{a_{i}}%
}_{\substack{=1\\\text{(since }a_{i}=0\text{)}}}\right)  \left(
\prod_{\substack{i\in S;\\a_{i}>0}}t_{i}^{a_{i}}\right)  \left(
\prod_{\substack{i\in\left[  n\right]  \setminus S;\\a_{i}=0}%
}\underbrace{w^{-a_{i}}}_{\substack{=1\\\text{(since }a_{i}=0\text{)}%
}}\right)  \left(  \prod_{\substack{i\in\left[  n\right]  \setminus
S;\\a_{i}<0}}w^{-a_{i}}\right) \\
&  =\underbrace{\left(  \prod_{\substack{i\in S;\\a_{i}=0}}1\right)  }%
_{=1}\left(  \prod_{\substack{i\in S;\\a_{i}>0}}t_{i}^{a_{i}}\right)
\underbrace{\left(  \prod_{\substack{i\in\left[  n\right]  \setminus
S;\\a_{i}=0}}1\right)  }_{=1}\left(  \prod_{\substack{i\in\left[  n\right]
\setminus S;\\a_{i}<0}}w^{-a_{i}}\right) \\
&  =\left(  \prod_{\substack{i\in S;\\a_{i}>0}}t_{i}^{a_{i}}\right)  \left(
\prod_{\substack{i\in\left[  n\right]  \setminus S;\\a_{i}<0}}w^{-a_{i}%
}\right)  ,
\end{align*}
we obtain $B\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}\right)
=\left(  \prod_{i\in S}t_{i}^{a_{i}}\right)  \left(  \prod_{i\in\left[
n\right]  \setminus S}w^{-a_{i}}\right)  $. This proves Proposition
\ref{prop.pathless.BSwd} \textbf{(a)}.
\end{verlong}

\textbf{(b)} We must show that there exists a unique continuous $\mathbf{k}%
$-linear map $B_{S}:\mathcal{Q}_{S}\rightarrow\mathcal{T}_{S}\left[  \left[
w\right]  \right]  $ satisfying%
\begin{equation}
\left(
\begin{array}
[c]{l}%
B_{S}\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}\right)  =\left(
\prod_{i\in S}t_{i}^{a_{i}}\right)  \left(  \prod_{i\in\left[  n\right]
\setminus S}w^{-a_{i}}\right) \\
\ \ \ \ \ \ \ \ \ \ \text{for each }S\text{-adequate }\left(  a_{1}%
,a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}%
\end{array}
\right)  . \label{pf.prop.pathless.BSwd.b.cond}%
\end{equation}
The uniqueness of such a map is clear (because the elements of $\mathcal{Q}%
_{S}$ are infinite $\mathbf{k}$-linear combinations of the Laurent monomials
$q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ for $S$-adequate $n$-tuples
$\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}$; but the formula
(\ref{pf.prop.pathless.BSwd.b.cond}) uniquely determines the value of $B_{S}$
on such a $\mathbf{k}$-linear combination). Thus, it remains to prove its existence.

For each $f\in\mathcal{Q}_{S}$, we have $B\left(  f\right)  \in\mathcal{T}%
_{S}\left[  \left[  w\right]  \right]  $\ \ \ \ \footnote{\textit{Proof.} Let
$f\in\mathcal{Q}_{S}$. We must show that $B\left(  f\right)  \in
\mathcal{T}_{S}\left[  \left[  w\right]  \right]  $. Since the map $B$ is
$\mathbf{k}$-linear and continuous, we can WLOG assume that $f$ is a Laurent
monomial of the form $q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ for some
$S$-adequate $n$-tuple $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in
\mathbb{Z}^{n}$ (because $f$ is always an infinite $\mathbf{k}$-linear
combination of such Laurent monomials). Assume this. Consider this $\left(
a_{1},a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}$.
\par
Thus, $f=q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$. Applying the map $B$
to both sides of this equality, we obtain%
\begin{align*}
B\left(  f\right)   &  =B\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}%
^{a_{n}}\right)  =\left(  \prod_{i\in S}t_{i}^{a_{i}}\right)  \left(
\prod_{i\in\left[  n\right]  \setminus S}w^{-a_{i}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{prop.pathless.BSwd}
\textbf{(a)}}\right) \\
&  \in\mathcal{T}_{S}\left[  \left[  w\right]  \right]  .
\end{align*}
This is precisely what we wanted to show.}. Hence, we can define a map
$\widetilde{B_{S}}:\mathcal{Q}_{S}\rightarrow\mathcal{T}_{S}\left[  \left[
w\right]  \right]  $ by%
\[
\widetilde{B_{S}}\left(  f\right)  =B\left(  f\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }f\in\mathcal{Q}_{S}.
\]
This map $\widetilde{B_{S}}$ is a restriction of the map $B$; hence, it is a
continuous $\mathbf{k}$-linear map (since $B$ is a continuous $\mathbf{k}%
$-linear map). Furthermore, it satisfies%
\begin{align*}
\widetilde{B_{S}}\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}%
}\right)   &  =B\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\widetilde{B_{S}%
}\right) \\
&  =\left(  \prod_{i\in S}t_{i}^{a_{i}}\right)  \left(  \prod_{i\in\left[
n\right]  \setminus S}w^{-a_{i}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition \ref{prop.pathless.BSwd} \textbf{(a)}}\right)
\end{align*}
for each $S$-adequate $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in
\mathbb{Z}^{n}$. Hence, $\widetilde{B_{S}}$ is a continuous $\mathbf{k}%
$-linear map $B_{S}:\mathcal{Q}_{S}\rightarrow\mathcal{T}_{S}\left[  \left[
w\right]  \right]  $ satisfying (\ref{pf.prop.pathless.BSwd.b.cond}). Thus,
the existence of such a map $B_{S}$ is proven. As we have explained, this
completes the proof of Proposition \ref{prop.pathless.BSwd} \textbf{(b)}.
\end{proof}

\begin{proposition}
\label{prop.pathless.cd}Let $S$ be a subset of $\left[  n-1\right]  $. Then,
the diagrams%
\begin{equation}%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%\calX_S \arinj[d] \ar[r]^{A_S} & \calQ_S \arinj[d] \ar[r]^-{B_S}
%& \calT_S\left[\left[w\right]\right] \arinj[d] \\
%\calX\ar[r]_{A} & \calQ\ar[r]_-{B} & \calT\left[\left[w\right]\right]
%}} }%
%BeginExpansion
\xymatrix{
\calX_S \arinj[d] \ar[r]^{A_S} & \calQ_S \arinj[d] \ar[r]^-{B_S}
& \calT_S\left[\left[w\right]\right] \arinj[d] \\
\calX\ar[r]_{A} & \calQ\ar[r]_-{B} & \calT\left[\left[w\right]\right]
}
%EndExpansion
\label{eq.prop.pathless.cd.1}%
\end{equation}
and%
\begin{equation}%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%\calX_S \arinj[d] \ar[r]^{D_S} & \calT^\prime_S \arinj[d] \ar[r]^-{E_S}
%& \calT^\prime_S\left[\left[w\right]\right] \arinj[d] \\
%\calX\ar[r]_{D} & \calT^\prime\ar[r]_-{E} & \calT^\prime\left[\left
%[w\right]\right]
%}} }%
%BeginExpansion
\xymatrix{
\calX_S \arinj[d] \ar[r]^{D_S} & \calT^\prime_S \arinj[d] \ar[r]^-{E_S}
& \calT^\prime_S\left[\left[w\right]\right] \arinj[d] \\
\calX\ar[r]_{D} & \calT^\prime\ar[r]_-{E} & \calT^\prime\left[\left
[w\right]\right]
}
%EndExpansion
\label{eq.prop.pathless.cd.2}%
\end{equation}
(where the vertical arrows are the obvious inclusion maps) are commutative.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.pathless.cd}.]The commutativity of the left
square of (\ref{eq.prop.pathless.cd.1}) is obvious\footnote{\textquotedblleft
Obvious\textquotedblright\ in the following sense: You want to prove that a
diagram of the form%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%\mathcal{A}_1 \ar[r]^{f_1} \ar[d]_{f_2} & \mathcal{A}_2 \ar[d]^{f_3} \\
%\mathcal{A}_3 \ar[r]_{f_4} & \mathcal{A}_4
%}}}%
%BeginExpansion
\xymatrix{
\mathcal{A}_1 \ar[r]^{f_1} \ar[d]_{f_2} & \mathcal{A}_2 \ar[d]^{f_3} \\
\mathcal{A}_3 \ar[r]_{f_4} & \mathcal{A}_4
}%
%EndExpansion
\]
is commutative, where $\mathcal{A}_{1},\mathcal{A}_{2},\mathcal{A}%
_{3},\mathcal{A}_{4}$ are four $\mathbf{k}$-algebras and $f_{1},f_{2}%
,f_{3},f_{4}$ are four $\mathbf{k}$-algebra homomorphisms. (In our concrete
case, $\mathcal{A}_{1}=\mathcal{X}_{S}$, $\mathcal{A}_{2}=\mathcal{Q}_{S}$,
$\mathcal{A}_{3}=\mathcal{X}$, $\mathcal{A}_{4}=\mathcal{Q}$, $f_{1}=A_{S}$
and $f_{4}=A$, whereas $f_{2}$ and $f_{3}$ are the inclusion maps
$\mathcal{X}_{S}\rightarrow\mathcal{X}$ and $\mathcal{Q}_{S}\rightarrow
\mathcal{Q}$.) In order to prove this commutativity, it suffices to show that
it holds \textbf{on a generating set} of the $\mathbf{k}$-algebra
$\mathcal{A}_{1}$. In other words, it suffices to pick some generating set
$\mathfrak{G}$ of the $\mathbf{k}$-algebra $\mathcal{A}_{1}$ and show that all
$g\in\mathfrak{G}$ satisfy $\left(  f_{3}\circ f_{1}\right)  \left(  g\right)
=\left(  f_{4}\circ f_{2}\right)  \left(  g\right)  $. (In our concrete case,
it is most reasonable to pick $\mathfrak{G}=\left\{  x_{i,j}\ \mid\ \left(
i,j\right)  \in\mathfrak{P}_{S}\right\}  $. The proof then becomes completely
clear.)}. So is the commutativity of each of the two squares of
(\ref{eq.prop.pathless.cd.2})\footnote{for similar reasons}. It thus remains
to prove the commutativity of the right square of (\ref{eq.prop.pathless.cd.1}%
). In other words, we must show that $B_{S}\left(  p\right)  =B\left(
p\right)  $ for each $p\in\mathcal{Q}_{S}$.

So fix $p\in\mathcal{Q}_{S}$. Since both maps $B_{S}$ and $B$ are continuous
and $\mathbf{k}$-linear, we can WLOG assume that $p$ is a Laurent monomial of
the form $q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ for an $S$-adequate
$n$-tuple $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}$ (since
the elements of $\mathcal{Q}_{S}$ are infinite $\mathbf{k}$-linear
combinations of Laurent monomials of this form). Assume this, and fix this
$\left(  a_{1},a_{2},\ldots,a_{n}\right)  $.

From $p=q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$, we obtain
\[
B\left(  p\right)  =B\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}%
}\right)  =\left(  \prod_{i\in S}t_{i}^{a_{i}}\right)  \left(  \prod
_{i\in\left[  n\right]  \setminus S}w^{-a_{i}}\right)
\]
(by Proposition \ref{prop.pathless.BSwd} \textbf{(a)}). Comparing this with%
\begin{align*}
B_{S}\left(  p\right)   &  =B_{S}\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }p=q_{1}^{a_{1}%
}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}\right) \\
&  =\left(  \prod_{i\in S}t_{i}^{a_{i}}\right)  \left(  \prod_{i\in\left[
n\right]  \setminus S}w^{-a_{i}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }B_{S}\right)  ,
\end{align*}
we obtain $B_{S}\left(  p\right)  =B\left(  p\right)  $. This proves the
commutativity of the right square of (\ref{eq.prop.pathless.cd.1}). The proof
of Proposition \ref{prop.pathless.cd} is thus complete.
\end{proof}

\begin{proposition}
\label{prop.pathless.BS-alg}Let $S$ be a subset of $\left[  n-1\right]  $.
Then, $B_{S}:\mathcal{Q}_{S}\rightarrow\mathcal{T}_{S}\left[  \left[
w\right]  \right]  $ is a continuous $\mathbf{k}$-algebra homomorphism.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.pathless.BS-alg}.]We merely need to show that
$B_{S}$ is a $\mathbf{k}$-algebra homomorphism. To this purpose, by linearity,
we only need to prove that $B_{S}\left(  1\right)  =1$ and $B_{S}\left(
\mathfrak{mn}\right)  =B_{S}\left(  \mathfrak{m}\right)  B_{S}\left(
\mathfrak{n}\right)  $ for any two Laurent monomials $\mathfrak{m}$ and
$\mathfrak{n}$ of the form $q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$
for $S$-adequate $n$-tuples $\left(  a_{1},a_{2},\ldots,a_{n}\right)
\in\mathbb{Z}^{n}$ (since the elements of $\mathcal{Q}_{S}$ are infinite
$\mathbf{k}$-linear combinations of Laurent monomials of this form). This is
easy and left to the reader.
\end{proof}

\begin{proposition}
\label{prop.pathless.AB}Let $S$ be a subset of $\left[  n-1\right]  $. Let
$\left(  i,j\right)  \in\mathfrak{P}_{S}$. Then,%
\[
\left(  E_{S}\circ D_{S}\right)  \left(  x_{i,j}\right)  =\left(  B_{S}\circ
A_{S}\right)  \left(  x_{i,j}\right)  .
\]

\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.pathless.AB}.]From $\left(  i,j\right)
\in\mathfrak{P}_{S}$, we obtain $i\in S$ and $j\in\left[  n\right]  \setminus
S$. Thus, the definition of $B_{S}$ reveals that $B_{S}\left(  q_{i}\right)
=t_{i}$ and $B_{S}\left(  q_{j}^{-1}\right)  =w$.

Proposition \ref{prop.pathless.BS-alg} shows that $B_{S}:\mathcal{Q}%
_{S}\rightarrow\mathcal{T}_{S}\left[  \left[  w\right]  \right]  $ is a
continuous $\mathbf{k}$-algebra homomorphism. But the definition of $A_{S}$
yields
\[
A_{S}\left(  x_{i,j}\right)  =-\dfrac{q_{i}+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}%
}=-\dfrac{q_{i}+\beta+\alpha q_{j}^{-1}}{1-q_{i}q_{j}^{-1}}.
\]
Applying the map $B_{S}$ to both sides of this equality, we find%
\begin{align*}
B_{S}\left(  A_{S}\left(  x_{i,j}\right)  \right)   &  =B_{S}\left(
-\dfrac{q_{i}+\beta+\alpha q_{j}^{-1}}{1-q_{i}q_{j}^{-1}}\right)
=-\dfrac{B_{S}\left(  q_{i}\right)  +\beta+\alpha B_{S}\left(  q_{j}%
^{-1}\right)  }{1-B_{S}\left(  q_{i}\right)  B_{S}\left(  q_{j}^{-1}\right)
}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }B_{S}\text{ is a }\mathbf{k}\text{-algebra homomorphism, and
thus}\\
\text{respects sums, products and fractions (as long as}\\
\text{the denominators of the fractions are invertible)}%
\end{array}
\right) \\
&  =-\dfrac{t_{i}+\beta+\alpha w}{1-t_{i}w}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }B_{S}\left(  q_{i}\right)  =t_{i}\text{ and }B_{S}\left(
q_{j}^{-1}\right)  =w\right)  .
\end{align*}
Comparing this with%
\[
\left(  E_{S}\circ D_{S}\right)  \left(  x_{i,j}\right)  =E_{S}\left(
\underbrace{D_{S}\left(  x_{i,j}\right)  }_{=t_{i}}\right)  =E_{S}\left(
t_{i}\right)  =-\dfrac{t_{i}+\beta+\alpha w}{1-t_{i}w},
\]
we obtain $B_{S}\left(  A_{S}\left(  x_{i,j}\right)  \right)  =\left(
E_{S}\circ D_{S}\right)  \left(  x_{i,j}\right)  $. Thus, $\left(  E_{S}\circ
D_{S}\right)  \left(  x_{i,j}\right)  =B_{S}\left(  A_{S}\left(
x_{i,j}\right)  \right)  =\left(  B_{S}\circ A_{S}\right)  \left(
x_{i,j}\right)  $. This proves Proposition \ref{prop.pathless.AB}.
\end{proof}

\begin{proposition}
\label{prop.pathless.S}Let $\mathfrak{m}\in\mathfrak{M}$ be a pathless monomial.

\textbf{(a)} There exists a subset $S$ of $\left[  n-1\right]  $ such that
$\mathfrak{m}$ is $S$-friendly.

\textbf{(b)} Let $S$ be such a subset. Then, $\mathfrak{m}\in\mathcal{X}_{S}$
and $\left(  E\circ D\right)  \left(  \mathfrak{m}\right)  =\left(  B_{S}\circ
A_{S}\right)  \left(  \mathfrak{m}\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.pathless.S}.]\textbf{(a)} Write $\mathfrak{m}$
in the form $\mathfrak{m}=\prod_{\substack{\left(  i,j\right)  \in\left[
n\right]  ^{2};\\i<j}}x_{i,j}^{a_{i,j}}$. For each $i\in\left[  n-1\right]  $,
define a $b_{i}\in\mathbb{N}$ by $b_{i}=\sum_{j=i+1}^{n}a_{i,j}$. Define a
subset $S$ of $\left[  n-1\right]  $ by $S=\left\{  i\in\left[  n-1\right]
\ \mid\ b_{i}>0\right\}  $. Then $\mathfrak{m}$ is $S$%
-friendly\footnote{\textit{Proof.} We need to show that every indeterminate
$x_{i,j}$ that appears in $\mathfrak{m}$ satisfies $i\in S$ and $j\notin S$.
\par
Indeed, assume the contrary. Thus, some indeterminate $x_{i,j}$ that appears
in $\mathfrak{m}$ does \textbf{not} satisfy $i\in S$ and $j\notin S$. Fix such
an indeterminate $x_{i,j}$, and denote it by $x_{u,v}$. Thus, $x_{u,v}$ is an
indeterminate that appears in $\mathfrak{m}$ but does \textbf{not} satisfy
$u\in S$ and $v\notin S$. Therefore, we have either $u\notin S$ or $v\in S$
(or both).
\par
We have $1\leq u<v\leq n$ (since the indeterminate $x_{u,v}$ exists) and thus
$u\in\left[  n-1\right]  $. The definition of $b_{u}$ yields $b_{u}%
=\sum_{j=u+1}^{n}a_{u,j}$. But $v\geq u+1$ (since $u<v$). Hence, $a_{u,v}$ is
an addend of the sum $\sum_{j=u+1}^{n}a_{u,j}$. Hence, $\sum_{j=u+1}%
^{n}a_{u,j}\geq a_{u,v}$. But $a_{u,v}>0$ (since the indeterminate $x_{u,v}$
appears in $\mathfrak{m}$). Hence, $b_{u}=\sum_{j=u+1}^{n}a_{u,j}\geq
a_{u,v}>0$. Therefore, $u\in S$ (by the definition of $S$). Hence, $u\notin S$
cannot hold. Therefore, $v\in S$ (since we know that we have either $u\notin
S$ or $v\in S$). In other words, $v\in\left[  n-1\right]  $ and $b_{v}>0$ (by
the definition of $S$). But the definition of $b_{v}$ yields $b_{v}%
=\sum_{j=v+1}^{n}a_{v,j}=\sum_{w=v+1}^{n}a_{v,w}$. Hence, $\sum_{w=v+1}%
^{n}a_{v,w}=b_{v}>0$. Hence, there exists some $w\in\left\{  v+1,v+2,\ldots
,n\right\}  $ such that $a_{v,w}>0$. Fix such a $w$.
\par
We have $v<w$ (since $w\in\left\{  v+1,v+2,\ldots,n\right\}  $), hence
$u<v<w$. Thus, $\left(  u,v\right)  \neq\left(  v,w\right)  $. Moreover, the
indeterminate $x_{v,w}$ appears in $\mathfrak{m}$ (since $a_{v,w}>0$). Thus,
both indeterminates $x_{u,v}$ and $x_{v,w}$ appear in $\mathfrak{m}$. Hence,
$x_{u,v}x_{v,w}\mid\mathfrak{m}$ (since $\left(  u,v\right)  \neq\left(
v,w\right)  $).
\par
But the monomial $\mathfrak{m}$ is pathless. In other words, there exists no
triple $\left(  i,j,k\right)  \in\left[  n\right]  ^{3}$ satisfying $i<j<k$
and $x_{i,j}x_{j,k}\mid\mathfrak{m}$. This contradicts the fact that $\left(
u,v,w\right)  $ is such a triple (since $u<v<w$ and $x_{u,v}x_{v,w}%
\mid\mathfrak{m}$). This contradiction completes our proof.}. This proves
Proposition \ref{prop.pathless.S} \textbf{(a)}.

