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\newcommand{\ZZ}{\mathbb Z}
\newcommand{\NN}{\mathbb N}
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\newcommand{\spr}{\operatorname{spr}}

\title{Splitting a Polynomial into Linear Factors after an Injective Ring Extension}
\author{GPT-5.5, with prompting by Darij Grinberg}
\date{\today}

\begin{document}

\maketitle

\begin{quote}
\textbf{Manifest.} {This is an answer to \url{https://mathoverflow.net/questions/429134}, generated by GPT-5.5 with significant shepherding and editing. I still intend to write up this argument in my own words and possibly extend it to an expository paper due to its many connections to algebraic combinatorics.}\footnote{This work is in the public domain. \\ I thank Victor Reiner and Brendon Rhoades for help with finding precedents in the literature. --Darij}
\end{quote}

All rings in this note are commutative and unital, and all ring homomorphisms preserve the unity.

A small notational warning: if $x_1,\ldots,x_N$ are indeterminates, then
$\ZZ[x_1,\ldots,x_N]$ denotes a polynomial ring. If $u_1,\ldots,u_N$ are elements of some ring $B$, then
$\ZZ[u_1,\ldots,u_N]\subseteq B$ denotes the subalgebra generated by these elements. Such a subalgebra need not be a polynomial ring unless this is proved.

\section{\label{sec:ABC}The universal splitting map is split}

Fix a positive integer $m$. Set
\[
        B_m=\ZZ[a_1,b_1,a_2,b_2,\ldots,a_m,b_m],
\]
the polynomial ring in the $2m$ indeterminates $a_1,b_1,\ldots,a_m,b_m$ over $\ZZ$.
For each subset $I\subseteq [m]:=\{1,2,\ldots,m\}$, define the monomial
\[
        z_I=\prod_{i\in I}a_i\prod_{i\notin I}b_i\in B_m .
\]
For $0\leq r\leq m$, define
\[
        E_r=\sum_{\substack{I\subseteq [m];\\ |I|=r}} z_I .
\]
For example,
\begin{align*}
E_0 &= z_\varnothing = b_1 b_2 \cdots b_m; \\
E_1 &= \sum_{i=1}^m z_{\left\{i\right\}}
= a_1 b_2 b_3 \cdots b_m + b_1 a_2 b_3 \cdots b_m + b_1 b_2 a_3 \cdots b_m + \cdots + b_1 b_2 \cdots b_{m-1} a_m; \\
E_2 &= \sum_{1\leq i<j\leq m} z_{\left\{i,j\right\}}
= \sum_{1\leq i<j\leq m} b_1 b_2 \cdots b_{i-1} a_i b_{i+1} b_{i+2} \cdots b_{j-1} a_j b_{j+1} b_{j+2} \cdots b_m; \\
\cdots; \\
E_m &= z_{[m]} = a_1 a_2 \cdots a_m.
\end{align*}

We observe the following:
\begin{proposition}\label{prop:hg-vieta}
We have
\begin{align}
        \sum_{j=0}^m E_jX^j
        =
        \prod_{r=1}^m (a_rX+b_r)
        \qquad \text{ in the univariate polynomial ring } B_m[X].
        \label{prop:hg-vieta:1}
\end{align}
\end{proposition}

\begin{proof}
This is a homogeneous form of Vi\`ete's theorem.
The proof is easy:
Multiply the right hand side and collect terms with like powers of $X$.
\end{proof}

% In other words, Proposition~\ref{prop:hg-vieta} says that
% \begin{align}
        % \prod_{i=1}^m(a_iX+b_i)=\sum_{r=0}^m E_rX^r .
        % \label{prop:hg-vieta:2}
% \end{align}

Let
\[
        C_m=\ZZ[E_0,E_1,\ldots,E_m]\subseteq B_m .
\]
At this point, $C_m$ means the subalgebra of $B_m$ generated by
$E_0,E_1,\ldots,E_m$. Lemma~\ref{lem:alg-independent} will show that this subalgebra is actually a polynomial ring over $\ZZ$ on these generators.

The main claim is the following.

\begin{theorem}[splitting theorem]\label{thm:splitting}
The inclusion
\[
        C_m\hookrightarrow B_m
\]
is split as a $C_m$-module homomorphism. Equivalently, there is a $C_m$-linear map
\[
        \rho:B_m\to C_m
\]
whose restriction to $C_m$ is the identity.
\end{theorem}

The proof of this theorem will be delivered at the end of Subsection~\ref{subsec:split-pf}.

\subsection{Algebraic independence of the \texorpdfstring{$E_r$}{Er}'s}

\begin{lemma}\label{lem:alg-independent}
The elements $E_0,E_1,\ldots,E_m$ are algebraically independent over $\ZZ$. Hence
\[
        C_m=\ZZ[E_0,E_1,\ldots,E_m]
\]
is a polynomial ring over $\ZZ$ on the displayed generators.
\end{lemma}

\begin{proof}
We will use the theory of monomial orders (\cite[\S 2.2]{CoxLittleOShea2025}).
If a monomial order has been chosen in a polynomial ring, then $\LT(f)$ denotes the leading monomial of a nonzero polynomial $f$, not including its coefficient. For a monomial $M$ and a variable $x$, we write $\nu_x(M)$ for the exponent of $x$ in $M$.

Choose any monomial order on $B_m$. For each $r$, write
\[
        \LT(E_r)=z_{I_r}
\]
for some $r$-element subset $I_r\subseteq [m]$.

We claim that
\[
        I_0\subset I_1\subset\cdots\subset I_m .
\]
Indeed, suppose that $I_r\not\subseteq I_{r+1}$ for some $0\leq r<m$. Choose
\[
        x\in I_r\setminus I_{r+1},
        \qquad
        y\in I_{r+1}\setminus I_r.
\]
Put
\[
        I_r'=(I_r\setminus\{x\})\cup\{y\},
        \qquad
        I_{r+1}'=(I_{r+1}\setminus\{y\})\cup\{x\}.
\]
Then
\[
        z_{I_r}z_{I_{r+1}}=z_{I_r'}z_{I_{r+1}'}.
\]
But $z_{I_r}>z_{I_r'}$ and $z_{I_{r+1}}>z_{I_{r+1}'}$ by the definition of leading term. Multiplicativity of the monomial order gives
\[
        z_{I_r}z_{I_{r+1}}>z_{I_r'}z_{I_{r+1}'},
\]
a contradiction to the previous equality. This proves the claim.

