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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 2.0, \today}
\maketitle

\begin{abstract}
\textbf{Abstract.} Fix a commutative ring $\mathbf{k}$, an element $\beta
\in\mathbf{k}$ 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 $\mathcal{J}_{\beta}$
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)  $ for $1\leq i<j<k\leq
n$. The quotient algebra $\mathcal{X}/\mathcal{J}_{\beta}$ (at least for a
certain universal choice of $\mathbf{k}$ and $\beta$) 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 natural question is to find
a combinatorial basis of this quotient algebra. One can define the
\textit{pathless monomials}, i.e., the monomials in $\mathcal{X}$ that have 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 indeed span the $\mathbf{k}$-module
$\mathcal{X}/\mathcal{J}_{\beta}$; however, they turn out (in general) to be
$\mathbf{k}$-linearly dependent. More combinatorially: Reducing a given
monomial in $\mathcal{X}$ modulo the ideal $\mathcal{J}_{\beta}$ 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)  $ 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 $t_{i}$. For a certain class of monomials $\mathfrak{m}$ (those
corresponding to \textquotedblleft noncrossing trees\textquotedblright), they
have shown that whatever result one gets by reducing $\mathfrak{m}$ modulo
$\mathcal{J}_{\beta}$, the image of this result under $D$ is independent on
the choices made in the reduction process. M\'{e}sz\'{a}ros has conjectured
that this property holds not only for this class of monomials, but for any
polynomial $p\in\mathcal{X}$. We prove this result, in the following slightly
stronger form: If $p\in\mathcal{X}$, and if $q\in\mathcal{X}$ is a
$\mathbf{k}$-linear combination of pathless monomials satisfying $p\equiv
q\operatorname{mod}\mathcal{J}_{\beta}$, then $D\left(  q\right)  $ does not
depend on $q$ (as long as $\beta$ and $p$ are fixed).

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

\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
a number $\beta\in\mathbb{Q}$, 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 $\dfrac{x_{i,k}\left(  x_{i,j}+x_{j,k}%
+\beta\right)  }{x_{i,j}x_{j,k}}\mathfrak{m}$ in $p$.

Thus, each move modifies the polynomial, replacing a monomial by a sum of
three monomials (or two, if $\beta=0$). 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 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 interest in the reduction rule
$\mathfrak{m}\mapsto\dfrac{x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)
}{x_{i,j}x_{j,k}}\mathfrak{m}$ originates 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). 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}. Recent work on Grothendieck polynomials 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 somewhat greater generality
than all previously published sources. 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 $\mathcal{J}_{\beta}$ 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)  $ 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 $\mathcal{J}_{\beta}$. 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 $\mathcal{J}_{\beta}$, then the image of $q$ under the substitution
$x_{i,j}\mapsto t_{i}$ does not depend on $q$ (but only on $\beta$ 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 $\mathcal{J}_{\beta}$).

After the proof of Theorem~\ref{thm.t-red.unique}, we shall also outline an
answer (Proposition \ref{prop.forkless.basis}) to the (easier) question of
finding a basis for the quotient ring $\mathcal{X}/\mathcal{J}_{\beta}$. This
basis will be obtained using an explicitly given Gr\"{o}bner basis of the
ideal $\mathcal{J}_{\beta}$. Finally, we shall state a generalization (Theorem
\ref{conj.Jba.indep}) of Theorem~\ref{thm.t-red.unique} to an ideal
$\mathcal{J}_{\beta,\alpha}$ that \textquotedblleft deforms\textquotedblright%
\ $\mathcal{J}_{\beta}$; a proof of this generalization is left to the next
version of this preprint.

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
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.

\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.)

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 $\beta\in\mathbf{k}$. Let $\mathcal{J}_{\beta}$ 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)  $ 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 $\beta\in\mathbf{k}$ and $p\in\mathcal{X}$.
Then, there exists a pathless polynomial $q\in\mathcal{X}$ such that $p\equiv
q\operatorname{mod}\mathcal{J}_{\beta}$.
\end{proposition}

In general, this $q$ is not unique.\footnote{For instance, if $\mathbf{k}%
=\mathbb{Z}$, $\beta=1$ 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 $x_{1,2}%
x_{2,3}x_{3,4}\equiv q\operatorname{mod}\mathcal{J}_{\beta}$, but they are not
identical.}

We shall roughly outline a proof of Proposition \ref{prop.path-red.span} now;
a more detailed writeup of this proof can be found in Section
\ref{sect.span-detail} below.

\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
$\widetilde{\mathfrak{m}}=\mathfrak{m}\cdot\dfrac{x_{i,k}\left(
x_{i,j}+x_{j,k}+\beta\right)  }{x_{i,j}x_{j,k}}$, which is congruent to
$\mathfrak{m}$ modulo $\mathcal{J}_{\beta}$ but has the property that all
monomials appearing in it have a smaller weight than $\mathfrak{m}$. This
gives rise to a recursive algorithm for reducing a polynomial modulo the ideal
$\mathcal{J}_{\beta}$. 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 $\mathcal{J}_{\beta}$ 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 $\mathcal{X}/\mathcal{J}_{\beta}$ for a certain choice of
$\mathbf{k}$ and $\beta$ (namely, for the choice where $\mathbf{k}$ is a
univariate polynomial ring over $\mathbb{Q}$, and $\beta$ is the indeterminate
in $\mathbf{k}$). 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 \S 7]{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 was conjectured
by Karola M\'{e}sz\'{a}ros in a 2015 talk at MIT:

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

\section{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 three $\mathbf{k}$-algebras
$\mathcal{Q}^{\prime}$, $\mathcal{Q}$ and $\mathcal{T}$ and three $\mathbf{k}%
$-linear maps $A$, $B$ and $C$ (with $A$ and $C$ being $\mathbf{k}$-algebra
homomorphisms) forming a diagram%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%\calX\ar[r]^{A} & \calQ^\prime\ar[r]^{B} & \calQ\ar[r]^{C} & \calT}}}%
%BeginExpansion
\xymatrix{
\calX\ar[r]^{A} & \calQ^\prime\ar[r]^{B} & \calQ\ar[r]^{C} & \calT}%
%EndExpansion
.
\]
We shall eventually show that:

\begin{itemize}
\item (Proposition \ref{prop.A.killsJ} below) the homomorphism $A$ annihilates
the ideal $\mathcal{J}_{1}$, but

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

These two facts will allow us to prove Theorem~\ref{thm.t-red.unique} in the
case when $\beta=1$ (this is Lemma \ref{lem.t-red.unique.b=1} below). From
this case, we shall then escalate to a somewhat more general case: Namely, we
shall prove Theorem~\ref{thm.t-red.unique} in the case when $\beta$ is
regular\footnote{See Definition \ref{def.reg} for the meaning of
\textquotedblleft regular\textquotedblright.} (Proposition
\ref{prop.t-red.unique.b=reg}). This latter case already covers the situation
studied in \cite[\S 5]{EscMes15} (indeed, $\beta$ is a polynomial
indeterminate over $\mathbf{k}=\mathbb{Q}$ in this case). Finally, using this
case as a stepping stone, we shall obtain a proof of
Theorem~\ref{thm.t-red.unique} in full generality.

Let us define the notion of a regular element of a commutative ring:

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

\subsection{The algebra $\mathcal{Q}^{\prime}$ of \textquotedblleft
zero-sum\textquotedblright\ power series}

\begin{noncompile}
We want to define a ring $\mathcal{Q}^{\prime}$ of formal power series over
$\mathbf{k}$ in the indeterminates $q_{1}/q_{2},q_{2}/q_{3},\ldots
,q_{n-1}/q_{n}$. In order to define it formally, we proceed as follows:
\end{noncompile}

\begin{definition}
\label{def.Q'}\textbf{(a)} Set $\mathcal{Q}^{\prime}=\mathbf{k}\left[  \left[
r_{1},r_{2},\ldots,r_{n-1}\right]  \right]  $. (This is the ring of power
series in the $n-1$ indeterminates $r_{1},r_{2},\ldots,r_{n-1}$ over
$\mathbf{k}$.)

The $\mathbf{k}$-algebra $\mathcal{Q}^{\prime}$ has a topology: the product
topology, defined by regarding it as a direct product of many copies of
$\mathbf{k}$ (thus regarding each power series as the family of its
coefficients). (Each copy of $\mathbf{k}$ corresponds to a monomial.) Thus,
$\mathcal{Q}^{\prime}$ becomes a topological $\mathbf{k}$-algebra. Every ring
of power series in this note will be endowed with a topology in this very way.

\textbf{(b)} For each $i\in\left[  n\right]  $, we define a monomial $q_{i}$
in the indeterminates $r_{1},r_{2},\ldots,r_{n-1}$ by $q_{i}=r_{i}%
r_{i+1}\cdots r_{n-1}$. (In particular, $q_{n}=1$.)

Each of the elements $q_{1},q_{2},\ldots,q_{n}$ of $\mathcal{Q}^{\prime}$ is
regular (since it is a monomial), and they satisfy $q_{n}\mid q_{n-1}%
\mid\cdots\mid q_{1}$. 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 $1\leq i\leq j\leq n$.
\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-1}^{a_{1}+a_{2}+\cdots+a_{n-1}}
\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-1}^{b_{n-1}}=q_{1}^{b_{1}}q_{2}%
^{b_{2}-b_{1}}q_{3}^{b_{3}-b_{2}}\cdots q_{n-1}^{b_{n-1}-b_{n-2}}%
q_{n}^{-b_{n-1}} \label{eq.Q'.reindex2}%
\end{equation}
for all $\left(  b_{1},b_{2},\ldots,b_{n-1}\right)  \in\mathbb{Z}^{n-1}$.

\begin{definition}
Let $\mathfrak{Z}$ denote the set of all $n$-tuples $\left(  a_{1}%
,a_{2},\ldots,a_{n}\right)  \in\mathbb{Z}^{n}$ satisfying $a_{1}+a_{2}%
+\cdots+a_{n}=0$ and $a_{1}+a_{2}+\cdots+a_{i}\geq0$ for all $i\in\left\{
1,2,\ldots,n\right\}  $.
\end{definition}

The topological $\mathbf{k}$-module $\mathcal{Q}^{\prime}$ has 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, $\left(  r_{1}^{b_{1}}r_{2}^{b_{2}}\cdots r_{n-1}^{b_{n-1}%
}\right)  _{\left(  b_{1},b_{2},\ldots,b_{n-1}\right)  \in\mathbb{N}^{n-1}}$
is a topological basis of the topological $\mathbf{k}$-module $\mathcal{Q}%
^{\prime}$, because each power series can be uniquely represented as an
infinite $\mathbf{k}$-linear combination of all the monomials.} $\left(
q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}\right)  _{\left(  a_{1}%
,a_{2},\ldots,a_{n}\right)  \in\mathfrak{Z}}$. Indeed, this is just a
reindexing of the standard topological basis \newline$\left(  r_{1}^{b_{1}%
}r_{2}^{b_{2}}\cdots r_{n-1}^{b_{n-1}}\right)  _{\left(  b_{1},b_{2}%
,\ldots,b_{n-1}\right)  \in\mathbb{N}^{n-1}}$ via the bijection%
\begin{align*}
\mathfrak{Z}  &  \rightarrow\mathbb{N}^{n-1},\\
\left(  a_{1},a_{2},\ldots,a_{n}\right)   &  \mapsto\left(  a_{1},a_{1}%
+a_{2},a_{1}+a_{2}+a_{3},\ldots,a_{1}+a_{2}+\cdots+a_{n-1}\right)
\end{align*}
(because of the equality (\ref{eq.Q'.reindex})).

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

\begin{definition}
Define a $\mathbf{k}$-algebra homomorphism $A:\mathcal{X}\rightarrow
\mathcal{Q}^{\prime}$ by%
\[
A\left(  x_{i,j}\right)  =\dfrac{q_{i}/q_{j}}{1-q_{i}/q_{j}}=\sum_{k>0}\left(
q_{i}/q_{j}\right)  ^{k}\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  i,j\right)
\in\left[  n\right]  ^{2}\text{ satisfying }i<j.
\]
Notice that this is well-defined, since $q_{i}/q_{j}=r_{i}r_{i+1}\cdots
r_{j-1}\in\mathcal{Q}^{\prime}$ is homogeneous of positive degree and thus can
be substituted into a power series.
\end{definition}

\begin{proposition}
\label{prop.A.killsJ}We have $A\left(  \mathcal{J}_{1}\right)  =0$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.A.killsJ}.]The ideal $\mathcal{J}_{1}$ 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}+1\right)  $ for $\left(  i,j,k\right)
\in\left[  n\right]  ^{3}$ satisfying $i<j<k$. Thus, it suffices to show that
$A\left(  x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+1\right)  \right)  =0$
for all triples $\left(  i,j,k\right)  \in\left[  n\right]  ^{3}$ satisfying
$i<j<k$. So let us fix such a triple.

Set $y=q_{i}/q_{j}$ and $z=q_{j}/q_{k}$. Both $y$ and $z$ are homogeneous of
positive degree, and thus we can freely divide by power series such as $1-y$
and $1-z$.

The definition of $A$ yields $A\left(  x_{i,j}\right)  =\dfrac{q_{i}/q_{j}%
}{1-q_{i}/q_{j}}=\dfrac{y}{1-y}$ (since $q_{i}/q_{j}=y$). Similarly, $A\left(
x_{j,k}\right)  =\dfrac{z}{1-z}$.

The definition of $A$ yields $A\left(  x_{i,k}\right)  =\dfrac{q_{i}/q_{k}%
}{1-q_{i}/q_{k}}=\dfrac{yz}{1-yz}$ (since $q_{i}/q_{k}=\underbrace{\left(
q_{i}/q_{j}\right)  }_{=y}\underbrace{\left(  q_{j}/q_{k}\right)  }_{=z}=yz$).