\textbf{(b)} We know that $\mathfrak{m}\in\mathcal{X}_{S}$ (since
$\mathfrak{m}$ is $S$-friendly). Now, we shall show that $\left(  E\circ
D\right)  \mid_{\mathcal{X}_{S}}=B_{S}\circ A_{S}$ (if we regard $B_{S}\circ
A_{S}$ as a map to $\mathcal{T}\left[  \left[  w\right]  \right]  $ and regard
$E\circ D$ as a map to $\mathcal{T}\left[  \left[  w\right]  \right]  $).

The map $\left(  E\circ D\right)  \mid_{\mathcal{X}_{S}}$ is a $\mathbf{k}%
$-algebra homomorphism (since $D$ and $E$ are $\mathbf{k}$-algebra
homomorphisms), and the map $B_{S}\circ A_{S}$ is a $\mathbf{k}$-algebra
homomorphism (since both $B_{S}$ and $A_{S}$ are $\mathbf{k}$-algebra
homomorphisms\footnote{Here we are using Proposition
\ref{prop.pathless.BS-alg}.}). Hence, we are trying to prove that two
$\mathbf{k}$-algebra homomorphisms are equal (namely, the homomorphisms
$\left(  E\circ D\right)  \mid_{\mathcal{X}_{S}}$ and $B_{S}\circ A_{S}$). It
is clearly enough to prove this on the generating family $\left(
x_{i,j}\right)  _{\left(  i,j\right)  \in\mathfrak{P}_{S}}$ of the
$\mathbf{k}$-algebra $\mathcal{X}_{S}$. In other words, it is enough to prove
that $\left(  \left(  E\circ D\right)  \mid_{\mathcal{X}_{S}}\right)  \left(
x_{i,j}\right)  =\left(  B_{S}\circ A_{S}\right)  \left(  x_{i,j}\right)  $
for each $\left(  i,j\right)  \in\mathfrak{P}_{S}$.

So let us fix some $\left(  i,j\right)  \in\mathfrak{P}_{S}$. Proposition
\ref{prop.pathless.cd} shows that the diagram (\ref{eq.prop.pathless.cd.2}) is
commutative. Thus, $\left(  E\circ D\right)  \mid_{\mathcal{X}_{S}}=E_{S}\circ
D_{S}$ (provided that we regard $E_{S}\circ D_{S}$ as a map to $\mathcal{T}%
^{\prime}\left[  \left[  w\right]  \right]  $), and thus%
\[
\left(  \left(  E\circ D\right)  \mid_{\mathcal{X}_{S}}\right)  \left(
x_{i,j}\right)  =\left(  E_{S}\circ D_{S}\right)  \left(  x_{i,j}\right)
=\left(  B_{S}\circ A_{S}\right)  \left(  x_{i,j}\right)
\]
(by Proposition \ref{prop.pathless.AB}).

This completes our proof of $\left(  E\circ D\right)  \mid_{\mathcal{X}_{S}%
}=B_{S}\circ A_{S}$. Now, from $\mathfrak{m}\in\mathcal{X}_{S}$, we obtain
$\left(  E\circ D\right)  \left(  \mathfrak{m}\right)  =\underbrace{\left(
\left(  E\circ D\right)  \mid_{\mathcal{X}_{S}}\right)  }_{=B_{S}\circ A_{S}%
}\left(  \mathfrak{m}\right)  =\left(  B_{S}\circ A_{S}\right)  \left(
\mathfrak{m}\right)  $. This completes the proof of Proposition
\ref{prop.pathless.S} \textbf{(b)}.
\end{proof}

\subsection{$\left(  E\circ D\right)  \left(  q\right)  =\left(  B\circ
A\right)  \left(  q\right)  $ for pathless $q$}

\begin{corollary}
\label{cor.pathless.D}Let $q\in\mathcal{X}$ be pathless. Then, $\left(  E\circ
D\right)  \left(  q\right)  =\left(  B\circ A\right)  \left(  q\right)  $.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.pathless.D}.]The polynomial $q$ is pathless,
i.e., is a $\mathbf{k}$-linear combination of pathless monomials. Hence, we
WLOG assume that $q$ is a pathless monomial $\mathfrak{m}$ (since both maps
$E\circ D$ and $B\circ A$ are $\mathbf{k}$-linear). Consider this
$\mathfrak{m}$.

Proposition \ref{prop.pathless.S} \textbf{(a)} shows that there exists a
subset $S$ of $\left[  n-1\right]  $ such that $\mathfrak{m}$ is $S$-friendly.
Consider this $S$.

Proposition \ref{prop.pathless.S} \textbf{(b)} yields $\mathfrak{m}%
\in\mathcal{X}_{S}$ and $\left(  E\circ D\right)  \left(  \mathfrak{m}\right)
=\left(  B_{S}\circ A_{S}\right)  \left(  \mathfrak{m}\right)  $. But the
commutativity of the diagram (\ref{eq.prop.pathless.cd.1}) in Proposition
\ref{prop.pathless.cd} shows that $B_{S}\circ A_{S}=\left(  B\circ A\right)
\mid_{\mathcal{X}_{S}}$ (provided that we regard $B_{S}\circ A_{S}$ as a map
to $\mathcal{T}\left[  \left[  w\right]  \right]  $). Hence,
\[
\underbrace{\left(  B_{S}\circ A_{S}\right)  }_{=\left(  B\circ A\right)
\mid_{\mathcal{X}_{S}}}\left(  \mathfrak{m}\right)  =\left(  \left(  B\circ
A\right)  \mid_{\mathcal{X}_{S}}\right)  \left(  \mathfrak{m}\right)  =\left(
B\circ A\right)  \left(  \mathfrak{m}\right)  .
\]
Thus, $\left(  E\circ D\right)  \left(  \mathfrak{m}\right)  =\left(
B_{S}\circ A_{S}\right)  \left(  \mathfrak{m}\right)  =\left(  B\circ
A\right)  \left(  \mathfrak{m}\right)  $. Since $q=\mathfrak{m}$, this
rewrites as $\left(  E\circ D\right)  \left(  q\right)  =\left(  B\circ
A\right)  \left(  q\right)  $. This proves Corollary \ref{cor.pathless.D}.
\end{proof}

\subsection{Proof of Theorem~\ref{thm.t-red.unique}}

\begin{lemma}
\label{lem.t-red.unique.0}Let $p\in\mathcal{X}$ be a pathless polynomial such
that $p\in\mathcal{J}$. Then, $D\left(  p\right)  =0$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.t-red.unique.0}.]We have $A\left(  \underbrace{p}%
_{\in\mathcal{J}}\right)  \in A\left(  \mathcal{J}\right)  =0$ (by Proposition
\ref{prop.A.killsJ}); thus, $A\left(  p\right)  =0$. But Corollary
\ref{cor.pathless.D} (applied to $q=p$) yields
\[
\left(  E\circ D\right)  \left(  p\right)  =\left(  B\circ A\right)  \left(
p\right)  =B\left(  \underbrace{A\left(  p\right)  }_{=0}\right)  =B\left(
0\right)  =0
\]
(since the map $B$ is $\mathbf{k}$-linear). Thus, $E\left(  D\left(  p\right)
\right)  =\left(  E\circ D\right)  \left(  p\right)  =0$. Since the
$\mathbf{k}$-linear map $E$ is injective (by Proposition \ref{prop.E.inj}), we
thus conclude that $D\left(  p\right)  =0$. This proves Lemma
\ref{lem.t-red.unique.0}.
\end{proof}

We are now ready to prove Theorem~\ref{thm.t-red.unique}:

\begin{proof}
[Proof of Theorem~\ref{thm.t-red.unique}.]We need to prove that $D\left(
q\right)  $ does not depend on the choice of $q$. In other words, we need to
prove that if $f$ and $g$ are two pathless polynomials $q\in\mathcal{X}$ such
that $p\equiv q\operatorname{mod}\mathcal{J}$, then $D\left(  f\right)
=D\left(  g\right)  $.

So let $f$ and $g$ be two pathless polynomials $q\in\mathcal{X}$ such that
$p\equiv q\operatorname{mod}\mathcal{J}$. Thus, $p\equiv f\operatorname{mod}%
\mathcal{J}$ and $p\equiv g\operatorname{mod}\mathcal{J}$. Hence, $f\equiv
p\equiv g\operatorname{mod}\mathcal{J}$, so that $f-g\in\mathcal{J}$. Also,
the polynomial $f-g\in\mathcal{X}$ is pathless (since it is the difference of
the two pathless polynomials $f$ and $g$). Thus, Lemma
\ref{lem.t-red.unique.0} (applied to $f-g$ instead of $p$) shows that
$D\left(  f-g\right)  =0$. Thus, $0=D\left(  f-g\right)  =D\left(  f\right)
-D\left(  g\right)  $ (since $D$ is a $\mathbf{k}$-algebra homomorphism). In
other words, $D\left(  f\right)  =D\left(  g\right)  $. This proves
Theorem~\ref{thm.t-red.unique}.
\end{proof}

\subsection{\label{sect.symmetry}Appendix: A symmetric description of
$\mathcal{J}$}

In this section, let us give a different description of $\mathcal{J}$ that
reveals a symmetry inherent in the setting. First, we introduce auxiliary
polynomials. So far, we have only been considering indeterminates $x_{i,j}$
corresponding to pairs $\left(  i,j\right)  \in\left[  n\right]  ^{2}$
satisfying $i<j$. We shall now also define $x_{i,j}$ for pairs $\left(
i,j\right)  \in\left[  n\right]  ^{2}$ satisfying $i>j$; but these $x_{i,j}$
will not be new indeterminates, but rather will be polynomials in
$\mathcal{X}$:

\begin{definition}
\label{def.symmetry.xji}\textbf{(a)} Let $\left(  i,j\right)  \in\left[
n\right]  ^{2}$ be a pair satisfying $i>j$. Then, we define an element
$x_{i,j}\in\mathcal{X}$ by $x_{i,j}=-\beta-x_{j,i}$.

Thus, an element $x_{i,j}\in\mathcal{X}$ is defined for any pair $\left(
i,j\right)  $ of two distinct elements of $\left[  n\right]  $.

\textbf{(b)} For any three distinct elements $i,j,k$ of $\left[  n\right]  $,
we define a polynomial $J_{i,j,k}\in\mathcal{X}$ by%
\[
J_{i,j,k}=x_{i,j}x_{j,k}+x_{j,k}x_{k,i}+x_{k,i}x_{i,j}+\beta\left(
x_{i,j}+x_{j,k}+x_{k,i}\right)  +\beta^{2}-\alpha.
\]

\end{definition}

\begin{proposition}
\label{prop.symmetry}The ideal $\mathcal{J}$ of $\mathcal{X}$ is generated by
all polynomials $J_{i,j,k}$ for $i,j,k$ being three distinct elements of
$\left[  n\right]  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.symmetry}.]If $\left(  u_{p}\right)  _{p\in
P}$ is any family of elements of $\mathcal{X}$, then $\left\langle
u_{p}\right\rangle _{p\in P}$ shall mean the ideal of $\mathcal{X}$ generated
by this family $\left(  u_{p}\right)  _{p\in P}$. Thus, we need to prove that
$\mathcal{J}=\left\langle J_{i,j,k}\right\rangle _{i,j,k\text{ are three
distinct elements of }\left[  n\right]  }$.

We know (from the definition of $\mathcal{J}$) that%
\begin{equation}
\mathcal{J}=\left\langle x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}%
+\beta\right)  -\alpha\right\rangle _{\left(  i,j,k\right)  \in\left[
n\right]  ^{3}\text{ satisfying }i<j<k}. \label{pf.prop.symmetry.J=1}%
\end{equation}


\begin{vershort}
But for each $\left(  i,j,k\right)  \in\left[  n\right]  ^{3}$ satisfying
$i<j<k$, a straightforward computation reveals that%
\[
x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  -\alpha
=J_{i,j,k}.
\]

\end{vershort}

\begin{verlong}
But for each $\left(  i,j,k\right)  \in\left[  n\right]  ^{3}$ satisfying
$i<j<k$, we have%
\begin{equation}
x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  -\alpha=J_{i,j,k}
\label{pf.prop.symmetry.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.symmetry.2}):} Let $\left(
i,j,k\right)  \in\left[  n\right]  ^{3}$ be such that $i<j<k$. Then, the
definition of $J_{i,j,k}$ yields%
\begin{align*}
J_{i,j,k}  &  =x_{i,j}x_{j,k}+x_{j,k}x_{k,i}+x_{k,i}x_{i,j}+\beta\left(
x_{i,j}+x_{j,k}+x_{k,i}\right)  +\beta^{2}-\alpha\\
&  =x_{i,j}x_{j,k}+\underbrace{x_{k,i}}_{\substack{=-\beta-x_{i,k}\\\text{(by
the definition of }x_{k,i}\text{,}\\\text{since }k>i\text{)}}}\left(
x_{i,j}+x_{j,k}+\beta\right)  +\beta\left(  x_{i,j}+x_{j,k}\right)  +\beta
^{2}-\alpha\\
&  =x_{i,j}x_{j,k}+\left(  -\beta-x_{i,k}\right)  \left(  x_{i,j}%
+x_{j,k}+\beta\right)  +\beta\left(  x_{i,j}+x_{j,k}\right)  +\beta^{2}%
-\alpha\\
&  =x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  -\alpha.
\end{align*}
This proves (\ref{pf.prop.symmetry.2}).}.
\end{verlong}

\noindent Hence, (\ref{pf.prop.symmetry.J=1}) rewrites as follows:%
\begin{equation}
\mathcal{J}=\left\langle J_{i,j,k}\right\rangle _{\left(  i,j,k\right)
\in\left[  n\right]  ^{3}\text{ satisfying }i<j<k}.
\label{pf.prop.symmetry.J=3}%
\end{equation}


\begin{vershort}
On the other hand, the definition of $J_{i,j,k}$ shows that $J_{i,j,k}$ is
symmetric in its three arguments $i,j,k$; in other words, we have%
\[
J_{i,j,k}=J_{i,k,j}=J_{j,i,k}=J_{j,k,i}=J_{k,i,j}=J_{k,j,i}%
\]
whenever $i,j,k$ are three distinct elements of $\left[  n\right]  $. Thus,%
\[
\left\langle J_{i,j,k}\right\rangle _{i,j,k\text{ are three distinct elements
of }\left[  n\right]  }=\left\langle J_{i,j,k}\right\rangle _{\left(
i,j,k\right)  \in\left[  n\right]  ^{3}\text{ satisfying }i<j<k}.
\]

\end{vershort}

\begin{verlong}
On the other hand, the definition of $J_{i,j,k}$ shows that $J_{i,j,k}$ is
symmetric in its three arguments $i,j,k$; in other words, we have%
\[
J_{i,j,k}=J_{i,k,j}=J_{j,i,k}=J_{j,k,i}=J_{k,i,j}=J_{k,j,i}%
\]
whenever $i,j,k$ are three distinct elements of $\left[  n\right]  $. Hence,
the family \newline$\left(  J_{i,j,k}\right)  _{i,j,k\text{ are three distinct
elements of }\left[  n\right]  }$ contains the same elements as the family
\newline$\left(  J_{i,j,k}\right)  _{\left(  i,j,k\right)  \in\left[
n\right]  ^{3}\text{ satisfying }i<j<k}$, just repeated more often. Therefore,
the ideal generated by the former family equals the ideal generated by the
latter family. In other words,%
\[
\left\langle J_{i,j,k}\right\rangle _{i,j,k\text{ are three distinct elements
of }\left[  n\right]  }=\left\langle J_{i,j,k}\right\rangle _{\left(
i,j,k\right)  \in\left[  n\right]  ^{3}\text{ satisfying }i<j<k}.
\]

\end{verlong}

\noindent Comparing this with (\ref{pf.prop.symmetry.J=3}), we obtain
$\mathcal{J}=\left\langle J_{i,j,k}\right\rangle _{i,j,k\text{ are three
distinct elements of }\left[  n\right]  }$. This proves Proposition
\ref{prop.symmetry}.
\end{proof}

Proposition \ref{prop.symmetry} reveals a hidden symmetry in the definitions
of $\mathcal{X}$ and $\mathcal{J}$:

\begin{proposition}
\label{prop.symmetry2}Consider the symmetric group $S_{n}$ (that is, the group
of all permutations of $\left[  n\right]  $).