A chain with one $r$-element set for each $r$ is obtained by adding the elements of $[m]$ in some order. Relabeling the indices $1,2,\ldots,m$, we may therefore assume
\[
        I_r=\{1,2,\ldots,r\} \qquad \text{for all } 0 \leq r \leq m.
\]
Hence
\[
        \LT(E_r)=M_r:=a_1a_2\cdots a_r b_{r+1}b_{r+2}\cdots b_m .
\]
For $\mathbf k=(k_0,k_1,\ldots,k_m)\in\NN^{m+1}$, the leading monomial of
\[
        E^{\mathbf k}:=E_0^{k_0}E_1^{k_1}\cdots E_m^{k_m}
\]
is therefore
\[
        M^{\mathbf k}:=M_0^{k_0}M_1^{k_1}\cdots M_m^{k_m}.
\]
The exponents of the variables in $M^{\mathbf k}$ determine $\mathbf k$: namely,
\[
        k_0=\nu_{b_1}(M^{\mathbf k}),
\]
\[
        k_i=\nu_{a_i}(M^{\mathbf k})-\nu_{a_{i+1}}(M^{\mathbf k})
        \qquad(1\leq i<m),
\]
and
\[
        k_m=\nu_{a_m}(M^{\mathbf k}).
\]
Thus the monomials $M^{\mathbf k}$ are pairwise distinct. Therefore the elements $E^{\mathbf k}$ are $\ZZ$-linearly independent, since each has a distinct leading monomial with coefficient $1$. This proves algebraic independence.
\end{proof}

\begin{remark}\label{rem:specific-order-alg-indep}
The proof of Lemma~\ref{lem:alg-independent} was written so that it works with any monomial order. If one only wants a quick proof of algebraic independence, one can instead choose a specific lexicographic order, for example
\[
        a_1>a_2>\cdots>a_m>b_1>b_2>\cdots>b_m.
\]
Then
\[
        \LT(E_r)=a_1a_2\cdots a_r b_{r+1}b_{r+2}\cdots b_m
\]
immediately, and the distinct-leading-monomial argument for the products
$E_0^{k_0}\cdots E_m^{k_m}$ goes through without the preliminary chain argument.
\end{remark}

\subsection{The diagonal subring and chain monomials}

Give $B_m$ the $\ZZ^m$-grading determined by
\[
        \deg a_i=\deg b_i=\varepsilon_i,
\]
where $\varepsilon_1,\ldots,\varepsilon_m$ is the standard basis of $\ZZ^m$. Let
\[
        \delta=(1,1,\ldots,1)\in\ZZ^m.
\]
Each $z_I$ and each $E_r$ has multidegree $\delta$. Define the diagonal subring
\[
        A_m=\bigoplus_{n\geq0}(B_m)_{n\delta},
\]
where $(B_m)_\gamma$ denotes the $\gamma$-th degree component of $B_m$.
This is a subring of $B_m$, because the product of two homogeneous elements of multidegrees
$n\delta$ and $n'\delta$ is homogeneous of multidegree $(n+n')\delta$.

A monomial of $B_m$ is called \emph{diagonal of diagonal degree $n$} if it has multidegree $n\delta$. Equivalently, it has the form
\[
        \prod_{i=1}^m a_i^{h_i}b_i^{n-h_i}
\]
for a uniquely determined vector
\[
        h=(h_1,\ldots,h_m)\in\{0,1,\ldots,n\}^m .
\]
This vector $h$ is called the \emph{height vector} of the diagonal monomial, and its entries $h_i$ are called the \emph{heights} of the respective labels $i$. Thus
\[
        h_i=\nu_{a_i}(M)
\]
for a diagonal monomial $M$ of diagonal degree $n$ (where $\nu_x(M)$ denotes the exponent of a variable $x$ in $M$); equivalently, the exponent of $b_i$ in $M$ is $n-h_i$.
Clearly,
the diagonal monomials (of all diagonal degrees $n$) form a basis of the $\ZZ$-module $A_m$.

A \emph{chain monomial} will mean a product
\[
        z_{I_1}z_{I_2}\cdots z_{I_n}
\]
where the subsets $I_1,I_2,\ldots,I_n$ form a multichain under inclusion (i.e., any $i$ and $j$ satisfy $I_i \subseteq I_j$ or $I_j \subseteq I_i$). Since the ring is commutative, the order of the factors is immaterial. When convenient, we write such a monomial in the form $z_{I_1}z_{I_2}\cdots z_{I_n}$ with
\[
        I_n\subseteq I_{n-1}\subseteq\cdots\subseteq I_1.
\]
Note that $n=0$ is allowed.

\begin{lemma}\label{lem:diagonal-chain-span}
The subring $A_m$ is generated, as a $\ZZ$-algebra, by the $2^m$ monomials
\[
        z_I\qquad(I\subseteq [m]).
\]
Moreover, the chain monomials are precisely the ordinary monomial basis of $A_m$ written in this generating set. In particular, they form a $\ZZ$-basis of $A_m$.
\end{lemma}

\begin{proof}
Each $z_I$ has multidegree $\delta$, so the subring generated by the $z_I$'s is contained in $A_m$.

Conversely, consider a monomial of multidegree $n\delta$. Such a monomial has the form
\[
        \prod_{i=1}^m a_i^{h_i}b_i^{n-h_i}
\]
for a unique vector
\[
        h=(h_1,\ldots,h_m)\in\{0,1,\ldots,n\}^m .
\]
For each $s\in\{1,2,\ldots,n\}$, set
\[
        I_s(h)=\{i\in[m]\mid h_i\geq s\}.
\]
Then
\[
        I_n(h)\subseteq I_{n-1}(h)\subseteq\cdots\subseteq I_1(h)
\]
and
\[
        \prod_{s=1}^n z_{I_s(h)}
        =\prod_{i=1}^m a_i^{h_i}b_i^{n-h_i}.
\]
Thus every diagonal monomial is a chain monomial in the $z_I$'s. Conversely, every chain monomial is visibly a diagonal monomial.\footnote{And it is easy to see that distinct chains $\left(I_1, I_2, \ldots, I_n\right)$ lead to different chain monomials $z_{I_1} z_{I_2} \cdots z_{I_n}$: indeed, the length $n$ of the chain is the diagonal degree of the chain monomial, and each exponent $h_i = \nu_{a_i}(z_{I_1} z_{I_2} \cdots z_{I_n})$ is the number of subsets $I_1, I_2, \ldots, I_n$ that contain $i$, which (because of the chain condition $I_n \subseteq I_{n-1} \subseteq \cdots \subseteq I_1$) entails that the first $h_i$ sets $I_1, I_2, \ldots, I_{h_i}$ contain $i$ while the remaining $n - h_i$ sets $I_{h_i+1}, I_{h_i+2}, \ldots, I_n$ don't. Thus, from the monomial $z_{I_1} z_{I_2} \cdots z_{I_n}$, we can tell which $i$'s lie in which $I_j$'s.}
The ordinary monomials in the $a_i$'s and $b_i$'s form a $\ZZ$-basis of $B_m$, so the diagonal monomials form a $\ZZ$-basis of $A_m$.
Hence the chain monomials form a $\ZZ$-basis of $A_m$.
\end{proof}