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}+1\right)  \right) \\
&  =\underbrace{A\left(  x_{i,j}\right)  }_{=\dfrac{y}{1-y}}%
\underbrace{A\left(  x_{j,k}\right)  }_{=\dfrac{z}{1-z}}-\underbrace{A\left(
x_{i,k}\right)  }_{=\dfrac{yz}{1-yz}}\left(  \underbrace{A\left(
x_{i,j}\right)  }_{=\dfrac{y}{1-y}}+\underbrace{A\left(  x_{j,k}\right)
}_{=\dfrac{z}{1-z}}+1\right) \\
&  =\dfrac{y}{1-y}\dfrac{z}{1-z}-\dfrac{yz}{1-yz}\left(  \dfrac{y}{1-y}%
+\dfrac{z}{1-z}+1\right) \\
&  =0,
\end{align*}
qed.
\end{proof}

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

\begin{definition}
Let $\mathcal{Q}$ be the $\mathbf{k}$-algebra $\mathbf{k}\left[  \left[
q_{1},q_{2},\ldots,q_{n-1}\right]  \right]  $. This is the ring of formal
power series in the $n-1$ indeterminates $q_{1},q_{2},\ldots,q_{n-1}$ over
$\mathbf{k}$. These indeterminates have nothing to do with the monomials
$q_{i}$ from Definition \ref{def.Q'} \textbf{(a)} (but of course the same
names have been chosen for similarity).

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

\subsection{The continuous $\mathbf{k}$-linear map $B:\mathcal{Q}^{\prime
}\rightarrow\mathcal{Q}$}

Before we define our next map, let us show three simple lemmas:

\begin{lemma}
\label{lem.B.wd-1}Let $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in
\mathbb{Z}^{n}$ be such that $a_{1}+a_{2}+\cdots+a_{n}=0$. Let $N=\sum
_{i\in\left[  n\right]  }\max\left\{  a_{i},0\right\}  $. Then, $\left(
a_{1},a_{2},\ldots,a_{n}\right)  \in\left\{  -N,-N+1,\ldots,N\right\}  ^{n}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.B.wd-1}.]We have%
\[
0=a_{1}+a_{2}+\cdots+a_{n}=\sum_{i\in\left[  n\right]  }a_{i}=\sum
_{\substack{i\in\left[  n\right]  ;\\a_{i}\geq0}}a_{i}+\sum_{\substack{i\in
\left[  n\right]  ;\\a_{i}<0}}a_{i}%
\]
(since each $i\in\left[  n\right]  $ satisfies either $a_{i}\geq0$ or
$a_{i}<0$, but not both). Solving this equation for $\sum_{\substack{i\in
\left[  n\right]  ;\\a_{i}\geq0}}a_{i}$, we obtain%
\[
\sum_{\substack{i\in\left[  n\right]  ;\\a_{i}\geq0}}a_{i}=-\sum
_{\substack{i\in\left[  n\right]  ;\\a_{i}<0}}a_{i}=\sum_{\substack{i\in
\left[  n\right]  ;\\a_{i}<0}}\underbrace{\left(  -a_{i}\right)
}_{\substack{=\left\vert a_{i}\right\vert \\\text{(since }a_{i}<0\text{)}%
}}=\sum_{\substack{i\in\left[  n\right]  ;\\a_{i}<0}}\left\vert a_{i}%
\right\vert .
\]
But%
\begin{align*}
N  &  =\sum_{i\in\left[  n\right]  }\max\left\{  a_{i},0\right\}
=\sum_{\substack{i\in\left[  n\right]  ;\\a_{i}\geq0}}\underbrace{\max\left\{
a_{i},0\right\}  }_{\substack{=a_{i}\\\text{(since }a_{i}\geq0\text{)}}%
}+\sum_{\substack{i\in\left[  n\right]  ;\\a_{i}<0}}\underbrace{\max\left\{
a_{i},0\right\}  }_{\substack{=0\\\text{(since }a_{i}<0\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since each }i\in\left[  n\right]  \text{
satisfies either }a_{i}\geq0\text{ or }a_{i}<0\text{, but not both}\right) \\
&  =\sum_{\substack{i\in\left[  n\right]  ;\\a_{i}\geq0}}a_{i}%
+\underbrace{\sum_{\substack{i\in\left[  n\right]  ;\\a_{i}<0}}0}_{=0}%
=\sum_{\substack{i\in\left[  n\right]  ;\\a_{i}\geq0}}a_{i}=\sum
_{\substack{i\in\left[  n\right]  ;\\a_{i}<0}}\left\vert a_{i}\right\vert .
\end{align*}


Now, we claim that%
\begin{equation}
\left\vert a_{j}\right\vert \leq N\ \ \ \ \ \ \ \ \ \ \text{for each }%
j\in\left[  n\right]  . \label{pf.lem.B.wd-1.4}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.B.wd-1.4}):} Fix $j\in\left[  n\right]  $. We
want to prove (\ref{pf.lem.B.wd-1.4}). We are in one of the following two cases:

\textit{Case 1:} We have $a_{j}\geq0$.

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

Let us consider Case 1 first. In this case, we have $a_{j}\geq0$. Thus,
$a_{j}$ is an addend in the sum $\sum_{\substack{i\in\left[  n\right]
;\\a_{i}\geq0}}a_{i}$. Since this sum consists purely of nonnegative
integers\footnote{because $a_{i}$ is a nonnegative integer for each
$i\in\left[  n\right]  $ satisfying $a_{i}\geq0$}, we therefore conclude that
$a_{j}$ is at most as large as this sum. In other words, $a_{j}\leq
\sum_{\substack{i\in\left[  n\right]  ;\\a_{i}\geq0}}a_{i}$. In light of
$N=\sum_{\substack{i\in\left[  n\right]  ;\\a_{i}\geq0}}a_{i}$, this rewrites
as $a_{j}\leq N$. But $a_{j}\geq0$ and thus $\left\vert a_{j}\right\vert
=a_{j}\leq N$. Hence, (\ref{pf.lem.B.wd-1.4}) is proven in Case 1.

Let us now consider Case 2. In this case, we have $a_{j}<0$. Hence,
$\left\vert a_{j}\right\vert $ is an addend in the sum $\sum_{\substack{i\in
\left[  n\right]  ;\\a_{i}<0}}\left\vert a_{i}\right\vert $. Since this sum
consists purely of nonnegative integers\footnote{because $\left\vert
a_{i}\right\vert $ is a nonnegative integer for each $i\in\left[  n\right]  $
satisfying $a_{i}<0$}, we therefore conclude that its addend $\left\vert
a_{j}\right\vert $ is at most as large as the whole sum. In other words,
$\left\vert a_{j}\right\vert \leq\sum_{\substack{i\in\left[  n\right]
;\\a_{i}<0}}\left\vert a_{i}\right\vert $. In light of $N=\sum_{\substack{i\in
\left[  n\right]  ;\\a_{i}<0}}\left\vert a_{i}\right\vert $, this rewrites as
$\left\vert a_{j}\right\vert \leq N$. Hence, (\ref{pf.lem.B.wd-1.4}) is proven
in Case 2.

We have thus proven (\ref{pf.lem.B.wd-1.4}) in both Cases 1 and 2. Hence,
(\ref{pf.lem.B.wd-1.4}) is proven.]

From (\ref{pf.lem.B.wd-1.4}), we conclude that $a_{j}\in\left\{
-N,-N+1,\ldots,N\right\}  $ for each $j\in\left[  n\right]  $ (since $a_{j}%
\in\mathbb{Z}$). Hence, $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\left\{
-N,-N+1,\ldots,N\right\}  ^{n}$. This proves Lemma \ref{lem.B.wd-1}.
\end{proof}

\begin{lemma}
\label{lem.B.wd0}Let $b_{1},b_{2},\ldots,b_{n-1}$ be $n-1$ nonnegative
integers. Let $N=b_{1}+b_{2}+\cdots+b_{n-1}$. Let $\left(  a_{1},a_{2}%
,\ldots,a_{n}\right)  \in\mathfrak{Z}$ be such that each $i\in\left[
n-1\right]  $ satisfies $b_{i}=\max\left\{  a_{i},0\right\}  $. Then, $\left(
a_{1},a_{2},\ldots,a_{n}\right)  $ belongs to $\left\{  -N,-N+1,\ldots
,N\right\}  ^{n}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.B.wd0}.]From $\left(  a_{1},a_{2},\ldots
,a_{n}\right)  \in\mathfrak{Z}$, we obtain $a_{1}+a_{2}+\cdots+a_{n}=0$ and
$a_{1}+a_{2}+\cdots+a_{i}\geq0$ for all $i\in\left\{  1,2,\ldots,n\right\}  $.
From $a_{1}+a_{2}+\cdots+a_{i}\geq0$ for all $i\in\left\{  1,2,\ldots
,n\right\}  $, we obtain $a_{1}+a_{2}+\cdots+a_{n-1}\geq0$%
.\ \ \ \ \footnote{To be fully precise: The inequality $a_{1}+a_{2}%
+\cdots+a_{n-1}\geq0$ follows from $a_{1}+a_{2}+\cdots+a_{i}\geq0$ (applied to
$i=n-1$) when $n\geq2$. But when $n<2$, it follows from $a_{1}+a_{2}%
+\cdots+a_{n-1}=\left(  \text{empty sum}\right)  =0$.}

Now,%
\[
0=a_{1}+a_{2}+\cdots+a_{n}=\underbrace{\left(  a_{1}+a_{2}+\cdots
+a_{n-1}\right)  }_{\geq0}+a_{n}\geq0+a_{n}=a_{n},
\]
so that $a_{n}\leq0$ and therefore $\max\left\{  a_{n},0\right\}  =0$.

We know that each $i\in\left[  n-1\right]  $ satisfies $b_{i}=\max\left\{
a_{i},0\right\}  $. Hence, $\sum_{i\in\left[  n-1\right]  }b_{i}=\sum
_{i\in\left[  n-1\right]  }\max\left\{  a_{i},0\right\}  $. Thus,
\[
\sum_{i\in\left[  n-1\right]  }\max\left\{  a_{i},0\right\}  =\sum
_{i\in\left[  n-1\right]  }b_{i}=b_{1}+b_{2}+\cdots+b_{n-1}=N.
\]
Now,%
\[
\sum_{i\in\left[  n\right]  }\max\left\{  a_{i},0\right\}  =\underbrace{\sum
_{i\in\left[  n-1\right]  }\max\left\{  a_{i},0\right\}  }_{=N}%
+\underbrace{\max\left\{  a_{n},0\right\}  }_{=0}=N+0=N.
\]
In other words, $N=\sum_{i\in\left[  n\right]  }\max\left\{  a_{i},0\right\}
$. Hence, Lemma \ref{lem.B.wd-1} shows that $\left(  a_{1},a_{2},\ldots
,a_{n}\right)  \in\left\{  -N,-N+1,\ldots,N\right\}  ^{n}$. This proves Lemma
\ref{lem.B.wd0}.
\end{proof}

\begin{lemma}
\label{lem.B.wd1}Let $\mathfrak{m}$ be a monomial in the indeterminates
$q_{1},q_{2},\ldots,q_{n-1}$ (with nonnegative exponents). Then, there exist
only finitely many $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{Z}$
satisfying $\prod_{\substack{i\in\left[  n-1\right]  ;\\a_{i}>0}}q_{i}^{a_{i}%
}=\mathfrak{m}$. (Here, again, $q_{1},q_{2},\ldots,q_{n-1}$ are the
indeterminates of $\mathcal{Q}$, not the monomials $q_{i}$ from Definition
\ref{def.Q'} \textbf{(b)}.)
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.B.wd1}.]Write $\mathfrak{m}$ in the form
$\mathfrak{m}=\prod_{i\in\left[  n-1\right]  }q_{i}^{b_{i}}$. Let
$N=b_{1}+b_{2}+\cdots+b_{n-1}$. We want to prove that there exist only
finitely many $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{Z}$
satisfying $\prod_{\substack{i\in\left[  n-1\right]  ;\\a_{i}>0}}q_{i}^{a_{i}%
}=\mathfrak{m}$. We shall show that each such $\left(  a_{1},a_{2}%
,\ldots,a_{n}\right)  $ must belong to the set $\left\{  -N,-N+1,\ldots
,N\right\}  ^{n}$.

Indeed, let $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{Z}$ be such
that $\prod_{\substack{i\in\left[  n-1\right]  ;\\a_{i}>0}}q_{i}^{a_{i}%
}=\mathfrak{m}$. We must show that $\left(  a_{1},a_{2},\ldots,a_{n}\right)  $
belongs to $\left\{  -N,-N+1,\ldots,N\right\}  ^{n}$.

We have
\begin{align*}
\prod_{i\in\left[  n-1\right]  }q_{i}^{\max\left\{  a_{i},0\right\}  }  &
=\left(  \prod_{\substack{i\in\left[  n-1\right]  ;\\a_{i}>0}%
}\underbrace{q_{i}^{\max\left\{  a_{i},0\right\}  }}_{\substack{=q_{i}^{a_{i}%
}\\\text{(since }\max\left\{  a_{i},0\right\}  =a_{i}\\\text{(since }%
a_{i}>0\text{))}}}\right)  \left(  \prod_{\substack{i\in\left[  n-1\right]
;\\a_{i}\leq0}}\underbrace{q_{i}^{\max\left\{  a_{i},0\right\}  }%
}_{\substack{=q_{i}^{0}\\\text{(since }\max\left\{  a_{i},0\right\}
=0\\\text{(since }a_{i}\leq0\text{))}}}\right) \\
&  =\left(  \prod_{\substack{i\in\left[  n-1\right]  ;\\a_{i}>0}}q_{i}^{a_{i}%
}\right)  \left(  \prod_{\substack{i\in\left[  n-1\right]  ;\\a_{i}\leq
0}}\underbrace{q_{i}^{0}}_{=1}\right)  =\prod_{\substack{i\in\left[
n-1\right]  ;\\a_{i}>0}}q_{i}^{a_{i}}=\mathfrak{m}=\prod_{i\in\left[
n-1\right]  }q_{i}^{b_{i}}.
\end{align*}
Thus, $\prod_{i\in\left[  n-1\right]  }q_{i}^{b_{i}}=\prod_{i\in\left[
n-1\right]  }q_{i}^{\max\left\{  a_{i},0\right\}  }$. In other words, each
$i\in\left[  n-1\right]  $ satisfies $b_{i}=\max\left\{  a_{i},0\right\}  $.
Hence, $\left(  a_{1},a_{2},\ldots,a_{n}\right)  $ belongs to $\left\{
-N,-N+1,\ldots,N\right\}  ^{n}$ (by Lemma \ref{lem.B.wd0}).