\textbf{(a)} There is a unique action of the group $S_{n}$ on $\mathcal{X}$ by
$\mathbf{k}$-algebra automorphisms satisfying%
\begin{align*}
\sigma\cdot x_{i,j}=x_{\sigma\left(  i\right)  ,\sigma\left(  j\right)
}\ \ \ \ \ \ \ \ \ \  &  \text{for all }\sigma\in S_{n}\\
&  \text{and all pairs }\left(  i,j\right)  \text{ of distinct elements of
}\left[  n\right]  .
\end{align*}


\textbf{(b)} The ideal $\mathcal{J}$ is invariant under this action of $S_{n}%
$, and thus the quotient $\mathbf{k}$-algebra $\mathcal{X}/\mathcal{J}$
inherits this action of $S_{n}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.symmetry2}.]\textbf{(a)} Let $\mathcal{Y}$ be
the polynomial ring
\[
\mathbf{k}\left[  y_{i,j}\ \mid\ \left(  i,j\right)  \in\left[  n\right]
^{2}\text{ such that }i\neq j\right]  .
\]
This is a polynomial ring in $n\left(  n-1\right)  $ indeterminates $y_{i,j}$
over $\mathbf{k}$. The symmetric group $S_{n}$ acts on $\mathcal{Y}$ by
$\mathbf{k}$-algebra automorphisms; this action is defined by%
\begin{align*}
\sigma\cdot y_{i,j}=y_{\sigma\left(  i\right)  ,\sigma\left(  j\right)
}\ \ \ \ \ \ \ \ \ \  &  \text{for all }\sigma\in S_{n}\\
&  \text{and all pairs }\left(  i,j\right)  \text{ of distinct elements of
}\left[  n\right]  .
\end{align*}
(The well-definedness of this action follows easily from the universal
property of the polynomial ring $\mathcal{Y}$.)

Let $\phi:\mathcal{Y}\rightarrow\mathcal{X}$ be the unique $\mathbf{k}%
$-algebra homomorphism that sends each $y_{i,j}$ to $x_{i,j}$. This $\phi$ is
well-defined by the universal property of the polynomial ring $\mathcal{Y}$.
Also, $\phi$ is surjective, since the generators $x_{i,j}$ of $\mathcal{X}$
all belong to the image of $\phi$.

If $\left(  u_{p}\right)  _{p\in P}$ is any family of elements of
$\mathcal{Y}$, then $\left\langle u_{p}\right\rangle _{p\in P}$ shall mean the
ideal of $\mathcal{Y}$ generated by this family $\left(  u_{p}\right)  _{p\in
P}$. Define an ideal $\mathcal{K}$ of $\mathcal{Y}$ by%
\begin{equation}
\mathcal{K}=\left\langle y_{i,j}+y_{j,i}+\beta\right\rangle _{\left(
i,j\right)  \in\left[  n\right]  ^{2}\text{ such that }i\neq j}.
\label{pf.prop.symmetry2.defK}%
\end{equation}
Clearly, this ideal $\mathcal{K}$ is $S_{n}$-invariant. Hence, the quotient
algebra $\mathcal{Y}/\mathcal{K}$ inherits the $S_{n}$-action from
$\mathcal{Y}$.

We are going to show that $\mathcal{X}\cong\mathcal{Y}/\mathcal{K}$.

\begin{vershort}
Let $\pi:\mathcal{Y}\rightarrow\mathcal{Y}/\mathcal{K}$ be the canonical
projection; this is a surjective $\mathbf{k}$-algebra homomorphism. The
$\mathbf{k}$-algebra $\mathcal{Y}$ is generated by the $y_{i,j}$ for all
$\left(  i,j\right)  \in\left[  n\right]  ^{2}$ such that $i\neq j$. Hence,
the quotient algebra $\mathcal{Y}/\mathcal{K}$ is generated by their
projections $\pi\left(  y_{i,j}\right)  $.
\end{vershort}

\begin{verlong}
Let $\pi:\mathcal{Y}\rightarrow\mathcal{Y}/\mathcal{K}$ be the canonical
projection; this is a surjective $\mathbf{k}$-algebra homomorphism. The
$\mathbf{k}$-algebra $\mathcal{Y}$ is generated by the $y_{i,j}$ for all
$\left(  i,j\right)  \in\left[  n\right]  ^{2}$ such that $i\neq j$. Hence,
the quotient algebra $\mathcal{Y}/\mathcal{K}$ is generated by the projections
of these $y_{i,j}$. In other words, the quotient algebra $\mathcal{Y}%
/\mathcal{K}$ is generated by the $\pi\left(  y_{i,j}\right)  $ for all
$\left(  i,j\right)  \in\left[  n\right]  ^{2}$ such that $i\neq j$.
\end{verlong}

It is easy to see that $\phi\left(  \mathcal{K}\right)  =0$%
\ \ \ \ \footnote{\textit{Proof.} It is clearly sufficient to show that
$\phi\left(  y_{i,j}+y_{j,i}+\beta\right)  =0$ for each $\left(  i,j\right)
\in\left[  n\right]  ^{2}$ such that $i\neq j$ (because of
(\ref{pf.prop.symmetry2.defK})). So let us fix some $\left(  i,j\right)
\in\left[  n\right]  ^{2}$ such that $i\neq j$. We must prove that
$\phi\left(  y_{i,j}+y_{j,i}+\beta\right)  =0$.
\par
This statement is clearly symmetric in $i$ and $j$; thus, we WLOG assume that
$i\leq j$. Hence, $i<j$ (since $i\neq j$). The definition of $\phi$ yields
$\phi\left(  y_{i,j}\right)  =x_{i,j}$ and $\phi\left(  y_{j,i}\right)
=x_{j,i}=-\beta-x_{i,j}$ (by the definition of $x_{j,i}$, since $j>i$). Now,
$\phi$ is a $\mathbf{k}$-algebra homomorphism. Thus,
\[
\phi\left(  y_{i,j}+y_{j,i}+\beta\right)  =\underbrace{\phi\left(
y_{i,j}\right)  }_{=x_{i,j}}+\underbrace{\phi\left(  y_{j,i}\right)
}_{=-\beta-x_{i,j}}+\beta=x_{i,j}+\left(  -\beta-x_{i,j}\right)  +\beta=0.
\]
This completes our proof.}. Hence, the $\mathbf{k}$-algebra homomorphism
$\phi$ factors through the projection $\pi:\mathcal{Y}\rightarrow
\mathcal{Y}/\mathcal{K}$. More precisely: There exists a $\mathbf{k}$-algebra
homomorphism $\phi^{\prime}:\mathcal{Y}/\mathcal{K}\rightarrow\mathcal{X}$
satisfying $\phi=\phi^{\prime}\circ\pi$. Consider this $\phi^{\prime}$. Thus,
each $\left(  i,j\right)  \in\left[  n\right]  ^{2}$ such that $i\neq j$
satisfies%
\begin{equation}
\phi^{\prime}\left(  \pi\left(  y_{i,j}\right)  \right)  =\underbrace{\left(
\phi^{\prime}\circ\pi\right)  }_{=\phi}\left(  y_{i,j}\right)  =\phi\left(
y_{i,j}\right)  =x_{i,j} \label{pf.prop.symmetry2.phi'}%
\end{equation}
(by the definition of $\phi$).

Define a $\mathbf{k}$-algebra homomorphism $\zeta:\mathcal{X}\rightarrow
\mathcal{Y}/\mathcal{K}$ by requiring that%
\begin{equation}
\zeta\left(  x_{i,j}\right)  =\pi\left(  y_{i,j}\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }\left(  i,j\right)  \in\left[  n\right]
^{2}\text{ satisfying }i<j. \label{pf.prop.symmetry2.zeta}%
\end{equation}
(This is well-defined by the universal property of $\mathcal{X}$.) Then, it is
easy to see that%
\begin{equation}
\zeta\left(  x_{i,j}\right)  =\pi\left(  y_{i,j}\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }\left(  i,j\right)  \in\left[  n\right]
^{2}\text{ satisfying }i\neq j \label{pf.prop.symmetry2.zeta2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.symmetry2.zeta2}):} Let $\left(
i,j\right)  \in\left[  n\right]  ^{2}$ be such that $i\neq j$. We must prove
that $\zeta\left(  x_{i,j}\right)  =\pi\left(  y_{i,j}\right)  $.
\par
If $i<j$, then this follows immediately from (\ref{pf.prop.symmetry2.zeta}).
Thus, we WLOG assume that we don't have $i<j$. Hence, $i\geq j$, so that $i>j$
(since $i\neq j$). In other words, $j<i$. Thus, (\ref{pf.prop.symmetry2.zeta})
(applied to $\left(  j,i\right)  $ instead of $\left(  i,j\right)  $) shows
that $\zeta\left(  x_{j,i}\right)  =\pi\left(  y_{j,i}\right)  $.
\par
Notice that $y_{i,j}+y_{j,i}+\beta\in\mathcal{K}$ (by
(\ref{pf.prop.symmetry2.defK})), so that $\pi\left(  y_{i,j}+y_{j,i}%
+\beta\right)  =0$ (since $\pi$ is the canonical projection $\mathcal{Y}%
\rightarrow\mathcal{Y}/\mathcal{K}$).
\par
But the definition of $x_{i,j}$ yields $x_{i,j}=-\beta-x_{j,i}$ (since $i>j$).
Applying the map $\zeta$ to both sides of this equality, we obtain%
\begin{align*}
\zeta\left(  x_{i,j}\right)   &  =\zeta\left(  -\beta-x_{j,i}\right)
=-\beta-\underbrace{\zeta\left(  x_{j,i}\right)  }_{=\pi\left(  y_{j,i}%
\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\zeta\text{ is a
}\mathbf{k}\text{-algebra homomorphism}\right) \\
&  =-\beta-\pi\left(  y_{j,i}\right)  .
\end{align*}
On the other hand, $\pi$ is a $\mathbf{k}$-algebra homomorphism, so that
$\pi\left(  y_{i,j}+y_{j,i}+\beta\right)  =\pi\left(  y_{i,j}\right)
+\pi\left(  y_{j,i}\right)  +\beta$. Thus,%
\[
\pi\left(  y_{i,j}\right)  +\pi\left(  y_{j,i}\right)  +\beta=\pi\left(
y_{i,j}+y_{j,i}+\beta\right)  =0.
\]
Hence, $\pi\left(  y_{i,j}\right)  =-\beta-\pi\left(  y_{j,i}\right)  $.
Comparing this with $\zeta\left(  x_{i,j}\right)  =-\beta-\pi\left(
y_{j,i}\right)  $, we obtain $\zeta\left(  x_{i,j}\right)  =\pi\left(
y_{i,j}\right)  $. This completes our proof of (\ref{pf.prop.symmetry2.zeta2}%
).}.

\begin{vershort}
The equality (\ref{pf.prop.symmetry2.phi'}) shows that the $\mathbf{k}%
$-algebra homomorphism $\phi^{\prime}:\mathcal{Y}/\mathcal{K}\rightarrow
\mathcal{X}$ sends the generators $\pi\left(  y_{i,j}\right)  $ of
$\mathcal{Y}/\mathcal{K}$ to the respective generators $x_{i,j}$ of
$\mathcal{X}$. The equality (\ref{pf.prop.symmetry2.zeta2}) shows that the
$\mathbf{k}$-algebra homomorphism $\zeta:\mathcal{X}\rightarrow\mathcal{Y}%
/\mathcal{K}$ sends the generators $x_{i,j}$ of $\mathcal{X}$ back to the
respective generators $\pi\left(  y_{i,j}\right)  $ of $\mathcal{Y}%
/\mathcal{K}$. Hence, these two $\mathbf{k}$-algebra homomorphisms
$\phi^{\prime}:\mathcal{Y}/\mathcal{K}\rightarrow\mathcal{X}$ and
$\zeta:\mathcal{X}\rightarrow\mathcal{Y}/\mathcal{K}$ are mutually inverse.
Thus, $\phi^{\prime}$ is a $\mathbf{k}$-algebra isomorphism.
\end{vershort}

\begin{verlong}
Now, $\zeta\circ\phi^{\prime}=\operatorname*{id}$%
\ \ \ \ \footnote{\textit{Proof.} Both $\zeta\circ\phi^{\prime}$ and
$\operatorname*{id}$ are $\mathbf{k}$-algebra homomorphisms (since $\zeta$ and
$\phi^{\prime}$ are $\mathbf{k}$-algebra homomorphisms). Thus, we want to
prove the equality of two $\mathbf{k}$-algebra homomorphisms. It therefore
suffices to check their equality on the generators $\pi\left(  y_{i,j}\right)
$ of the $\mathbf{k}$-algebra $\mathcal{Y}/\mathcal{K}$ (since we know that
the $\mathbf{k}$-algebra $\mathcal{Y}/\mathcal{K}$ is generated by the
$\pi\left(  y_{i,j}\right)  $ for all $\left(  i,j\right)  \in\left[
n\right]  ^{2}$ such that $i\neq j$). In other words, it suffices to check
that $\left(  \zeta\circ\phi^{\prime}\right)  \left(  \pi\left(
y_{i,j}\right)  \right)  =\pi\left(  y_{i,j}\right)  $ for each $\left(
i,j\right)  \in\left[  n\right]  ^{2}$ such that $i\neq j$.
\par
But this is easy: For each $\left(  i,j\right)  \in\left[  n\right]  ^{2}$
such that $i\neq j$, we have%
\[
\left(  \zeta\circ\phi^{\prime}\right)  \left(  \pi\left(  y_{i,j}\right)
\right)  =\zeta\left(  \underbrace{\phi^{\prime}\left(  \pi\left(
y_{i,j}\right)  \right)  }_{\substack{=x_{i,j}\\\text{(by
(\ref{pf.prop.symmetry2.phi'}))}}}\right)  =\zeta\left(  x_{i,j}\right)
=\pi\left(  y_{i,j}\right)
\]
(by (\ref{pf.prop.symmetry2.zeta2})). Thus, $\zeta\circ\phi^{\prime
}=\operatorname*{id}$ is proven.} and $\phi^{\prime}\circ\zeta
=\operatorname*{id}$\ \ \ \ \footnote{The proof of this is analogous to the
proof of $\zeta\circ\phi^{\prime}=\operatorname*{id}$ (but instead of the
generators $\pi\left(  y_{i,j}\right)  $ of the $\mathbf{k}$-algebra
$\mathcal{Y}/\mathcal{K}$, we now must use the generators $x_{i,j}$ of the
$\mathbf{k}$-algebra $\mathcal{X}$).}. Hence, the maps $\phi^{\prime}$ and
$\zeta$ are mutually inverse. Thus, the $\mathbf{k}$-algebra homomorphism
$\phi^{\prime}$ is invertible, and thus is a $\mathbf{k}$-algebra isomorphism.
\end{verlong}

\noindent Hence, $\mathcal{X}\cong\mathcal{Y}/\mathcal{K}$ as $\mathbf{k}%
$-algebras. Therefore, the $S_{n}$-action on $\mathcal{Y}/\mathcal{K}$ can be
transported to $\mathcal{X}$. The result is an action of the group $S_{n}$ on
$\mathcal{X}$ by $\mathbf{k}$-algebra automorphisms satisfying%
\begin{align*}
\sigma\cdot x_{i,j}=x_{\sigma\left(  i\right)  ,\sigma\left(  j\right)
}\ \ \ \ \ \ \ \ \ \  &  \text{for all }\sigma\in S_{n}\\
&  \text{and all pairs }\left(  i,j\right)  \text{ of distinct elements of
}\left[  n\right]  .
\end{align*}
Moreover, this is clearly the only such action (because any $\mathbf{k}%
$-algebra automorphism of $\mathcal{X}$ is determined by its action on the
generators $x_{i,j}$). This proves Proposition \ref{prop.symmetry2}
\textbf{(a)}.

\textbf{(b)} If $i,j,k$ are three distinct elements of $\left[  n\right]  $,
then $\sigma\cdot J_{i,j,k}=J_{\sigma\left(  i\right)  ,\sigma\left(
j\right)  ,\sigma\left(  k\right)  }$ for each $\sigma\in S_{n}$ (as follows
easily from the definitions of the elements involved). Hence, the action of
$S_{n}$ on $\mathcal{X}$ permutes the family $\left(  J_{i,j,k}\right)
_{i,j,k\text{ are three distinct elements of }\left[  n\right]  }$. Thus, the
ideal generated by this family is $S_{n}$-invariant. Since this ideal is
$\mathcal{J}$ (by Proposition \ref{prop.symmetry}), we have thus shown that
$\mathcal{J}$ is $S_{n}$-invariant. Hence, the quotient $\mathbf{k}$-algebra
$\mathcal{X}/\mathcal{J}$ inherits an $S_{n}$-action from $\mathcal{X}$.
Proposition \ref{prop.symmetry2} \textbf{(b)} is thus proven.
\end{proof}

\section{Forkless polynomials and a basis of $\mathcal{X}/\mathcal{J}$}

\subsection{Statements}

We have thus answered one of the major questions about the ideal $\mathcal{J}%
$; but we have begged perhaps the most obvious one: Can we find a basis of the
$\mathbf{k}$-module $\mathcal{X}/\mathcal{J}$? This turns out to be much
simpler than the above; the key is to use a different strategy. Instead of
reducing polynomials to pathless polynomials, we shall reduce them to
\textit{forkless} polynomials, defined as follows:

\begin{definition}
A monomial $\mathfrak{m}\in\mathfrak{M}$ is said to be \textit{forkless} if
there exists no triple $\left(  i,j,k\right)  \in\left[  n\right]  ^{3}$
satisfying $i<j<k$ and $x_{i,j}x_{i,k}\mid\mathfrak{m}$ (as monomials).

A polynomial $p\in\mathcal{X}$ is said to be \textit{forkless} if it is a
$\mathbf{k}$-linear combination of forkless monomials.
\end{definition}

The following characterization of forkless polynomials is rather obvious:

\begin{proposition}
Let $\mathfrak{m}\in\mathfrak{M}$. Then, the monomial $\mathfrak{m}$ is
forkless if and only if there exist a map $f:\left[  n-1\right]
\rightarrow\left[  n\right]  $ and a map $g:\left[  n-1\right]  \rightarrow
\mathbb{N}$ such that
\[
\left(  f\left(  i\right)  >i\text{ for each }i\in\left[  n-1\right]  \right)
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \mathfrak{m}=\prod
_{i\in\left[  n-1\right]  }x_{i,f\left(  i\right)  }^{g\left(  i\right)  }.
\]

\end{proposition}

Now, we claim the following:

\begin{theorem}
\label{thm.forkless.span-uni}Let $p\in\mathcal{X}$. Then, there exists a
\textbf{unique} forkless polynomial $q\in\mathcal{X}$ such that $p\equiv
q\operatorname{mod}\mathcal{J}$.
\end{theorem}

\begin{proposition}
\label{prop.forkless.basis}The projections of the forkless monomials
$\mathfrak{m}\in\mathfrak{M}$ onto the quotient ring $\mathcal{X}/\mathcal{J}$
form a basis of the $\mathbf{k}$-module $\mathcal{X}/\mathcal{J}$.
\end{proposition}

\subsection{A reminder on Gr\"{o}bner bases}

Theorem \ref{thm.forkless.span-uni} and Proposition \ref{prop.forkless.basis}
can be proven using the theory of Gr\"{o}bner bases. See, e.g.,
\cite{BecWei98} for an introduction. Let us outline the argument. We shall use
the following concepts:

\begin{definition}
\label{def.groebner.all}Let $\Xi$ be a set of indeterminates. Let
$\mathcal{X}_{\Xi}$ be the polynomial ring $\mathbf{k}\left[  \xi\ \mid
\ \xi\in\Xi\right]  $ over $\mathbf{k}$ in these indeterminates. Let
$\mathfrak{M}_{\Xi}$ be the set of all monomials in these indeterminates
(i.e., the free abelian monoid on the set $\Xi$). (For example, if
$\Xi=\left\{  x_{i,j}\ \mid\ \left(  i,j\right)  \in\left[  n\right]
^{2}\text{ satisfying }i<j\right\}  $, then $\mathcal{X}_{\Xi}=\mathcal{X}$
and $\mathfrak{M}_{\Xi}=\mathfrak{M}$.)