We shall also use the elementary straightening identity
\begin{equation}\label{eq:straightening-z}
        z_Iz_J=z_{I\cap J}z_{I\cup J}
        \qquad(I,J\subseteq[m]),
\end{equation}
which holds directly in $A_m$, since both sides have the same exponent of each variable $a_i$ and $b_i$.

\begin{remark}[Hibi-ring context, not used in the proof]\label{rem:hibi-context}
One standard way to package Lemma~\ref{lem:diagonal-chain-span} and \eqref{eq:straightening-z} is by introducing the polynomial ring
\[
        T_m=\ZZ[y_I\mid I\subseteq[m]]
\]
and the ideal $J_m$ generated by
\[
        y_Iy_J-y_{I\cap J}y_{I\cup J}
        \qquad(I,J\subseteq[m]).
\]
Then the map
\[
        T_m/J_m\longrightarrow A_m,
        \qquad y_I\longmapsto z_I,
\]
is an isomorphism. This is the Hibi ring of the Boolean lattice $2^{[m]}$; see Hibi \cite{Hibi1987}. We shall not need this presentation below.
\end{remark}

\subsection{A bijection between monomials and pairs of permutations and tuples}

Recall that the symmetric group $S_m$ consists of the permutations of $[m] := \{1,2,\ldots,m\}$.
We write permutations in one-line notation; for example, $31524$ or $(3,1,5,2,4)$ is a permutation in $S_5$. (We shall occasionally omit commas and parentheses in this way.)
In particular, $12\cdots m$ is the identity permutation in $S_m$.

For a permutation $\pi=(\pi_1,\ldots,\pi_m)\in S_m$, set
\[
        \Des(\pi)=\{s\in\{1,\ldots,m-1\}\mid \pi_s>\pi_{s+1}\},
        \qquad
        \des(\pi)=|\Des(\pi)|.
\]
For $0\leq j\leq m$, write
\[
        P_j^\pi=\{\pi_1,\pi_2,\ldots,\pi_j\},
\]
so that $P_0^\pi=\varnothing$ and $P_m^\pi=[m]$. Define the Garsia--Stanton descent monomial
\[
        u_\pi=\prod_{s\in\Des(\pi)} z_{P_s^\pi}\in A_m .
\]
Thus $u_{12\cdots m}=1$.

Given $\pi\in S_m$ and $\mathbf i=(i_0,i_1,\ldots,i_m)\in\NN^{m+1}$, put
\begin{equation}\label{eq:M-pi-i}
        M_{\pi,\mathbf i}
        =u_\pi\prod_{j=0}^m \left(z_{P_j^\pi}\right)^{i_j}.
\end{equation}
This is a chain monomial, of diagonal degree
\[
        \des(\pi)+i_0+i_1+\cdots+i_m .
\]

If $h=(h_1,\ldots,h_m)\in\ZZ^m$, then the \emph{sorting permutation} of $h$, denoted
$\spr(h)$, will mean the permutation in $S_m$ obtained by listing the labels
$1,2,\ldots,m$ in decreasing order of their heights $h_i$, breaking ties
in increasing order of the labels. That is, $\spr(h)$ is the unique permutation
$\pi \in S_m$ satisfying
\begin{align}
        h_{\pi_1}\geq h_{\pi_2}\geq\cdots\geq h_{\pi_m}
\label{eq.spr.1}
\end{align}
and
\begin{align}
\text{ if $h_{\pi_s}=h_{\pi_{s+1}}$, then }
\pi_s<\pi_{s+1}.
\label{eq.spr.2}
\end{align}
(This is related to the notion of standardization in combinatorics,
but not literally the same; we use the name ``sorting permutation'' to avoid confusion.)


\begin{lemma}[height vector of the distinguished chain monomial]\label{lem:spr-of-M}
Let $\pi\in S_m$ and $\mathbf i=(i_0,i_1,\ldots,i_m)\in\NN^{m+1}$. Let $h$ be the height vector of the chain monomial $M_{\pi,\mathbf i}$. Then
\[
        \spr(h)=\pi .
\]
\end{lemma}

\begin{proof}
We analyze the height vector $h = (h_1, h_2, \ldots, h_m)$ of $M_{\pi,\mathbf i}$.
For $1\leq s\leq m$, put
\[
        d_s^\pi=\#\{q\in\Des(\pi)\mid q\geq s\},
        \qquad
        g_s=i_s+i_{s+1}+\cdots+i_m .
\]
The factor $u_\pi$ contributes $d_s^\pi$ to the height of the label $\pi_s$, while the prefix factors
\[
        \prod_{j=0}^m \left(z_{P_j^\pi}\right)^{i_j}
\]
contribute $g_s$ to the height of the same label. Hence
\begin{equation}\label{eq:height-of-M-pi-i}
        h_{\pi_s}=d_s^\pi+g_s
        \qquad(1\leq s\leq m).
\end{equation}
The sequence $g_1,g_2,\ldots,g_m$ is weakly decreasing. The sequence
$d_1^\pi,d_2^\pi,\ldots,d_m^\pi$ is also weakly decreasing; more precisely,
\[
        d_s^\pi=d_{s+1}^\pi+1 \quad\text{if } s\in\Des(\pi),
        \qquad
        d_s^\pi=d_{s+1}^\pi \quad\text{if } s\notin\Des(\pi).
\]
Thus \eqref{eq:height-of-M-pi-i} gives
\[
        h_{\pi_1}\geq h_{\pi_2}\geq\cdots\geq h_{\pi_m}.
\]
Moreover, if $s\in\Des(\pi)$, then $d_s^\pi=d_{s+1}^\pi+1$, whence
\[
        h_{\pi_s}>h_{\pi_{s+1}}.
\]
If equality $h_{\pi_s}=h_{\pi_{s+1}}$ occurs, then therefore $s\notin\Des(\pi)$, so $\pi_s<\pi_{s+1}$. This is exactly the rule defining the sorting permutation: list the labels in decreasing height, and break equal-height ties increasingly. Hence $\spr(h)=\pi$.
\end{proof}

\begin{lemma}[labelled-antichain bijection]\label{lem:ppartition-bijection}
The map
\[
        (\pi,\mathbf i)\longmapsto M_{\pi,\mathbf i}
\]
is a bijection from
\[
        \{(\pi,\mathbf i)\mid \pi\in S_m,\  \mathbf i\in\NN^{m+1}\}
\]
to the set of all chain monomials in $A_m$.
More precisely, for each $n\geq0$, it restricts to a bijection from the pairs satisfying
\[
        \des(\pi)+i_0+i_1+\cdots+i_m=n
\]
to the chain monomials of diagonal degree $n$.
\end{lemma}

\begin{proof}
\textit{Well-definedness.}
For a fixed pair $(\pi,\mathbf i)$, the sets
\[
        P_0^\pi\subseteq P_1^\pi\subseteq\cdots\subseteq P_m^\pi
\]
form a chain. The additional factors appearing in $u_\pi$ are among the same sets $P_s^\pi$. Hence $M_{\pi,\mathbf i}$ is a chain monomial. Its diagonal degree is
\[
        \des(\pi)+i_0+i_1+\cdots+i_m,
\]
since each $z_I$ has diagonal degree $1$.