Now, forget that we fixed $\left(  a_{1},a_{2},\ldots,a_{n}\right)  $. We thus
have shown that each $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in
\mathfrak{Z}$ satisfying $\prod_{\substack{i\in\left[  n-1\right]  ;\\a_{i}%
>0}}q_{i}^{a_{i}}=\mathfrak{m}$ must belong to the set $\left\{
-N,-N+1,\ldots,N\right\}  ^{n}$. Therefore, there exist only finitely many
such $\left(  a_{1},a_{2},\ldots,a_{n}\right)  $ (because the set $\left\{
-N,-N+1,\ldots,N\right\}  ^{n}$ is finite). This proves Lemma \ref{lem.B.wd1}.
\end{proof}

\begin{definition}
We define a continuous $\mathbf{k}$-linear map $B:\mathcal{Q}^{\prime
}\rightarrow\mathcal{Q}$ by setting%
\[
B\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}\right)  =\prod
_{\substack{i\in\left[  n-1\right]  ;\\a_{i}>0}}q_{i}^{a_{i}}%
\ \ \ \ \ \ \ \ \ \ \text{for each }\left(  a_{1},a_{2},\ldots,a_{n}\right)
\in\mathfrak{Z}.
\]
This is well-defined, since $\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\right)  _{\left(  a_{1},a_{2},\ldots,a_{n}\right)
\in\mathfrak{Z}}$ is a topological basis of $\mathcal{Q}^{\prime}$, 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 monomials).
\end{definition}

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

\subsection{The algebra $\mathcal{T}$ of power series}

\begin{definition}
We define a topological $\mathbf{k}$-algebra $\mathcal{T}$ by $\mathcal{T}%
=\mathbf{k}\left[  \left[  t_{1},t_{2},\ldots,t_{n-1}\right]  \right]  $ (with
the usual topology). Again, this is simply a ring of formal power series over
$\mathbf{k}$.

We shall regard $\mathcal{T}^{\prime}$ as a $\mathbf{k}$-subalgebra of
$\mathcal{T}$ (in the obvious way). Thus, $D:\mathcal{X}\rightarrow
\mathcal{T}^{\prime}$ becomes a $\mathbf{k}$-algebra homomorphism
$\mathcal{X}\rightarrow\mathcal{T}$.
\end{definition}

\subsection{The continuous $\mathbf{k}$-algebra homomorphism $C:\mathcal{Q}%
\rightarrow\mathcal{T}$}

\begin{definition}
We define a continuous $\mathbf{k}$-algebra homomorphism $C:\mathcal{Q}%
\rightarrow\mathcal{T}$ by%
\[
C\left(  q_{i}\right)  =\dfrac{t_{i}}{1+t_{i}}\ \ \ \ \ \ \ \ \ \ \text{for
each }i\in\left[  n-1\right]  .
\]
This is well-defined, because for each $i\in\left[  n-1\right]  $, the power
series $\dfrac{t_{i}}{1+t_{i}}$ has constant term $0$ and thus can be
substituted into power series.
\end{definition}

Thus, we have defined the following spaces and maps between them:%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%\calX\ar[r]^{A} & \calQ^\prime\ar[r]^{B} & \calQ\ar[r]^{C} & \calT}}}%
%BeginExpansion
\xymatrix{
\calX\ar[r]^{A} & \calQ^\prime\ar[r]^{B} & \calQ\ar[r]^{C} & \calT}%
%EndExpansion
.
\]


It is worth reminding ourselves that $A$ and $C$ 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 composition $C\circ B\circ A$ 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 $\left(  a_{1},a_{2},\ldots,a_{n}\right)
\in\mathfrak{Z}$ 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)  $. Let
$\mathfrak{Z}_{S}$ denote the set of all $S$-adequate $n$-tuples $\left(
a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{Z}$. We let $\mathcal{Q}%
_{S}^{\prime}$ denote the subset of $\mathcal{Q}^{\prime}$ consisting of all
infinite $\mathbf{k}$-linear combinations of the monomials $q_{1}^{a_{1}}%
q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ for $S$-adequate $n$-tuples $\left(
a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{Z}$. It is easy to see that
$\mathcal{Q}_{S}^{\prime}$ is a topological $\mathbf{k}$-subalgebra of
$\mathcal{Q}^{\prime}$ (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 here, but rather monomials defined
by $q_{i}=r_{i}r_{i+1}\cdots r_{n-1}$. But their products $q_{1}^{a_{1}}%
q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ are monomials. Explicitly, they can be
rewritten as products of the $r_{1},r_{2},\ldots,r_{n-1}$ using
(\ref{eq.Q'.reindex}). Thus, it is easy to see that the elements of
$\mathcal{Q}_{S}^{\prime}$ are the infinite $\mathbf{k}$-linear combinations
of the monomials $r_{1}^{b_{1}}r_{2}^{b_{2}}\cdots r_{n-1}^{b_{n-1}}$ for all
$\left(  b_{1},b_{2},\ldots,b_{n-1}\right)  \in\mathbb{N}^{n-1}$ 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 $b_{0}=0$ and $b_{n}=0$. But we won't need this characterization.)

\textbf{(e)} We let $\mathcal{Q}_{S}$ denote the topological $\mathbf{k}%
$-algebra $\mathbf{k}\left[  \left[  q_{i}\ \mid\ i\in S\right]  \right]  $.
This is a topological subalgebra of $\mathcal{Q}$.

\textbf{(f)} 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 $\mathcal{T}$.

\textbf{(g)} We define a $\mathbf{k}$-algebra homomorphism $A_{S}%
:\mathcal{X}_{S}\rightarrow\mathcal{Q}_{S}^{\prime}$ by%
\[
A_{S}\left(  x_{i,j}\right)  =\dfrac{q_{i}/q_{j}}{1-q_{i}/q_{j}}=\sum
_{k>0}\left(  q_{i}/q_{j}\right)  ^{k}\ \ \ \ \ \ \ \ \ \ \text{for all
}\left(  i,j\right)  \in\mathfrak{P}_{S}.
\]
This is easily seen to be well-defined (because for each $\left(  i,j\right)
\in\mathfrak{P}_{S}$ and $k>0$, 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
and belongs to $\mathfrak{Z}$, and therefore the monomial $\left(  q_{i}%
/q_{j}\right)  ^{k}$ is in $\mathcal{Q}_{S}^{\prime}$).

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

\textbf{(i)} We define a continuous $\mathbf{k}$-algebra homomorphism
$C_{S}:\mathcal{Q}_{S}\rightarrow\mathcal{T}_{S}$ by%
\[
C_{S}\left(  q_{i}\right)  =\dfrac{t_{i}}{1+t_{i}}%
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in S.
\]
This is well-defined, because for each $i\in S$, the power series
$\dfrac{t_{i}}{1+t_{i}}$ has constant term $0$ and thus can be substituted
into power series.
\end{definition}

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

\textbf{(a)} We have $B\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}%
}\right)  =\prod_{i\in S}q_{i}^{a_{i}}$ for each $S$-adequate $n$-tuple
$\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{Z}$.

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

\begin{proof}
[Proof of Proposition \ref{prop.pathless.BSwd}.]\textbf{(a)} Let $\left(
a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{Z}$ be an $S$-adequate
$n$-tuple. We must show that $B\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\right)  =\prod_{i\in S}q_{i}^{a_{i}}$.

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 $a_{i}>0$). In particular, each $i\in\left[  n-1\right]  $
satisfying $a_{i}>0$ must belong to $S$.

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

\textbf{(b)} We must show that there exists a unique continuous $\mathbf{k}%
$-linear map $B_{S}:\mathcal{Q}_{S}^{\prime}\rightarrow\mathcal{Q}_{S}$
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)
=\prod_{i\in S}q_{i}^{a_{i}}\\
\ \ \ \ \ \ \ \ \ \ \text{for each }S\text{-adequate }\left(  a_{1}%
,a_{2},\ldots,a_{n}\right)  \in\mathfrak{Z}%
\end{array}
\right)  . \label{pf.prop.pathless.BSwd.b.cond}%
\end{equation}
The uniqueness of such a map is clear (because the elements of $\mathcal{Q}%
_{S}^{\prime}$ are infinite $\mathbf{k}$-linear combinations of the monomials
$q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ for $S$-adequate $n$-tuples
$\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{Z}$; 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 $f\in\mathcal{Q}_{S}^{\prime}$, we have $B\left(  f\right)
\in\mathcal{Q}_{S}$\ \ \ \ \footnote{\textit{Proof.} Let $f\in\mathcal{Q}%
_{S}^{\prime}$. We must show that $B\left(  f\right)  \in\mathcal{Q}_{S}$.
Since the map $B$ is $\mathbf{k}$-linear and continuous, we can WLOG assume
that $f$ is a monomial of the form $q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}$ for some $S$-adequate $n$-tuple $\left(  a_{1},a_{2}%
,\ldots,a_{n}\right)  \in\mathfrak{Z}$ (because $f$ is always an infinite
$\mathbf{k}$-linear combination of such monomials). Assume this. Consider this
$\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{Z}$.
\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)  =\prod_{i\in S}q_{i}^{a_{i}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by Proposition \ref{prop.pathless.BSwd} \textbf{(a)}}\right) \\
&  \in\mathcal{Q}_{S}.
\end{align*}
This is precisely what we wanted to show.}. Hence, we can define a map
$\widetilde{B_{S}}:\mathcal{Q}_{S}^{\prime}\rightarrow\mathcal{Q}_{S}$ by%
\[
\widetilde{B_{S}}\left(  f\right)  =B\left(  f\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }f\in\mathcal{Q}_{S}^{\prime}.
\]
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) \\
&  =\prod_{i\in S}q_{i}^{a_{i}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition \ref{prop.pathless.BSwd} \textbf{(a)}}\right)
\end{align*}
for each $S$-adequate $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in
\mathfrak{Z}$. Hence, $\widetilde{B_{S}}$ is a continuous $\mathbf{k}$-linear
map $B_{S}:\mathcal{Q}_{S}^{\prime}\rightarrow\mathcal{Q}_{S}$ 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 diagram%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%\calX_S \arinj[d] \ar[r]^{A_S} & \calQ^\prime_S \arinj[d] \ar[r]^{B_S}
%& \calQ_S \arinj[d] \ar[r]^{C_S}
%& \calT_S \arinj[d] \\
%\calX\ar[r]_{A} & \calQ^\prime\ar[r]_{B} & \calQ\ar[r]_{C} & \calT}}}%
%BeginExpansion
\xymatrix{
\calX_S \arinj[d] \ar[r]^{A_S} & \calQ^\prime_S \arinj[d] \ar[r]^{B_S}
& \calQ_S \arinj[d] \ar[r]^{C_S}
& \calT_S \arinj[d] \\
\calX\ar[r]_{A} & \calQ^\prime\ar[r]_{B} & \calQ\ar[r]_{C} & \calT}%
%EndExpansion
\]
is commutative.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.pathless.cd}.]The commutativity of the left
square 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, $\mathcal{A}_{1}=\mathcal{X}_{S}$, $\mathcal{A}_{2}=\mathcal{Q}%
_{S}^{\prime}$, $\mathcal{A}_{3}=\mathcal{X}$, $\mathcal{A}_{4}=\mathcal{Q}%
^{\prime}$, $f_{1}=A_{S}$ and $f_{4}=A$, whereas $f_{2}$ and $f_{3}$ are the
inclusion maps $\mathcal{X}_{S}\rightarrow\mathcal{X}$ and $\mathcal{Q}%
_{S}^{\prime}\rightarrow\mathcal{Q}^{\prime}$.) 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 the right square\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 Hausdorff topological $\mathbf{k}$-algebras and
$f_{1},f_{2},f_{3},f_{4}$ are four continuous $\mathbf{k}$-algebra
homomorphisms. (In our concrete case, $\mathcal{A}_{1}=\mathcal{Q}_{S}$,
$\mathcal{A}_{2}=\mathcal{T}_{S}$, $\mathcal{A}_{3}=\mathcal{Q}$,
$\mathcal{A}_{4}=\mathcal{T}$, $f_{1}=C_{S}$ and $f_{4}=C$, whereas $f_{2}$
and $f_{3}$ are the inclusion maps $\mathcal{Q}_{S}\rightarrow\mathcal{Q}$ and
$\mathcal{T}_{S}\rightarrow\mathcal{T}$.) In order to prove this
commutativity, it suffices to show that it holds \textbf{on a topological
generating set} of the $\mathbf{k}$-algebra $\mathcal{A}_{1}$. (A
\textit{topological generating set} of a topological $\mathbf{k}$-algebra
$\mathcal{A}$ means a subset $\mathfrak{G}$ of $\mathcal{A}$ such that the
$\mathbf{k}$-subalgebra of $\mathcal{A}$ generated by $\mathfrak{G}$ is dense
in $\mathcal{A}$.) In other words, it suffices to pick some topological
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\{
q_{i}\ \mid\ i\in S\right\}  $. The proof then becomes completely clear.)}. It
thus remains to prove the commutativity of the middle square. In other words,
we must show that $B_{S}\left(  p\right)  =B\left(  p\right)  $ for each
$p\in\mathcal{Q}_{S}^{\prime}$.

So fix $p\in\mathcal{Q}_{S}^{\prime}$. Since both maps $B_{S}$ and $B$ are
continuous and $\mathbf{k}$-linear, we can WLOG assume that $p$ is a monomial
of the form $q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ for an
$S$-adequate $n$-tuple $\left(  a_{1},a_{2},\ldots,a_{n}\right)
\in\mathfrak{Z}$ (since the elements of $\mathcal{Q}_{S}^{\prime}$ are
infinite $\mathbf{k}$-linear combinations of 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
\[
B\left(  p\right)  =B\left(  q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}%
}\right)  =\prod_{i\in S}q_{i}^{a_{i}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition \ref{prop.pathless.BSwd} \textbf{(a)}}\right)  .
\]
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) \\
&  =\prod_{i\in S}q_{i}^{a_{i}}\ \ \ \ \ \ \ \ \ \ \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 middle square. 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, $B_{S}:\mathcal{Q}_{S}^{\prime}\rightarrow\mathcal{Q}_{S}$ 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 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 $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{Z}$ (since
the elements of $\mathcal{Q}_{S}^{\prime}$ are infinite $\mathbf{k}$-linear
combinations of 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,%
\[
\left(  B_{S}\circ A_{S}\right)  \left(  x_{i,j}\right)  =\dfrac{q_{i}%
}{1-q_{i}}.
\]

\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 $j\in\left[  n\right]  \setminus
S$, so that $j\notin S$. From this, it becomes clear that $B_{S}\left(
q_{i}/q_{j}\right)  =q_{i}$ (by the definition of $B_{S}$).