\textbf{(a)} A \textit{term order} on $\mathfrak{M}_{\Xi}$ is a total order on
the set $\mathfrak{M}_{\Xi}$ that satisfies the following conditions:

\begin{itemize}
\item Each $\mathfrak{m}\in\mathfrak{M}_{\Xi}$ satisfies $1\leq\mathfrak{m}$
(where $1$ is the trivial monomial in $\mathfrak{M}_{\Xi}$).

\item If $\mathfrak{m}$, $\mathfrak{u}$ and $\mathfrak{v}$ are three elements
of $\mathfrak{M}_{\Xi}$ satisfying $\mathfrak{u}\leq\mathfrak{v}$, then
$\mathfrak{mu}\leq\mathfrak{mv}$.
\end{itemize}

\textbf{(b)} If we are given a total order on the set $\Xi$, then we
canonically obtain a term order on $\mathfrak{M}_{\Xi}$ defined as follows:
For two monomials $\mathfrak{m}=\prod_{\xi\in\Xi}\xi^{m_{\xi}}$ and
$\mathfrak{n}=\prod_{\xi\in\Xi}\xi^{n_{\xi}}$ in $\mathfrak{M}_{\Xi}$, we set
$\mathfrak{m}\leq\mathfrak{n}$ if and only if either $\mathfrak{m}%
=\mathfrak{n}$ or the largest $\xi\in\Xi$ for which $m_{\xi}$ and $n_{\xi}$
differ satisfies $m_{\xi}<n_{\xi}$. This term order is called the
\textit{inverse lexicographical order on the set }$\mathfrak{M}_{\Xi}$
\textit{determined by the given total order on }$\Xi$.

\textbf{(c)} Two monomials $\mathfrak{m}=\prod_{\xi\in\Xi}\xi^{m_{\xi}}$ and
$\mathfrak{n}=\prod_{\xi\in\Xi}\xi^{n_{\xi}}$ in $\mathfrak{M}_{\Xi}$ are said
to be \textit{non-disjoint} if there exists some $\xi\in\Xi$ satisfying
$m_{\xi}>0$ and $n_{\xi}>0$. Otherwise, $\mathfrak{m}$ and $\mathfrak{n}$ are
said to be \textit{disjoint}.

From now on, let us assume that some term order on $\mathfrak{M}_{\Xi}$ has
been chosen. The next definitions will all rely on this term order.

\textbf{(d)} If $f\in\mathcal{X}_{\Xi}$ is a nonzero polynomial, then the
\textit{head term} of $f$ denotes the largest $\mathfrak{m}\in\mathfrak{M}%
_{\Xi}$ such that the coefficient of $\mathfrak{m}$ in $f$ is nonzero. This
head term will be denoted by $\operatorname*{HT}\left(  f\right)  $.
Furthermore, if $f\in\mathcal{X}_{\Xi}$ is a nonzero polynomial, then the
\textit{head coefficient} of $f$ is defined to be the coefficient of
$\operatorname*{HT}\left(  f\right)  $ in $f$; this coefficient will be
denoted by $\operatorname*{HC}\left(  f\right)  $.

\textbf{(e)} A nonzero polynomial $f\in\mathcal{X}_{\Xi}$ is said to be
\textit{monic} if its head coefficient $\operatorname*{HC}\left(  f\right)  $
is $1$.

\textbf{(f)} If $\mathfrak{m}=\prod_{\xi\in\Xi}\xi^{m_{\xi}}$ and
$\mathfrak{n}=\prod_{\xi\in\Xi}\xi^{n_{\xi}}$ are two monomials in
$\mathfrak{M}_{\Xi}$, then the \textit{lowest common multiple}
$\operatorname{lcm}\left(  \mathfrak{m},\mathfrak{n}\right)  $ of
$\mathfrak{m}$ and $\mathfrak{n}$ is defined to be the monomial $\prod_{\xi
\in\Xi}\xi^{\max\left\{  m_{\xi},n_{\xi}\right\}  }$. (Thus,
$\operatorname{lcm}\left(  \mathfrak{m},\mathfrak{n}\right)  =\mathfrak{mn}$
if and only if $\mathfrak{m}$ and $\mathfrak{n}$ are disjoint.)

\textbf{(g)} If $g_{1}$ and $g_{2}$ are two monic polynomials in
$\mathcal{X}_{\Xi}$, then the \textit{S-polynomial of }$g_{1}$\textit{ and
}$g_{2}$ is defined to be the polynomial $\mathfrak{s}_{1}g_{1}-\mathfrak{s}%
_{2}g_{2}$, where $\mathfrak{s}_{1}$ and $\mathfrak{s}_{2}$ are the unique two
monomials satisfying $\mathfrak{s}_{1}\operatorname*{HT}\left(  g_{1}\right)
=\mathfrak{s}_{2}\operatorname*{HT}\left(  g_{2}\right)  =\operatorname{lcm}%
\left(  \operatorname*{HT}\left(  g_{1}\right)  ,\operatorname*{HT}\left(
g_{2}\right)  \right)  $. This S-polynomial is denoted by
$\operatorname*{spol}\left(  g_{1},g_{2}\right)  $.

From now on, let $G$ be a subset of $\mathcal{X}_{\Xi}$ that consists of monic polynomials.

\textbf{(h)} We define a binary relation $\underset{G}{\longrightarrow}$ on
the set $\mathcal{X}_{\Xi}$ as follows: For two polynomials $f$ and $g$ in
$\mathcal{X}_{\Xi}$, we set $f\underset{G}{\longrightarrow}g$ (and say that
$f$ \textit{reduces to }$g$ \textit{modulo }$G$) if there exists some $p\in G$
and some monomials $\mathfrak{t}\in\mathfrak{M}_{\Xi}$ and $\mathfrak{s}%
\in\mathfrak{M}_{\Xi}$ with the following properties:

\begin{itemize}
\item The coefficient of $\mathfrak{t}$ in $f$ is $\neq0$.

\item We have $\mathfrak{s}\cdot\operatorname*{HT}\left(  p\right)
=\mathfrak{t}$.

\item If $a$ is the coefficient of $\mathfrak{t}$ in $f$, then $g=f-a\cdot
\mathfrak{s}\cdot p$.
\end{itemize}

\textbf{(i)} We let $\overset{\ast}{\underset{G}{\longrightarrow}}$ denote the
reflexive-and-transitive closure of the relation $\underset{G}{\longrightarrow
}$.

\textbf{(j)} We say that a monomial $\mathfrak{m}\in\mathfrak{M}_{\Xi}$ is
$G$\textit{-reduced} if it is not divisible by the head term of any element of
$G$.

We say that a polynomial $q\in\mathcal{X}_{\Xi}$ is $G$\textit{-reduced} if
$q$ is a $\mathbf{k}$-linear combination of $G$-reduced monomials.

\textbf{(k)} Let $\mathcal{I}$ be an ideal of $\mathcal{X}_{\Xi}$. The set $G$
is said to be a \textit{Gr\"{o}bner basis} of the ideal $\mathcal{I}$ if and
only if the set $G$ generates $\mathcal{I}$ and has the following two
equivalent properties:

\begin{itemize}
\item For each $p\in\mathcal{X}_{\Xi}$, there is a unique $G$-reduced
$q\in\mathcal{X}_{\Xi}$ such that $p\overset{\ast
}{\underset{G}{\longrightarrow}}q$.

\item For each $p\in\mathcal{I}$, we have $p\overset{\ast
}{\underset{G}{\longrightarrow}}0$.
\end{itemize}
\end{definition}

The definition we just gave is modelled after the definitions in \cite[Chapter
5]{BecWei98}; however, there are several minor differences:

\begin{itemize}
\item We use the word \textquotedblleft monomial\textquotedblright\ in the
same meaning as \cite[Chapter 5]{BecWei98} use the word \textquotedblleft
term\textquotedblright\ (but not in the same meaning as \cite[Chapter
5]{BecWei98} use the word \textquotedblleft monomial\textquotedblright).

\item We allow $\mathbf{k}$ to be a commutative ring, whereas \cite[Chapter
5]{BecWei98} require $\mathbf{k}$ to be a field. This leads to some
complications in the theory of Gr\"{o}bner bases; in particular, not every
ideal has a Gr\"{o}bner basis anymore. However, everything \textbf{we} are
going to use about Gr\"{o}bner bases in this paper is still true in our
general setting.

\item We require the elements of the Gr\"{o}bner basis $G$ to be monic,
whereas \cite[Chapter 5]{BecWei98} merely assume them to be nonzero
polynomials. In this way, we are sacrificing some of the generality of
\cite[Chapter 5]{BecWei98} (a sacrifice necessary to ensure that things don't
go wrong when $\mathbf{k}$ is not a field). However, this is not a major loss
of generality, since in the situation of \cite[Chapter 5]{BecWei98} the
difference between monic polynomials and arbitrary nonzero polynomials is not
particularly large (we can scale any nonzero polynomial by a constant scalar
to obtain a monic polynomial, and so we can assume the polynomials to be monic
in most of the proofs).
\end{itemize}

The following fact is useful even if almost trivial:

\begin{lemma}
\label{lem.groebner.to0}Let $\Xi$, $\mathcal{X}_{\Xi}$ and $\mathfrak{M}_{\Xi
}$ be as in Definition \ref{def.groebner.all}. Let $G$ be a subset of
$\mathcal{X}_{\Xi}$ that consists of monic polynomials. Let $S$ be a finite
set. For each $s\in S$, let $g_{s}$ be an element of $G$, and let
$\mathfrak{s}_{s}\in\mathfrak{M}_{\Xi}$ and $a_{s}\in\mathbf{k}$ be arbitrary.
Assume that the monomials $\mathfrak{s}_{s}\operatorname*{HT}\left(
g_{s}\right)  $ for all $s\in S$ are distinct. Then, $\sum_{s\in S}%
a_{s}\mathfrak{s}_{s}g_{s}\overset{\ast}{\underset{G}{\longrightarrow}}0$.
\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.groebner.to0}.]See \cite{verlong}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.groebner.to0} (sketched).]We proceed by strong
induction on $\left\vert S\right\vert $. If $\left\vert S\right\vert =0$, then
Lemma \ref{lem.groebner.to0} is obvious\footnote{because in this case, we have
$\sum_{s\in S}w_{s}g_{s}=\left(  \text{empty sum}\right)  =0\overset{\ast
}{\underset{G}{\longrightarrow}}0$}. Hence, WLOG assume that $\left\vert
S\right\vert >0$.

Let $t$ be the element of $S$ with highest $\mathfrak{s}_{t}\operatorname*{HT}%
\left(  g_{t}\right)  $. Then, $t$ is the \textbf{unique} element of $S$ with
highest $\mathfrak{s}_{t}\operatorname*{HT}\left(  g_{t}\right)  $ (since the
monomials $\mathfrak{s}_{s}\operatorname*{HT}\left(  g_{s}\right)  $ for all
$s\in S$ are distinct). By the induction hypothesis, we have $\sum_{s\in
S\setminus\left\{  t\right\}  }a_{s}\mathfrak{s}_{s}g_{s}\overset{\ast
}{\underset{G}{\longrightarrow}}0$. If $a_{t}=0$, then this immediately
rewrites as $\sum_{s\in S}a_{s}\mathfrak{s}_{s}g_{s}\overset{\ast
}{\underset{G}{\longrightarrow}}0$, and so we are done. Hence, WLOG assume
that $a_{t}\neq0$. Hence, the head term of $\sum_{s\in S}a_{s}\mathfrak{s}%
_{s}g_{s}$ is $\mathfrak{s}_{t}\operatorname*{HT}\left(  g_{t}\right)  $, with
coefficient $a_{t}$ (since $t$ is the \textbf{unique} element of $S$ with
highest $\mathfrak{s}_{t}\operatorname*{HT}\left(  g_{t}\right)  $). Thus, we
know that:

\begin{itemize}
\item The coefficient of $\mathfrak{s}_{t}\operatorname*{HT}\left(
g_{t}\right)  $ is $\neq0$ (since $a_{t}\neq0$).

\item We have $\mathfrak{s}_{t}\cdot\operatorname*{HT}\left(  g_{t}\right)
=\mathfrak{s}_{t}\operatorname*{HT}\left(  g_{t}\right)  $.

\item If $a$ is the coefficient of $\mathfrak{s}_{t}\operatorname*{HT}\left(
g_{t}\right)  $ in $\sum_{s\in S}a_{s}\mathfrak{s}_{s}g_{s}$, then $\sum_{s\in
S\setminus\left\{  t\right\}  }a_{s}\mathfrak{s}_{s}g_{s}=\sum_{s\in S}%
a_{s}\mathfrak{s}_{s}g_{s}-a\cdot\mathfrak{s}_{t}\cdot g_{t}$ (because this
$a$ is $a_{t}$).
\end{itemize}

Hence, the definition of the relation $\underset{G}{\longrightarrow}$ yields
$\sum_{s\in S}a_{s}\mathfrak{s}_{s}g_{s}\underset{G}{\longrightarrow}%
\sum_{s\in S\setminus\left\{  t\right\}  }a_{s}\mathfrak{s}_{s}g_{s}$.
Combining this with $\sum_{s\in S\setminus\left\{  t\right\}  }a_{s}%
\mathfrak{s}_{s}g_{s}\overset{\ast}{\underset{G}{\longrightarrow}}0$, we
obtain $\sum_{s\in S}a_{s}\mathfrak{s}_{s}g_{s}\overset{\ast
}{\underset{G}{\longrightarrow}}0$. This completes the induction step, and
thus Lemma \ref{lem.groebner.to0} is proven.
\end{proof}
\end{verlong}

\begin{verlong}
The following criterion for a set to be a Gr\"{o}bner basis is well-known (it
is, in fact, the main ingredient in the proof of the correctness of
Buchberger's algorithm):

\begin{proposition}
\label{prop.groebner.buch0}Let $\Xi$, $\mathcal{X}_{\Xi}$ and $\mathfrak{M}%
_{\Xi}$ be as in Definition \ref{def.groebner.all}. Let $\mathcal{I}$ be an
ideal of $\mathcal{X}_{\Xi}$. Let $G$ be a subset of $\mathcal{X}_{\Xi}$ that
consists of monic polynomials. Assume that the set $G$ generates $\mathcal{I}%
$. Then, $G$ is a Gr\"{o}bner basis of $\mathcal{I}$ if and only if it has the
following property:

\begin{itemize}
\item If $g_{1}$ and $g_{2}$ are any two elements of the set $G$, then
$\operatorname*{spol}\left(  g_{1},g_{2}\right)  \overset{\ast
}{\underset{G}{\longrightarrow}}0$.
\end{itemize}
\end{proposition}

Proofs of Proposition \ref{prop.groebner.buch0} (at least in the case when
$\mathbf{k}$ is a field) can be found in \cite[Theorem 5.48, (iii)
$\Longleftrightarrow$ (i)]{BecWei98}, \cite[Theorem 2.14]{EneHer12},
\cite[Theorem 1.1.33]{Graaf16}, \cite[V.3 \textbf{i)} $\Longleftrightarrow$
\textbf{ii)}]{MalBlo15}, \cite[Th\'{e}or\`{e}me (Buchberger) \textbf{(i)}
$\Longleftrightarrow$ \textbf{(ii)}]{Monass02}, and (in a slight variation) in
\cite[Chapter 2, \S 6, Theorem 6]{CoLiOs15}\footnote{Different sources state
slightly different versions of Proposition \ref{prop.groebner.buch0}. For
example, some texts require $\operatorname*{spol}\left(  g_{1},g_{2}\right)
\overset{\ast}{\underset{G}{\longrightarrow}}0$ not for any two elements
$g_{1}$ and $g_{2}$ of $G$, but only for any two \textbf{distinct} elements
$g_{1}$ and $g_{2}$ of $G$. However, this makes no difference, because if
$g_{1}$ and $g_{2}$ are equal, then $\operatorname*{spol}\left(  g_{1}%
,g_{2}\right)  =0\overset{\ast}{\underset{G}{\longrightarrow}}0$. Similarly,
other texts require $g_{1}<g_{2}$ (with respect to some chosen total order on
$G$); this also does not change much, since $\operatorname*{spol}\left(
g_{1},g_{2}\right)  =-\operatorname*{spol}\left(  g_{2},g_{1}\right)  $.}.

The following fact (known as \textquotedblleft Buchberger's first
criterion\textquotedblright) somewhat simplifies dealing with S-polynomials:

\begin{proposition}
\label{prop.groebner.buchS}Let $\Xi$, $\mathcal{X}_{\Xi}$ and $\mathfrak{M}%
_{\Xi}$ be as in Definition \ref{def.groebner.all}. Let $G$ be a subset of
$\mathcal{X}_{\Xi}$ that consists of monic polynomials. Let $g_{1}$ and
$g_{2}$ be two elements of the set $G$ such that the head terms of $g_{1}$ and
$g_{2}$ are disjoint. Then, $\operatorname*{spol}\left(  g_{1},g_{2}\right)
\overset{\ast}{\underset{G}{\longrightarrow}}0$.
\end{proposition}

Proposition \ref{prop.groebner.buchS} can be found in \cite[Lemma
5.66]{BecWei98}, \cite[Lemma 1.1.38]{Graaf16}, \cite[Theorem 4]{Grinbe17},
\cite[Proposition 2.15]{EneHer12}, \cite[V.6 \textbf{i)}]{MalBlo15} and
\cite[Lemme (in the section \textquotedblleft Am\'{e}liorations de
l'algorithme\textquotedblright)]{Monass02}\footnote{That said, the proof of
\cite[Lemme (in the section \textquotedblleft Am\'{e}liorations de
l'algorithme\textquotedblright)]{Monass02} is incorrect.}.