\textit{Surjectivity.}
Let a chain monomial of diagonal degree $n$ be given. By Lemma~\ref{lem:diagonal-chain-span}, it has a unique height vector
\[
        h=(h_1,\ldots,h_m)\in\{0,1,\ldots,n\}^m,
\]
where the monomial is
\[
        \prod_{i=1}^m a_i^{h_i}b_i^{n-h_i}.
\]
Let
\[
        \pi=\spr(h)
\]
be the sorting permutation of $h$. Thus $\pi$ lists the labels in decreasing
order of their heights, breaking ties increasingly:
\[
        h_{\pi_1}\geq h_{\pi_2}\geq\cdots\geq h_{\pi_m},
\]
and if $h_{\pi_s}=h_{\pi_{s+1}}$, then $\pi_s<\pi_{s+1}$.

For $1\leq s\leq m$, put
\[
        d_s^\pi=\#\{q\in\Des(\pi)\mid q\geq s\}.
\]
This sequence $(d_1^\pi, d_2^\pi, \ldots, d_m^\pi)$ is the ``descent staircase'' of $\pi$: it drops by $1$ exactly at those $s$ that are descents of $\pi$.
Now define
\[
        g_s=h_{\pi_s}-d_s^\pi\qquad(1\leq s\leq m).
\]
We claim that
\begin{equation}
        g_1\geq g_2\geq\cdots\geq g_m\geq0.
\label{pf.lem:ppartition-bijection.g-dec}
\end{equation}
Indeed, if $s\notin\Des(\pi)$, then $d_s^\pi=d_{s+1}^\pi$, and the inequality $g_s\geq g_{s+1}$ follows from $h_{\pi_s}\geq h_{\pi_{s+1}}$. If $s\in\Des(\pi)$, then $\pi_s>\pi_{s+1}$; by the tie-breaking rule this forces
$h_{\pi_s}>h_{\pi_{s+1}}$, whence
$h_{\pi_s}\geq h_{\pi_{s+1}}+1$. Since $d_s^\pi=d_{s+1}^\pi+1$ in this case, we again get $g_s\geq g_{s+1}$. Finally, from $d_m^\pi=0$, we obtain
$g_m=h_{\pi_m}\geq0$.
Thus, \eqref{pf.lem:ppartition-bijection.g-dec} is proved.

Set
\[
        i_s=g_s-g_{s+1}\qquad(1\leq s<m),
\]
\[
        i_m=g_m,
\]
and
\[
        i_0=n-\des(\pi)-g_1.
\]
The last integer $i_0$ is nonnegative, since $g_1=h_{\pi_1}-\des(\pi)$ and $h_{\pi_1}\leq n$. These formulas give a vector
$\mathbf i = (i_0, i_1, \ldots, i_m) \in\NN^{m+1}$,
since \eqref{pf.lem:ppartition-bijection.g-dec} yields that $i_1, i_2, \ldots, i_m$ are nonnegative as well.
Moreover, $i_0 + i_1 + \cdots + i_m = n-\des(\pi)$, whence $\des(\pi) + i_0 + i_1 + \cdots + i_m = n$; in other words, the monomial $M_{\pi,\mathbf i}$ has diagonal degree $n$.

For $1\leq s\leq m$, the height of $M_{\pi,\mathbf i}$ at the label $\pi_s$ is
\[
        d_s^\pi+i_s+i_{s+1}+\cdots+i_m.
\]
By the definition of the $i$'s, this equals $d_s^\pi+g_s=h_{\pi_s}$.
Hence $M_{\pi,\mathbf i}$ has the same height vector as the original chain monomial, and it also has diagonal degree $n$. These data determine a diagonal monomial, so $M_{\pi,\mathbf i}$ is the original chain monomial.
So our original chain monomial has the form $M_{\pi,\mathbf i}$ for some $\pi\in S_m$ and $\mathbf i\in\NN^{m+1}$.
This proves the surjectivity of the map in the lemma.

\textit{Injectivity.}
We show that the pair $(\pi,\mathbf i)$ can be recovered from the monomial $M_{\pi,\mathbf i}$. Let $h$ be its height vector, and let $n$ be its diagonal degree. By Lemma~\ref{lem:spr-of-M}, the sorting permutation of $h$ is $\pi$. Thus $\pi$ is recovered from the monomial.

Once $\pi$ is known, its descent staircase $d_s^\pi$ is known. Define
\[
        g_s=h_{\pi_s}-d_s^\pi\qquad(1\leq s\leq m).
\]
For the monomial $M_{\pi,\mathbf i}$, equation~\eqref{eq:height-of-M-pi-i} says precisely that
\[
        g_s=i_s+i_{s+1}+\cdots+i_m.
\]
Hence the successive differences recover
\[
        i_s=g_s-g_{s+1}\qquad(1\leq s<m),
        \qquad
        i_m=g_m.
\]
Finally,
\[
        i_0=n-\des(\pi)-g_1,
\]
because $n$ is the diagonal degree of $M_{\pi,\mathbf i}$ and $\des(\pi)+g_1=\des(\pi)+i_1+\cdots+i_m$ is the contribution of all factors other than $(z_{P_0^\pi})^{i_0}$. Thus all entries of $\mathbf i$ are recovered. Therefore two different pairs $(\pi,\mathbf i)$ cannot give the same chain monomial.
This proves the injectivity of the map in the lemma.
\end{proof}

