\documentclass[12pt]{article} \usepackage{e-jc} \specs{P3.3}{24(3)}{2017} \usepackage{amsmath, amsthm, amssymb} \usepackage{graphicx} \usepackage[colorlinks=true,citecolor=black,linkcolor=black,urlcolor=blue]{hyperref} \newcommand{\doi}[1]{\href{http://dx.doi.org/#1}{\texttt{doi:#1}}} \newcommand{\arxiv}[1]{\href{http://arxiv.org/abs/#1}{\texttt{arXiv:#1}}} \newtheorem{Theorem}{Theorem}[section] \newtheorem{Proposition}[Theorem]{Proposition} \newtheorem{Corollary}[Theorem]{Corollary} \newtheorem{Lemma}[Theorem]{Lemma} \newtheorem{Example}[Theorem]{Example} \theoremstyle{definition} \newtheorem{Remark}[Theorem]{Remark} \newtheorem{Definition}[Theorem]{Definition} \newtheorem{Conjecture}[Theorem]{Conjecture} \newtheorem{Problem}[Theorem]{Problem} \newcommand{\twolines}[2]{{\scriptstyle #1 \atop \scriptstyle \vphantom{\sum}#2}} \def\Sym{\operatorname{\mathsf{Sym}}} \def\DSym{\operatorname{\mathsf{DSym}}} \def\QSym{\operatorname{\mathsf{QSym}}} \def\DQsym{\operatorname{\mathsf{DQSym}}} \title{\bf Ideals and quotients of diagonally \\ quasi-symmetric functions} \author{Shu Xiao Li\\ \small Department of Mathematics and Statistics\\[-0.8ex] \small York University\\[-0.8ex] \small Toronto, Canada\\ \small\tt lishu3@yorku.ca} \date{\dateline{Dec 1, 2016}{Jun 29, 2017}{Jul 14, 2017}\\ \small Mathematics Subject Classifications: 05E05} \allowdisplaybreaks \begin{document} \maketitle \begin{abstract} In 2004, J.-C. Aval, F. Bergeron and N. Bergeron studied the algebra of diagonally quasi-symmetric functions $\operatorname{\mathsf{DQSym}}$ in the ring $\mathbb{Q}[\mathbf{x},\mathbf{y}]$ with two sets of variables. They made conjectures on the structure of the quotient $\mathbb{Q}[\mathbf{x},\mathbf{y}]/\langle\operatorname{\mathsf{DQSym}}^+\rangle$, which is a quasi-symmetric analogue of the diagonal harmonic polynomials. In this paper, we construct a Hilbert basis for this quotient when there are infinitely many variables i.e. $\mathbf{x}=x_1,x_2,\dots$ and $\mathbf{y}=y_1,y_2,\dots$. Then we apply this construction to the case where there are finitely many variables, and compute the second column of its Hilbert matrix. \end{abstract} %\setcounter{tocdepth}{3} \section{Introduction} In the polynomial ring $\mathbb{Q}[\mathbf{x}_n]=\mathbb{Q}[x_1,\dots,x_n]$ with $n$ variables, the ring of symmetric polynomials (cf. \cite{Mac, Sagan}), $\Sym_n$, is the subspace of invariants under the symmetric group $S_n$ action \[ \sigma\cdot f(x_1,x_2,\dots,x_n)=f(x_{\sigma(1)},x_{\sigma(2)},\dots,x_{\sigma(n)}). \] The quotient space $\mathbb{Q}[\mathbf{x}_n]/\langle \Sym_n^+\rangle$ over the ideal generated by symmetric polynomials with no constant term is thus called the coinvariant space of symmetric group. Classic results by Artin \cite{Artin} and Steinberg \cite{Steinberg} asserts that this quotient forms an $S_n$-module that is isomorphic to the left regular representation. Moreover, considering the natural scalar product \[ \langle f,g\rangle = \big(f(\partial x_1,\dots,\partial x_n)(g(x_1,\dots,x_n))\big)(0,0,\dots, 0), \] this quotient is equal to the orthogonal complement of $\Sym_n$. In particular, the coinvariant space is killed by Laplacian operator $\Delta=\partial x_1^2+\cdots+\partial x_n^2$. Hence, it is also known as the harmonic space. One can show that $\{h_k(x_k,\dots,x_n):1\leq k\leq n\}$ forms a Gr\"{o}bner basis of $\langle \Sym_n^+\rangle$ with respect to the usual order $x_1>\cdots >x_n$, where $h_k$ is the complete homogeneous basis of degree $k$. As a result, the dimension of $\mathbb{Q}[\mathbf{x}_n]/\langle \Sym_n^+\rangle$ is $n!$. One generalization is the diagonal harmonic space. In the context of $\mathbb{Q}[\mathbf{x}_n,\mathbf{y}_n]=\mathbb{Q}[x_1,\dots,x_n,y_1,\dots,y_n]$, the diagonally symmetric functions, $\DSym_n$, is the space of invariants under the diagonal action of $S_n$ \[ \sigma\cdot f(x_1,\dots,x_n,y_1,\dots,y_n)=f(x_{\sigma(1)},\dots,x_{\sigma(n)},y_{\sigma(1)},\dots,y_{\sigma(n)}). \] The diagonal harmonics, $\mathbb{Q}[\mathbf{x}_n,\mathbf{y}_n]/\langle \DSym_n^+\rangle$, was studied by Garsia and Haiman \cite{GH, Haiman} where it was used to prove the $n!