We can combine Proposition \ref{prop.groebner.buch0} with Proposition
\ref{prop.groebner.buchS} to obtain the following fact:
\end{verlong}

\begin{vershort}
One of Buchberger's celebrated results is the following proposition, which
allows us to verify that a given finite set $G$ is a Gr\"{o}bner basis of an
ideal $\mathcal{I}$ using a finite computation:
\end{vershort}

\begin{proposition}
\label{prop.groebner.buch1}Let $\Xi$, $\mathcal{X}_{\Xi}$ and $\mathfrak{M}%
_{\Xi}$ be as in Definition \ref{def.groebner.all}. Let $\mathcal{I}$ be an
ideal of $\mathcal{X}_{\Xi}$. Let $G$ be a subset of $\mathcal{X}_{\Xi}$ that
consists of monic polynomials. Assume that the set $G$ generates $\mathcal{I}%
$. Then, $G$ is a Gr\"{o}bner basis of $\mathcal{I}$ if and only if it has the
following property:

\begin{itemize}
\item If $g_{1}$ and $g_{2}$ are two elements of the set $G$ such that the
head terms of $g_{1}$ and $g_{2}$ are non-disjoint, then $\operatorname*{spol}%
\left(  g_{1},g_{2}\right)  \overset{\ast}{\underset{G}{\longrightarrow}}0$.
\end{itemize}
\end{proposition}

\begin{vershort}
Proposition \ref{prop.groebner.buch1} appears (at least in the case when
$\mathbf{k}$ is a field) in \cite[Theorem 5.68, (ii) $\Longleftrightarrow$
(i)]{BecWei98} and \cite[Conclusion after the proof of Lemma 1.1.38]{Graaf16}.
\end{vershort}

\begin{verlong}
Proposition \ref{prop.groebner.buch1} follows trivially from Proposition
\ref{prop.groebner.buch0} after recalling Proposition
\ref{prop.groebner.buchS}. Explicitly, Proposition \ref{prop.groebner.buch1}
appears (at least in the case when $\mathbf{k}$ is a field) in \cite[Theorem
5.68, (ii) $\Longleftrightarrow$ (i)]{BecWei98} and \cite[Conclusion after the
proof of Lemma 1.1.38]{Graaf16}.
\end{verlong}

We shall also use the following simple fact, known as the \textquotedblleft
Macaulay-Buchberger basis theorem\textquotedblright:

\begin{proposition}
\label{prop.groebner.standardbasis}Let $\Xi$, $\mathcal{X}_{\Xi}$ and
$\mathfrak{M}_{\Xi}$ be as in Definition \ref{def.groebner.all}. Let
$\mathcal{I}$ be an ideal of $\mathcal{X}_{\Xi}$. Let $G$ be a Gr\"{o}bner
basis of $\mathcal{I}$. The projections of the $G$-reduced monomials onto the
quotient ring $\mathcal{X}_{\Xi}/\mathcal{I}$ form a basis of the $\mathbf{k}%
$-module $\mathcal{X}_{\Xi}/\mathcal{I}$.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.groebner.standardbasis}.]See \cite[Chapter 5,
\S 3, Proposition 1 and Proposition 4]{CoLiOs15} or \cite[Th\'{e}or\`{e}me in
the section \textquotedblleft Espaces quotients\textquotedblright]{Monass02}
or \cite[Theorem 1.2.6]{Sturmf08} or \cite{verlong}.
\end{proof}
\end{vershort}

\begin{verlong}
Proposition \ref{prop.groebner.standardbasis} is easy to prove; it also
appears (in the case when $\mathbf{k}$ is a field) in various texts (e.g.,
\cite[Chapter 5, \S 3, Proposition 1 and Proposition 4]{CoLiOs15},
\cite[Th\'{e}or\`{e}me in the section \textquotedblleft Espaces
quotients\textquotedblright]{Monass02} or \cite[Theorem 1.2.6]{Sturmf08}).

For the sake of completeness, let us also outline its proof here:

\begin{proof}
[Proof of Proposition \ref{prop.groebner.standardbasis} (sketched).]For each
polynomial $p\in\mathcal{X}_{\Xi}$, we let $\overline{p}$ denote the
projection of $p$ onto the quotient ring $\mathcal{X}_{\Xi}/\mathcal{I}$. Let
$\mathfrak{M}_{\operatorname*{red}}$ be the set of all $G$-reduced monomials.
Thus, we must prove that the family $\left(  \overline{\mathfrak{m}}\right)
_{\mathfrak{m}\in\mathfrak{M}_{\operatorname*{red}}}$ is a basis of the
$\mathbf{k}$-module $\mathcal{X}_{\Xi}/\mathcal{I}$.

First, let us prove that this family is $\mathbf{k}$-linearly independent.
Indeed, let $\left(  \lambda_{\mathfrak{m}}\right)  _{\mathfrak{m}%
\in\mathfrak{M}_{\operatorname*{red}}}$ be a family of elements of
$\mathbf{k}$ such that all but finitely many $\mathfrak{m}\in\mathfrak{M}%
_{\operatorname*{red}}$ satisfy $\lambda_{\mathfrak{m}}=0$, and such that
$\sum_{\mathfrak{m}\in\mathfrak{M}_{\operatorname*{red}}}\lambda
_{\mathfrak{m}}\overline{\mathfrak{m}}=0$.

Now, let $f=\sum_{\mathfrak{m}\in\mathfrak{M}_{\operatorname*{red}}}%
\lambda_{\mathfrak{m}}\mathfrak{m}$. Thus, $\overline{f}=\overline
{\sum_{\mathfrak{m}\in\mathfrak{M}_{\operatorname*{red}}}\lambda
_{\mathfrak{m}}\mathfrak{m}}=\sum_{\mathfrak{m}\in\mathfrak{M}%
_{\operatorname*{red}}}\lambda_{\mathfrak{m}}\overline{\mathfrak{m}}=0$, so
that $f\in\mathcal{I}$. Assume (for the sake of contradiction) that $f\neq0$.

The definition of a Gr\"{o}bner basis shows that for each $p\in\mathcal{I}$,
we have $p\overset{\ast}{\underset{G}{\longrightarrow}}0$. Applying this to
$p=f$, we obtain $f\overset{\ast}{\underset{G}{\longrightarrow}}0$. Since
$f\neq0$, we thus conclude that there exists some $g\in\mathcal{X}_{\Xi}$ such
that $f\underset{G}{\longrightarrow}g$ and $g\overset{\ast
}{\underset{G}{\longrightarrow}}0$ (because the relation $\overset{\ast
}{\underset{G}{\longrightarrow}}$ is the reflexive-and-transitive closure of
the relation $\underset{G}{\longrightarrow}$). Consider this $g$.

We have $f\underset{G}{\longrightarrow}g$. By the definition of the relation
$\underset{G}{\longrightarrow}$, this means that there exists some $p\in G$
and some monomials $\mathfrak{t}\in\mathfrak{M}_{\Xi}$ and $\mathfrak{s}%
\in\mathfrak{M}_{\Xi}$ with the following properties:

\begin{itemize}
\item The coefficient of $\mathfrak{t}$ in $f$ is $\neq0$.

\item We have $\mathfrak{s}\cdot\operatorname*{HT}\left(  p\right)
=\mathfrak{t}$.

\item If $a$ is the coefficient of $\mathfrak{t}$ in $f$, then $g=f-a\cdot
\mathfrak{s}\cdot p$.
\end{itemize}

The coefficient of $\mathfrak{t}$ in $f$ is $\neq0$. In other words, the
monomial $\mathfrak{t}$ appears in $f$. Hence, $\mathfrak{t}\in\mathfrak{M}%
_{\operatorname*{red}}$ (since all monomials appearing in $f$ belong to
$\mathfrak{M}_{\operatorname*{red}}$ (by the definition of $f$)). In other
words, the monomial $\mathfrak{t}$ is $G$-reduced. In other words, the
monomial $\mathfrak{t}$ is not divisible by the head term of any element of
$G$ (by the definition of \textquotedblleft$G$-reduced\textquotedblright). But
this contradicts the fact that $\mathfrak{t}$ is divisible by
$\operatorname*{HT}\left(  p\right)  $ (since $\mathfrak{s}\cdot
\operatorname*{HT}\left(  p\right)  =\mathfrak{t}$), because
$\operatorname*{HT}\left(  p\right)  $ is the head term of the element $p$ of
$G$. This contradiction shows that our assumption (that $f\neq0$) was wrong.
Hence, we have $f=0$.

Thus, $\sum_{\mathfrak{m}\in\mathfrak{M}_{\operatorname*{red}}}\lambda
_{\mathfrak{m}}\mathfrak{m}=f=0$. Hence, $\lambda_{\mathfrak{m}}=0$ for each
$\mathfrak{m}\in\mathfrak{M}_{\operatorname*{red}}$ (since the monomials
$\mathfrak{m}\in\mathfrak{M}_{\operatorname*{red}}$ are $\mathbf{k}$-linearly
independent). In other words, $\left(  \lambda_{\mathfrak{m}}\right)
_{\mathfrak{m}\in\mathfrak{M}_{\operatorname*{red}}}=\left(  0\right)
_{\mathfrak{m}\in\mathfrak{M}_{\operatorname*{red}}}$.

Now, forget that we fixed $\left(  \lambda_{\mathfrak{m}}\right)
_{\mathfrak{m}\in\mathfrak{M}_{\operatorname*{red}}}$. We thus have shown that
if $\left(  \lambda_{\mathfrak{m}}\right)  _{\mathfrak{m}\in\mathfrak{M}%
_{\operatorname*{red}}}$ is a family of elements of $\mathbf{k}$ such that all
but finitely many $\mathfrak{m}\in\mathfrak{M}_{\operatorname*{red}}$ satisfy
$\lambda_{\mathfrak{m}}=0$, and such that $\sum_{\mathfrak{m}\in
\mathfrak{M}_{\operatorname*{red}}}\lambda_{\mathfrak{m}}\overline
{\mathfrak{m}}=0$, then $\left(  \lambda_{\mathfrak{m}}\right)  _{\mathfrak{m}%
\in\mathfrak{M}_{\operatorname*{red}}}=\left(  0\right)  _{\mathfrak{m}%
\in\mathfrak{M}_{\operatorname*{red}}}$. In other words, the family $\left(
\overline{\mathfrak{m}}\right)  _{\mathfrak{m}\in\mathfrak{M}%
_{\operatorname*{red}}}$ is $\mathbf{k}$-linearly independent.

Next, let us prove that this family spans the $\mathbf{k}$-module
$\mathcal{X}_{\Xi}/\mathcal{I}$. Indeed, let $r\in\mathcal{X}_{\Xi
}/\mathcal{I}$ be arbitrary.

It is easy to see that if two elements $f$ and $g$ of $\mathcal{X}_{\Xi}$
satisfy $f\underset{G}{\longrightarrow}g$, then $\overline{f}=\overline{g}$
(because $f\underset{G}{\longrightarrow}g$ implies that $g=f-a\cdot
\mathfrak{s}\cdot p$ for some $a\in\mathbf{k}$, some $\mathfrak{s}%
\in\mathfrak{M}_{\Xi}$ and some $p\in G$, but this shows that $f-g=a\cdot
\mathfrak{s}\cdot\underbrace{p}_{\in G\subseteq\mathcal{I}}\in a\cdot
\mathfrak{s}\cdot\mathcal{I}\subseteq\mathcal{I}$). Hence, if two elements $f$
and $g$ of $\mathcal{X}_{\Xi}$ satisfy $f\overset{\ast
}{\underset{G}{\longrightarrow}}g$, then%
\begin{equation}
\overline{f}=\overline{g} \label{pf.prop.groebner.standardbasis.barf=barg}%
\end{equation}
(because the relation $\overset{\ast}{\underset{G}{\longrightarrow}}$ is the
reflexive-and-transitive closure of the relation $\underset{G}{\longrightarrow
}$).

We have $r\in\mathcal{X}_{\Xi}/\mathcal{I}$. Thus, we can write $r$ as
$r=\overline{f}$ for some $f\in\mathcal{X}_{\Xi}$. Consider this $f$.

The definition of a Gr\"{o}bner basis shows that for each $p\in\mathcal{X}%
_{\Xi}$, there is a unique $G$-reduced $q\in\mathcal{X}_{\Xi}$ such that
$p\overset{\ast}{\underset{G}{\longrightarrow}}q$. Applying this to $p=f$, we
conclude that there is a unique $G$-reduced $q\in\mathcal{X}_{\Xi}$ such that
$f\overset{\ast}{\underset{G}{\longrightarrow}}q$. Consider this $q$. Applying
(\ref{pf.prop.groebner.standardbasis.barf=barg}) to $g=q$, we obtain
$\overline{f}=\overline{q}$.

But the polynomial $q$ is $G$-reduced. In other words, $q$ is a $\mathbf{k}%
$-linear combination of $G$-reduced monomials. In other words, $q$ is a
$\mathbf{k}$-linear combination of the family $\left(  \mathfrak{m}\right)
_{\mathfrak{m}\in\mathfrak{M}_{\operatorname*{red}}}$. Hence, $\overline{q}$
is a $\mathbf{k}$-linear combination of the family $\left(  \overline
{\mathfrak{m}}\right)  _{\mathfrak{m}\in\mathfrak{M}_{\operatorname*{red}}}$.
In view of $r=\overline{f}=\overline{q}$, this shows that $r$ is a
$\mathbf{k}$-linear combination of the family $\left(  \overline{\mathfrak{m}%
}\right)  _{\mathfrak{m}\in\mathfrak{M}_{\operatorname*{red}}}$.

Now, forget that we fixed $r$. We thus have shown that each $r\in
\mathcal{X}_{\Xi}/\mathcal{I}$ is a $\mathbf{k}$-linear combination of the
family $\left(  \overline{\mathfrak{m}}\right)  _{\mathfrak{m}\in
\mathfrak{M}_{\operatorname*{red}}}$. In other words, the family $\left(
\overline{\mathfrak{m}}\right)  _{\mathfrak{m}\in\mathfrak{M}%
_{\operatorname*{red}}}$ spans the $\mathbf{k}$-module $\mathcal{X}_{\Xi
}/\mathcal{I}$. Since we also know that this family is $\mathbf{k}$-linearly
independent, we can thus conclude that $\left(  \overline{\mathfrak{m}%
}\right)  _{\mathfrak{m}\in\mathfrak{M}_{\operatorname*{red}}}$ is a basis of
the $\mathbf{k}$-module $\mathcal{X}_{\Xi}/\mathcal{I}$. This proves
Proposition \ref{prop.groebner.standardbasis}.
\end{proof}
\end{verlong}

\subsection{The proofs}

The main workhorse of the proofs is the following fact:

\begin{proposition}
\label{prop.forkless.groebner}Consider the inverse lexicographical order on
the set $\mathfrak{M}$ of monomials determined by%
\begin{align*}
&  x_{1,2}>x_{1,3}>\cdots>x_{1,n}\\
&  >x_{2,3}>x_{2,4}>\cdots>x_{2,n}\\
&  >\cdots\\
&  >x_{n-1,n}.
\end{align*}
Then, the set%
\begin{equation}
\left\{  x_{i,k}x_{i,j}-x_{i,j}x_{j,k}+x_{i,k}x_{j,k}+\beta x_{i,k}%
+\alpha\ \mid\ \left(  i,j,k\right)  \in\left[  n\right]  ^{3}\text{
satisfying }i<j<k\right\}  \label{eq.prop.forkless.groebner.family}%
\end{equation}
is a Gr\"{o}bner basis of the ideal $\mathcal{J}$ of $\mathcal{X}$ (with
respect to this order).
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.forkless.groebner} (sketched).]The elements
$x_{i,k}x_{i,j}-x_{i,j}x_{j,k}+x_{i,k}x_{j,k}+\beta x_{i,k}+\alpha$ of the set
(\ref{eq.prop.forkless.groebner.family}) differ from the designated generators
$x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  -\alpha$ of the
ideal $\mathcal{J}$ merely by a factor of $-1$ (indeed, $x_{i,k}%
x_{i,j}-x_{i,j}x_{j,k}+x_{i,k}x_{j,k}+\beta x_{i,k}+\alpha=\left(  -1\right)
\left(  x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)
-\alpha\right)  $). Thus, they generate the ideal $\mathcal{J}$. Hence, in
order to prove that they form a Gr\"{o}bner basis of $\mathcal{J}$, we merely
need to show the following claim:

\begin{statement}
\textit{Claim 1:} Let $g_{1}$ and $g_{2}$ be two elements of the set
(\ref{eq.prop.forkless.groebner.family}) such that the head terms of $g_{1}$
and $g_{2}$ are non-disjoint. Then, $\operatorname*{spol}\left(  g_{1}%
,g_{2}\right)  \overset{\ast}{\underset{G}{\longrightarrow}}0$, where $G$ is
the set (\ref{eq.prop.forkless.groebner.family}).
\end{statement}

(Indeed, proving Claim 1 is sufficient because of Proposition
\ref{prop.groebner.buch1}.)

In order to prove Claim 1, we fix two elements $g_{1}$ and $g_{2}$ of the set
(\ref{eq.prop.forkless.groebner.family}) such that the head terms of $g_{1}$
and $g_{2}$ are non-disjoint. Thus,%
\[
g_{1}=x_{i_{1},k_{1}}x_{i_{1},j_{1}}-x_{i_{1},j_{1}}x_{j_{1},k_{1}}%
+x_{i_{1},k_{1}}x_{j_{1},k_{1}}+\beta x_{i_{1},k_{1}}+\alpha
\]
for some $\left(  i_{1},j_{1},k_{1}\right)  \in\left[  n\right]  ^{3}$
satisfying $i_{1}<j_{1}<k_{1}$, and%
\[
g_{2}=x_{i_{2},k_{2}}x_{i_{2},j_{2}}-x_{i_{2},j_{2}}x_{j_{2},k_{2}}%
+x_{i_{2},k_{2}}x_{j_{2},k_{2}}+\beta x_{i_{2},k_{2}}+\alpha
\]
for some $\left(  i_{2},j_{2},k_{2}\right)  \in\left[  n\right]  ^{3}$
satisfying $i_{2}<j_{2}<k_{2}$. Since the head terms $x_{i_{1},k_{1}}%
x_{i_{1},j_{1}}$ and $x_{i_{2},k_{2}}x_{i_{2},j_{2}}$ of $g_{1}$ and $g_{2}$
are non-disjoint, we must have $i_{1}=i_{2}$. Furthermore, one of $j_{1}$ and
$k_{1}$ must equal one of $j_{2}$ and $k_{2}$ (for the same reason). Thus,
there are at most four distinct integers among $i_{1},i_{2},j_{1},j_{2}%
,k_{1},k_{2}$.

We can now finish off Claim 1 by straightforward computations, after
distinguishing several cases based upon which of the numbers $j_{1}$ and
$k_{1}$ equal which of the numbers $j_{2}$ and $k_{2}$. We WLOG assume that
$\left(  i_{1},j_{1},k_{1}\right)  \neq\left(  i_{2},j_{2},k_{2}\right)  $
(since otherwise, it is clear that $\operatorname*{spol}\left(  g_{1}%
,g_{2}\right)  =0\overset{\ast}{\underset{G}{\longrightarrow}}0$). Thus, there
are \textbf{exactly} four distinct integers among $i_{1},i_{2},j_{1}%
,j_{2},k_{1},k_{2}$ (since $i_{1}=i_{2}$, since $i_{1}<j_{1}<k_{1}$ and
$i_{2}<j_{2}<k_{2}$, and since one of $j_{1}$ and $k_{1}$ equals one of
$j_{2}$ and $k_{2}$). Let us denote these four integers by $a,b,c,d$ in
increasing order (so that $a<b<c<d$). Hence, $i_{1}=a$ (since $i_{1}%
<j_{1}<k_{1}$ and $i_{2}<j_{2}<k_{2}$), whereas the two pairs $\left(
j_{1},k_{1}\right)  $ and $\left(  j_{2},k_{2}\right)  $ are two of the three
pairs $\left(  b,c\right)  $, $\left(  b,d\right)  $ and $\left(  c,d\right)
$ (for the same reason). Hence, $g_{1}$ and $g_{2}$ are two of the three
polynomials%
\begin{align*}
&  x_{a,c}x_{a,b}-x_{a,b}x_{b,c}+x_{a,c}x_{b,c}+\beta x_{a,c}+\alpha,\\
&  x_{a,d}x_{a,b}-x_{a,b}x_{b,d}+x_{a,d}x_{b,d}+\beta x_{a,d}+\alpha,\\
&  x_{a,d}x_{a,c}-x_{a,c}x_{c,d}+x_{a,d}x_{c,d}+\beta x_{a,d}+\alpha.
\end{align*}
It thus remains to verify that $\operatorname*{spol}\left(  g_{1}%
,g_{2}\right)  \overset{\ast}{\underset{G}{\longrightarrow}}0$.