\begin{remark}[Stanley's fundamental theorem on $P$-partitions]\label{rem:ppartition-context}
Lemma~\ref{lem:ppartition-bijection} is precisely the antichain case of Stanley's fundamental theorem on $P$-partitions \cite[Lemma 3.15.3]{StanleyEC1}. A function $h:[m]\to\{0,1,\ldots,n\}$ is a $P$-partition for the labelled antichain on $[m]$, since there are no order relations. Sorting $h$ gives a unique permutation $\pi$; the tie-breaking rule produces the strict inequalities at descents of $\pi$. Subtracting the descent staircase $d_s^\pi$ leaves an ordinary partition $g_1\geq\cdots\geq g_m\geq0$, whose successive differences are the numbers $i_1,\ldots,i_m$; the number $i_0$ records the unused vertical room up to $n$.
\end{remark}

\begin{remark}[Eulerian identity from the bijection]\label{rem:eulerian-from-bijection}
Comparing cardinalities in Lemma~\ref{lem:ppartition-bijection} for fixed diagonal degree $n$ gives Worpitzky's identity
\[
        (n+1)^m
        =
        \sum_{\pi\in S_m}
        \binom{n-\des(\pi)+m}{m},
\]
where the binomial coefficient is interpreted as $0$ when $n-\des(\pi)<0$. Indeed, the left hand side counts height vectors
$h\in\{0,1,\ldots,n\}^m$, while for a fixed $\pi$ the number of
$(i_0,\ldots,i_m)\in\NN^{m+1}$ satisfying
$i_0+\cdots+i_m=n-\des(\pi)$ is the displayed binomial coefficient. Equivalently,
\[
        \sum_{n\geq0}(n+1)^m t^n
        =
        \frac{\sum_{\pi\in S_m}t^{\des(\pi)}}{(1-t)^{m+1}}.
\]
Conversely, if one already knows this identity, then in Lemma~\ref{lem:ppartition-bijection} it would be enough to prove either surjectivity or injectivity for each fixed $n$; the equality of the finite cardinalities would force the other property by the pigeonhole principle.
\end{remark}



\subsection{Triangularity and freeness}

Our next goal is to show that the chain monomials $M_{\pi, \mathbf i}$ are
the leading terms of the polynomials $u_\pi E_{\mathbf i}$ with respect
to a certain partial order on diagonal monomials. We shall now define this order.
It is useful to keep in mind that this order is \textbf{not a monomial order}.
It is only an order on the basis of diagonal chain monomials of a fixed
diagonal degree; it does not respect multiplication.

The \emph{decreasing rearrangement} of any $m$-tuple
$h =(h_1,\ldots,h_m) \in \ZZ^m$ is defined to be the $m$-tuple
$h^\downarrow=(h^\downarrow_1\geq h^\downarrow_2\geq\cdots\geq h^\downarrow_m)$
that contains the same entries as $h$ but sorted in weakly decreasing
order. For instance, $(2, 5, 2, 7)^\downarrow = (7, 5, 2, 2)$.
We note the following simple fact:

\begin{lemma} \label{lem:spr-and-rear-inj}
Let $h=(h_1,\ldots,h_m)$ and $k=(k_1,\ldots,k_m)$ be two $m$-tuples
in $\ZZ^m$.
If $\spr(h) = \spr(k)$ and $h^\downarrow = k^\downarrow$,
then $h = k$.
\end{lemma}

\begin{proof}
Let $\pi = \spr(h) = \spr(k)$.
Then, from \eqref{eq.spr.1}, we see that
$h^\downarrow = (h_{\pi_1}, h_{\pi_2}, \ldots, h_{\pi_m})$.
Likewise,
$k^\downarrow = (k_{\pi_1}, k_{\pi_2}, \ldots, k_{\pi_m})$.
Thus, from $h^\downarrow = k^\downarrow$, we obtain
$(h_{\pi_1}, h_{\pi_2}, \ldots, h_{\pi_m})
= (k_{\pi_1}, k_{\pi_2}, \ldots, k_{\pi_m})$.
Since $\pi$ is a permutation of $[m]$, this gives $h_i=k_i$ for every label $i$. Thus $h=k$.
\end{proof}

Let $h=(h_1,\ldots,h_m)$ and $k=(k_1,\ldots,k_m)$ be two height vectors with
entries in $\{0,1,\ldots,n\}$.  Let
\[
        h^\downarrow=(h^\downarrow_1\geq h^\downarrow_2\geq\cdots\geq h^\downarrow_m)
        \qquad \text{and} \qquad
        k^\downarrow=(k^\downarrow_1\geq k^\downarrow_2\geq\cdots\geq k^\downarrow_m)
\]
be their decreasing rearrangements.  We say that
$h^\downarrow$ is \emph{dominated by} $k^\downarrow$ if
\[
        h^\downarrow_1+\cdots+h^\downarrow_q
        \leq
        k^\downarrow_1+\cdots+k^\downarrow_q
        \qquad\text{for every }q\in\{1,2,\ldots,m\}.
\]
% Since all height vectors considered here have the same diagonal degree
% (not quite!), the case $q=m$ is always an equality.

We define the \emph{dominance--lex order} on diagonal monomials of diagonal
degree $n$ as follows. First compare their decreasingly sorted height vectors
by dominance. If these sorted height vectors are unequal, then the monomial
whose sorted height vector strictly dominates the other is declared to be
larger. If the decreasingly sorted height vectors are equal, break ties by the
sorting permutation: the height vector whose sorting permutation is
lexicographically smaller is declared to be larger. Explicitly, two distinct
diagonal monomials $\mathbf m$ and $\mathbf n$, with respective height vectors
$h$ and $k$, satisfy $\mathbf m > \mathbf n$ if and only if either
\[
h^\downarrow \text{ strictly dominates } k^\downarrow\text{ (i.e., dominates it and is not equal to it)},
\]
or else
\[
h^\downarrow = k^\downarrow
\quad\text{and}\quad
\spr(h) < \spr(k)
\]
(where the $<$ sign refers to lexicographic order of one-line notations).
This is a partial order\footnote{It is easily seen to be asymmetric
and transitive.},
because dominance is only a partial order. Nevertheless, the leading term
statements below make sense: a monomial is a leading monomial of a polynomial
if it is the unique maximal monomial among the monomials appearing in that
polynomial.
Not every polynomial has a leading monomial, but if it does, then this
monomial is unique.