$ conjecture and Macdonald positivity. In particular, its dimension turns out to be $(n+1)^{n-1}$. More interesting results and applications can be found in \cite{BBGHT, BGHT, Haglund}. The ring of quasi-symmetric functions, $\QSym$, was introduced by Gessel \cite{Gessel} as generating function for Stanley's $P$-partitions \cite{Stanley}. It soon shows great importance in algebraic combinatorics e.g. \cite{ABS, GKLLRT}. In our context, $\QSym_n$ can be defined as the space of invariants in $\mathbb{Q}[\mathbf{x}_n]$ under the $S_n$-action of Hivert \[ \sigma\cdot \left(x_{i_1}^{a_1}\cdots x_{i_k}^{a_k}\right)=x_{j_1}^{a_1}\cdots x_{j_k}^{a_k} \] where $i_1<\cdots \ell(\tilde{\alpha})-\ell(\tilde{\gamma})$ and $\tilde{\alpha}_{i(j+\ell(\tilde{\alpha})-\ell(\tilde{\gamma}))}=\tilde{\gamma}_{ij}$ for all $j\geq 1$. We write $\tilde{\alpha}= \big( {0 \atop 0} \big) \tilde{\beta}$ if $\tilde{\alpha}_{11}=\tilde{\alpha}_{21}=0$ and $\tilde{\alpha}_{i(j+1)}=\tilde{\beta}_{ij}$ for all $j\geq2$. Note that for each generalized bicomposition $\tilde{\alpha}$ that is not a bicomposition, there is a unique way to decompose it into $\tilde{\alpha}=\tilde{\beta} \big( {0 \atop 0} \big) \gamma$ for some generalized bicomposition $\tilde{\beta}$ and bicomposition $\gamma$. \subsection{Diagonally quasi-symmetric functions} The algebra of diagonally quasi-symmetric functions, $\DQsym$, is a subalgebra of $\mathbb{Q}[[\mathbf{x},\mathbf{y}]]$ spanned by monomials indexed by bicompositions $$M_{\alpha}=\sum_{i_1<\dots0$, \begin{equation}\tag{3.1}\label{eq:3.1} G_{\tilde{\epsilon}}=G_{\tilde{\alpha}\beta}-x_{k+1}G_{\tilde{\alpha} \big( {\beta_{11}-1 \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \ \beta_{21} \ \ \, \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}. \end{equation} If $\beta_{11}=0$, \begin{equation}\tag{3.2}\label{eq:3.2} G_{\tilde{\epsilon}}=G_{\tilde{\alpha}\beta}-y_{k+1}G_{\tilde{\alpha} \big( { \ \ 0 \ \ \ \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \beta_{21}-1 \ \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}. \end{equation} Inductively, $\{G_{\tilde{\epsilon}}\}$ is defined for all generalized bicomposition $\tilde{\epsilon}$. Clearly $G_{\tilde{\epsilon}}$ is homogeneous in degree $|\tilde{\epsilon}|$. Hence, we have a notion of leading monomial of $G_{\tilde{\epsilon}}$, denoted by $LM(G_{\tilde{\epsilon}})$ with respect to the lexicographic order with $x_1>y_1>x_2>y_2>\cdots$. To show that $\{G_{\tilde{\epsilon}}\}$ form a basis, it suffices to prove the leading monomial of $G_{\tilde{\epsilon}}$ is $\mathbf{X}^{\tilde{\epsilon}}$. \begin{Lemma}\label{lemma:3.1} Let $\tilde{\alpha}= \big( {a \atop b} \big) \tilde{\beta}$ be a generalized bicomposition, \begin{enumerate} \item if $a=b=0$, then $G_{\tilde{\alpha}}=G_{\tilde{\beta}}(x_2,x_3,\dots,y_2,y_3,\dots)$, \item if $a>0$, then $G_{\tilde{\alpha}}=x_1G_{ \big( {a-1 \atop b} \big) \tilde{\beta}}+P(x_2,x_3,\dots,y_2,y_3,\dots)$, \item if $a=0$ and $b>0$, then $G_{\tilde{\alpha}}=y_1G_{ \big( {0 \atop b-1} \big) \tilde{\beta}}+P(x_2,x_3,\dots,y_2,y_3,\dots)$ \end{enumerate} for some $P\in\mathbb{Q}[[\mathbf{x},\mathbf{y}]]$. \end{Lemma} \begin{proof} We prove by induction on the length of $\tilde{\alpha}$. \begin{enumerate} \item If $\tilde{\alpha}= \big( {0 \atop 0} \big) $, then $G_{\tilde{\alpha}}=1$ and we are done. \item If $\tilde{\beta}=\beta$ is a bicomposition, \begin{enumerate} \item\label{2a} if $a=b=0$ and $\beta$ non-zero, \begin{enumerate} \item if $\beta_{11}\geq1$ and $\beta_{11}+\beta_{21}\geq 2$, using (\ref{eq:2.1}) and (\ref{eq:3.1}), we get \begin{align*} G_{\tilde{\alpha}}&=G_{\beta}-x_1G_{ \big( {\beta_{11}-1 \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \ \beta_{21} \ \ \, \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ &=F_{\beta}-x_1F_{ \big( {\beta_{11}-1 \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \ \beta_{21} \ \ \, \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ &=F_{\beta}(x_2,x_3,\dots,y_2,y_3,\dots)=G_{\beta}(x_2,x_3,\dots,y_2,y_3,\dots) \end{align*} and the lemma follows. \item if $\beta_{11}=1$ and $\beta_{21}=0$, using (\ref{eq:2.2}), (\ref{eq:3.1}) and