Proposition \ref{prop.pathless.BS-alg} shows that $B_{S}:\mathcal{Q}%
_{S}^{\prime}\rightarrow\mathcal{Q}_{S}$ is a continuous $\mathbf{k}$-algebra
homomorphism. Now,%
\begin{align*}
\left(  B_{S}\circ A_{S}\right)  \left(  x_{i,j}\right)   &  =B_{S}\left(
\underbrace{A_{S}\left(  x_{i,j}\right)  }_{=\dfrac{q_{i}/q_{j}}{1-q_{i}%
/q_{j}}}\right)  =B_{S}\left(  \dfrac{q_{i}/q_{j}}{1-q_{i}/q_{j}}\right)
=\dfrac{B_{S}\left(  q_{i}/q_{j}\right)  }{1-B_{S}\left(  q_{i}/q_{j}\right)
}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }B_{S}\text{ is a continuous }\mathbf{k}\text{-algebra
homomorphism,}\\
\text{and thus commutes with any power series}%
\end{array}
\right) \\
&  =\dfrac{q_{i}}{1-q_{i}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
B_{S}\left(  q_{i}/q_{j}\right)  =q_{i}\right)  .
\end{align*}
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 $D\left(  \mathfrak{m}\right)  =\left(  C_{S}\circ 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 $D\mid_{\mathcal{X}%
_{S}}=C_{S}\circ B_{S}\circ A_{S}$ (if we regard $C_{S}\circ B_{S}\circ A_{S}$
as a map to $\mathcal{T}$).

The map $D\mid_{\mathcal{X}_{S}}$ is a $\mathbf{k}$-algebra homomorphism
(since $D$ is a $\mathbf{k}$-algebra homomorphism), and the map $C_{S}\circ
B_{S}\circ A_{S}$ is a $\mathbf{k}$-algebra homomorphism (since all of $C_{S}%
$, $B_{S}$ 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 $D\mid_{\mathcal{X}_{S}}$ and $C_{S}\circ B_{S}\circ A_{S}$). 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(  D\mid_{\mathcal{X}_{S}}\right)  \left(  x_{i,j}\right)  =\left(
C_{S}\circ 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 $\left(  i,j\right)  \in\mathfrak{P}_{S}$. Then, $\left(
D\mid_{\mathcal{X}_{S}}\right)  \left(  x_{i,j}\right)  =D\left(
x_{i,j}\right)  =t_{i}$ (by the definition of $D$). Comparing this with%
\begin{align*}
\left(  C_{S}\circ B_{S}\circ A_{S}\right)  \left(  x_{i,j}\right)   &
=C_{S}\left(  \underbrace{\left(  B_{S}\circ A_{S}\right)  \left(
x_{i,j}\right)  }_{\substack{=\dfrac{q_{i}}{1-q_{i}}\\\text{(by Proposition
\ref{prop.pathless.AB})}}}\right)  =C_{S}\left(  \dfrac{q_{i}}{1-q_{i}%
}\right)  =\dfrac{C_{S}\left(  q_{i}\right)  }{1-C_{S}\left(  q_{i}\right)
}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because the map }C_{S}\text{ is a continuous }\mathbf{k}\text{-algebra}%
\\
\text{homomorphism, and thus}\\
\text{commutes with any power series}%
\end{array}
\right) \\
&  =\dfrac{\left(  \dfrac{t_{i}}{1+t_{i}}\right)  }{1-\dfrac{t_{i}}{1+t_{i}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }C_{S}\left(  q_{i}\right)
=\dfrac{t_{i}}{1+t_{i}}\right) \\
&  =t_{i},
\end{align*}
we obtain $\left(  D\mid_{\mathcal{X}_{S}}\right)  \left(  x_{i,j}\right)
=\left(  C_{S}\circ B_{S}\circ A_{S}\right)  \left(  x_{i,j}\right)  $.

This completes our proof of $D\mid_{\mathcal{X}_{S}}=C_{S}\circ B_{S}\circ
A_{S}$. Now, from $\mathfrak{m}\in\mathcal{X}_{S}$, we obtain $D\left(
\mathfrak{m}\right)  =\underbrace{\left(  D\mid_{\mathcal{X}_{S}}\right)
}_{=C_{S}\circ B_{S}\circ A_{S}}\left(  \mathfrak{m}\right)  =\left(
C_{S}\circ 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{$D\left(  q\right)  =\left(  C\circ B\circ A\right)  \left(
q\right)  $ for pathless $q$}

\begin{corollary}
\label{cor.pathless.D}Let $q\in\mathcal{X}$ be pathless. Then, $D\left(
q\right)  =\left(  C\circ 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
$D$ and $C\circ B\circ A$ 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 $D\left(  \mathfrak{m}\right)  =\left(  C_{S}\circ
B_{S}\circ A_{S}\right)  \left(  \mathfrak{m}\right)  $. But the commutativity
of the diagram in Proposition \ref{prop.pathless.cd} shows that $C_{S}\circ
B_{S}\circ A_{S}=\left(  C\circ B\circ A\right)  \mid_{\mathcal{X}_{S}}$
(provided that we regard $C_{S}\circ B_{S}\circ A_{S}$ as a map to
$\mathcal{T}$). Hence,
\[
\underbrace{\left(  C_{S}\circ B_{S}\circ A_{S}\right)  }_{=\left(  C\circ
B\circ A\right)  \mid_{\mathcal{X}_{S}}}\left(  \mathfrak{m}\right)  =\left(
\left(  C\circ B\circ A\right)  \mid_{\mathcal{X}_{S}}\right)  \left(
\mathfrak{m}\right)  =\left(  C\circ B\circ A\right)  \left(  \mathfrak{m}%
\right)  .
\]
Thus, $D\left(  \mathfrak{m}\right)  =\left(  C_{S}\circ B_{S}\circ
A_{S}\right)  \left(  \mathfrak{m}\right)  =\left(  C\circ B\circ A\right)
\left(  \mathfrak{m}\right)  $. Since $q=\mathfrak{m}$, this rewrites as
$D\left(  q\right)  =\left(  C\circ 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} for $\beta=1$}

\begin{lemma}
\label{lem.t-red.unique.b=1.0}Let $p\in\mathcal{X}$ be a pathless polynomial
such that $p\in\mathcal{J}_{1}$. Then, $D\left(  p\right)  =0$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.t-red.unique.b=1.0}.]We have $A\left(  \underbrace{p}%
_{\in\mathcal{J}_{1}}\right)  \in A\left(  \mathcal{J}_{1}\right)  =0$ (by
Proposition \ref{prop.A.killsJ}); thus, $A\left(  p\right)  =0$. But Corollary
\ref{cor.pathless.D} (applied to $q=p$) yields
\[
D\left(  p\right)  =\left(  C\circ B\circ A\right)  \left(  p\right)  =\left(
C\circ B\right)  \left(  \underbrace{A\left(  p\right)  }_{=0}\right)
=\left(  C\circ B\right)  \left(  0\right)  =0
\]
(since the map $C\circ B$ is $\mathbf{k}$-linear). This proves Lemma
\ref{lem.t-red.unique.b=1.0}.
\end{proof}

We are now ready to prove Theorem~\ref{thm.t-red.unique} in the case $\beta=1$:

\begin{lemma}
\label{lem.t-red.unique.b=1}Let $p\in\mathcal{X}$. Consider any pathless
polynomial $q\in\mathcal{X}$ such that $p\equiv q\operatorname{mod}%
\mathcal{J}_{1}$. Then, $D\left(  q\right)  $ does not depend on the choice of
$q$ (but merely on the choice of $p$).
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.t-red.unique.b=1}.]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 $p\equiv q\operatorname{mod}\mathcal{J}_{1}$, then $D\left(  f\right)
=D\left(  g\right)  $.

So let $f$ and $g$ be two pathless polynomials $q\in\mathcal{X}$ such that
$p\equiv q\operatorname{mod}\mathcal{J}_{1}$. Thus, $p\equiv
f\operatorname{mod}\mathcal{J}_{1}$ and $p\equiv g\operatorname{mod}%
\mathcal{J}_{1}$. Hence, $f\equiv p\equiv g\operatorname{mod}\mathcal{J}_{1}$,
so that $f-g\in\mathcal{J}_{1}$. 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.b=1.0} (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 Lemma \ref{lem.t-red.unique.b=1}.
\end{proof}

\subsection{Proof of Theorem~\ref{thm.t-red.unique} for $\beta$ regular}

Let us now state a simple lemma:

\begin{lemma}
\label{lem.b=reg.inj}Let $\beta$ be a regular element of $\mathbf{k}$. Define
a $\mathbf{k}$-algebra homomorphism $G:\mathcal{T}^{\prime}\rightarrow
\mathcal{T}^{\prime}$ by%
\[
G\left(  t_{i}\right)  =\beta t_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }%
i\in\left[  n-1\right]  .
\]
This map $G$ is injective.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.b=reg.inj}.]Let $\mathfrak{T}$ denote the set of all
monomials in the indeterminates $t_{1},t_{2},\ldots,t_{n-1}$. It is easy to
see that%
\begin{equation}
G\left(  \mathfrak{t}\right)  =\beta^{\deg\mathfrak{t}}\mathfrak{t}%
\ \ \ \ \ \ \ \ \ \ \text{for each monomial }\mathfrak{t}\in\mathfrak{T}.
\label{pf.lem.b=reg.inj.G(t)=}%
\end{equation}


Now, let $f\in\operatorname*{Ker}G$. Then, $f\in\mathcal{T}^{\prime}$. Hence,
we can write $f$ in the form $f=\sum_{\mathfrak{t}\in\mathfrak{T}}%
\lambda_{\mathfrak{t}}\mathfrak{t}$ for some family $\left(  \lambda
_{\mathfrak{t}}\right)  _{\mathfrak{t}\in\mathfrak{T}}\in\mathbf{k}%
^{\mathfrak{T}}$ (since every polynomial in $\mathcal{T}^{\prime}$ is a
$\mathbf{k}$-linear combination of the monomials $\mathfrak{t}\in\mathfrak{T}%
$). Consider this family $\left(  \lambda_{\mathfrak{t}}\right)
_{\mathfrak{t}\in\mathfrak{T}}$.

From $f\in\operatorname*{Ker}G$, we obtain $G\left(  f\right)  =0$. Thus,%
\begin{align*}
0  &  =G\left(  f\right)  =G\left(  \sum_{\mathfrak{t}\in\mathfrak{T}}%
\lambda_{\mathfrak{t}}\mathfrak{t}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }f=\sum_{\mathfrak{t}\in\mathfrak{T}}\lambda_{\mathfrak{t}%
}\mathfrak{t}\right) \\
&  =\sum_{\mathfrak{t}\in\mathfrak{T}}\lambda_{\mathfrak{t}}%
\underbrace{G\left(  \mathfrak{t}\right)  }_{\substack{=\beta^{\deg
\mathfrak{t}}\mathfrak{t}\\\text{(by (\ref{pf.lem.b=reg.inj.G(t)=}))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }G\text{ is }\mathbf{k}%
\text{-linear}\right) \\
&  =\sum_{\mathfrak{t}\in\mathfrak{T}}\lambda_{\mathfrak{t}}\beta
^{\deg\mathfrak{t}}\mathfrak{t}=\sum_{\mathfrak{t}\in\mathfrak{T}}\beta
^{\deg\mathfrak{t}}\lambda_{\mathfrak{t}}\mathfrak{t}.
\end{align*}
In other words, $\sum_{\mathfrak{t}\in\mathfrak{T}}\beta^{\deg\mathfrak{t}%
}\lambda_{\mathfrak{t}}\mathfrak{t}=0$. Since the monomials $\mathfrak{t}%
\in\mathfrak{T}$ in $\mathcal{T}^{\prime}$ are $\mathbf{k}$-linearly
independent, we thus obtain
\begin{equation}
\beta^{\deg\mathfrak{t}}\lambda_{\mathfrak{t}}=0\ \ \ \ \ \ \ \ \ \ \text{for
each }\mathfrak{t}\in\mathfrak{T}. \label{pf.lem.b=reg.inj.2}%
\end{equation}


Thus, we can easily obtain $\lambda_{\mathfrak{t}}=0$ for each $\mathfrak{t}%
\in\mathfrak{T}$\ \ \ \ \footnote{\textit{Proof.} Let $\mathfrak{t}%
\in\mathfrak{T}$. From (\ref{pf.lem.b=reg.inj.2}), we obtain $\beta
^{\deg\mathfrak{t}}\lambda_{\mathfrak{t}}=0$.
\par
Recall that the element $\beta$ of $\mathbf{k}$ is regular. Hence, its power
$\beta^{\deg\mathfrak{t}}$ is regular as well. Thus, from $\beta
^{\deg\mathfrak{t}}\lambda_{\mathfrak{t}}=0$, we obtain $\lambda
_{\mathfrak{t}}=0$. Qed.}. Hence, $\sum_{\mathfrak{t}\in\mathfrak{T}%
}\underbrace{\lambda_{\mathfrak{t}}}_{=0}\mathfrak{t}=\sum_{\mathfrak{t}%
\in\mathfrak{T}}0\mathfrak{t}=0$, so that $f=\sum_{\mathfrak{t}\in
\mathfrak{T}}\lambda_{\mathfrak{t}}\mathfrak{t}=0$.