For $\mathbf i=(i_0,i_1,\ldots,i_m)\in\NN^{m+1}$, put
\[
        E_{\mathbf i}=E_0^{i_0}E_1^{i_1}\cdots E_m^{i_m}.
\]
Also put
\[
        g_s=i_s+i_{s+1}+\cdots+i_m
        \qquad(1\leq s\leq m).
\]
Thus
\[
        g_1\geq g_2\geq\cdots\geq g_m\geq0.
\]

\begin{lemma}[dominance for terms of $E_{\mathbf i}$]\label{lem:E-i-dominance}
Let $w=(w_1,\ldots,w_m)$ be the height vector of a diagonal monomial appearing in
$E_{\mathbf i}$.  Then
\[
        w^\downarrow
        \text{ is dominated by }
        (g_1,g_2,\ldots,g_m).
\]
Moreover, for any permutation $\rho\in S_m$, the term
\[
        \prod_{j=0}^m \left(z_{P_j^\rho}\right)^{i_j}
\]
of $E_{\mathbf i}$ has height vector $w$ given by
\[
        w_{\rho_s}=g_s\qquad(1\leq s\leq m).
\]
\end{lemma}

\begin{proof}
Recall that $E_{\mathbf i} = E_0^{i_0} E_1^{i_1} \cdots E_m^{i_m}$,
where each $E_j$ is the sum of the $z_J$'s over all $j$-element subsets
$J \subseteq [m]$.
Thus, expanding this product, we obtain a sum of terms, each of which
is a product of such $z_J$'s. To be more specific:
A term of $E_{\mathbf i}$ is obtained by choosing, for each of the $i_j$
copies of $E_j$, a $j$-element subset $J\subseteq[m]$, and then
multiplying the $z_J$'s corresponding to all the chosen $J$'s.
In the height vector of this term, the height $w_x$ is
then the total number of chosen subsets $J$ that contain $x$.

Let $S$ be any $q$-element subset of $[m]$.  Each chosen $j$-element subset
meets $S$ in at most $\min(q,j)$ elements.  Hence
\[
        \sum_{x\in S}w_x
        \leq \sum_{j=0}^m i_j\min(q,j)
        =g_1+g_2+\cdots+g_q .
\]
Taking the maximum over all $q$-element subsets $S$ gives the desired
dominance inequality.

The final assertion follows directly from the definition of the prefixes
$P_j^\rho=\{\rho_1,\ldots,\rho_j\}$: the label $\rho_s$ lies in $P_j^\rho$
exactly when $j\geq s$, so it is counted $i_s+i_{s+1}+\cdots+i_m=g_s$
times.
\end{proof}

\begin{remark}
Lemma~\ref{lem:E-i-dominance} says, in particular, that the leading term of
$E_{\mathbf i}$ itself is
\[
        \prod_{j=0}^m\left(z_{\{1,2,\ldots,j\}}\right)^{i_j},
\]
with respect to the dominance--lex order.\footnote{We have not nailed
down the leading coefficient yet. (It is $1$, and this is a particular case
of Lemma~\ref{lem:unique-top-contribution} below.)}
  Indeed, the maximal sorted height
vector is $(g_1,\ldots,g_m)$, and among all terms with this sorted height
vector the lexicographically smallest sorting permutation is $12\cdots m$.
This is the usual triangularity underlying the fundamental theorem on
symmetric polynomials (see, e.g., \cite[\S 4.3, Proposition 5]{Bosch2018}),
in homogenized form.  The point of the next lemmas is
that, after multiplying by $u_\pi$, the relevant prefix order becomes
$\pi_1,\pi_2,\ldots,\pi_m$ rather than $1,2,\ldots,m$.
\end{remark}

\begin{lemma}[dominance after multiplying by $u_\pi$]\label{lem:dominance-after-u}
Let $\pi\in S_m$ and $\mathbf i=(i_0,\ldots,i_m)\in\NN^{m+1}$.  Let $h^*$ be
the height vector of
\[
        M_{\pi,\mathbf i}=u_\pi\prod_{j=0}^m\left(z_{P_j^\pi}\right)^{i_j}.
\]
Let $h$ be the height vector of any diagonal monomial appearing in
\[
        u_\pi E_{\mathbf i}.
\]
Then $h^\downarrow$ is dominated by $(h^*)^\downarrow$.
\end{lemma}

\begin{proof}
Put
\[
        d_s=d_s^\pi=\#\{q\in\Des(\pi)\mid q\geq s\},
        \qquad
        g_s=i_s+i_{s+1}+\cdots+i_m
        \qquad(1\leq s\leq m).
\]
The factor $u_\pi$ contributes height $d_s$ to the label $\pi_s$.  Hence,
the height vector $h^*$ satisfies
\[
        h^*_{\pi_s}=d_s+g_s\qquad(1\leq s\leq m),
\]
and $\spr(h^*)=\pi$ by Lemma~\ref{lem:spr-of-M}.  In particular,
\[
        (h^*)^\downarrow=(d_1+g_1,d_2+g_2,\ldots,d_m+g_m)
\]
because $d_1 \geq d_2 \geq \cdots \geq d_m$ and
$g_1 \geq g_2 \geq \cdots \geq g_m$.

Now consider the diagonal monomial in $u_\pi E_{\mathbf i}$
whose height vector is $h$.
Let $w = (w_1, w_2, \ldots, w_m)$ be the height vector of
the corresponding monomial in $E_{\mathbf i}$.
Thus,
\[
        h_x=d_x^{\pi,\mathrm{lab}}+w_x,
        \qquad\text{where }d_{\pi_s}^{\pi,\mathrm{lab}}=d_s
\]
(since $(d_1^{\pi,\mathrm{lab}}, d_2^{\pi,\mathrm{lab}},
\ldots, d_m^{\pi,\mathrm{lab}})$ is the height vector of
$u_\pi$).
Let $S$ be a $q$-element subset of $[m]$.  Since
$d_1\geq d_2\geq\cdots\geq d_m$, the largest possible sum of the
$d$-contributions over $q$ labels is $d_1+\cdots+d_q$.  Hence
\begin{align}
        \sum_{x\in S}d_x^{\pi,\mathrm{lab}}
        \leq d_1+\cdots+d_q .
\label{pf.lem:dominance-after-u.d-ineq}
\end{align}
By Lemma~\ref{lem:E-i-dominance}, applied in its proof to the same subset
$S$, we also have
\begin{align}
        \sum_{x\in S}w_x
        \leq g_1+\cdots+g_q .
\label{pf.lem:dominance-after-u.g-ineq}
\end{align}
Adding these inequalities gives
\begin{align}
        \sum_{x\in S}h_x
        \leq \sum_{s=1}^q(d_s+g_s)
        =\sum_{s=1}^q(h^*)^\downarrow_s .
\label{pf.lem:dominance-after-u.d+g-ineq}
\end{align}
Taking the maximum over all $q$-element subsets $S$ proves the dominance
claim.
% This is the special rearrangement inequality needed here: sorted
% $d$ plus sorted $w$ dominates the sorted sum.
\end{proof}

\begin{lemma}[the equality case]\label{lem:dominance-equality-case}
Keep the notation of Lemma~\ref{lem:dominance-after-u}.  Assume that
$h^\downarrow=(h^*)^\downarrow$.  Then
\[
        \spr(h)\geq_{\mathrm{lex}}\pi .
\]
Moreover, if $\spr(h)=\pi$, then $h=h^*$.
\end{lemma}