induction on $\ell(\tilde{\beta})$, we get \begin{align*} G_{\tilde{\alpha}}&=G_{\beta}-x_1G_{ \big( {0 \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop 0 \ \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ &=G_{\beta}-x_1G_{\big( {\beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}(x_2,x_3,\dots,y_2,y_3,\dots)\\ &=F_{\beta}-x_1F_{\big( {\beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}(x_2,x_3,\dots,y_2,y_3,\dots)\\ &=F_{\beta}(x_2,x_3,\dots,y_2,y_3,\dots)=G_{\beta}(x_2,x_3,\dots,y_2,y_3,\dots) \end{align*} and the lemma follows. \item if $\beta_{11}=0$ and $\beta_{21}\geq 2$, using (\ref{eq:2.3}) and (\ref{eq:3.2}), we get \begin{align*} G_{\tilde{\alpha}}&=G_{\beta}-y_1G_{ \big( { \ \ 0 \ \ \ \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \beta_{21}-1 \ \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ &=F_{\beta}-y_1F_{ \big( { \ \ 0 \ \ \ \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \beta_{21}-1 \ \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ &=F_{\beta}(x_2,x_3,\dots,y_2,y_3,\dots)=G_{\beta}(x_2,x_3,\dots,y_2,y_3,\dots) \end{align*} and the lemma follows. \item if $\beta_{11}=0$ and $\beta_{21}=1$, using (\ref{eq:2.4}), (\ref{eq:3.2}) and induction on $\ell(\tilde{\beta})$, we get \begin{align*} G_{\tilde{\alpha}}&=G_{\beta}-y_1G_{ \big( {0 \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop 0 \ \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ &=G_{\beta}-y_1G_{\big( {\beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}(x_2,x_3,\dots,y_2,y_3,\dots)\\ &=F_{\beta}-y_1F_{\big( {\beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}(x_2,x_3,\dots,y_2,y_3,\dots)\\ &=F_{\beta}(x_2,x_3,\dots,y_2,y_3,\dots)=G_{\beta}(x_2,x_3,\dots,y_2,y_3,\dots) \end{align*} and the lemma follows. \end{enumerate} \item if $a\geq1$ and $a+b\geq 2$, by definition $G_{\tilde{\alpha}}=F_{\tilde{\alpha}}$. Using (\ref{eq:2.1}), we get $$G_{\tilde{\alpha}}=F_{\tilde{\alpha}}=x_1F_{ \big( {a-1 \atop b} \big) \beta}+F_{\tilde{\alpha}}(x_2,x_3,\dots,y_2,y_3,\dots)$$ and the lemma follows, with $P=F_{\tilde{\alpha}}(x_2,x_3,\dots,y_2,y_3,\dots)$. \item if $a=1$ and $b=0$, by definition $G_{\tilde{\alpha}}=F_{\tilde{\alpha}}$. Using (\ref{eq:2.2}) and (\ref{2a}). we get \begin{align*} G_{\tilde{\alpha}}=F_{\tilde{\alpha}}&=x_1F_{\beta}(x_2,x_3,\dots,y_2,y_3,\dots)+F_{\tilde{\alpha}}(x_2,x_3,\dots,y_2,y_3,\dots)\\ &=x_1G_{ \big( {0 \atop 0} \big) \beta}+F_{\tilde{\alpha}}(x_2,x_3,\dots,y_2,y_3,\dots) \end{align*} and the lemma follows with $P=F_{\tilde{\alpha}}(x_2,x_3,\dots,y_2,y_3,\dots)$. \item if $a=0$ and $b\geq2$, by definition $G_{\tilde{\alpha}}=F_{\tilde{\alpha}}$. Using (\ref{eq:2.3}), we get $$G_{\tilde{\alpha}}=F_{\tilde{\alpha}}=y_1F_{ \big( {a \atop b-1} \big) \beta}+F_{\tilde{\alpha}}(x_2,x_3,\dots,y_2,y_3,\dots)$$ and the lemma follows, with $P=F_{\tilde{\alpha}}(x_2,x_3,\dots,y_2,y_3,\dots)$. \item if $a=0$ and $b=1$, by definition $G_{\tilde{\alpha}}=F_{\tilde{\alpha}}$. Using (\ref{eq:2.4}) and (\ref{2a}). we get \begin{align*} G_{\tilde{\alpha}}=F_{\tilde{\alpha}}&=y_1F_{\beta}(x_2,x_3,\dots,y_2,y_3,\dots)+F_{\tilde{\alpha}}(x_2,x_3,\dots,y_2,y_3,\dots)\\ &=y_1G_{ \big( {0 \atop 0} \big) \beta}+F_{\tilde{\alpha}}(x_2,x_3,\dots,y_2,y_3,\dots) \end{align*} and the lemma follows with $P=F_{\tilde{\alpha}}(x_2,x_3,\dots,y_2,y_3,\dots)$. \end{enumerate} \item In the general case, let $\tilde{\alpha}=\tilde{\gamma} \big( {0 \atop 0} \big) \beta$ where $\beta$ is a non-empty bicomposition and $k=\ell(\tilde{\alpha})-\ell(\beta)-1$. We prove by induction on $k$. If $k=1$, then we are back in case (\ref{2a}) above. Hence, we assume $k>1$ and $\tilde{\gamma}= \big( {a \atop b} \big) \tilde{\mu}$. \begin{enumerate} \item If $a=b=0$, \begin{enumerate} \item if $\beta_{11}\geq1$, by induction and (\ref{eq:3.1}), we have \begin{align*} G_{\tilde{\alpha}}=&G_{ \big( {0 \atop 0} \big) \tilde{\mu} \big( {0 \atop 0} \big) \beta}=G_{ \big( {0 \atop 0} \big) \tilde{\mu}\beta}-x_kG_{ \big( {0 \atop 0} \big) \tilde{\mu} \big( {\beta_{11}-1 \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \ \beta_{21} \ \ \, \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ =&G_{\tilde{\mu}\beta}(x_2,x_3,\dots,y_2,y_3,\dots)\\ &-x_{(k-1)+1}G_{\tilde{\mu} \big( {\beta_{11}-1 \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \ \beta_{21} \ \ \, \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}(x_2,x_3,\dots,y_2,y_3,\dots)\\ =&G_{\tilde{\mu} \big( {0 \atop 0} \big) \beta}(x_2,x_3,\dots,y_2,y_3,\dots) \end{align*} and the lemma follows. \item if $\beta_{11}=0$, by induction and (\ref{eq:3.2}), we have \begin{align*} G_{\tilde{\alpha}}=&G_{ \big( {0 \atop 0} \big) \tilde{\mu} \big( {0 \atop 0} \big) \beta}=G_{ \big( {0 \atop 0} \big) \tilde{\mu}\beta}-y_kG_{ \big( {0 \atop 0} \big) \tilde{\mu} \big( { \ \ 0 \ \ \ \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \beta_{21}-1 \ \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ \\ =&G_{\tilde{\mu}\beta}(x_2,x_3,\dots,y_2,y_3,\dots)\\ &-y_{(k-1)+1}G_{\tilde{\mu} \big( { \ \ 0 \ \ \ \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \beta_{21}-1 \ \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}(x_2,x_3,\dots,y_2,y_3,\dots)\\ =&G_{\tilde{\mu} \big( {0 \atop 0} \big) \beta}(x_2,x_3,\dots,y_2,y_3,\dots) \end{align*} and the lemma follows. \end{enumerate} \item If $a\geq1$, \begin{enumerate} \item if $\beta_{11}\geq1$, by induction and (\ref{eq:3.1}), we have \begin{align*} G_{\tilde{\alpha}}=&G_{ \big( {a \atop b} \big) \tilde{\mu} \big( {0 \atop 0} \big) \beta}=G_{ \big( {a \atop b} \big) \tilde{\mu}\beta}-x_kG_{ \big( {a \atop b} \big) \tilde{\mu} \big( {\beta_{11}-1 \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \ \beta_{21} \ \ \, \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ =&x_1G_{ \big( {a-1 \atop b} \big) \tilde{\mu}\beta}+P_1(x_2,x_3,\dots,y_2,y_3,\dots)\\ &-x_k\Big(x_1G_{ \big( {a-1 \atop b} \big) \tilde{\mu} \big( {\beta_{11}-1 \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \ \beta_{21} \ \ \, \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ &+P_2(x_2,x_3,\dots,y_2,y_3,\dots)\Big)\\ =&x_1\left(G_{ \big( {a-1 \atop b} \big) \tilde{\mu}\beta}-x_kG_{ \big( {a-1 \atop b} \big) \tilde{\mu} \big( {\beta_{11}-1 \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \ \beta_{21} \ \ \, \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\right)\\ &+P(x_2,x_3,\dots,y_2,y_3,\dots)\\ =&x_1G_{ \big( {a-1 \atop b} \big) \tilde{\mu} \big( {0 \atop 0} \big) \beta}+P(x_2,x_3,\dots,y_2,y_3,\dots) \end{align*} and the lemma follows with $P=P_1-x_kP_2$. \item if $\beta_{11}=0$, by induction and (\ref{eq:3.2}), we have \begin{align*} G_{\tilde{\alpha}}=&G_{ \big( {a \atop b} \big) \tilde{\mu} \big( {0 \atop 0} \big) \beta}=G_{ \big( {a \atop b} \big) \tilde{\mu}\beta}-y_kG_{ \big( {a \atop b} \big) \tilde{\mu} \big( { \ \ 0 \ \ \ \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \beta_{21}-1 \ \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ =&x_1G_{ \big( {a-1 \atop b} \big) \tilde{\mu}\beta}+P_1(x_2,x_3,\dots,y_2,y_3,\dots)\\ &-y_k\Big(x_1G_{ \big( {a-1 \atop b} \big) \tilde{\mu} \big( { \ \ 0 \ \ \ \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \beta_{21}-1 \ \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ &+P_2(x_2,x_3,\dots,y_2,y_3,\dots)\Big)\\ =&x_1\left(G_{ \big( {a-1 \atop b} \big) \tilde{\mu}\beta}-y_kG_{ \big( {a-1 \atop b} \big) \tilde{\mu} \big( { \ \ 0 \ \ \ \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \beta_{21}-1 \ \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\right)\\ &+P(x_2,x_3,\dots,y_2,y_3,\dots)\\ =&x_1G_{ \big( {a-1 \atop b} \big) \tilde{\mu} \big( {0 \atop 0} \big) \beta}+P(x_2,x_3,\dots,y_2,y_3,\dots) \end{align*} and the lemma follows with $P=P_1-y_kP_2$. \end{enumerate} \item If $a=0$ and $b\geq1$, \begin{enumerate} \item if $\beta_{11}\geq1$, by induction and (\ref{eq:3.1}), we have \begin{align*} G_{\tilde{\alpha}}=&G_{ \big( {0 \atop b} \big) \tilde{\mu} \big( {0 \atop 0} \big) \beta}=G_{ \big( {0 \atop b} \big) \tilde{\mu}\beta}-x_kG_{ \big( {0 \atop b} \big) \tilde{\mu} \big( {\beta_{11}-1 \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \ \beta_{21} \ \ \, \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ =&y_1G_{ \big( {0 \atop