Let us do this. Set%
\begin{align*}
u_{1}  &  =x_{a,c}x_{a,b}-x_{a,b}x_{b,c}+x_{a,c}x_{b,c}+\beta x_{a,c}%
+\alpha;\\
u_{2}  &  =x_{a,d}x_{a,b}-x_{a,b}x_{b,d}+x_{a,d}x_{b,d}+\beta x_{a,d}%
+\alpha;\\
u_{3}  &  =x_{a,d}x_{a,c}-x_{a,c}x_{c,d}+x_{a,d}x_{c,d}+\beta x_{a,d}%
+\alpha;\\
u_{4}  &  =x_{b,c}x_{b,d}-x_{b,c}x_{c,d}+x_{b,d}x_{c,d}+\beta x_{b,d}+\alpha.
\end{align*}
All four polynomials $u_{1},u_{2},u_{3},u_{4}$ belong to $G$. We shall prove
that $\operatorname*{spol}\left(  g_{1},g_{2}\right)  \overset{\ast
}{\underset{G}{\longrightarrow}}0$ whenever $g_{1}$ and $g_{2}$ are two of the
three polynomials $u_{1},u_{2},u_{3}$. In other words, we shall prove that
$\operatorname*{spol}\left(  u_{1},u_{2}\right)  \overset{\ast
}{\underset{G}{\longrightarrow}}0$, $\operatorname*{spol}\left(  u_{1}%
,u_{3}\right)  \overset{\ast}{\underset{G}{\longrightarrow}}0$ and
$\operatorname*{spol}\left(  u_{2},u_{3}\right)  \overset{\ast
}{\underset{G}{\longrightarrow}}0$.

Start with the neat identity%
\[
u_{1}\left(  x_{a,d}-x_{b,d}\right)  -u_{2}\left(  x_{a,c}-x_{b,c}\right)
-u_{3}\left(  x_{b,c}-x_{b,d}\right)  +u_{4}\left(  x_{a,c}-x_{a,d}\right)
=0.
\]
Expanding and bringing $6$ of the $8$ addends on the right hand side, we
obtain%
\[
x_{a,d}u_{1}-x_{a,c}u_{2}=-x_{b,c}u_{2}-x_{a,c}u_{4}+x_{b,d}u_{1}+x_{b,c}%
u_{3}+x_{a,d}u_{4}-x_{b,d}u_{3}.
\]
Since the monomials
\[
x_{b,c}\operatorname*{HT}\left(  u_{2}\right)  ,x_{a,c}\operatorname*{HT}%
\left(  u_{4}\right)  ,x_{b,d}\operatorname*{HT}\left(  u_{1}\right)
,x_{b,c}\operatorname*{HT}\left(  u_{3}\right)  ,x_{a,d}\operatorname*{HT}%
\left(  u_{4}\right)  ,x_{b,d}\operatorname*{HT}\left(  u_{3}\right)
\]
are distinct, we thus conclude that $x_{a,d}u_{1}-x_{a,c}u_{2}\overset{\ast
}{\underset{G}{\longrightarrow}}0$ (by Lemma \ref{lem.groebner.to0}). In other
words, $\operatorname*{spol}\left(  u_{1},u_{2}\right)  \overset{\ast
}{\underset{G}{\longrightarrow}}0$ (since $\operatorname*{spol}\left(
u_{1},u_{2}\right)  =x_{a,d}u_{1}-x_{a,c}u_{2}$).

Next, observe the identity%
\[
x_{a,d}u_{1}-x_{a,b}u_{3}=\beta u_{3}-\beta u_{2}-x_{a,b}u_{4}-x_{b,c}%
u_{2}+x_{b,c}u_{3}+x_{a,d}u_{4}+x_{c,d}u_{1}-x_{c,d}u_{2}.
\]
Since the monomials
\begin{align*}
&  \operatorname*{HT}\left(  u_{3}\right)  ,\operatorname*{HT}\left(
u_{2}\right)  ,x_{a,b}\operatorname*{HT}\left(  u_{4}\right)  ,x_{b,c}%
\operatorname*{HT}\left(  u_{2}\right)  ,x_{b,c}\operatorname*{HT}\left(
u_{3}\right)  ,\\
&  x_{a,d}\operatorname*{HT}\left(  u_{4}\right)  ,x_{c,d}\operatorname*{HT}%
\left(  u_{1}\right)  ,x_{c,d}\operatorname*{HT}\left(  u_{2}\right)
\end{align*}
are distinct, we can conclude that $x_{a,d}u_{1}-x_{a,b}u_{3}\overset{\ast
}{\underset{G}{\longrightarrow}}0$ (by Lemma \ref{lem.groebner.to0}). In other
words, $\operatorname*{spol}\left(  u_{1},u_{3}\right)  \overset{\ast
}{\underset{G}{\longrightarrow}}0$.

Finally, the identity we need for $\operatorname*{spol}\left(  u_{2}%
,u_{3}\right)  \overset{\ast}{\underset{G}{\longrightarrow}}0$ is%
\[
x_{a,c}u_{2}-x_{a,b}u_{3}=\beta u_{3}-\beta u_{2}-x_{a,b}u_{4}+x_{a,c}%
u_{4}-x_{b,d}u_{1}+x_{c,d}u_{1}+x_{b,d}u_{3}-x_{c,d}u_{2}.
\]
The same distinctness argument works here.

We have thus proven Claim 1. Thus, Proposition \ref{prop.forkless.groebner} is proven.
\end{proof}

\begin{remark}
Proposition \ref{prop.forkless.groebner} can be generalized somewhat. Namely,
instead of requiring the total order on $\mathfrak{M}$ to be inverse
lexicographic, it suffices to assume that we are given \textbf{any} term order
on $\mathfrak{M}$ satisfying the following condition: For every $\left(
i,j,k\right)  \in\left[  n\right]  ^{3}$ satisfying $i<j<k$, we have
$x_{i,k}>x_{j,k}$ and $x_{i,j}>x_{j,k}$.

In fact, this condition ensures that the head term of the polynomial
$x_{i,k}x_{i,j}-x_{i,j}x_{j,k}+x_{i,k}x_{j,k}+\beta x_{i,k}+\alpha$ (for
$\left(  i,j,k\right)  \in\left[  n\right]  ^{3}$ satisfying $i<j<k$) is
$x_{i,k}x_{i,j}$; but this is all that was needed from our term order to make
the above proof of Proposition \ref{prop.forkless.groebner} valid.
\end{remark}

\begin{proof}
[Proof of Proposition \ref{prop.forkless.basis} (sketched).]Let $G$ be the set
(\ref{eq.prop.forkless.groebner.family}). Then, Proposition
\ref{prop.forkless.groebner} shows that $G$ is a Gr\"{o}bner basis of the
ideal $\mathcal{J}$ of $\mathcal{X}$ (where $\mathfrak{M}$ is endowed with the
term order defined in Proposition \ref{prop.forkless.groebner}). Hence,
Proposition \ref{prop.groebner.standardbasis} (applied to $\Xi=\left\{
x_{i,j}\ \mid\ \left(  i,j\right)  \in\left[  n\right]  ^{2}\text{ satisfying
}i<j\right\}  $, $\mathcal{X}_{\Xi}=\mathcal{X}$, $\mathfrak{M}_{\Xi
}=\mathfrak{M}$ and $\mathcal{I}=\mathcal{J}$) shows that the projections of
the $G$-reduced monomials onto the quotient ring $\mathcal{X}/\mathcal{J}$
form a basis of the $\mathbf{k}$-module $\mathcal{X}/\mathcal{J}$. Since the
$G$-reduced monomials are precisely the forkless monomials, this yields
Proposition \ref{prop.forkless.basis}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.forkless.span-uni} (sketched).]Theorem
\ref{thm.forkless.span-uni} is merely a restatement of Proposition
\ref{prop.forkless.basis}.
\end{proof}

\begin{remark}
\label{rmk.forkless.span-uni.existence}Let us notice that the
\textquotedblleft existence\textquotedblright\ part of Theorem
\ref{thm.forkless.span-uni} can also be proven similarly to how we proved
Proposition \ref{prop.path-red.span}. This time, we need to define a different
notion of \textquotedblleft weight\textquotedblright: Instead of defining the
weight of a monomial $\mathfrak{m}=\prod_{\substack{\left(  i,j\right)
\in\left[  n\right]  ^{2};\\i<j}}x_{i,j}^{a_{i,j}}$ to be
$\operatorname*{weight}\mathfrak{m}=\sum_{\substack{\left(  i,j\right)
\in\left[  n\right]  ^{2};\\i<j}}a_{i,j}\left(  n-j+i\right)  $, we now must
define it to be $\operatorname*{weight}\mathfrak{m}=\sum_{\substack{\left(
i,j\right)  \in\left[  n\right]  ^{2};\\i<j}}a_{i,j}\left(  j-i\right)  $.
\end{remark}

\begin{question}
Is there a similarly simple argument for the \textquotedblleft
uniqueness\textquotedblright\ part?
\end{question}

\subsection{Dimensions}

The $\mathbf{k}$-module $\mathcal{X}/\mathcal{J}$ is free of infinite rank
whenever $n\geq2$; indeed, the basis given in Proposition
\ref{prop.forkless.basis} is infinite (for any $k\in\mathbb{N}$, the monomial
$x_{1,2}^{k}$ is forkless). However, $\mathcal{X}/\mathcal{J}$ can be equipped
with a filtration, whose filtered parts are of finite rank. Namely, recall
that the polynomial ring $\mathcal{X}$ is graded (by total degree) and thus
filtered; this filtration is then inherited by its quotient ring
$\mathcal{X}/\mathcal{J}$. For each $k\in\mathbb{N}$, we let $\left(
\mathcal{X}/\mathcal{J}\right)  _{\leq k}$ denote the $k$-th part of the
filtration on $\mathcal{X}/\mathcal{J}$ (that is, the projection onto
$\mathcal{X}/\mathcal{J}$ of all polynomials $p\in\mathcal{X}$ of total degree
$\leq k$). A moment of thought reveals that the basis of $\mathcal{X}%
/\mathcal{J}$ given in Proposition \ref{prop.forkless.basis} is a filtered
basis: For each $k\in\mathbb{N}$, the projections of the forkless monomials
$\mathfrak{m}\in\mathfrak{M}$ of total degree $\leq k$ onto the quotient ring
$\mathcal{X}/\mathcal{J}$ form a basis of the $\mathbf{k}$-module $\left(
\mathcal{X}/\mathcal{J}\right)  _{\leq k}$. This basis is a finite basis, and
so its size is a nonnegative integer. What is this integer?

Of course, it suffices to count the forkless monomials $\mathfrak{m}%
\in\mathfrak{M}$ of total degree $k$ for each $k\in\mathbb{N}$. This is a
relatively easy counting problem using some classical results
\cite[Proposition 1.3.7 and Proposition 1.3.10]{Stanley-EC1}; the answer is
the following:

\begin{proposition}
\label{prop.forkless.count}For each $k\in\mathbb{N}$, let $f_{n,k}$ be the
number of forkless monomials $\mathfrak{m}\in\mathfrak{M}$ of total degree
$k$. Then,%
\[
\sum_{k\in\mathbb{N}}f_{n,k}t^{k}=\dfrac{\left(  1+0t\right)  \left(
1+1t\right)  \cdots\left(  1+\left(  n-2\right)  t\right)  }{\left(
1-t\right)  ^{n-1}}%
\]
(as formal power series in $\mathbb{Z}\left[  \left[  t\right]  \right]  $).
\end{proposition}

When $\alpha=0$ and $\beta=0$, the ideal $\mathcal{J}$ of $\mathcal{X}$ is
homogeneous. Thus, in this case, the quotient $\mathbf{k}$-algebra
$\mathcal{X}/\mathcal{J}$ inherits not only a filtration, but also a grading
from $\mathcal{X}$; its Hilbert series (with respect to this grading) is the
power series $\sum_{k\in\mathbb{N}}f_{n,k}t^{k}$ of Proposition
\ref{prop.forkless.count}.

\begin{vershort}
\begin{question}
The filtration on $\mathcal{X}/\mathcal{J}$ considered above is not the only
natural one. Another is the filtration by \textquotedblleft
weight\textquotedblright\ (as in the proof of Proposition
\ref{prop.path-red.span}), or by the alternative notion of \textquotedblleft
weight\textquotedblright\ mentioned in Remark
\ref{rmk.forkless.span-uni.existence}. What are the analogues of Proposition
\ref{prop.forkless.count} for these filtrations?
\end{question}
\end{vershort}

\begin{verlong}
\begin{question}
The filtration on $\mathcal{X}/\mathcal{J}$ considered above is not the only
natural one. Another is the filtration by \textquotedblleft
weight\textquotedblright\ (as in Definition \ref{def.span-detail.weight}), or
by the alternative notion of \textquotedblleft weight\textquotedblright%
\ mentioned in Remark \ref{rmk.forkless.span-uni.existence}. What are the
analogues of Proposition \ref{prop.forkless.count} for these filtrations?
\end{question}
\end{verlong}

\section{Further questions}

Let us finally indicate some further directions of research not mentioned so far.

\subsection{\label{subsect.Qrat}The kernel of $A$}

\begin{question}
\label{quest.Ainj}\textbf{(a)} Is $\mathcal{J}$ the kernel of the map
$A:\mathcal{X}\rightarrow\mathcal{Q}$ from Definition \ref{def.A}?

\textbf{(b)} Consider the polynomial ring $\mathbf{k}\left[  \widetilde{q}%
_{1},\widetilde{q}_{2},\ldots,\widetilde{q}_{n}\right]  $ in $n$
indeterminates $\widetilde{q}_{1},\widetilde{q}_{2},\ldots,\widetilde{q}_{n}$
over $\mathbf{k}$. Let $\mathcal{Q}_{\operatorname*{rat}}$ denote the
localization of this polynomial ring at the multiplicative subset generated by
all differences of the form $\widetilde{q}_{i}-\widetilde{q}_{j}$ (for $1\leq
i<j\leq n$). Then, the morphism $A:\mathcal{X}\rightarrow\mathcal{Q}$ factors
through a $\mathbf{k}$-algebra homomorphism $\widetilde{A}:\mathcal{X}%
\rightarrow\mathcal{Q}_{\operatorname*{rat}}$ which sends each $x_{i,j}$ to
$-\dfrac{\widetilde{q}_{i}+\beta+\alpha/\widetilde{q}_{j}}{1-\widetilde{q}%
_{i}/\widetilde{q}_{j}}=-\dfrac{\widetilde{q}_{i}\widetilde{q}_{j}%
+\beta\widetilde{q}_{j}+\alpha}{\widetilde{q}_{j}-\widetilde{q}_{i}}%
\in\mathcal{Q}_{\operatorname*{rat}}$. Is $\mathcal{J}$ the kernel of this
latter homomorphism $\widetilde{A}$ ?
\end{question}

Parts \textbf{(a)} and \textbf{(b)} of Question \ref{quest.Ainj} are
equivalent, since the canonical $\mathbf{k}$-algebra homomorphism
$\mathcal{Q}_{\operatorname*{rat}}\rightarrow\mathcal{Q}$ is injective. This
question is interesting partly because a positive answer to part \textbf{(b)}
would provide a realization of $\mathcal{X}/\mathcal{J}$ as a subalgebra of a
localized polynomial ring in (only) $n$ indeterminates. This subalgebra would
probably not be the whole $\mathcal{Q}_{\operatorname*{rat}}$.

(Perhaps it can be shown -- by some kind of multidimensional residues -- that
$A$ maps the forkless monomials in $\mathcal{X}$ to linearly independent
elements of $\mathcal{Q}$. Such a proof would then immediately yield positive
answers to parts \textbf{(a)} and \textbf{(b)} of Question \ref{quest.Ainj} as
well as an alternative proof of Theorem \ref{thm.forkless.span-uni}.)

An approach to Question \ref{quest.Ainj} \textbf{(b)} might begin with finding
a basis of the $\mathbf{k}$-module $\mathcal{Q}_{\operatorname*{rat}}$. It
turns out that such a basis is rather easy to construct:

\begin{proposition}
\label{prop.Qrat.basis}In $\mathcal{Q}_{\operatorname*{rat}}$, consider the
family of all elements of the form $\prod_{i=1}^{n}g_{i}$, where each $g_{i}$
has either the form $\dfrac{1}{\left(  \widetilde{q}_{i}-\widetilde{q}%
_{j}\right)  ^{m}}$ for some $j\in\left\{  i+1,i+2,\ldots,n\right\}  $ and
$m>0$ or the form $\widetilde{q}_{i}^{k}$ for some $k\in\mathbb{N}$. This
family is a basis of the $\mathbf{k}$-module $\mathcal{Q}_{\operatorname*{rat}%
}$.
\end{proposition}

Notice that this family is similar to the forkless monomials in Proposition
\ref{prop.forkless.basis}, but it is \textquotedblleft
larger\textquotedblright\ (if we would allow the $g_{i}$ to have the form
$\widetilde{q}_{i}^{k}$ only for $k=0$, then we would obtain a restricted
family that would be in an obvious bijection with the forkless monomials).

Proposition \ref{prop.Qrat.basis} is closely related to results by Horiuchi
and Terao (\cite{HorTer03} and \cite{Terao02}); indeed, if $\mathbf{k}$ is a
field, then $\mathcal{Q}_{\operatorname*{rat}}$ can be regarded as the ring of
regular functions on the complement of the braid arrangement in $\mathbf{k}%
^{n}$, and such functions are what they have studied (although usually not the
whole $\mathcal{Q}_{\operatorname*{rat}}$). Notice however that they worked
only over fields $\mathbf{k}$ of characteristic $0$.