\begin{proof}
Let
\[
        \sigma=\spr(h).
\]
For $q\in\{1,2,\ldots,m\}$, set
\[
        S_q=\{\sigma_1,\ldots,\sigma_q\}.
\]
This is a $q$-element subset on which the sum of the entries of $h$ is
maximal.  Since $h^\downarrow=(h^*)^\downarrow$, the dominance inequality
\eqref{pf.lem:dominance-after-u.d+g-ineq} in
Lemma~\ref{lem:dominance-after-u} is an equality for $S=S_q$.  The two
inequalities \eqref{pf.lem:dominance-after-u.d-ineq}
and \eqref{pf.lem:dominance-after-u.g-ineq}
that were added in the proof of that lemma therefore must both
be equalities.  In particular,
\begin{equation}\label{eq:d-equality-for-Sq}
        \sum_{x\in S_q}d_x^{\pi,\mathrm{lab}}
        =d_1+d_2+\cdots+d_q
        \qquad(1\leq q\leq m).
\end{equation}

We now interpret this condition.  Decompose the word $\pi_1,\ldots,\pi_m$ into
its increasing runs\footnote{The \emph{increasing runs} of a word are
its inclusion-maximal increasing contiguous subsequences. For instance,
the increasing runs of the word $425816397$ are $4$, $258$, $16$, $39$ and $7$.}:
\[
        R_1\mid R_2\mid\cdots\mid R_t .
\]
The cuts occur exactly at the descents of $\pi$.  The sequence
$d_1,d_2,\ldots,d_m$ is constant on each run and strictly decreases when we
move from one run to the next.  Thus the labels in $R_1$ have the largest
$d$-value, the labels in $R_2$ have the next largest $d$-value, and so on.
Equation~\eqref{eq:d-equality-for-Sq}, for all $q$, says exactly that the
first $q$ entries of $\sigma$ always choose $q$ labels with largest possible
$d$-sum.  Consequently, $\sigma$ must list all labels of $R_1$ before all
labels of $R_2$, all labels of $R_2$ before all labels of $R_3$, and so on.
Indeed, if a label from a later run occurred before a label from an earlier
run, then for a suitable $q$ the set $S_q$ would contain the former but not
the latter; replacing the former by the latter would strictly increase the
left hand side of~\eqref{eq:d-equality-for-Sq}, a contradiction.

Thus $\sigma$ is obtained from $\pi$ by permuting entries only inside the
increasing runs of $\pi$.  Each such run is already written in increasing
order, and increasing order is lexicographically smallest among all orderings
of the same set of labels.  Therefore
\[
        \sigma\geq_{\mathrm{lex}}\pi .
\]

Finally, if $\spr(h)=\pi$, then $h$ and $h^*$ have the same decreasing
rearrangement and the same sorting permutation.
Thus, Lemma~\ref{lem:spr-and-rear-inj} yields $h = h^*$.
% This determines the height
% vector: for every $s$, the entry $h_{\pi_s}$ is the $s$-th entry of
% $h^\downarrow$, and the entry $h^*_{\pi_s}$ is the $s$-th entry of
% $(h^*)^\downarrow$.  Hence $h_{\pi_s}=h^*_{\pi_s}$ for all $s$, so $h=h^*$.
\end{proof}

\begin{lemma}[coefficient of the top contribution]\label{lem:unique-top-contribution}
Let $\pi\in S_m$ and $\mathbf i\in\NN^{m+1}$.  In the expansion of
$u_\pi E_{\mathbf i}$, the monomial $M_{\pi,\mathbf i}$
appears with coefficient $1$.
\end{lemma}

\begin{proof}
We must show that the coefficient of the monomial
\[
        M_{\pi,\mathbf i}
        =
        u_\pi\prod_{j=0}^m \left(z_{P_j^\pi}\right)^{i_j}
\]
in \(u_\pi E_{\mathbf i}\) is \(1\). Since multiplication by the monomial
\(u_\pi\) sends distinct monomials to distinct monomials, it is enough to
show that the coefficient of
\[
        \prod_{j=0}^m \left(z_{P_j^\pi}\right)^{i_j}
\]
in \(E_{\mathbf i}\) is \(1\).

For this proof only, we equip the polynomial ring $B_m$ with the
lexicographic monomial order corresponding to
\[
        a_{\pi_1}>a_{\pi_2}>\cdots>a_{\pi_m}>
        b_{\pi_1}>b_{\pi_2}>\cdots>b_{\pi_m}.
\]
We claim that, for each \(j\), the leading monomial of \(E_j\) is
\(z_{P_j^\pi}\), with coefficient \(1\). Indeed, among all \(j\)-element
subsets \(J\subseteq [m]\), the subset \(P_j^\pi=\{\pi_1,\ldots,\pi_j\}\)
is lexicographically first in the following sense: if \(J\neq P_j^\pi\), and
\(t\) is the first index such that \(\pi_t\) belongs to exactly one of
\(J\) and \(P_j^\pi\), then \(t\leq j\), and \(\pi_t\in P_j^\pi\setminus J\).
Thus \(z_{P_j^\pi}\) has exponent \(1\) on the variable \(a_{\pi_t}\), whereas
\(z_J\) has exponent \(0\) on this variable. Hence
\[
        z_{P_j^\pi}>z_J
\]
in our lexicographic order.

Therefore the polynomial $E_j$ has leading monomial
\[
        \LT(E_j)=z_{P_j^\pi}
        \qquad(0\leq j\leq m),
\]
with leading coefficient \(1\). Since leading monomials and leading
coefficients multiply in a monomial order, the product
\[
        E_{\mathbf i}=E_0^{i_0}E_1^{i_1}\cdots E_m^{i_m}
\]
has leading monomial
\[
        \prod_{j=0}^m \left(z_{P_j^\pi}\right)^{i_j}
\]
with coefficient \(1\).

Thus, as we have explained above,
the coefficient of \(M_{\pi,\mathbf i}\) in \(u_\pi E_{\mathbf i}\) is
also \(1\).
\end{proof}

% \begin{proof}
% We must show that the coefficient of the monomial
% $M_{\pi,\mathbf i} = u_\pi\prod_{j=0}^m \left(z_{P_j^\pi}\right)^{i_j}$
% in $u_\pi E_{\mathbf i}$ is $1$
% (since we already know that this monomial appears in $u_\pi E_{\mathbf i}$,
% coming from the prefix choice).