b-1} \big) \tilde{\mu}\beta}+P_1(x_2,x_3,\dots,y_2,y_3,\dots)\\ &-x_k\Big(y_1G_{ \big( {0 \atop b-1} \big) \tilde{\mu} \big( {\beta_{11}-1 \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \ \beta_{21} \ \ \, \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ &+P_2(x_2,x_3,\dots,y_2,y_3,\dots)\Big)\\ =&y_1\left(G_{ \big( {0 \atop b-1} \big) \tilde{\mu}\beta}-x_kG_{ \big( {0 \atop b-1} \big) \tilde{\mu} \big( {\beta_{11}-1 \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \ \beta_{21} \ \ \, \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\right)\\ &+P(x_2,x_3,\dots,y_2,y_3,\dots)\\ =&y_1G_{ \big( {0 \atop b-1} \big) \tilde{\mu} \big( {0 \atop 0} \big) \beta}+P(x_2,x_3,\dots,y_2,y_3,\dots) \end{align*} and the lemma follows with $P=P_1-x_kP_2$. \item if $\beta_{11}=0$, by induction and (\ref{eq:3.2}), we have \begin{align*} G_{\tilde{\alpha}}=&G_{ \big( {0 \atop b} \big) \tilde{\mu} \big( {0 \atop 0} \big) \beta}=G_{ \big( {0 \atop b} \big) \tilde{\mu}\beta}-y_kG_{ \big( {0 \atop b} \big) \tilde{\mu} \big( { \ \ 0 \ \ \ \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \beta_{21}-1 \ \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ =&y_1G_{ \big( {0 \atop b-1} \big) \tilde{\mu}\beta}+P_1(x_2,x_3,\dots,y_2,y_3,\dots)\\ &-y_k\Big(y_1G_{ \big( {0 \atop b-1} \big) \tilde{\mu} \big( { \ \ 0 \ \ \ \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \beta_{21}-1 \ \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\\ &+P_2(x_2,x_3,\dots,y_2,y_3,\dots)\Big)\\ =&y_1\left(G_{ \big( {0 \atop b-1} \big) \tilde{\mu}\beta}-y_kG_{ \big( {0 \atop b-1} \big) \tilde{\mu} \big( { \ \ 0 \ \ \ \ \beta_{12} \ \cdots \ \beta_{1\ell(\beta)} \atop \beta_{21}-1 \ \beta_{22} \ \cdots \ \beta_{2\ell(\beta)}} \big)}\right)\\ &+P(x_2,x_3,\dots,y_2,y_3,\dots)\\ =&y_1G_{ \big( {0 \atop b-1} \big) \tilde{\mu} \big( {0 \atop 0} \big) \beta}+P(x_2,x_3,\dots,y_2,y_3,\dots) \end{align*} and the lemma follows with $P=P_1-y_kP_2$.\qedhere \end{enumerate} \end{enumerate} \end{enumerate} \end{proof} \begin{Corollary} Let $\tilde{\epsilon}$ be a generalized bicomposition, then the leading monomial of $G_{\tilde{\epsilon}}$ is $\mathbf{X}^{\tilde{\epsilon}}$. Hence, $\{G_{\tilde{\alpha}}\}$ forms a Hilbert basis for $R$. \end{Corollary} \begin{proof} We prove by induction on $\ell(\tilde{\epsilon})$ and $|\tilde{\epsilon}|$. If $\tilde{\epsilon}=\big( {0 \atop 0} \big)$, by definition $G_{\tilde{\epsilon}}=1=X^{\tilde{\epsilon}}$. Otherwise, let $\tilde{\epsilon}=\big( {a \atop b} \big)\tilde{\beta}$. \begin{enumerate} \item If $a=b=0$ and $\tilde{\beta}$ non-zero, by induction on $\ell(\tilde{\epsilon})$ and Lemma \ref{lemma:3.1}, we have $$LM(G_{\tilde{\epsilon}})=LM(G_{\tilde{\beta}}(x_2,x_3,\dots,y_2,y_3,\dots))=(x_2,x_3,\dots,y_2,y_3,\dots)^{\tilde{\beta}}=\mathbf{X}^{\tilde{\epsilon}}.$$ \item If $a\geq1$, by induction on $|\tilde{\epsilon}|$ and Lemma \ref{lemma:3.1}, we have $$LM(G_{\tilde{\epsilon}})=LM\left(x_1G_{\big( {a-1 \atop b} \big)\tilde{\beta}}\right)=\mathbf{X}^{\tilde{\epsilon}}.$$ \item If $a=0$ and $b\geq1$, by induction on $|\tilde{\epsilon}|$ and Lemma \ref{lemma:3.1}, we have \[LM(G_{\tilde{\epsilon}})=LM\left(y_1G_{\big( {0 \atop b-1} \big)\tilde{\beta}}\right)=\mathbf{X}^{\tilde{\epsilon}}.\qedhere\] \end{enumerate} \end{proof} \section{The Hilbert Basis} The set $\{x^{\tilde{\alpha}}F_{\beta}\}$ is a spanning set of the ideal $I$. For each $\tilde{\alpha}$ and $\beta$, we write $x^{\tilde{\alpha}}F_{\beta}$ in terms of the $G$ basis by the following rules. (1) We reorder the product $x^{\tilde{\alpha}}F_{\beta}$ as $\cdots\left( x_2^{\tilde{\alpha}_{21}}\left( y_2^{\tilde{\alpha}_{22}}\left( x_1^{\tilde{\alpha}_{11}}\left( y_1^{\tilde{\alpha}_{21}}F_{\beta}\right) \right) \right) \right) $. (2) We reduce the above product recursively using (\ref{eq:3.1}) \begin{equation}\tag{4.1} x_iG_{\tilde{\gamma}}=x_iG_{ \big( {\cdots \ \tilde{\gamma}_{1i} \ \cdots \atop \cdots \ \tilde{\gamma}_{2i} \ \cdots}\big)}=G_{ \big( {\cdots \ \tilde{\gamma}_{1i}+1 \ \cdots \atop \cdots \ \ \tilde{\gamma}_{2i} \ \ \cdots}\big)}-G_{ \big( {\cdots \ 0 \ \tilde{\gamma}_{1i}+1 \ \cdots \atop \cdots \ 0 \ \ \tilde{\gamma}_{2i} \ \ \cdots}\big)}; \end{equation} or using (\ref{eq:3.2}) when $\tilde{\gamma}_{1i}=0$ for some i, \begin{equation}\tag{4.2} y_iG_{\tilde{\gamma}}=y_iG_{ \big( {\cdots \ \tilde{\gamma}_{1i} \ \cdots \atop \cdots \ \tilde{\gamma}_{2i} \ \cdots}\big)}=G_{ \big( {\cdots \ \ \tilde{\gamma}_{1i} \ \ \cdots \atop \cdots \ \tilde{\gamma}_{2i}+1 \ \cdots}\big)}-G_{ \big( {\cdots \ 0 \ \ \tilde{\gamma}_{1i} \ \ \cdots \atop \cdots \ 0 \ \tilde{\gamma}_{2i}+1 \ \cdots}\big)}. \end{equation} (3) When $\tilde{\gamma}_{1i}=a>0$, we reduce $y_iG_{\tilde{\gamma}}$ as \begin{align*}\tag{4.3} y_1G_{\tilde{\gamma}}&=y_iG_{ \big( {\cdots \ \ a \ \ \cdots \atop \cdots \ \tilde{\gamma}_{2i} \ \cdots}\big)}=y_i\left( G_{ \big( {\cdots \ 0 \ \ a \ \ \cdots \atop \cdots \ 0 \ \tilde{\gamma}_{2i} \ \cdots}\big)} + x_iG_{ \big( {\cdots \ a-1 \ \cdots \atop \cdots \ \tilde{\gamma}_{2i} \ \cdots}\big)} \right) \\ &=y_iG_{ \big( {\cdots \ 0 \ \ a \ \ \cdots \atop \cdots \ 0 \ \tilde{\gamma}_{2i} \ \cdots}\big)} + x_i\left( y_iG_{ \big( {\cdots \ a-1 \ \cdots \atop \cdots \ \tilde{\gamma}_{2i} \ \cdots}\big)}\right) =\cdots\\ &=\sum_{k=0}^{a-1}x_i^k\left( y_iG_{ \big( {\cdots \ 0 \ a-k \ \cdots \atop \cdots \ 0 \ \tilde{\gamma}_{2i} \ \cdots}\big)}\right) + x_i^a\left( y_i G_{ \big( {\cdots \ \ 0 \ \ \cdots \atop \cdots \ \tilde{\gamma}_{2i} \ \cdots}\big)}\right). \end{align*} The $``\cdots"$ above means $\tilde{\gamma}_{11} \ \cdots \ \tilde{\gamma}_{1(i-1)}$, $\tilde{\gamma}_{1(i+1)} \ \cdots \ \tilde{\gamma}_{1\ell(\tilde{\gamma})}$, $\tilde{\gamma}_{21} \ \cdots \ \tilde{\gamma}_{2(i-1)}$ or $\tilde{\gamma}_{2(i+1)} \ \cdots \ \tilde{\gamma}_{1\ell(\tilde{\gamma})}$ with respect to their positions in the generalized bicomposition. For example, \begin{align*} &y_1F_{\big( {1 \atop 0} \big)}=y_1\left( G_{\big( {0 \ 1 \atop 0 \ 0} \big)} + x_1G_{\big( {0 \atop 0} \big)}\right) = y_1G_{\big( {0 \ 1 \atop 0 \ 0} \big)}+x_1y_1G_{\big( {0 \atop 0} \big)}\\ =&G_{\big( {0 \ 1 \atop 1 \ 0} \big)} - G_{\big( {0 \ 0 \ 1 \atop 0 \ 1 \ 0} \big)}+x_1\left( G_{\big( {0 \atop 1} \big)}-G_{\big( {0 \ 0 \atop 0 \ 1} \big)}\right)\\ =&G_{\big( {0 \ 1 \atop 1 \ 0} \big)} - G_{\big( {0 \ 0 \ 1 \atop 0 \ 1 \ 0} \big)}+G_{\big( {1 \atop 1} \big)} - G_{\big( {0 \ 1 \atop 0 \ 1} \big)} - G_{\big( {1 \ 0 \atop 0 \ 1} \big)} + G_{\big( {0 \ 1 \ 0 \atop 0 \ 0 \ 1} \big)}. \end{align*} For each of the above rule, we choose one $G_{\tilde{\eta}}$ as leading basis element. We define a function $\phi$ from $\left(\{x_i\}\times\{G_{\tilde{\gamma}}\}\right)\cup\left(\{y_i\}\times\{G_{\tilde{\gamma}}\}\right)$ to $\{G_{\tilde{\gamma}}\}$ as follows. In the case of rules (4.1), (4.2), we choose $\phi\left( x_i,G_{\tilde{\gamma}}\right) =G_{ \big( {\cdots \ 0 \ \tilde{\gamma}_{1i}+1 \ \cdots \atop \cdots \ 0 \ \ \tilde{\gamma}_{2i} \ \ \cdots}\big)}$ and $\phi\left( y_i,G_{\tilde{\gamma}}\right) =G_{ \big( {\cdots \ 0 \ \ \tilde{\gamma}_{1i} \ \ \cdots \atop \cdots \ 0 \ \tilde{\gamma}_{2i}+1 \ \cdots}\big)}$. In the case of rule (4.3), we choose $\phi\left( y_i,G_{\tilde{\gamma}}\right) =\phi\left( y_i,G_{ \big( {\cdots \ 0 \ \ a \ \ \cdots \atop \cdots \ 0 \ \tilde{\gamma}_{2i} \ \cdots}\big)}\right) =G_{ \big( {\cdots \ 0 \ 0 \ \ a \ \ \cdots \atop \cdots \ 0 \ 1 \ \tilde{\gamma}_{2i} \ \cdots}\big)}$. In the other words, at each step of the expansion, we choose the lexicographically smallest $\tilde{\eta}$ such that $G_{\tilde{\eta}}$ appears as a term in the expansion. \begin{Lemma} The above process of choosing is invertible, i.e. $\phi$ is injective. \end{Lemma} \begin{proof} Since each time we multiply $x_i$ or $y_i$, the chosen term contains a $\binom00$ at position $i$. Combining this fact with the rule that we have to multiply $y_i$ before $x_i$, we have the following inverse function. Let $i$ be the largest number that $\left( \tilde{\gamma}_{1i},\tilde{\gamma}_{2i}\right) =(0,0)$ and $00$, then $\phi^{-1}\left( G_{ \big( {\cdots \ 0 \ \tilde{\gamma}_{1(i+1)} \ \cdots \atop \cdots \ 0 \ \tilde{\gamma}_{2(i+1)} \ \cdots}\big)}\right) =x_iG_{ \big( {\cdots \ \tilde{\gamma}_{1(i+1)}-1 \ \cdots \atop \cdots \ \ \tilde{\gamma}_{2(i+1)} \ \ \cdots}\big)}$. (2) If $\tilde{\gamma}_{1(i+1)}=0$ and, $\tilde{\gamma}_{1(i+2)}=0$ or $\tilde{\gamma}_{2(i+1)}>1$, then $\phi^{-1}\left( G_{ \big( {\cdots \ 0 \ \tilde{\gamma}_{1(i+1)} \ \cdots \atop \cdots \ 0 \ \tilde{\gamma}_{2(i+1)} \ \cdots}\big)}\right) =y_iG_{ \big( {\cdots \ \ \tilde{\gamma}_{1(i+1)} \ \ \cdots \atop \cdots \ \tilde{\gamma}_{2(i+1)}-1 \ \cdots}\big)}$. (3) If $\tilde{\gamma}_{1(i+1)}=0$, $\tilde{\gamma}_{2(i+1)}=1$ and $\tilde{\gamma}_{1(i+2)}>0$, then $\phi^{-1}\left( G_{ \big( {\cdots \ 0 \ 0 \ \tilde{\gamma}_{1(i+2)} \ \cdots \atop \cdots \ 0 \ 1 \ \tilde{\gamma}_{2(i+2)} \ \cdots}\big)}\right) =y_iG_{ \big( {\cdots \ \tilde{\gamma}_{1(i+2)} \ \cdots \atop \cdots \ \tilde{\gamma}_{2(i+2)} \ \cdots}\big)}$. \end{proof} Then, we can construct a map $\Phi:\{X^{\tilde{\alpha}}F_{\beta}:|\beta|\geq 1\}\to\{G_{\tilde{\gamma}}\}$ that is defined by ``composing" $\phi$ with itself $(|\tilde{\alpha}|-1)$ times. By the above Lemma, we also have $\Phi$ is injective. For simplicity, we define $\phi^{-1}(G_{\tilde{\gamma}})$ (or $\Phi^{-1}(G_{\tilde{\gamma}})$) to be $X^{\tilde{\alpha}}G_{\tilde{\beta}}$ (or $X^{\tilde{\alpha}}F_{\beta}$) if $\phi(X^{\tilde{\alpha}}G_{\tilde{\beta}})=G_{\tilde{\gamma}}$ (or $\Phi(X^{\tilde{\alpha}}F_{\beta})=G_{\tilde{\gamma}}$ respectively). \begin{Lemma} In the expansion of $X^{\tilde{\alpha}}F_{\beta}$ in the $G$ basis using the rules above, the term $\Phi(X^{\tilde{\alpha}}F_{\beta})$ appears only once. In particular, it has coefficients $1$ or $-1$. \end{Lemma} \begin{proof} We begin with the claim that if $\tilde{\mu}\neq\tilde{\nu}$, then $\phi(x_iG_{\tilde{\mu}})$ and $\phi(y_iG_{\tilde{\mu}})$ do not appear in the expansion of $x_iG_{\tilde{\nu}}$ and $y_iG_{\tilde{\nu}}$ respectively. Let $k$ be the smallest integer such that $(\tilde{\mu}_{k1},\tilde{\mu}_{k2})\neq(\tilde{\nu}_{k1},\tilde{\nu}_{k2})$. In rules (4.1), (4.2) and (4.3), for all $G_{\tilde{\gamma}}$ in the expansion of $x_iG_{\tilde{\mu}}$ or $y_iG_{\tilde{\mu}}$, the first $i-1$ columns of $\tilde{\gamma}$ is the same as that of $\tilde{\mu}$. Hence, the claim follows if $ki$ before $y_i$, which violates our rule. And the same condition applies to $\nu$. With this restriction, it is easy to check that the claim holds. If $k>i$, in both cases, if we choose any term in the expansion that is not $\phi(x_iG_{\tilde{\nu}})$ or $\phi(y_iG_{\tilde{\nu}})$, then the $i$ or $i+1$ column of its index must be different from that of $\phi(x_iG_{\tilde{\mu}})$ or $\phi(y_iG_{\tilde{\mu}})$. If we choose $\phi(x_iG_{\tilde{\nu}})$ or $\phi(y_iG_{\tilde{\nu}})$, we also have $\phi(x_iG_{\tilde{\mu}})\neq\phi(x_iG_{\tilde{\nu}})$ and $\phi(y_iG_{\tilde{\mu}})\phi(y_iG_{\tilde{\nu}})$ because $\mu\neq\nu$. Since each term in the expansion of $X^{\tilde{\alpha}}F_{\beta}$ corresponds to a sequence of choice using rules (4.1), (4.2) or (43), if at some point, we choose a term that is different from the choice in $\Phi$, then a recursive use of the claim asserts that $\Phi(X^{\tilde{\alpha}}F_{\beta})$ will not appear again. \end{proof} We now define an order $(<_G)$ on the set of generalized bicompositions as follows \begin{enumerate} \item If $\tilde{\alpha}$ and $\tilde{\beta}$ are bicompositions, then $\tilde{\alpha}<_G\tilde{\beta}$ if $\tilde{\alpha}<_{lex}\tilde{\beta}$. \item If $\tilde{\alpha}$ is a bicomposition and $\tilde{\beta}$ is not, then $\tilde{\alpha}<_G\tilde{\beta}$. \item If $\tilde{\alpha}=\tilde{\mu}\big( {0 \atop 0} \big)\alpha'$, $\tilde{\beta}=\tilde{\nu}\big( {0 \atop 0} \big)\beta'$ where $\alpha'$ and $\beta'$ are bicompositions, let $u=\ell(\tilde{\alpha})-\ell(\alpha')-1$, $v=\ell(\tilde{\alpha})-\ell(\beta')-1$, then $\tilde{\alpha}<_G\tilde{\beta}$ if \begin{enumerate} \item $u0$ and $\beta'_{11}=0$, or \item $u=v$, $\alpha'_{11}>0$, $\beta'_{11}>0$ (or $\alpha'_{11}=0$, $\beta'_{11}=0$) and $\overleftarrow{\phi}(G_{\tilde{\alpha}})<_G\overleftarrow{\phi}(G_{\tilde{\beta}})$ where we define $\overleftarrow{\phi}(G_{\tilde{\delta}})$ to be $\tilde{\gamma}$ if $\phi(x_iG_{\tilde{\gamma}})=G_{\tilde{\delta}}$ or $\phi(y_iG_{\tilde{\gamma}})=G_{\tilde{\delta}}$ for some $i$. \end{enumerate} \end{enumerate} \begin{Lemma}\label{lem:4.3} The order defined above is a total order on the set of generalized bicompositions such that if $G_{\tilde{\gamma}}=\Phi(X^{\tilde{\alpha}}F_{\beta})$, then for all $G_{\tilde{\delta}}$ that appears in the expansion of $X^{\tilde{\alpha}}F_{\beta}$, we have $\tilde{\gamma}\geq_G\tilde{\delta}$. \end{Lemma} \begin{proof} Clearly this is a total order. If $\tilde{\alpha}<\tilde{\beta}$ by (1) or (2), then $\tilde{\beta}$ cannot appear in the expansion of $\Phi^{-1}(\tilde{\alpha})=\tilde{\alpha}$. If $\tilde{\alpha}<\tilde{\beta}$ by (3a), that means $\phi^{-1}(\tilde{\alpha})=x_{u+1}G_{\tilde{\gamma}}$ or $y_{u+1}G_{\tilde{\gamma}}$ for some $\tilde{\gamma}$. Hence, $\tilde{\beta}$ cannot appear in the expansion of $\Phi^{-1}(\tilde{\alpha})$ because $\tilde{\beta}_{(v+1)1}=\tilde{\beta}_{(v+1)2}=0$ cannot be created. If $\tilde{\alpha}<\tilde{\beta}$ by (3b), that means $\phi^{-1}(\tilde{\alpha})=x_{u+1}G_{\tilde{\gamma}}$ for some $\tilde{\gamma}$. Hence, $\tilde{\beta}$ cannot appear in the expansion of $\Phi^{-1}(\tilde{\alpha})$ because it is not in that of $x_{u+1}G_{\tilde{\delta}}$ for any $\tilde{\delta}$. \end{proof} With this ordering, there is a unique leading $G_{\tilde{\delta}}$ for each expansion of $X^{\tilde{\alpha}}F_{\beta}$. \begin{Theorem}\label{thm:4.4} The set $A=\{G_{\tilde{\alpha}} \mid G_{\tilde{\alpha}}\notin \text{Img}(\Phi)\}$ forms a Hilbert basis for the quotient space R/I. \end{Theorem} \begin{proof} For any polynomial $p$ in $R$, we write $p$ in terms of the $G$ basis with $<_G$ order. For each term $G_{\tilde{\alpha}}\in Img(\Phi)$, we subtract $p$ by $\Phi^{-1}(G_{\tilde{\alpha}})\in I$ and $G_{\tilde{\alpha}}$ is cancelled. If we repeat this process (possibly countably many times), we can express $p$ as a series of $A$. \end{proof} \section{Finitely many variables case} In the case that there are finitely many variables, $R_n=\mathbb{Q}[x_1,\dots,x_n,y_1,\dots,y_n]$, the above constructions of $\DQsym(x_1,\dots,x_n,y_1,\dots,y_n)$, the $F$, $G$ bases and the ideal $I_n=<\DQsym^+(x_1,\dots,x_n,y_1,\dots,y_n)>$ remain the same by taking $x_i=y_i=0$ for $i>n$. In this case, $LM(G_{\tilde{\alpha}})=X^{\tilde{\alpha}}$ whenever $\ell(\tilde{\alpha})\leq n$ and hence $\{G_{\tilde{\alpha}}:\ell(\tilde{\alpha})\leq n\}$ spans $R_n$. Let $R_n^{i,j}$ be the span of $\{X^{\tilde{\alpha}}:\ell(\tilde{\alpha})\leq n,\sum_k\tilde{\alpha}_{1k}=i, \sum_k\tilde{\alpha}_{2k}=j\}$. Since $I_n$ is bihomogeneous in $\mathbf{x}$ and $\mathbf{y}$, $\displaystyle I_n=\bigoplus_{i,j}I_n^{i,j}$ where $I_n^{i,j}=I_n\cap R_n^{i,j}$, and $\displaystyle R_n/I_n=\bigoplus_{i,j}V_n^{i,j}$ where $V_n^{i,j}=R_n/I_n\cap R_n^{i,j}$. The Hilbert matrix corresponding to $R_n/I_n$ is the matrix $M_n(i,j)=\dim(V_n^{i-1,j-1})$. The goal of this section is to compute the second column of the Hilbert matrix. The proof is slight generalization of the one in \cite{ABB}. \begin{Lemma}\label{lem:5.1} The set $\{G_{\tilde{\alpha}} \mid G_{\tilde{\alpha}}\notin Img(\Phi), \ell(\tilde{\alpha})\leq n\}$ spans the quotient $R_n/I_n$. \end{Lemma} \begin{proof} Among all $\tilde{\alpha}$ such that $G_{\tilde{\alpha}}\in Img(\Phi)$, $\ell(\tilde{\alpha})\leq n$ and $G_{\tilde{\alpha}}$ cannot be reduced to $0$, let $\tilde{\beta}$ be the smallest one with respect to the $<_G$ order. Then, \begin{align*} G_{\tilde{\beta}}&=G_{\tilde{\beta}}-\Phi^{-1}(G_{\tilde{\beta}})+\Phi^{-1}(G_{\tilde{\beta}})\\ &\equiv G_{\tilde{\beta}}-\Phi^{-1}(G_{\tilde{\beta}}) \mod I_n \end{align*} But since $G_{\tilde{\beta}}$ is the leading term in $\Phi^{-1}(G_{\tilde{\beta}})$, terms in $G_{\tilde{\beta}}-\Phi^{-1}(G_{\tilde{\beta}})$ are strictly smaller than $G_{\tilde{\beta}}$, and thus they reduce to $0$. This contradicts to our assumption on $\tilde{\beta}$. \end{proof} Let $B_n$ be the set of generalized bicompositions $\{\tilde{\alpha}\}$ such that $\displaystyle\sum_{i=1}^{k}(\tilde{\alpha}_{1i}+\tilde{\alpha}_{2i})\ell(\tilde{\epsilon})$, since $\tilde{\alpha}\in A_n$, we have $\sum_{i\geq k}(\tilde{\alpha}_{1i}+\tilde{\alpha}_{2i})