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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 $1\leq i0}\left(
q_{i}/q_{j}\right) ^{k}\ \ \ \ \ \ \ \ \ \ \text{for all }\left( i,j\right)
\in\left[ n\right] ^{2}\text{ satisfying }i0}}q_{i}^{a_{i}%
}=\mathfrak{m}$. (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 $i0}\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};\\i0\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 $1\leq u0$ (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 $v0$). 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 $i0\\\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 $\mathfrak{m}=\prod_{\substack{\left(
i,j\right) \in\left[ n\right] ^{2};\\ii\geq1$. 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) j\\\text{(since }j0}+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 $ii\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
$\Xi=\left\{ x_{i,j}\ \mid\ \left( i,j\right) \in\left[ n\right]
^{2}\text{ satisfying }i0$ 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}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 = 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. = 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
}ix_{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