Now, forget that we fixed $f$. We thus have shown that $f=0$ for each
$f\in\operatorname*{Ker}G$. In other words, $\operatorname*{Ker}G=0$. Hence,
the map $G$ is injective (since $G$ is $\mathbf{k}$-linear). This proves Lemma
\ref{lem.b=reg.inj}.
\end{proof}

The following lemma generalizes Lemma \ref{lem.t-red.unique.b=1.0}:

\begin{lemma}
\label{lem.t-red.unique.b=reg-rest}Let $\beta$ be a regular element of
$\mathbf{k}$. Let $p\in\mathcal{X}$ be a pathless polynomial such that
$p\in\mathcal{J}_{\beta}$. Then, $D\left(  p\right)  =0$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.t-red.unique.b=reg-rest}.]Define a $\mathbf{k}%
$-algebra homomorphism $F:\mathcal{X}\rightarrow\mathcal{X}$ by%
\[
F\left(  x_{i,j}\right)  =\beta x_{i,j}\ \ \ \ \ \ \ \ \ \ \text{for all
}\left(  i,j\right)  \in\left[  n\right]  ^{2}\text{ satisfying }i<j.
\]


Define a $\mathbf{k}$-algebra homomorphism $G:\mathcal{T}^{\prime}%
\rightarrow\mathcal{T}^{\prime}$ by%
\[
G\left(  t_{i}\right)  =\beta t_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }%
i\in\left[  n-1\right]  .
\]
Lemma \ref{lem.b=reg.inj} shows that this map $G$ is injective. Hence,
$\operatorname*{Ker}G=0$.

A trivial computation shows that the diagram%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%\calX\ar[r]^{D} \ar[d]_{F} & \calT^\prime\ar[d]^{G} \\
%\calX\ar[r]_{D} & \calT^{\prime}
%}}}%
%BeginExpansion
\xymatrix{
\calX\ar[r]^{D} \ar[d]_{F} & \calT^\prime\ar[d]^{G} \\
\calX\ar[r]_{D} & \calT^{\prime}
}%
%EndExpansion
\]
is commutative. In other words, $D\circ F=G\circ D$.

It is easy to check that $F\left(  \mathcal{J}_{\beta}\right)  \subseteq
\mathcal{J}_{1}$. (Indeed, the $\mathbf{k}$-algebra homomorphism $F$ sends
each generator $x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  $
of $\mathcal{J}_{\beta}$ to%
\[
\left(  \beta x_{i,j}\right)  \left(  \beta x_{j,k}\right)  -\left(  \beta
x_{i,k}\right)  \left(  \beta x_{i,j}+\beta x_{j,k}+\beta\right)  =\beta
^{2}\underbrace{\left(  x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}%
+1\right)  \right)  }_{\substack{\in\mathcal{J}_{1}\\\text{(since the ideal
}\mathcal{J}_{1}\text{ is generated}\\\text{by polynomials like this)}}%
}\in\mathcal{J}_{1}.
\]
)

It is easy to see (using the definition of $F$) that $F\left(  \mathfrak{m}%
\right)  =\beta^{\deg\mathfrak{m}}\mathfrak{m}$ for each monomial
$\mathfrak{m}\in\mathfrak{M}$. Hence, the map $F$ sends each pathless monomial
$\mathfrak{m}\in\mathfrak{M}$ to a scalar multiple of a pathless monomial
(namely, to $\beta^{\deg\mathfrak{m}}\mathfrak{m}$). Thus, the map $F$ sends
pathless polynomials to pathless polynomials. Therefore, $F\left(  p\right)
\in\mathcal{X}$ is pathless (since $p\in\mathcal{X}$ is pathless). Also,
$F\left(  \underbrace{p}_{\in\mathcal{J}_{\beta}}\right)  \in F\left(
\mathcal{J}_{\beta}\right)  \subseteq\mathcal{J}_{1}$. Thus, Lemma
\ref{lem.t-red.unique.b=1.0} (applied to $F\left(  p\right)  $ instead of $p$)
shows that $D\left(  F\left(  p\right)  \right)  =0$. Since%
\[
D\left(  F\left(  p\right)  \right)  =\underbrace{\left(  D\circ F\right)
}_{=G\circ D}\left(  p\right)  =\left(  G\circ D\right)  \left(  p\right)
=G\left(  D\left(  p\right)  \right)  ,
\]
this rewrites as $G\left(  D\left(  p\right)  \right)  =0$. Hence, $D\left(
p\right)  \in\operatorname*{Ker}G=0$, so that $D\left(  p\right)  =0$. This
proves Lemma \ref{lem.t-red.unique.b=reg-rest}.
\end{proof}

As a consequence, we obtain a further particular case of
Theorem~\ref{thm.t-red.unique}:

\begin{proposition}
\label{prop.t-red.unique.b=reg}Let $\beta$ be a regular element of
$\mathbf{k}$. Let $p\in\mathcal{X}$. Consider any pathless polynomial
$q\in\mathcal{X}$ such that $p\equiv q\operatorname{mod}\mathcal{J}_{\beta}$.
Then, $D\left(  q\right)  $ does not depend on the choice of $q$ (but merely
on the choice of $\beta$ and $p$).
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.t-red.unique.b=reg}.]Proposition
\ref{prop.t-red.unique.b=reg} follows from Lemma
\ref{lem.t-red.unique.b=reg-rest} in the same way as we derived Lemma
\ref{lem.t-red.unique.b=1} from Lemma \ref{lem.t-red.unique.b=1.0} (with the
only difference that $\mathcal{J}_{1}$ is replaced by $\mathcal{J}_{\beta}$
throughout the proof).
\end{proof}

\subsection{Proof of Theorem~\ref{thm.t-red.unique} for all $\beta$}

Let us now prepare for the final steps of our way to
Theorem~\ref{thm.t-red.unique}.

\begin{definition}
\label{def.m}Fix any element $\beta\in\mathbf{k}$.

\textbf{(a)} Let $\mathbf{m}$ be the polynomial ring $\mathbf{k}\left[
b\right]  $ in one indeterminate $b$ over $\mathbf{k}$.

\textbf{(b)} Let $\pi:\mathbf{m}\rightarrow\mathbf{k}$ be the unique
$\mathbf{k}$-algebra homomorphism from $\mathbf{k}\left[  b\right]  $ to
$\mathbf{k}$ that sends $b$ to $\beta$. (This is well-defined by the universal
property of the polynomial ring $\mathbf{k}\left[  b\right]  =\mathbf{m}$.) In
other words, $\pi$ is the evaluation map that sends each polynomial
$f\in\mathbf{k}\left[  b\right]  $ to its evaluation $f\left(  \beta\right)  $
at $\beta$. It is easy to see that $\operatorname*{Ker}\pi=\left(
b-\beta\right)  \mathbf{m}$.

\textbf{(c)} Let $\mathcal{X}^{\left[  \mathbf{m}\right]  }$ be the polynomial
ring $\mathbf{m}\left[  x_{i,j}\ \mid\ \left(  i,j\right)  \in\left[
n\right]  ^{2}\text{ satisfying }i<j\right]  $. In other words, $\mathcal{X}%
^{\left[  \mathbf{m}\right]  }$ is the $\mathbf{m}$-algebra that is defined in
the same way as $\mathcal{X}$ but using the base ring $\mathbf{m}$ instead of
$\mathbf{k}$.

\textbf{(d)} Let $\mathcal{T}^{\prime\left[  \mathbf{m}\right]  }$ be the
polynomial ring $\mathbf{m}\left[  t_{1},t_{2},\ldots,t_{n-1}\right]  $. In
other words, $\mathcal{T}^{\prime\left[  \mathbf{m}\right]  }$ is the
$\mathbf{m}$-algebra that is defined in the same way as $\mathcal{T}^{\prime}$
but using the base ring $\mathbf{m}$ instead of $\mathbf{k}$.

\textbf{(e)} Let $D^{\left[  \mathbf{m}\right]  }:\mathcal{X}^{\left[
\mathbf{m}\right]  }\rightarrow\mathcal{T}^{\prime\left[  \mathbf{m}\right]
}$ be the $\mathbf{m}$-algebra homomorphism defined in the same way as the
$\mathbf{k}$-algebra homomorphism $D:\mathcal{X}\rightarrow\mathcal{T}%
^{\prime}$ but using the base ring $\mathbf{m}$ instead of $\mathbf{k}$.

\textbf{(f)} The $\mathbf{k}$-algebra homomorphism $\pi:\mathbf{m}%
\rightarrow\mathbf{k}$ induces a $\mathbf{k}$-algebra homomorphism
\[
\mathbf{m}\left[  x_{i,j}\ \mid\ \left(  i,j\right)  \in\left[  n\right]
^{2}\text{ satisfying }i<j\right]  \rightarrow\mathbf{k}\left[  x_{i,j}%
\ \mid\ \left(  i,j\right)  \in\left[  n\right]  ^{2}\text{ satisfying
}i<j\right]
\]
which sends each $f\in\mathbf{m}$ to $\pi\left(  f\right)  $ while sending
each indeterminate $x_{i,j}$ to the corresponding indeterminate $x_{i,j}$.
This latter homomorphism will be denoted by $\pi^{\mathcal{X}}$. Notice that
$\pi^{\mathcal{X}}$ is a $\mathbf{k}$-algebra homomorphism from $\mathcal{X}%
^{\left[  \mathbf{m}\right]  }$ to $\mathcal{X}$ (since $\mathbf{m}\left[
x_{i,j}\ \mid\ \left(  i,j\right)  \in\left[  n\right]  ^{2}\text{ satisfying
}i<j\right]  =\mathcal{X}^{\left[  \mathbf{m}\right]  }$ and $\mathbf{k}%
\left[  x_{i,j}\ \mid\ \left(  i,j\right)  \in\left[  n\right]  ^{2}\text{
satisfying }i<j\right]  =\mathcal{X}$).

Explicitly speaking, the homomorphism $\pi^{\mathcal{X}}$ acts on a polynomial
$f\in\mathcal{X}^{\left[  \mathbf{m}\right]  }$ by applying the map $\pi$ to
each of its coefficients.

\textbf{(g)} The $\mathbf{k}$-algebra homomorphism $\pi:\mathbf{m}%
\rightarrow\mathbf{k}$ induces a $\mathbf{k}$-algebra homomorphism
\[
\mathbf{m}\left[  t_{1},t_{2},\ldots,t_{n-1}\right]  \rightarrow
\mathbf{k}\left[  t_{1},t_{2},\ldots,t_{n-1}\right]
\]
which sends each $f\in\mathbf{m}$ to $\pi\left(  f\right)  $ while sending
each indeterminate $t_{i}$ to the corresponding indeterminate $t_{i}$. This
latter homomorphism will be denoted by $\pi^{\mathcal{T}^{\prime}}$. Notice
that $\pi^{\mathcal{T}^{\prime}}$ is a $\mathbf{k}$-algebra homomorphism from
$\mathcal{T}^{\prime\left[  \mathbf{m}\right]  }$ to $\mathcal{T}^{\prime}$
(since $\mathbf{m}\left[  t_{1},t_{2},\ldots,t_{n-1}\right]  =\mathcal{T}%
^{\prime\left[  \mathbf{m}\right]  }$ and $\mathbf{k}\left[  t_{1}%
,t_{2},\ldots,t_{n-1}\right]  =\mathcal{T}^{\prime}$).

Explicitly speaking, the homomorphism $\pi^{\mathcal{T}^{\prime}}$ acts on a
polynomial $f\in\mathcal{T}^{\prime\left[  \mathbf{m}\right]  }$ by applying
the map $\pi$ to each of its coefficients.

\textbf{(h)} Let $\mathcal{J}_{b}^{\left[  \mathbf{m}\right]  }$ be the ideal
of $\mathcal{X}^{\left[  \mathbf{m}\right]  }$ defined in the same way as
$\mathcal{J}_{\beta}$ but using the base ring $\mathbf{m}$ and the element
$b\in\mathbf{m}$ instead of the base ring $\mathbf{k}$ and the element
$\beta\in\mathbf{k}$.
\end{definition}

We notice some simple facts:

\begin{lemma}
\label{lem.m.basics}\textbf{(a)} The element $b$ of $\mathbf{m}$ is regular.

\textbf{(b)} We have $\mathcal{J}_{\beta}=\pi^{\mathcal{X}}\left(
\mathcal{J}_{b}^{\left[  \mathbf{m}\right]  }\right)  $.

\textbf{(c)} For every pathless polynomial $f\in\mathcal{X}$, there exists a
pathless polynomial $g\in\mathcal{X}^{\left[  \mathbf{m}\right]  }$ satisfying
$f=\pi^{\mathcal{X}}\left(  g\right)  $.

\textbf{(d)} We have $\operatorname*{Ker}\left(  \pi^{\mathcal{X}}\right)
=\left(  b-\beta\right)  \mathcal{X}^{\left[  \mathbf{m}\right]  }$.

\textbf{(e)} We have $\pi^{\mathcal{T}^{\prime}}\circ D^{\left[
\mathbf{m}\right]  }=D\circ\pi^{\mathcal{X}}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.m.basics}.]\textbf{(a)} This is clear, since $b$ is
the indeterminate in the polynomial ring $\mathbf{k}\left[  b\right]  $.

\textbf{(b)} The $\mathbf{k}$-algebra homomorphism $\pi:\mathbf{m}%
\rightarrow\mathbf{k}$ is surjective (since the canonical inclusion map
$\iota:\mathbf{k}\rightarrow\mathbf{m}$ satisfies $\pi\circ\iota
=\operatorname*{id}$). Hence, the $\mathbf{k}$-algebra homomorphism
\newline$\pi^{\mathcal{X}}:\mathbf{m}\left[  x_{i,j}\ \mid\ \left(
i,j\right)  \in\left[  n\right]  ^{2}\text{ satisfying }i<j\right]
\rightarrow\mathbf{k}\left[  x_{i,j}\ \mid\ \left(  i,j\right)  \in\left[
n\right]  ^{2}\text{ satisfying }i<j\right]  $ induced by $\pi$ is surjective
as well. Since $\mathbf{m}\left[  x_{i,j}\ \mid\ \left(  i,j\right)
\in\left[  n\right]  ^{2}\text{ satisfying }i<j\right]  =\mathcal{X}^{\left[
\mathbf{m}\right]  }$ and $\mathbf{k}\left[  x_{i,j}\ \mid\ \left(
i,j\right)  \in\left[  n\right]  ^{2}\text{ satisfying }i<j\right]
=\mathcal{X}$, this rewrites as follows: The $\mathbf{k}$-algebra homomorphism
$\pi^{\mathcal{X}}:\mathcal{X}^{\left[  \mathbf{m}\right]  }\rightarrow
\mathcal{X}$ is surjective\footnote{In fact, slightly more holds: Let
$\iota:\mathbf{k}\rightarrow\mathbf{m}$ be the canonical inclusion map. Let
$\iota^{\mathcal{X}}$ denote the $\mathbf{k}$-algebra homomorphism
$\mathbf{k}\left[  x_{i,j}\ \mid\ \left(  i,j\right)  \in\left[  n\right]
^{2}\text{ satisfying }i<j\right]  \rightarrow\mathbf{m}\left[  x_{i,j}%
\ \mid\ \left(  i,j\right)  \in\left[  n\right]  ^{2}\text{ satisfying
}i<j\right]  $ induced by this $\mathbf{k}$-algebra homomorphism $\iota$.
Then, from $\pi\circ\iota=\operatorname*{id}$, we easily obtain $\pi
^{\mathcal{X}}\circ\iota^{\mathcal{X}}=\operatorname*{id}$ (by functoriality).
Hence, $\pi^{\mathcal{X}}$ is surjective.}. Hence, $\pi^{\mathcal{X}}\left(
\mathcal{X}^{\left[  \mathbf{m}\right]  }\right)  =\mathcal{X}$.