Let us only briefly hint to how Proposition \ref{prop.Qrat.basis} is proven;
the details shall be deferred to future work. We can construct $\mathcal{Q}%
_{\operatorname*{rat}}$ recursively: For any $n>0$, we can first construct the
$\mathbf{k}$-algebra $\mathcal{Q}_{\operatorname*{rat},n-1}$ defined as the
localization of the polynomial ring $\mathbf{k}\left[  \widetilde{q}%
_{2},\widetilde{q}_{3},\ldots,\widetilde{q}_{n}\right]  $ at the
multiplicative subset generated by all differences of the form $\widetilde{q}%
_{i}-\widetilde{q}_{j}$ (for $2\leq i<j\leq n$); then, $\mathcal{Q}%
_{\operatorname*{rat}}$ is isomorphic to the localization of the polynomial
ring $\mathcal{Q}_{\operatorname*{rat},n-1}\left[  \widetilde{q}_{1}\right]  $
at the multiplicative subset generated by all differences of the form
$\widetilde{q}_{1}-\widetilde{q_{j}}$ (for $2\leq j\leq n$). Thus, Proposition
\ref{prop.Qrat.basis} can be proven by induction over $n$, using the following fact:

\begin{proposition}
\label{prop.Qrat.indstep}Let $A$ be a commutative ring. Let $f_{1}%
,f_{2},\ldots,f_{n}$ be $n$ elements of $A$. Assume that for each $1\leq
i<j\leq n$, the element $f_{i}-f_{j}$ of $A$ is invertible. Let $B$ be the
localization of the polynomial ring $A\left[  x\right]  $ at the
multiplicative subset generated by all differences of the form $x-f_{j}$ (for
$1\leq j\leq n$). Then, $B$ is a free $A$-module, with a basis consisting of
the following elements:

\begin{itemize}
\item all elements of the form $\dfrac{1}{\left(  x-f_{j}\right)  ^{m}}$ for
$j\in\left\{  1,2,\ldots,n\right\}  $ and $m>0$;

\item all elements of the form $x^{k}$ for $k\in\mathbb{N}$.
\end{itemize}
\end{proposition}

Proposition \ref{prop.Qrat.indstep} is essentially a form of partial fraction
decomposition, saying that any element of $B$ can be uniquely written as an
$A$-linear combination of elements of the form $\dfrac{1}{\left(
x-f_{j}\right)  ^{m}}$ for $j\in\left\{  1,2,\ldots,n\right\}  $ and $m>0$,
plus a polynomial in $A\left[  x\right]  $. This can be proven by thoroughly
analyzing the corresponding proof in the case when $A$ is a field; the
invertibility of the differences $f_{i}-f_{j}$ is actually what is needed here
(since it entails that the ideals $\left(  x-f_{j}\right)  A\left[  x\right]
$ of $A\left[  x\right]  $ for $j\in\left\{  1,2,\ldots,n\right\}  $ are
pairwise comaximal).

\subsection{Isomorphisms between $\mathcal{X}/\mathcal{J}$ for different
$\alpha,\beta$}

Let us now rename the ideal $\mathcal{J}$ as $\mathcal{J}_{\beta,\alpha}$, in
order to stress its dependence on $\beta$ and $\alpha$.

\begin{question}
When are the $\mathbf{k}$-algebras $\mathcal{X}/\mathcal{J}_{\beta,\alpha}$
for different choices of $\alpha$ and $\beta$ isomorphic?
\end{question}

For $n=2$, the answer is clearly \textquotedblleft always\textquotedblright,
because $\mathcal{X}/\mathcal{J}_{\beta,\alpha}$ does not depend on $\alpha$
and $\beta$ in this case (in fact, $\mathcal{J}_{\beta,\alpha}=0$ when $n=2$).
So the question only becomes interesting for $n\geq3$. The answer may well
depend on the base ring $\mathbf{k}$, and it is perhaps reasonable to assume
that $\mathbf{k}$ is a field here. It is easy to come up with an example where
the $\mathbf{k}$-algebras $\mathcal{X}/\mathcal{J}_{\beta,\alpha}$ for
different choices of $\alpha$ and $\beta$ are not isomorphic\footnote{For
example, let $\mathbf{k}$ be a field of characteristic $\neq2$, and let $n=3$.
Then, $\mathcal{X}/\mathcal{J}_{\beta,\alpha}$ is the quotient of the
polynomial ring $\mathcal{X}=\mathbf{k}\left[  x_{1,2},x_{1,3},x_{2,3}\right]
$ by the principal ideal generated by $x_{1,2}x_{2,3}-x_{1,3}\left(
x_{1,2}+x_{2,3}+\beta\right)  -\alpha$.
\par
We claim that any $\mathbf{k}$-algebra $\mathcal{X}/\mathcal{J}_{\beta,\alpha
}$ for $4\alpha=\beta^{2}$ is non-isomorphic to any $\mathbf{k}$-algebra
$\mathcal{X}/\mathcal{J}_{\beta,\alpha}$ for $4\alpha\neq\beta^{2}$.
\par
To see this, it suffices to show that the $\mathbf{k}$-algebra $\mathcal{X}%
/\mathcal{J}_{\beta,\alpha}$ has a \textquotedblleft$\mathbf{k}$-valued
singular point\textquotedblright\ (i.e., a $\mathbf{k}$-algebra homomorphism
$\varepsilon:\mathcal{X}/\mathcal{J}_{\beta,\alpha}\rightarrow\mathbf{k}$ such
that there exist three $\mathbf{k}$-linearly independent $\left(
\varepsilon,\varepsilon\right)  $-derivations $\mathcal{X}/\mathcal{J}%
_{\beta,\alpha}\rightarrow\mathbf{k}$, where an $\left(  \varepsilon
,\varepsilon\right)  $\textit{-derivation} means a $\mathbf{k}$-linear map
$\partial:\mathcal{X}/\mathcal{J}_{\beta,\alpha}\rightarrow\mathbf{k}$
satisfying $\partial\left(  fg\right)  =\partial\left(  f\right)
\varepsilon\left(  g\right)  +\varepsilon\left(  f\right)  \partial\left(
g\right)  $ for all $f,g$) if and only if $4\alpha=\beta^{2}$. But this is
easily verified.}. The following example should stress that isomorphisms
nevertheless can exist:

\begin{example}
\textbf{(a)} Let $\gamma\in\mathbf{k}$. The $\mathbf{k}$-algebra isomorphism
$\mathcal{X}\rightarrow\mathcal{X},\ x_{i,j}\mapsto\gamma-x_{i,j}$ descends to
a $\mathbf{k}$-algebra isomorphism $\mathcal{X}/\mathcal{J}_{\beta,\alpha
}\rightarrow\mathcal{X}/\mathcal{J}_{\beta+2\gamma,\alpha+\beta\gamma
+\gamma^{2}}$.

\textbf{(b)} Let $\rho\in\mathbf{k}$ be invertible. The $\mathbf{k}$-algebra
isomorphism $\mathcal{X}\rightarrow\mathcal{X},\ x_{i,j}\mapsto\rho x_{i,j}$
descends to a $\mathbf{k}$-algebra isomorphism $\mathcal{X}/\mathcal{J}%
_{\rho\beta,\rho^{2}\alpha}\rightarrow\mathcal{X}/\mathcal{J}_{\beta,\alpha}$.
\end{example}

\subsection{\label{subsect.arnold}A deformation of the Orlik-Terao algebra?}

We have already seen in Subsection \ref{subsect.Qrat} that the $\mathbf{k}%
$-algebra $\mathcal{X}/\mathcal{J}$ is closely connected to the localization
$\mathcal{Q}_{\operatorname*{rat}}$ from Question \ref{quest.Ainj}. In the
parlance of algebraic geometers, $\mathcal{Q}_{\operatorname*{rat}}$ is the
coordinate ring of the complement of the braid arrangement in $\mathbf{k}^{n}%
$. This complement has been the subject of a classical paper by Arnold
\cite{Arnold71}, which discussed its cohomology ring. Arnold's description of
this cohomology ring is remarkably similar to our definition of $\mathcal{X}%
/\mathcal{J}$ in the case when $\beta=0$ and $\alpha=0$. Namely, Arnold
considers the exterior (i.e., free anticommutative) algebra $A\left(
n\right)  $ in $\dbinom{n}{2}$ indeterminates $\omega_{i,j}$ for $1\leq
i<j\leq n$ up to the relations $\omega_{i,j}\omega_{j,k}+\omega_{j,k}%
\omega_{k,i}+\omega_{k,i}\omega_{i,j}=0$, where $\omega_{u,v}$ for $u>v$ is
defined to be a synonym for $\omega_{v,u}$. He gives a basis \cite[Corollary
3]{Arnold71} of this $\mathbf{k}$-module $A\left(  n\right)  $, which is
almost exactly the same as our basis of forkless monomials for $\mathcal{X}%
/\mathcal{J}$ (with the difference, of course, that his monomials are
squarefree because they live in an exterior algebra, and that his choice of
order is different).

Arnold's algebra $A\left(  n\right)  $ has since been significantly
generalized. Namely, to every matroid corresponds an \textit{Orlik-Solomon
algebra} \cite{CorEti01}; this recovers the algebra $A\left(  n\right)  $ when
the matroid is the graphical matroid of the complete graph $K_{n}$. Seeing
that the subdivision algebra $\mathcal{X}/\mathcal{J}$ can be viewed as a
commutative analogue of $A\left(  n\right)  $, we can thus ask for a similar
commutative analogue of an arbitrary Orlik-Solomon algebra.

Such an analogue, too, is known \cite{SchToh15}: it is the \textit{Orlik-Terao
algebra} of a finite family of vectors. This generalizes $\mathcal{X}%
/\mathcal{J}$ in the case when $\beta=0$ and $\alpha=0$. We may thus regard
$\mathcal{X}/\mathcal{J}$ as a deformation of a specific Orlik-Terao algebra,
and ask for a generalization:

\begin{question}
\textbf{(a)} Can an arbitrary Orlik-Terao algebra be deformed by two
parameters $\beta$ and $\alpha$, generalizing our $\mathcal{X}/\mathcal{J}$ ?
A deformation by one parameter $\hbar$ (which we suspect to correspond to our
$\mathcal{X}/\mathcal{J}$ for the braid arrangement with $\alpha=0$) has been
studied by McBreen and Proudfoot in \cite[\S A.2]{McbPro15} at least in the
case of a unimodular family of vectors.

\textbf{(b)} Does Theorem \ref{thm.t-red.unique} extend to Orlik-Terao algebras?
\end{question}

Note that our basis of forkless monomials for $\mathcal{X}/\mathcal{J}$ can be
regarded as an \textquotedblleft\textbf{nbc} basis\textquotedblright\ in the
sense of \cite{CorEti01} (except that our monomials are not required to be
squarefree). Indeed, if we totally order the monomials $x_{i,j}$ in such a way
that $x_{i,j}>x_{u,v}$ whenever $i<u$, then the broken circuits of the
graphical matroid of $K_{n}$ are precisely the sets of the form $\left\{
\left\{  i,j\right\}  ,\left\{  i,k\right\}  \right\}  $ for $i<j<k$; but
these correspond to the precise monomials $x_{i,j}x_{i,k}$ that a forkless
monomial cannot be divisible by. Proudfoot's and Speyer's \cite[Theorem
4]{ProSpe06} leads to a similar basis for Orlik-Terao algebras of arbitrary
hyperplane arrangements, and \cite[Theorem A.9]{McbPro15} extends this to its
one-parameter deformation for unimodular arrangements. We may still ask
similar questions about oriented matroids not coming from hyperplane
arrangements, and we may also ask for combinatorial proofs. Horiuchi's and
Terao's works \cite{HorTer03} and \cite{Terao02} seem relevant once again.

We end with an overview of algebras similar to $\mathcal{X}/\mathcal{J}$ that
have appeared in the literature, making no claims of completeness. See also
the last few paragraphs of the Introduction of \cite{Kirill16} for a history
of these algebras.

\begin{itemize}
\item As mentioned above, in \cite{Arnold71}, Arnold introduced the
noncommutative algebra $A\left(  n\right)  $ with anticommuting generators
$\omega_{i,j}$ (for $1\leq i<j\leq n$) and
relations\footnote{Anticommutativity of the generators means that
$\omega_{i,j}\omega_{u,v}=-\omega_{u,v}\omega_{i,j}$ for all $i<j$ and $u<v$,
and that $\omega_{i,j}^{2}=0$ for all $i<j$.}
\[
\omega_{i,j}\omega_{j,k}+\omega_{j,k}\omega_{i,k}+\omega_{i,k}\omega
_{i,j}=0\text{ (for }1\leq i<j<k\leq n\text{).}%
\]
This was probably the first algebra of this kind to be defined. Note that the
relations can be rewritten in the form%
\[
\omega_{i,j}\omega_{j,k}=-\omega_{i,k}\left(  \omega_{i,j}-\omega
_{j,k}\right)
\]
to reveal the similarity to the generators of $\mathcal{J}$, but Arnold's
algebra does not include the two \textquotedblleft deforming\textquotedblright%
\ parameters $\alpha$ and $\beta$ of our $\mathcal{X}/\mathcal{J}$. Arnold
showed that $A\left(  n\right)  $ is isomorphic to the (integer) cohomology
ring of the space $\left\{  \left(  z_{1},z_{2},\ldots,z_{n}\right)
\in\mathbb{C}^{n}\ \mid\ z_{1},z_{2},\ldots,z_{n}\text{ are distinct}\right\}
$, and found a $\mathbf{k}$-linear basis of $A\left(  n\right)  $. (Note that
he has been working with $\mathbf{k}=\mathbb{Z}$, but this clearly yields the
same results for all $\mathbf{k}$.)

This algebra $A\left(  n\right)  $ has later been generalized to the
Orlik-Solomon algebra of an arbitrary hyperplane arrangement, and more
generally of an arbitrary matroid (see, e.g., \cite{Yuzvin01} for an
exposition of the arrangement case); the algebra $A\left(  n\right)  $ is
recovered by taking the braid arrangement.

\item In \cite{VarGel87}, Gelfand and Varchenko have introduced a commutative
counterpart of the Orlik-Solomon algebra of a hyperplane arrangement $S$. This
algebra $P$ is generated by the constant function $1$ and the Heaviside
functions of the hyperplanes in the arrangement. If $S$ is the braid
arrangement, then this algebra $P$ is isomorphic to the commutative algebra
$P\left(  n\right)  $ with generators $x_{i,j}$ (for $1\leq i<j\leq n$) and
relations%
\begin{align*}
x_{i,j}^{2}  &  =x_{i,j}\text{ (for }1\leq i<j\leq n\text{), and}\\
x_{i,j}x_{j,k}\left(  x_{i,k}-1\right)  -\left(  x_{i,j}-1\right)  \left(
x_{j,k}-1\right)  x_{i,k}  &  =0\text{ (for }1\leq i<j<k\leq n\text{).}%
\end{align*}
(Actually, Gelfand and Varchenko require many more relations, corresponding to
all linear dependencies between the linear functions that determine the
hyperplanes of $S$; our relation $x_{i,j}x_{j,k}\left(  x_{i,k}-1\right)
-\left(  x_{i,j}-1\right)  \left(  x_{j,k}-1\right)  x_{i,k}=0$ corresponds to
the linear dependency $f_{i,j}+f_{j,k}-f_{i,k}=0$ between the linear functions
$f_{i,j}=x_{i}-x_{j}$, $f_{j,k}=x_{j}-x_{k}$ and $f_{i,k}=x_{i}-x_{k}$. We are
using the nontrivial fact that the above relations suffice.) The relation
$x_{i,j}x_{j,k}\left(  x_{i,k}-1\right)  -\left(  x_{i,j}-1\right)  \left(
x_{j,k}-1\right)  x_{i,k}=0$ rewrites as%
\[
x_{i,j}x_{j,k}=x_{i,k}\left(  x_{i,j}+x_{j,k}-1\right)  ,
\]
which is exactly one of the generators of $\mathcal{J}$ when $\beta=1$ and
$\alpha=0$. However, the additional relations $x_{i,j}^{2}=x_{i,j}$ make their
algebra $P\left(  n\right)  $ finite-dimensional as a $\mathbf{k}$-module
(unlike our $\mathcal{X}/\mathcal{J}$). Gelfand and Varchenko find a basis of
$P$ (at least in the case when $\mathbf{k}=\mathbb{C}$), defined in terms of
what they call \textquotedblleft open cycles\textquotedblright\ (and is
nowadays known as broken circuits). If $S$ is the braid arrangement, and if an
appropriate ordering of the hyperplanes is used, then this basis becomes
similar to our basis of forkless monomials (Proposition
\ref{prop.forkless.basis}), except that it only contains the squarefree
forkless monomials (as the $x_{i,j}^{2}=x_{i,j}$ relations render all other
monomials redundant).

This algebra $P$ can be straightforwardly generalized to arbitrary oriented matroids.

\item The Gelfand-Varchenko algebra $P\left(  n\right)  $ is filtered, and its
associated graded algebra is the commutative algebra $Q\left(  n\right)  $
with generators $x_{i,j}$ (for $1\leq i<j\leq n$) and relations
\begin{align*}
x_{i,j}^{2}  &  =0\text{ (for }1\leq i<j\leq n\text{), and}\\
x_{i,j}x_{j,k}  &  =x_{i,k}\left(  x_{i,j}+x_{j,k}\right)  \text{ (for }1\leq
i<j<k\leq n\text{).}%
\end{align*}
The latter relations are exactly the generators of $\mathcal{J}$ when
$\beta=0$ and $\alpha=0$. This connection is explored, e.g., in Moseley's
\cite{Mosele12} (although he imposes a much larger set of relations); again,
this $\mathbf{k}$-algebra is a finite-dimensional $\mathbf{k}$-module with a
\textquotedblleft broken circuit\textquotedblright\ basis. The algebra
$Q\left(  n\right)  $ also appears (as $A_{n}$) in Mathieu's \cite[\S 6]%
{Mathie95}.

Again, this generalizes to an arbitrary hyperplane arrangement, yielding what
is called its Artinian Orlik-Terao algebra (the algebra $W\left(
\mathcal{A}\right)  $ in \cite{OrlTer94}). But even the case of the type-A
braid arrangement is still the object of active research; see \cite{EarRei18}
(where $Q\left(  n\right)  $ appears as $\mathcal{U}^{n}$) and \cite[\S 8.2]%
{Berget18}\footnote{Note that \textquotedblleft$n$\textquotedblright\ should
be replaced by \textquotedblleft$n-1$\textquotedblright\ in \cite[Lemma
8.2]{Berget18}.} (where $Q\left(  n\right)  $ appears as $S_{\mathcal{A}%
^{\perp},-2}=\operatorname*{Sym}\left(  K\right)  /I_{\mathcal{A}^{\perp},-2}%
$, if we replace $n-1$ by $n$) for recent studies of the $S_{n}$-action on
$Q\left(  n\right)  $.