Let $J_{\beta}$ be the $\mathbb{Z}$-submodule of $\mathcal{X}$ spanned by all
elements of the form $x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}%
+\beta\right)  $ for $\left(  i,j,k\right)  \in\left[  n\right]  ^{3}$
satisfying $i<j<k$. Then, the ideal $\mathcal{J}_{\beta}$ of $\mathcal{X}$ is
generated by this $\mathbb{Z}$-submodule $J_{\beta}$ (since $\mathcal{J}%
_{\beta}$ is generated by all elements of this form). Thus, $\mathcal{J}%
_{\beta}=\mathcal{X}J_{\beta}$ (a product of two $\mathbb{Z}$-submodules of
$\mathcal{X}$).

Let $J_{b}$ be the $\mathbb{Z}$-submodule of $\mathcal{X}^{\left[
\mathbf{m}\right]  }$ spanned by all elements of the form $x_{i,j}%
x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+b\right)  $ for $\left(  i,j,k\right)
\in\left[  n\right]  ^{3}$ satisfying $i<j<k$. Then, the ideal $\mathcal{J}%
_{b}^{\left[  \mathbf{m}\right]  }$ of $\mathcal{X}^{\left[  \mathbf{m}%
\right]  }$ is generated by this $\mathbb{Z}$-submodule $J_{b}$ (since
$\mathcal{J}_{b}^{\left[  \mathbf{m}\right]  }$ is generated by all elements
of this form). Thus, $\mathcal{J}_{b}^{\left[  \mathbf{m}\right]
}=\mathcal{X}^{\left[  \mathbf{m}\right]  }J_{b}$ (a product of two
$\mathbb{Z}$-submodules of $\mathcal{X}^{\left[  \mathbf{m}\right]  }$).

For each $\left(  i,j,k\right)  \in\left[  n\right]  ^{3}$ satisfying $i<j<k$,
we have%
\begin{align*}
&  \pi^{\mathcal{X}}\left(  x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}%
+x_{j,k}+b\right)  \right) \\
&  =x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+\underbrace{\pi\left(
b\right)  }_{=\beta}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of the map }\pi^{\mathcal{X}}\right) \\
&  =x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  .
\end{align*}
Thus, the map $\pi^{\mathcal{X}}$ sends each of the designated generators of
the $\mathbb{Z}$-module $J_{b}$ to the corresponding generator of the
$\mathbb{Z}$-module $J_{\beta}$. Therefore, $\pi^{\mathcal{X}}\left(
J_{b}\right)  =J_{\beta}$ (since the map $\pi^{\mathcal{X}}$ is $\mathbb{Z}$-linear).

Now,%
\begin{align*}
\pi^{\mathcal{X}}\left(  \underbrace{\mathcal{J}_{b}^{\left[  \mathbf{m}%
\right]  }}_{=\mathcal{X}^{\left[  \mathbf{m}\right]  }J_{b}}\right)   &
=\pi^{\mathcal{X}}\left(  \mathcal{X}^{\left[  \mathbf{m}\right]  }%
J_{b}\right) \\
&  =\underbrace{\pi^{\mathcal{X}}\left(  \mathcal{X}^{\left[  \mathbf{m}%
\right]  }\right)  }_{=\mathcal{X}}\underbrace{\pi^{\mathcal{X}}\left(
J_{b}\right)  }_{=J_{\beta}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\pi^{\mathcal{X}}\text{ is a ring homomorphism}\right) \\
&  =\mathcal{X}J_{\beta}=\mathcal{J}_{\beta}.
\end{align*}
This proves Lemma \ref{lem.m.basics} \textbf{(b)}.

\textbf{(c)} For the purpose of this proof, we shall regard $\mathbf{k}$ as a
$\mathbf{k}$-subalgebra of $\mathbf{m}$ via the canonical inclusion
$\mathbf{k}\rightarrow\mathbf{m}$ (sending each $\lambda\in\mathbf{k}$ to the
constant polynomial $\lambda\in\mathbf{m}$). Then, $\mathcal{X}$ also becomes
a $\mathbf{k}$-subalgebra of $\mathcal{X}^{\left[  \mathbf{m}\right]  }$.
Moreover, it is fairly clear that $\pi^{\mathcal{X}}\left(  f\right)  =f$ for
each $f\in\mathcal{X}$ (because $\pi\left(  \lambda\right)  =\lambda$ for each
$\lambda\in\mathbf{k}$\textbf{).}

Now, let $f\in\mathcal{X}$ be a pathless polynomial. Then, $f$, when
considered as a polynomial in $\mathcal{X}^{\left[  \mathbf{m}\right]  }$, is
still a pathless polynomial. Moreover, $\pi^{\mathcal{X}}\left(  f\right)
=f$. Thus, there exists a pathless polynomial $g\in\mathcal{X}^{\left[
\mathbf{m}\right]  }$ satisfying $f=\pi^{\mathcal{X}}\left(  g\right)  $
(namely, $g=f$). This proves Lemma \ref{lem.m.basics} \textbf{(c)}.

\textbf{(d)} Let $p\in\operatorname*{Ker}\left(  \pi^{\mathcal{X}}\right)  $.
Thus, $p\in\mathcal{X}^{\left[  \mathbf{m}\right]  }$. Write $p$ in the form
$p=\sum_{\mathfrak{m}\in\mathfrak{M}}f_{\mathfrak{m}}\mathfrak{m}$ for some
family $\left(  f_{\mathfrak{m}}\right)  _{\mathfrak{m}\in\mathfrak{M}}%
\in\mathbf{m}^{\mathfrak{M}}$. (This is possible, since every polynomial in
$\mathcal{X}^{\left[  \mathbf{m}\right]  }$ is a unique $\mathbf{m}$-linear
combination of the monomials $\mathfrak{m}\in\mathfrak{M}$.) Applying the map
$\pi^{\mathcal{X}}$ to the equality $p=\sum_{\mathfrak{m}\in\mathfrak{M}%
}f_{\mathfrak{m}}\mathfrak{m}$, we obtain
\begin{equation}
\pi^{\mathcal{X}}\left(  p\right)  =\pi^{\mathcal{X}}\left(  \sum
_{\mathfrak{m}\in\mathfrak{M}}f_{\mathfrak{m}}\mathfrak{m}\right)
=\sum_{\mathfrak{m}\in\mathfrak{M}}\pi\left(  f_{\mathfrak{m}}\right)
\mathfrak{m} \label{pf.lem.m.basics.d.2}%
\end{equation}
(by the definition of $\pi^{\mathcal{X}}$).

Recall that $p\in\operatorname*{Ker}\left(  \pi^{\mathcal{X}}\right)  $. Thus,
$\pi^{\mathcal{X}}\left(  p\right)  =0$. In view of (\ref{pf.lem.m.basics.d.2}%
), this rewrites as $\sum_{\mathfrak{m}\in\mathfrak{M}}\pi\left(
f_{\mathfrak{m}}\right)  \mathfrak{m}=0$. Since the monomials $\mathfrak{m}%
\in\mathfrak{M}$ in $\mathcal{X}$ are $\mathbf{k}$-linearly independent, this
shows that $\pi\left(  f_{\mathfrak{m}}\right)  =0$ for each $\mathfrak{m}%
\in\mathfrak{M}$. Thus, for each $\mathfrak{m}\in\mathfrak{M}$, we have
$f_{\mathfrak{m}}\in\operatorname*{Ker}\pi=\left(  b-\beta\right)  \mathbf{m}%
$. Hence,%
\[
p=\sum_{\mathfrak{m}\in\mathfrak{M}}\underbrace{f_{\mathfrak{m}}}_{\in\left(
b-\beta\right)  \mathbf{m}}\mathfrak{m}\in\sum_{\mathfrak{m}\in\mathfrak{M}%
}\left(  b-\beta\right)  \underbrace{\mathbf{m}\mathfrak{m}}_{\subseteq
\mathcal{X}^{\left[  \mathbf{m}\right]  }}\subseteq\left(  b-\beta\right)
\mathcal{X}^{\left[  \mathbf{m}\right]  }.
\]


Now, forget that we fixed $p$. We thus have shown that $p\in\left(
b-\beta\right)  \mathcal{X}^{\left[  \mathbf{m}\right]  }$ for each
$p\in\operatorname*{Ker}\left(  \pi^{\mathcal{X}}\right)  $. In other words,
$\operatorname*{Ker}\left(  \pi^{\mathcal{X}}\right)  \subseteq\left(
b-\beta\right)  \mathcal{X}^{\left[  \mathbf{m}\right]  }$.

Observe that $b-\beta\in\left(  b-\beta\right)  \mathbf{m}=\operatorname*{Ker}%
\pi$, and thus $\pi\left(  b-\beta\right)  =0$.

On the other hand, let us regard $\mathbf{m}$ as a subring of the polynomial
ring $\mathcal{X}^{\left[  \mathbf{m}\right]  }$. Then, $\pi^{\mathcal{X}%
}\left(  f\right)  $ is well-defined for each $f\in\mathbf{m}$ (since
$f\in\mathbf{m}\subseteq\mathcal{X}^{\left[  \mathbf{m}\right]  }$).
Furthermore, each $f\in\mathbf{m}$ satisfies $\pi^{\mathcal{X}}\left(
f\right)  =\pi\left(  f\right)  $ (by the definition of $\pi^{\mathcal{X}}$).
Applying this to $f=b-\beta$, we obtain $\pi^{\mathcal{X}}\left(
b-\beta\right)  =\pi\left(  b-\beta\right)  =0$.

But $\pi^{\mathcal{X}}$ is a $\mathbf{k}$-algebra homomorphism. Hence,
\[
\pi^{\mathcal{X}}\left(  \left(  b-\beta\right)  \mathcal{X}^{\left[
\mathbf{m}\right]  }\right)  =\underbrace{\pi^{\mathcal{X}}\left(
b-\beta\right)  }_{=0}\pi^{\mathcal{X}}\left(  \mathcal{X}^{\left[
\mathbf{m}\right]  }\right)  =0\pi^{\mathcal{X}}\left(  \mathcal{X}^{\left[
\mathbf{m}\right]  }\right)  =0.
\]
Hence, $\left(  b-\beta\right)  \mathcal{X}^{\left[  \mathbf{m}\right]
}\subseteq\operatorname*{Ker}\left(  \pi^{\mathcal{X}}\right)  $. Combining
this with $\operatorname*{Ker}\left(  \pi^{\mathcal{X}}\right)  \subseteq
\left(  b-\beta\right)  \mathcal{X}^{\left[  \mathbf{m}\right]  }$, we obtain
$\operatorname*{Ker}\left(  \pi^{\mathcal{X}}\right)  =\left(  b-\beta\right)
\mathcal{X}^{\left[  \mathbf{m}\right]  }$. This proves Lemma
\ref{lem.m.basics} \textbf{(d)}.

\textbf{(e)} The definition of the map $D$ was canonical with respect to the
base ring. Thus, the map $D^{\left[  \mathbf{m}\right]  }$ (defined in the
same way as $D$, but using the base ring $\mathbf{m}$ instead of $\mathbf{k}$)
and the map $D$ fit into a commutative diagram%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%\calX^{\left[\mathbf{m}\right]} \ar[d]_{\pi^{\calX}} \ar[r]^{D^{\left
%[\mathbf{m}\right]}} & \calT^{\prime\left[\mathbf{m}\right]} \ar
%[d]^{\pi^{\calT^\prime}} \\
%\calX\ar[r]_{D} & \calT^\prime}}}%
%BeginExpansion
\xymatrix{
\calX^{\left[\mathbf{m}\right]} \ar[d]_{\pi^{\calX}} \ar[r]^{D^{\left
[\mathbf{m}\right]}} & \calT^{\prime\left[\mathbf{m}\right]} \ar
[d]^{\pi^{\calT^\prime}} \\
\calX\ar[r]_{D} & \calT^\prime}%
%EndExpansion
.
\]
The commutativity of this diagram shows that $\pi^{\mathcal{T}^{\prime}}\circ
D^{\left[  \mathbf{m}\right]  }=D\circ\pi^{\mathcal{X}}$. This proves Lemma
\ref{lem.m.basics} \textbf{(e)}.
\end{proof}

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

\begin{proof}
[Proof of Lemma \ref{lem.t-red.unique.rest}.]Lemma \ref{lem.m.basics}
\textbf{(c)} (applied to $f=p$) shows that there exists a pathless polynomial
$g\in\mathcal{X}^{\left[  \mathbf{m}\right]  }$ satisfying $p=\pi
^{\mathcal{X}}\left(  g\right)  $. Fix such a $g$, and denote it by $v$. Thus,
$v\in\mathcal{X}^{\left[  \mathbf{m}\right]  }$ is a pathless polynomial
satisfying $p=\pi^{\mathcal{X}}\left(  v\right)  $.

Lemma \ref{lem.m.basics} \textbf{(b)} shows that $\mathcal{J}_{\beta}%
=\pi^{\mathcal{X}}\left(  \mathcal{J}_{b}^{\left[  \mathbf{m}\right]
}\right)  $. Now, $p\in\mathcal{J}_{\beta}=\pi^{\mathcal{X}}\left(
\mathcal{J}_{b}^{\left[  \mathbf{m}\right]  }\right)  $. Hence, there exists
some $u\in\mathcal{J}_{b}^{\left[  \mathbf{m}\right]  }$ such that
$p=\pi^{\mathcal{X}}\left(  u\right)  $. Consider this $u$.

The map $\pi^{\mathcal{X}}$ is a $\mathbf{k}$-algebra homomorphism. Thus,
$\pi^{\mathcal{X}}\left(  u-v\right)  =\underbrace{\pi^{\mathcal{X}}\left(
u\right)  }_{=p}-\underbrace{\pi^{\mathcal{X}}\left(  v\right)  }_{=p}=p-p=0$.
Hence, $u-v\in\operatorname*{Ker}\left(  \pi^{\mathcal{X}}\right)  =\left(
b-\beta\right)  \mathcal{X}^{\left[  \mathbf{m}\right]  }$ (by Lemma
\ref{lem.m.basics} \textbf{(d)}). In other words, there exists some
$w\in\mathcal{X}^{\left[  \mathbf{m}\right]  }$ such that $u-v=\left(
b-\beta\right)  w$. Consider this $w$.

From $u-v=\left(  b-\beta\right)  w$, we obtain $u=v+\left(  b-\beta\right)
w$.

Proposition \ref{prop.path-red.span} (applied to $\mathbf{m}$, $b$,
$\mathcal{X}^{\left[  \mathbf{m}\right]  }$, $\mathcal{J}_{b}^{\left[
\mathbf{m}\right]  }$ and $w$ instead of $\mathbf{k}$, $\beta$, $\mathcal{X}$,
$\mathcal{J}_{\beta}$ and $p$) shows that there exists a pathless polynomial
$q\in\mathcal{X}^{\left[  \mathbf{m}\right]  }$ such that $w\equiv
q\operatorname{mod}\mathcal{J}_{b}^{\left[  \mathbf{m}\right]  }$. Consider
this $q$. We have%
\[
v+\left(  b-\beta\right)  \underbrace{q}_{\equiv w\operatorname{mod}%
\mathcal{J}_{b}^{\left[  \mathbf{m}\right]  }}\equiv v+\left(  b-\beta\right)
w=u\equiv0\operatorname{mod}\mathcal{J}_{b}^{\left[  \mathbf{m}\right]  }%
\]
(since $u\in\mathcal{J}_{b}^{\left[  \mathbf{m}\right]  }$). In other words,
$v+\left(  b-\beta\right)  q\in\mathcal{J}_{b}^{\left[  \mathbf{m}\right]  }$.

The element $b$ of $\mathbf{m}$ is regular (by Lemma \ref{lem.m.basics}
\textbf{(a)}). The polynomial $v+\left(  b-\beta\right)  q\in\mathcal{X}%
^{\left[  \mathbf{m}\right]  }$ is pathless (since it is an $\mathbf{m}%
$-linear combination of the two pathless polynomials $v$ and $q$) and
satisfies $v+\left(  b-\beta\right)  q\in\mathcal{J}_{b}^{\left[
\mathbf{m}\right]  }$ (as we have shown). Hence, Lemma
\ref{lem.t-red.unique.b=reg-rest} (applied to $\mathbf{m}$, $b$,
$\mathcal{X}^{\left[  \mathbf{m}\right]  }$, $\mathcal{J}_{b}^{\left[
\mathbf{m}\right]  }$, $\mathcal{T}^{\prime\left[  \mathbf{m}\right]  }$,
$D^{\left[  \mathbf{m}\right]  }$ and $v+\left(  b-\beta\right)  q$ instead of
$\mathbf{k}$, $\beta$, $\mathcal{X}$, $\mathcal{J}_{\beta}$, $\mathcal{T}%
^{\prime}$, $D$ and $p$) shows that $D^{\left[  \mathbf{m}\right]  }\left(
v+\left(  b-\beta\right)  q\right)  =0$.

But Lemma \ref{lem.m.basics} \textbf{(e)} yields $\pi^{\mathcal{T}^{\prime}%
}\circ D^{\left[  \mathbf{m}\right]  }=D\circ\pi^{\mathcal{X}}$. Hence,%
\begin{align*}
\underbrace{\left(  D\circ\pi^{\mathcal{X}}\right)  }_{=\pi^{\mathcal{T}%
^{\prime}}\circ D^{\left[  \mathbf{m}\right]  }}\left(  v+\left(
b-\beta\right)  q\right)   &  =\left(  \pi^{\mathcal{T}^{\prime}}\circ
D^{\left[  \mathbf{m}\right]  }\right)  \left(  v+\left(  b-\beta\right)
q\right) \\
&  =\pi^{\mathcal{T}^{\prime}}\left(  \underbrace{D^{\left[  \mathbf{m}%
\right]  }\left(  v+\left(  b-\beta\right)  q\right)  }_{=0}\right)
=\pi^{\mathcal{T}^{\prime}}\left(  0\right)  =0.
\end{align*}
Thus,
\begin{align}
0  &  =\left(  D\circ\pi^{\mathcal{X}}\right)  \left(  v+\left(
b-\beta\right)  q\right)  =D\left(  \underbrace{\pi^{\mathcal{X}}\left(
v+\left(  b-\beta\right)  q\right)  }_{\substack{=\pi^{\mathcal{X}}\left(
v\right)  +\pi^{\mathcal{X}}\left(  \left(  b-\beta\right)  q\right)
\\\text{(since }\pi^{\mathcal{X}}\text{ is a }\mathbf{k}\text{-algebra
homomorphism)}}}\right) \nonumber\\
&  =D\left(  \pi^{\mathcal{X}}\left(  v\right)  +\pi^{\mathcal{X}}\left(
\left(  b-\beta\right)  q\right)  \right)  .
\label{pf.lem.t-red.unique.rest.5}%
\end{align}


But $\left(  b-\beta\right)  \underbrace{q}_{\in\mathcal{X}^{\left[
\mathbf{m}\right]  }}\in\left(  b-\beta\right)  \mathcal{X}^{\left[
\mathbf{m}\right]  }=\operatorname*{Ker}\left(  \pi^{\mathcal{X}}\right)  $
and thus $\pi^{\mathcal{X}}\left(  \left(  b-\beta\right)  q\right)  =0$.
Thus, (\ref{pf.lem.t-red.unique.rest.5}) becomes%
\[
0=D\left(  \pi^{\mathcal{X}}\left(  v\right)  +\underbrace{\pi^{\mathcal{X}%
}\left(  \left(  b-\beta\right)  q\right)  }_{=0}\right)  =D\left(
\underbrace{\pi^{\mathcal{X}}\left(  v\right)  }_{=p}\right)  =D\left(
p\right)  .
\]
This proves Lemma \ref{lem.t-red.unique.rest}.
\end{proof}

\begin{vershort}
\begin{proof}
[Proof of Theorem \ref{thm.t-red.unique}.]Theorem \ref{thm.t-red.unique}
follows from Lemma \ref{lem.t-red.unique.rest} in the same way as we derived
Lemma \ref{lem.t-red.unique.b=1} from Lemma \ref{lem.t-red.unique.b=1.0} (with
the only difference that $\mathcal{J}_{1}$ is replaced by $\mathcal{J}_{\beta
}$ throughout the proof).
\end{proof}
\end{vershort}

\begin{verlong}
\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 $p\equiv q\operatorname{mod}\mathcal{J}_{\beta}$, then $D\left(
f\right)  =D\left(  g\right)  $.

So let $f$ and $g$ be two pathless polynomials $q\in\mathcal{X}$ such that
$p\equiv q\operatorname{mod}\mathcal{J}_{\beta}$. Thus, $p\equiv
f\operatorname{mod}\mathcal{J}_{\beta}$ and $p\equiv g\operatorname{mod}%
\mathcal{J}_{\beta}$. Hence, $f\equiv p\equiv g\operatorname{mod}%
\mathcal{J}_{\beta}$, so that $f-g\in\mathcal{J}_{\beta}$. 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.rest}
(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 Lemma \ref{lem.t-red.unique.b=1}.
\end{proof}
\end{verlong}

\subsection{\label{sect.span-detail}Appendix: Detailed proof of Proposition
\ref{prop.path-red.span}}

Let us now 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 $\mathfrak{m\in M}$ 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, $\mathfrak{m}\in\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}%
_{\beta}$.
\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}_{\beta}%
\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 $\mathfrak{m}%
\in\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta}$.

If $\mathfrak{m}$ is pathless, then this is obvious\footnote{\textit{Proof.}
Assume that $\mathfrak{m}$ is pathless. We must then show that $\mathfrak{m}%
\in\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta}$.
\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, $\mathfrak{m}\in\mathcal{X}_{\operatorname*{pathless}}\subseteq
\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta}$, 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 $x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)
\in\mathcal{J}_{\beta}$ (since $x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}%
+x_{j,k}+\beta\right)  $ is one of the designated generators of the ideal
$\mathcal{J}_{\beta}$). Thus, \newline$x_{i,j}x_{j,k}\equiv x_{i,k}\left(
x_{i,j}+x_{j,k}+\beta\right)  \operatorname{mod}\mathcal{J}_{\beta}$. 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)  \operatorname{mod}\mathcal{J}_{\beta}%
}\mathfrak{n}\equiv x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\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}\operatorname{mod}\mathcal{J}_{\beta}.
\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}+\mathcal{J}_{\beta}. \label{pf.lem.span-detail.1.3}%
\end{equation}
We shall now analyze the three monomials $x_{i,k}x_{i,j}\mathfrak{n}$,
$x_{i,k}x_{j,k}\mathfrak{n}$ and $x_{i,k}\mathfrak{n}$ 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
\[
x_{i,k}x_{i,j}\mathfrak{n}\in\mathcal{X}_{\operatorname*{pathless}%
}+\mathcal{J}_{\beta}.
\]


\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
\[
x_{i,k}x_{j,k}\mathfrak{n}\in\mathcal{X}_{\operatorname*{pathless}%
}+\mathcal{J}_{\beta}.
\]


\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
\[
x_{i,k}\mathfrak{n}\in\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}%
_{\beta}.
\]

\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}_{\beta}}%
+\underbrace{x_{i,k}x_{j,k}\mathfrak{n}}_{\in\mathcal{X}%
_{\operatorname*{pathless}}+\mathcal{J}_{\beta}}+\beta\underbrace{x_{i,k}%
\mathfrak{n}}_{\in\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta}%
}+\underbrace{\mathcal{J}_{\beta}}_{\subseteq\mathcal{X}%
_{\operatorname*{pathless}}+\mathcal{J}_{\beta}}\\
&  \subseteq\left(  \mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta
}\right)  +\left(  \mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta
}\right)  +\beta\left(  \mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}%
_{\beta}\right)  +\left(  \mathcal{X}_{\operatorname*{pathless}}%
+\mathcal{J}_{\beta}\right) \\
&  \subseteq\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta}%
\end{align*}
(since $\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta}$ 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 $\mathfrak{m}\in\mathcal{X}_{\operatorname*{pathless}%
}+\mathcal{J}_{\beta}$. 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 $\mathfrak{m}\in\mathcal{X}_{\operatorname*{pathless}}%
+\mathcal{J}_{\beta}$ for each $\mathfrak{m}\in\mathfrak{M}$. In other words,
$\mathfrak{M}\subseteq\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}%
_{\beta}$ (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 $\mathfrak{M}%
\subseteq\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta}$, we
obtain $\operatorname*{span}\mathfrak{M}\subseteq\operatorname*{span}\left(
\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta}\right)
=\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta}$ (since
$\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta}$ is a $\mathbf{k}%
$-submodule of $\mathcal{X}$). Hence, $\mathcal{X}=\operatorname*{span}%
\mathfrak{M}\subseteq\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}%
_{\beta}$.

\begin{noncompile}
In other words, $\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta}$
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,
$\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta}$ contains a
spanning set of $\mathcal{X}$ as a subset (namely, the spanning set
$\mathfrak{M}$). Therefore, $\mathcal{X}_{\operatorname*{pathless}%
}+\mathcal{J}_{\beta}$ must contain the whole $\mathcal{X}$ as a subset (since
$\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta}$ is a $\mathbf{k}%
$-submodule of $\mathcal{X}$). In other words, $\mathcal{X}\subseteq
\mathcal{X}_{\operatorname*{pathless}}+\mathcal{J}_{\beta}$.
\end{noncompile}

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

We have $p=u+\underbrace{v}_{\in\mathcal{J}_{\beta}}\in u+\mathcal{J}_{\beta}%
$. In other words, $p\equiv u\operatorname{mod}\mathcal{J}_{\beta}$. 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 $p\equiv q\operatorname{mod}%
\mathcal{J}_{\beta}$ (namely, $q=u$). This proves Proposition
\ref{prop.path-red.span}.
\end{proof}

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

\subsection{Statements}

We have thus answered one of the major questions about the ideal
$\mathcal{J}_{\beta}$; but we have begged perhaps the most obvious one: Can we
find a basis of the $\mathbf{k}$-module $\mathcal{X}/\mathcal{J}_{\beta}$?
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 $\beta\in\mathbf{k}$ and $p\in\mathcal{X}$.
Then, there exists a \textbf{unique} forkless polynomial $q\in\mathcal{X}$
such that $p\equiv q\operatorname{mod}\mathcal{J}_{\beta}$.
\end{theorem}

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

\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{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}

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[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:

\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}

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, (iii) $\Longleftrightarrow$ (i)]{BecWei98} and \cite[Conclusion after
the proof of Lemma 1.1.38]{Graaf16}.

We shall also use the following simple fact:

\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}

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}).