\item In \cite{OrlTer94}, Orlik and Terao assign an algebra $\mathbf{K}\left[
\alpha_{\mathcal{A}}^{-1}\right]  $ to any hyperplane arrangement
$\mathcal{A}$ in a finite-dimensional $\mathbf{K}$-vector space $V$, where
$\mathbf{K}$ is any field. Nowadays known as the (big) Orlik-Terao algebra, it
is simply the $\mathbf{K}$-subalgebra of the ring of rational functions on $V$
generated by the reciprocals of the linear forms whose kernels are the
hyperplanes of $\mathcal{A}$. When $\mathcal{A}$ is the braid arrangement,
this $\mathbf{K}$-algebra is isomorphic to the commutative $\mathbf{K}%
$-algebra $R\left(  n\right)  $ with generators $x_{i,j}$ (for $1\leq i<j\leq
n$) and relations%
\[
x_{i,j}x_{j,k}=x_{i,k}\left(  x_{i,j}+x_{j,k}\right)  \text{ (for }1\leq
i<j<k\leq n\text{).}%
\]
The relations here are exactly the generators of $\mathcal{J}$ when $\beta=0$
and $\alpha=0$. Unlike the previous algebras, this one is no longer
finite-dimensional, thus being the closest one so far to $\mathcal{X}%
/\mathcal{J}$. A basis of the (big) Orlik-Terao algebra has been found by
Proudfoot and Speyer in \cite{ProSpe06}.

\item Kirillov, in \cite[\S 4]{Kirill97}, introduces a noncommutative algebra
$\mathcal{G}_{n}$ with generators $\left[  i,j\right]  $ (for $1\leq i<j\leq
n$) and relations%
\begin{align}
\left[  i,j\right]  \left[  j,k\right]   &  =\left[  j,k\right]  \left[
i,k\right]  +\left[  i,k\right]  \left[  i,j\right]  \text{ (for }1\leq
i<j<k\leq n\text{);}\label{eq.context.Gn.rel1}\\
\left[  j,k\right]  \left[  i,j\right]   &  =\left[  i,k\right]  \left[
j,k\right]  +\left[  i,j\right]  \left[  i,k\right]  \text{ (for }1\leq
i<j<k\leq n\text{);}\label{eq.context.Gn.rel2}\\
\left[  i,j\right]  \left[  k,l\right]   &  =\left[  k,l\right]  \left[
i,j\right]  \text{ (for }i<j\text{ and }k<l\text{ with }\left\{  i,j\right\}
\cap\left\{  k,l\right\}  =\varnothing\text{).}\nonumber
\end{align}
Note that the abelianization of this $\mathcal{G}_{n}$ is $R\left(  n\right)
$. Kirillov states (without proof) a basis of $\mathcal{G}_{n}$ in
\cite[Theorem 4.3]{Kirill97}\footnote{Note that the \textquotedblleft%
$Z_{2}=\left\{  \varnothing,\left[  12\right]  \right\}  $\textquotedblright%
\ in \cite[Theorem 4.3]{Kirill97} is a typo, and should instead be
\textquotedblleft$Z_{2}=\left\{  \varnothing,\left[  12\right]  ,\left[
12\right]  ^{2},\left[  13\right]  ^{3},\ldots\right\}  $\textquotedblright.},
which (under abelianization) transforms into the basis from our Proposition
\ref{prop.forkless.basis}.

\item In \cite[Definition 10.5]{Kirill97}, Kirillov goes on to deform the
algebra $\mathcal{G}_{n}$, replacing (\ref{eq.context.Gn.rel1}) and
(\ref{eq.context.Gn.rel2}) by%
\begin{align*}
\left[  i,j\right]  \left[  j,k\right]   &  =\left[  j,k\right]  \left[
i,k\right]  +\left[  i,k\right]  \left[  i,j\right]  +\beta\left[  i,k\right]
\text{ (for }1\leq i<j<k\leq n\text{);}\\
\left[  j,k\right]  \left[  i,j\right]   &  =\left[  i,k\right]  \left[
j,k\right]  +\left[  i,j\right]  \left[  i,k\right]  +\beta\left[  i,k\right]
\text{ (for }1\leq i<j<k\leq n\text{),}%
\end{align*}
where $\beta\in\mathbf{k}$ is fixed. He denotes this algebra by $\mathcal{L}%
_{n,\beta}$, but leaves its properties to further study.

\item The quasi-classical Yang-Baxter algebra $\mathcal{B}\left(
A_{n}\right)  $ is the noncommutative $\mathbf{k}$-algebra with generators
$x_{i,j}$ (for $1\leq i<j\leq n$) and relations%
\begin{align*}
x_{i,j}x_{j,k} &  =x_{i,k}x_{i,j}+x_{j,k}x_{i,k}+\beta x_{i,k}\text{ (for
}1\leq i<j<k\leq n\text{);}\\
x_{i,j}x_{k,l} &  =x_{k,l}x_{i,j}\text{ (for }i<j\text{ and }k<l\text{ with
}\left\{  i,j\right\}  \cap\left\{  k,l\right\}  =\varnothing\text{).}%
\end{align*}
Note that this is not the same as $\mathcal{L}_{n,\beta}$, since the second
relation of $\mathcal{L}_{n,\beta}$ is missing here. This algebra
$\mathcal{B}\left(  A_{n}\right)  $ was also introduced by Kirillov (according
to \cite{Meszar09}).

\item M\'{e}sz\'{a}ros, in \cite{Meszar09}, studies the abelianization of
$\mathcal{B}\left(  A_{n}\right)  $; this is also the abelianization of
$\mathcal{L}_{n,\beta}$. This is the $\mathbf{k}$-algebra $\mathcal{X}%
/\mathcal{J}$ for $\alpha=0$.

\item In \cite[Definition 5.1]{Kirill16}, Kirillov starts with two parameters
$\alpha,\beta\in\mathbf{k}$ and defines the noncommutative $\mathbf{k}%
$-algebra $\widehat{\operatorname*{ACYB}}_{n}\left(  \alpha,\beta\right)  $,
which has generators $x_{i,j}$ (for $1\leq i<j\leq n$) and relations%
\begin{align*}
x_{i,j}x_{j,k} &  =x_{i,k}x_{i,j}+x_{j,k}x_{i,k}+\beta x_{i,k}+\alpha\text{
(for }1\leq i<j<k\leq n\text{);}\\
x_{i,j}x_{k,l} &  =x_{k,l}x_{i,j}\text{ (for }i<j\text{ and }k<l\text{ with
}\left\{  i,j\right\}  \cap\left\{  k,l\right\}  =\varnothing\text{).}%
\end{align*}
This algebra deforms $\mathcal{B}\left(  A_{n}\right)  $; its abelianization
is our $\mathcal{X}/\mathcal{J}$.
\end{itemize}

\begin{question}
Which of these algebras satisfy an analogue of Theorem~\ref{thm.t-red.unique}?
\end{question}

\begin{noncompile}
The following is obsolete, written back when I was not aware of Orlik-Terao algebras:

\begin{question}
Let $\mathcal{M}$ be a matroid on ground set $E$. Let $\mathcal{X}_{E}$ be the
polynomial ring $\mathbf{k}\left[  x_{e}\ \mid\ e\in E\right]  $. Let
$\partial:\mathcal{X}_{E}\rightarrow\mathcal{X}_{E}$ be the derivation of
$\mathcal{X}_{E}$ that sends each indeterminate $x_{e}$ to $1$. Let
$\mathcal{J}_{\mathcal{M}}$ be the ideal of $\mathcal{X}_{E}$ generated by all
elements of the form $\partial\left(  \prod_{e\in C}x_{e}\right)  $ where $C$
is a circuit of $\mathcal{M}$.

Let $\partial:\mathcal{X}_{E}\rightarrow\mathcal{X}_{E}$ be the derivation of
$\mathcal{X}_{E}$ that sends each indeterminate $x_{e}$ to $1$. Let
$\mathcal{J}_{\mathcal{M}}$ be the ideal of $\mathcal{X}_{E}$ generated by all
elements of the form $\partial\left(  \prod_{e\in C}x_{e}\right)  $ where $C$
is a circuit of $\mathcal{M}$.

\textbf{(a)} What can be said about the $\mathbf{k}$-algebra $\mathcal{X}%
_{E}/\mathcal{J}_{\mathcal{M}}$ ?

\textbf{(b)} It does not seem that $\mathcal{X}_{E}/\mathcal{J}_{\mathcal{M}}$
is a direct generalization of $\mathcal{X}/\mathcal{J}$ (for $\beta=0$ and
$\alpha=0$). Indeed, when $\mathcal{M}$ is the graphical matroid of the
complete graph $K_{n}$, the $\mathbf{k}$-algebra $\mathcal{X}_{E}%
/\mathcal{J}_{\mathcal{M}}$ has%
\begin{align*}
\text{generators}\ \ \ \ \ \ \ \ \ \  &  x_{i,j}\ \ \ \ \ \ \ \ \ \ \text{for
all }1\leq i<j\leq n\ \ \ \ \ \ \ \ \ \ \text{and}\\
\text{relations}\ \ \ \ \ \ \ \ \ \  &  x_{i,j}x_{j,k}+x_{i,k}x_{i,j}%
+x_{i,k}x_{j,k}=0\ \ \ \ \ \ \ \ \ \ \text{for all }1\leq i<j<k\leq n,
\end{align*}
whereas the $\mathbf{k}$-algebra $\mathcal{X}/\mathcal{J}$ (for $\beta=0$ and
$\alpha=0$) has%
\begin{align*}
\text{generators}\ \ \ \ \ \ \ \ \ \  &  x_{i,j}\ \ \ \ \ \ \ \ \ \ \text{for
all }1\leq i<j\leq n\ \ \ \ \ \ \ \ \ \ \text{and}\\
\text{relations}\ \ \ \ \ \ \ \ \ \  &  x_{i,j}x_{j,k}-x_{i,k}x_{i,j}%
-x_{i,k}x_{j,k}=0\ \ \ \ \ \ \ \ \ \ \text{for all }1\leq i<j<k\leq n.
\end{align*}
This suggests that we might want to tweak the definition of $\mathcal{J}%
_{\mathcal{M}}$ by introducing signs. Perhaps an oriented matroid structure on
$\mathcal{M}$ would help with that?

\textbf{(c)} Our basis of forkless monomials for $\mathcal{X}/\mathcal{J}$ can
be regarded as an \textquotedblleft\textbf{nbc} basis\textquotedblright\ in
the sense of \cite{CorEti01} (except that our monomials are not required to be
squarefree). Indeed, if we totally order the monomials $x_{i,j}$ in such a way
that $x_{i,j}>x_{u,v}$ whenever $i<u$, then the broken circuits of the
graphical matroid of $K_{n}$ are precisely the sets of the form $\left\{
\left\{  i,j\right\}  ,\left\{  i,k\right\}  \right\}  $ for $i<j<k$; but
these correspond to the precise monomials $x_{i,j}x_{i,k}$ that a forkless
monomial cannot be divisible by. Does this extend to the general case of a
matroid (or oriented matroid)?
\end{question}
\end{noncompile}

\subsection{Final questions}

Finally, we pose two lateral but (in our view) equally interesting questions
about $\mathcal{X}/\mathcal{J}$.

The first question, suggested by a referee, concerns the geometric background
of M\'{e}sz\'{a}ros's work. As mentioned in the Introduction, the algebra
$\mathcal{X}/\mathcal{J}$ generalizes M\'{e}sz\'{a}ros's \textquotedblleft
subdivision algebra\textquotedblright\ $\mathcal{S}\left(  A_{n}\right)  $
from \cite{Meszar09}. The latter owes its name to a geometric interpretation
of the relations $x_{i,j}x_{j,k}=x_{i,k}x_{i,j}+x_{i,k}x_{j,k}-x_{i,k}$ that
hold in $\mathcal{X}/\mathcal{J}$ when $\beta=-1$ and $\alpha=0$. For example,
if we consider the standard basis $\left(  e_{1},e_{2},\ldots,e_{n}\right)  $
of $\mathbb{R}^{n}$, then the cone $\left\langle e_{i}-e_{j},e_{j}%
-e_{k}\right\rangle _{+}$ (where $\left\langle u_{1},u_{2},\ldots
,u_{p}\right\rangle _{+}$ means the cone spanned by $p$ vectors $u_{1}%
,u_{2},\ldots,u_{p}$) is the union of the two cones $\left\langle e_{i}%
-e_{k},e_{i}-e_{j}\right\rangle _{+}$ and $\left\langle e_{i}-e_{k}%
,e_{j}-e_{k}\right\rangle _{+}$, while the intersection of the latter two
cones is $\left\langle e_{i}-e_{k}\right\rangle _{+}$. Thus, the indicator
functions of these cones satisfy%
\[
\mathbf{1}_{\left\langle e_{i}-e_{j},e_{j}-e_{k}\right\rangle _{+}}%
=\mathbf{1}_{\left\langle e_{i}-e_{k},e_{i}-e_{j}\right\rangle _{+}%
}+\mathbf{1}_{\left\langle e_{i}-e_{k},e_{j}-e_{k}\right\rangle _{+}%
}-\mathbf{1}_{\left\langle e_{i}-e_{k}\right\rangle _{+}}%
\]
(with $\mathbf{1}_{P}$ denoting the indicator function of a polyhedron $P$),
which is reminiscent of our relation $x_{i,j}x_{j,k}=x_{i,k}x_{i,j}%
+x_{i,k}x_{j,k}-x_{i,k}$. As M\'{e}sz\'{a}ros showed in \cite{Meszar09}, this
similarity can be used in studying root polytopes.

The cones $\left\langle e_{i}-e_{j},e_{j}-e_{k}\right\rangle _{+}$,
$\left\langle e_{i}-e_{k},e_{i}-e_{j}\right\rangle _{+}$, $\left\langle
e_{i}-e_{k},e_{j}-e_{k}\right\rangle _{+}$ and $\left\langle e_{i}%
-e_{k}\right\rangle _{+}$ are examples of \textit{permutohedral cones}, and
there is a rich theory of relations between their characteristic functions;
see \cite{Early18a} and \cite{Early18b}.

\begin{question}
Can such a geometric interpretation be given for the relations $x_{i,j}%
x_{j,k}=x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  +\alpha$ in
$\mathcal{X}/\mathcal{J}$ for general $\alpha$ and $\beta$, or at least for
values other than $\beta=-1$ and $\alpha=0$ ?
\end{question}

The relation $x_{i,j}x_{j,k}=x_{i,k}\left(  x_{i,j}+x_{j,k}-1\right)  $ in
$\mathcal{X}/\mathcal{J}$ for $\alpha=0$ and $\beta=-1$ also is connected to
MacMahon's Partition Analysis; see \cite[Fact 1.9]{AnPaRi01} for an equality
that not only follows its structure, but also to some extent anticipates our
homomorphism $A$ used above. Furthermore, MacMahon's $\Omega_{\geq}$ operator
(see, e.g., \cite[Definition 1.1]{AnPaRi01}) is rather similar to our linear
map $B$, in that it filters a Laurent series by removing all addends with
negative exponents. This connection, which I was alerted to by Richard
Stanley, is yet to be explored.

The last question, entirely out of left field, asks for a connection to the
notion of Rota-Baxter algebras (see, e.g., \cite{Guo09} for a survey):

\begin{question}
Is there anything to the superficial similarity \cite{Grinbe18} of the
relation $x_{i,j}x_{j,k}=x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  $ with
the axiom of a Rota-Baxter algebra?
\end{question}

\begin{noncompile}
The Rota-Baxter trail (unfinished, probably nonsense!).

There is an uncanny similarity between the definition of the ideal
$\mathcal{J}$ (at least for $\alpha=0$) and the concept of a
\textit{Rota-Baxter algebra} (see, e.g., \cite{Guo09} for an introduction).
Recall that a \textit{Rota-Baxter algebra of weight }$\beta$ means a
$\mathbf{k}$-algebra $R$ equipped with a $\mathbf{k}$-linear map
$P:R\rightarrow R$ (called its \textit{Rota-Baxter operator}) that satisfies%
\begin{equation}
P\left(  a\right)  P\left(  b\right)  =P\left(  P\left(  a\right)  b\right)
+P\left(  aP\left(  b\right)  \right)  +\beta P\left(  ab\right)
\label{eq.rota-baxter.def}%
\end{equation}
for all $a,b\in R$.

\begin{question}
Can the similarity between the equality (\ref{eq.rota-baxter.def}) and the
relations $\overline{x_{i,j}}\overline{x_{j,k}}=\overline{x_{i,k}}%
\overline{x_{i,j}}+\overline{x_{i,k}}\overline{x_{j,k}}+\beta\overline
{x_{i,k}}$ of the $\mathbf{k}$-algebra $\mathcal{X}/\mathcal{J}$ (for
$\alpha=0$) be concretized?
\end{question}

Let us give a partial answer. Consider a Rota-Baxter algebra $R$ of weight
$\beta$. For any monomial $\mathfrak{m}\in\mathfrak{M}$, we define a
$\mathbf{k}$-linear map $P_{\mathfrak{m}}:R^{\otimes n}\rightarrow R$ as
follows: It suffices to define $P_{\mathfrak{m}}\left(  r_{1}\otimes
r_{2}\otimes\cdots\otimes r_{n}\right)  $ for any $r_{1},r_{2},\ldots,r_{n}\in
R$. Consider the \textquotedblleft formal word\textquotedblright\ $r_{1}%
r_{2}\cdots r_{n}$ (this is not the product of $r_{1},r_{2},\ldots,r_{n}$, but
just a word with $n$ letters). Write the monomial $\mathfrak{m}$ as a product
$x_{i_{1},j_{1}}x_{i_{2},j_{2}}\cdots x_{i_{k},j_{k}}$. Now, we transform the
word $r_{1}r_{2}\cdots r_{n}$ as follows:

\begin{itemize}
\item First, for each $p\in\left\{  1,2,\ldots,k\right\}  $, we add an opening
parenthesis in front of the letter $r_{i}$, and a closing parenthesis after
the letter $r_{j-1}$.

\item Finally, we add a letter $P$ before each opening parenthesis, and
evaluate the resulting word as a product of elements of $R$.
\end{itemize}

For example, if $n=4$ and $\mathfrak{m}=x_{1,3}x_{2,3}x_{2,4}$, then
$P_{\mathfrak{m}}\left(  r_{1}\otimes r_{2}\otimes r_{3}\otimes r_{4}\right)
=P(r_{1}P(P(r_{2}))r_{3})r_{4}$.

Thus, a $\mathbf{k}$-linear map $P_{\mathfrak{m}}:R^{\otimes n}\rightarrow R$
is defined for every monomial $\mathfrak{m}\in\mathfrak{M}$. By linearity, we
thus define a $\mathbf{k}$-linear map $P_{f}:R^{\otimes n}\rightarrow R$ for
each polynomial $f\in\mathcal{X}$.

\begin{question}
Is it true that $P_{f}=0$ whenever $f\in\mathcal{J}$ and $\alpha=0$ ?
\end{question}

We suspect that the answer is positive. If so, then there is a canonical map
from $\mathcal{X}/\mathcal{J}$ (for $\alpha=0$) to the $n$-th component of the
operad of Rota-Baxter algebras. This map, however, is not injective, and it
will often happen that distinct monomials $\mathfrak{m}$ lead to the same map
$P_{\mathfrak{m}}$; this suggests that this is perhaps not the most natural construction.
\end{noncompile}

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\end{document}