\subsection{The proofs}

The main workhorse of the proofs is the following fact:

\begin{proposition}
\label{prop.forkless.groebner}Let $\beta\in\mathbf{k}$ and $p\in\mathcal{X}$.
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}%
\ \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 $\mathcal{J}_{\beta}$ of $\mathcal{X}$
(with respect to this order).
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.forkless.groebner} (sketched).]The elements
$x_{i,k}x_{i,j}-x_{i,j}x_{j,k}+x_{i,k}x_{j,k}+\beta x_{i,k}$ 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)  $ of the ideal
$\mathcal{J}_{\beta}$ 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}=\left(  -1\right)  \left(
x_{i,j}x_{j,k}-x_{i,k}\left(  x_{i,j}+x_{j,k}+\beta\right)  \right)  $). Thus,
they generate the ideal $\mathcal{J}_{\beta}$. Hence, in order to prove that
they form a Gr\"{o}bner basis of $\mathcal{J}_{\beta}$, 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,%
\[
g_{1}=x_{i_{1},k_{1}}x_{i_{1},j_{1}}-x_{i_{1},j_{1}}x_{j_{1},k_{1}}%
+x_{i_{1},k_{1}}x_{j_{1},k_{1}}+\beta x_{i_{1},k_{1}}%
\]
for some $\left(  i_{1},j_{1},k_{1}\right)  \in\left[  n\right]  ^{3}$
satisfying $i_{1}<j_{1}<k_{1}$, and%
\[
g_{2}=x_{i_{2},k_{2}}x_{i_{2},j_{2}}-x_{i_{2},j_{2}}x_{j_{2},k_{2}}%
+x_{i_{2},k_{2}}x_{j_{2},k_{2}}+\beta x_{i_{2},k_{2}}.
\]
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\footnote{This integer $b$ has nothing to do with the
indeterminate $b$ from Definition \ref{def.m}.} (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},\\
&  x_{a,d}x_{a,b}-x_{a,b}x_{b,d}+x_{a,d}x_{b,d}+\beta x_{a,d},\\
&  x_{a,d}x_{a,c}-x_{a,c}x_{c,d}+x_{a,d}x_{c,d}+\beta x_{a,d}.
\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. Actually, let's work in a slightly more general setting: Fix
$\alpha\in\mathbf{k}$. Let $G_{\alpha}$ be the set
\[
\left\{  x_{i,k}x_{i,j}-x_{i,j}x_{j,k}+x_{i,k}x_{j,k}+\beta x_{i,k}%
+\alpha\ \mid\ \left(  i,j,k\right)  \in\left[  n\right]  ^{3}\text{
satisfying }i<j<k\right\}  .
\]
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*}
We shall prove that $\operatorname*{spol}\left(  g_{1},g_{2}\right)
\overset{\ast}{\underset{G_{\alpha}}{\longrightarrow}}0$ whenever $g_{1}$ and
$g_{2}$ are two of the three polynomials $u_{1},u_{2},u_{3}$. This shall yield
our above claim (that $\operatorname*{spol}\left(  g_{1},g_{2}\right)
\overset{\ast}{\underset{G}{\longrightarrow}}0$) by setting $\alpha=0$. (The
additional generality that the parameter $\alpha$ gives us will help us prove
the more general Proposition \ref{prop.forkless.groebner.Jba} below.)

So we need to show that $\operatorname*{spol}\left(  u_{1},u_{2}\right)
\overset{\ast}{\underset{G_{\alpha}}{\longrightarrow}}0$,
$\operatorname*{spol}\left(  u_{1},u_{3}\right)  \overset{\ast
}{\underset{G_{\alpha}}{\longrightarrow}}0$ and $\operatorname*{spol}\left(
u_{2},u_{3}\right)  \overset{\ast}{\underset{G_{\alpha}}{\longrightarrow}}0$.

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 $x_{a,d}u_{1}-x_{a,c}u_{2}\overset{\ast
}{\underset{G_{\alpha}}{\longrightarrow}}0$ (by Lemma \ref{lem.groebner.to0}).
In other words, $\operatorname*{spol}\left(  u_{1},u_{2}\right)
\overset{\ast}{\underset{G_{\alpha}}{\longrightarrow}}0$ (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 $x_{a,d}u_{1}-x_{a,b}u_{3}\overset{\ast
}{\underset{G_{\alpha}}{\longrightarrow}}0$ (by Lemma \ref{lem.groebner.to0}).
In other words, $\operatorname*{spol}\left(  u_{1},u_{3}\right)
\overset{\ast}{\underset{G_{\alpha}}{\longrightarrow}}0$.

Finally, the identity we need for $\operatorname*{spol}\left(  u_{2}%
,u_{3}\right)  \overset{\ast}{\underset{G_{\alpha}}{\longrightarrow}}0$ 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.

\begin{noncompile}
Once this is assumed, we merely need to show that the set
(\ref{eq.prop.forkless.groebner.family}) is a Gr\"{o}bner basis \textbf{in the
case }$n=4$. This can be confirmed by a simple computation in the
\texttt{SageMath} computer algebra system \cite{SageMath}:
%TCIMACRO{\TeXButton{sage-out}{\begin{verbatim}
%sage: R.<x12, x13, x14, x23, x24, x34, b> = PolynomialRing(QQ, order="lex")
%sage: I = R.ideal([x12*x23-x13*(x12+x23+b), x23*x34-x24*(x23+x34+b),
%....:              x13*x34-x14*(x13+x34+b), x12*x24-x14*(x12+x24+b)])
%sage: I.groebner_basis()
%[x12*x13 - x12*x23 + x13*x23 + x13*b,
%x12*x14 - x12*x24 + x14*x24 + x14*b,
%x13*x14 - x13*x34 + x14*x34 + x14*b,
%x23*x24 - x23*x34 + x24*x34 + x24*b]
%\end{verbatim}}}%
%BeginExpansion
\begin{verbatim}
sage: R.<x12, x13, x14, x23, x24, x34, b> = PolynomialRing(QQ, order="lex")
sage: I = R.ideal([x12*x23-x13*(x12+x23+b), x23*x34-x24*(x23+x34+b),
....:              x13*x34-x14*(x13+x34+b), x12*x24-x14*(x12+x24+b)])
sage: I.groebner_basis()
[x12*x13 - x12*x23 + x13*x23 + x13*b,
x12*x14 - x12*x24 + x14*x24 + x14*b,
x13*x14 - x13*x34 + x14*x34 + x14*b,
x23*x24 - x23*x34 + x24*x34 + x24*b]
\end{verbatim}%
%EndExpansion


(Here, we have assumed that $\mathbf{k}=\mathbb{Q}$ for the sake of
simplicity, but the computations of the S-polynomials $\operatorname*{spol}%
\left(  g_{1},g_{2}\right)  $ are universal.)
\end{noncompile}

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

\begin{proof}
[Proof of Proposition \ref{prop.forkless.basis} (sketched).]Let $G_{\beta}$ be
the set (\ref{eq.prop.forkless.groebner.family}). Then, Proposition
\ref{prop.forkless.groebner} shows that $G_{\beta}$ is a Gr\"{o}bner basis of
the ideal $\mathcal{J}_{\beta}$ 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\}  $, $\mathcal{X}_{\Xi}=\mathcal{X}$, $\mathfrak{M}_{\Xi
}=\mathfrak{M}$, $\mathcal{I}=\mathcal{J}_{\beta}$ and $G=G_{\beta}$) shows
that the projections of the $G_{\beta}$-reduced monomials onto the quotient
ring $\mathcal{X}/\mathcal{J}_{\beta}$ form a basis of the $\mathbf{k}$-module
$\mathcal{X}/\mathcal{J}_{\beta}$. Since the $G_{\beta}$-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}

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}.\footnote{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)  $.} Is there a
similarly simple argument for the \textquotedblleft
uniqueness\textquotedblright\ part?

\section{A generalization?}

The ideals $\mathcal{J}_{\beta}$ of the $\mathbf{k}$-algebra $\mathcal{X}$ can
be \textquotedblleft deformed\textquotedblright\ by introducing a second
parameter $\alpha\in\mathbf{k}$, leading to the following definition:

\begin{definition}
Let $\beta\in\mathbf{k}$ and $\alpha\in\mathbf{k}$. Let $\mathcal{J}%
_{\beta,\alpha}$ 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 idea of this definition 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 $\mathcal{X}/\mathcal{J}_{\beta,\alpha}$, which he calls
the \textquotedblleft associative quasi-classical Yang--Baxter algebra of
weight $\left(  \alpha,\beta\right)  $\textquotedblright). The ideal
$\mathcal{J}_{\beta}$ is a particular case: $\mathcal{J}_{\beta}%
=\mathcal{J}_{\beta,0}$.

Similarly to Proposition \ref{prop.path-red.span}, we can prove the following:

\begin{proposition}
Let $\beta\in\mathbf{k}$, $\alpha\in\mathbf{k}$ and $p\in\mathcal{X}$. Then,
there exists a pathless polynomial $q\in\mathcal{X}$ such that $p\equiv
q\operatorname{mod}\mathcal{J}_{\beta,\alpha}$.
\end{proposition}

An analogue to Theorem \ref{thm.t-red.unique} also holds:

\begin{theorem}
\label{conj.Jba.indep}Let $\beta\in\mathbf{k}$, $\alpha\in\mathbf{k}$ and
$p\in\mathcal{X}$. Consider any pathless polynomial $q\in\mathcal{X}$ such
that $p\equiv q\operatorname{mod}\mathcal{J}_{\beta,\alpha}$. Then, $D\left(
q\right)  $ does not depend on the choice of $q$ (but merely on the choice of
$\beta$, $\alpha$ and $p$).
\end{theorem}

We have a proof of Theorem \ref{conj.Jba.indep}; this proof will appear in a
future version (\texttt{arXiv:1704.00839v3}) of this preprint. Notice that
Theorem \ref{conj.Jba.indep} prompts a further question: All four enumerative
questions posed in \cite[Exercise A22]{Stanle15} can be extended to the
\textquotedblleft deformed\textquotedblright\ case (by replacing the
\textquotedblleft reduction rule\textquotedblright\ $x_{ij}x_{jk}\rightarrow
x_{ik}\left(  x_{ij}+x_{jk}+\beta\right)  $ by $x_{ij}x_{jk}\rightarrow
x_{ik}\left(  x_{ij}+x_{jk}+\beta\right)  +\alpha$); the results are
deformations of Catalan and Narayana numbers. Which deformations?

The ideals $\mathcal{J}_{\beta,\alpha}$ also satisfy the following analogues
of Proposition \ref{prop.forkless.groebner} and Proposition
\ref{prop.forkless.basis}:

\begin{proposition}
\label{prop.forkless.groebner.Jba}Let $\beta\in\mathbf{k}$, $\alpha
\in\mathbf{k}$ and $p\in\mathcal{X}$. 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%
\[
\left\{  x_{i,k}x_{i,j}-x_{i,j}x_{j,k}+x_{i,k}x_{j,k}+\beta x_{i,k}%
+\alpha\ \mid\ \left(  i,j,k\right)  \in\left[  n\right]  ^{3}\text{
satisfying }i<j<k\right\}
\]
is a Gr\"{o}bner basis of the ideal $\mathcal{J}_{\beta,\alpha}$ of
$\mathcal{X}$ (with respect to this order).
\end{proposition}

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

These two propositions can be proven similarly to Proposition
\ref{prop.forkless.groebner} and Proposition \ref{prop.forkless.basis}.

\begin{thebibliography}{99999999}                                                                                         %


\bibitem[BecWei98]{BecWei98}Thomas Becker, Volker Weispfennig,
\textit{Gr\"{o}bner Bases: A Computational Approach to Commutative Algebra},
Corrected 2nd printing, Springer 1998.

\bibitem[CoLiOs15]{CoLiOs15}David A. Cox, John Little, Donal O'Shea,
\textit{Ideals, Varieties, and Algorithms}, Undergraduate Texts in
Mathematics, 4th edition, Springer 2015.

\bibitem[EneHer12]{EneHer12}Viviana Ene, J\"{u}rgen Herzog,
\textit{Gr\"{o}bner Bases in Commutative Algebra}, Graduate Studies in
Mathematics \#130, AMS 2012.

\bibitem[EscMes15]{EscMes15}\href{https://arxiv.org/abs/1502.03997v3}{Laura
Escobar, Karola M\'{e}sz\'{a}ros, \textit{Subword complexes via triangulations
of root polytopes}, arXiv:1502.03997v3.}

\bibitem[Graaf16]{Graaf16}Willem de Graaf, \textit{Computational Algebra},
lecture notes, version 21 September 2016.\newline\url{http://www.science.unitn.it/~degraaf/compalg/notes.pdf}

\bibitem[Kirill16]{Kirill16}Anatol N. Kirillov, \textit{On Some Quadratic
Algebras I }$\dfrac{1}{2}$\textit{: Combinatorics of Dunkl and Gaudin
Elements, Schubert, Grothendieck, Fuss-Catalan, Universal Tutte and Reduced
Polynomials}, SIGMA \textbf{12} (2016), 002, 172 pages,
\href{https://arxiv.org/abs/1502.00426v3}{arXiv:1502.00426v3}.

\bibitem[MalBlo15]{MalBlo15}Philippe Malbos, Thomas Blossier,
\textit{Alg\'{e}bre appliqu\'{e}e: Introduction aux bases de Gr\"{o}bner et
\`{a} leurs applications}, lecture notes, version 4 May 2015.\newline\url{http://math.univ-lyon1.fr/~malbos/Ens/algapip12.pdf}

\bibitem[Meszar09]{Meszar09}%
\href{http://www.ams.org/journals/tran/2011-363-08/S0002-9947-2011-05265-7/}{Karola
M\'{e}sz\'{a}ros, \textit{Root polytopes, triangulations, and the subdivision
algebra. I}, Trans. Amer. Math. Soc. 363 (2011), pp. 4359--4382.}\newline See
also \href{https://arxiv.org/abs/0904.2194v3}{arXiv:0904.2194v3} for a
preprint version.

\bibitem[MesDiz17]{MesDiz17}\href{https://arxiv.org/abs/1705.02418v3}{Karola
M\'{e}sz\'{a}ros, Avery St. Dizier, \textit{From generalized permutahedra to
Grothendieck polynomials via flow polytopes}, arXiv:1705.02418v3}.

\bibitem[Monass02]{Monass02}Denis Monasse, \textit{Introduction aux bases de
Gr\"{o}bner: th\'{e}orie et pratique}, lecture notes, version 19 November
2002.\newline\url{http://denis.monasse.free.fr/denis/articles/grobner.pdf}

\bibitem[SageMath]{SageMath}\href{http://www.sagemath.org}{The Sage
Developers, \textit{SageMath, the Sage Mathematics Software System (Version
7.6)}, 2017.}

\bibitem[Stanle15]{Stanle15}Richard P. Stanley, \textit{Catalan Numbers},
Cambridge University Press 2015.

\bibitem[Sturmf08]{Sturmf08}Bernd Sturmfels, \textit{Algorithms in Invariant
Theory}, 2nd edition, Springer 2008.
\end{thebibliography}


\end{document}