% Clearly, it suffices to prove that the coefficient of the monomial
% $\prod_{j=0}^m \left(z_{P_j^\pi}\right)^{i_j}$ in $E_{\mathbf i}$ is $1$
% (since multiplication by the monomial $u_\pi$ merely shifts the
% coefficients).
% But this can be done using monomial orders:
% Equip (just for this specific proof)
% the polynomial ring $B_m$ with the monomial order that is
% lexicographic with respect to the following ordering of
% variables:
% \[
% a_{\pi_1} > a_{\pi_2} > \cdots > a_{\pi_m}
% > b_{\pi_1} > b_{\pi_2} > \cdots > b_{\pi_m} .
% \]
% Then, each $E_j$ has leading term $z_{P_j^\pi}$ with
% coefficient $1$. Since leading terms and leading coefficients
% multiply, this implies that the product
% $E_{\mathbf i} = \prod_{j=0}^m E_j^{i_j}$ has leading term
% $\prod_{j=0}^m \left(z_{P_j^\pi}\right)^{i_j}$ with
% coefficient $1$.
% Thus the lemma follows.
% \end{proof}

% \begin{proof}[Old proof of Lemma \ref{lem:unique-top-contribution}.]
% Let $h^*$ be the height vector of $M_{\pi,\mathbf i}$.  Consider an arbitrary
% choice of terms from the factors $E_j$ in the product $u_\pi E_{\mathbf i}$,
% and assume that this choice produces a term whose height vector is $h^*$.
% We shall show that this arbitrary choice must be the prefix choice
% (i.e., the choice of $z_{P_j^\pi}$ from each copy of $E_j$).

% More explicitly, for each copy of $E_j$, the chosen term has the form $z_J$
% for some $j$-element subset $J$ of $[m]$.  Let $\mathcal J$ be the multiset of
% all subsets $J$ chosen in this way, over all copies of all the $E_j$'s.  Define
% a vector $w=(w_1,\ldots,w_m)$ by
% \[
        % w_x=\#\{J\in\mathcal J\mid x\in J\}
        % \qquad(x\in[m])
% \]
% (counted, of course, with multiplicity).
% Thus $w_x$ is the contribution of the chosen $E$-terms to the height at the
% label $x$.  If $d_x^{\pi,\mathrm{lab}}$ denotes the contribution of $u_\pi$
% to the height at $x$, then our assumption says
% \begin{equation}\label{pf.lem:unique-top.hstar-as-d-plus-w}
        % h_x^*=d_x^{\pi,\mathrm{lab}}+w_x
        % \qquad(x\in[m]).
% \end{equation}

% We first determine how much the vector $w$ contributes on each prefix
% $P_q^\pi$.  By Lemma~\ref{lem:spr-of-M}, we have $\spr(h^*)=\pi$.  Hence, if
% we sort the labels $1,2,\ldots,m$ by decreasing $h^*$-height, breaking ties
% in increasing order of the labels, then the sorted list is
% \[
        % \pi_1,\pi_2,\ldots,\pi_m .
% \]
% Consequently, for each $q\in\{0,1,\ldots,m\}$, the $q$ largest entries of
% $h^*$ are $h^*_{\pi_1}, h^*_{\pi_2}, \ldots, h^*_{\pi_q}$,
% that is, the $h^*_x$ for $x \in P_q^\pi$.  Therefore
% \begin{align}
        % \sum_{x\in P_q^\pi} h_x^*
        % &= \sum_{s=1}^q h^*_{\pi_s}        \notag\\
        % &= \sum_{s=1}^q \left(d_s+g_s\right),
        % \label{pf.lem:unique-top.hstar-prefix-sum}
% \end{align}
% where the last equality is the formula for the height vector of
% $M_{\pi,\mathbf i}$.
% On the other hand,
% \[
        % \sum_{x\in P_q^\pi} d_x^{\pi,\mathrm{lab}}
        % =d_1+d_2+\cdots+d_q .
% \]
% Subtracting this equality from~\eqref{pf.lem:unique-top.hstar-prefix-sum},
% and using~\eqref{pf.lem:unique-top.hstar-as-d-plus-w}, we obtain
% \begin{equation}\label{pf.lem:unique-top.w-prefix-sum}
        % \sum_{x\in P_q^\pi} w_x
        % =g_1+g_2+\cdots+g_q
        % \qquad(0\leq q\leq m).
% \end{equation}

% Now compare this equality with the elementary upper bound obtained from the
% chosen subsets.  Fix $q$.  For each $J\in\mathcal J$, we have
% \[
        % |J\cap P_q^\pi|\leq \min(q,|J|).
% \]
% Adding these inequalities over all $J\in\mathcal J$ gives
% \begin{align*}
        % \sum_{x\in P_q^\pi} w_x
        % &=\sum_{J\in\mathcal J}|J\cap P_q^\pi|  \\
        % &\leq \sum_{J\in\mathcal J}\min(q,|J|)
         % = g_1+g_2+\cdots+g_q .
% \end{align*}
% The last equality is the definition of the numbers $g_s$: a chosen subset of
% size $j$ contributes $1$ to each of $g_1,\ldots,g_j$ and $0$ to the later
% $g_s$'s, and there are $i_j$ chosen subsets of size $j$.
% But the left hand side is already equal to $g_1+\cdots+g_q$ by
% \eqref{pf.lem:unique-top.w-prefix-sum}.  Thus equality holds in each of the
% individual inequalities
% \[
        % |J\cap P_q^\pi|\leq \min(q,|J|)
        % \qquad(J\in\mathcal J)
% \]
% for this fixed $q$.  Since $q$ was arbitrary, equality holds for every
% $q\in\{0,1,\ldots,m\}$ and every $J\in\mathcal J$.

% Finally, take any chosen subset $J\in\mathcal J$, and let $j=|J|$.  Applying
% the preceding conclusion with $q=j$, we get
% \[
        % |J\cap P_j^\pi|=\min(j,j)=j.
% \]
% Both $J$ and $P_j^\pi$ have cardinality $j$, so this forces
% $J=P_j^\pi$.  Thus every copy of $E_j$ contributes the prefix term
% $z_{P_j^\pi}$, for every $j$.  This is precisely the prefix choice producing
% $M_{\pi,\mathbf i}$, and it is the only choice that can do so.  Hence the
% coefficient of $M_{\pi,\mathbf i}$ in $u_\pi E_{\mathbf i}$ is $1$.

% \end{proof}

% \begin{remark}[alternative proof of the coefficient]\label{rem:unique-top-symmetric-functions}
% The coefficient-$1$ part of Lemma~\ref{lem:unique-top-contribution} can also be viewed as a homogenized form of the classical triangularity of elementary symmetric functions.
% Let
% \[
        % \lambda=(m^{i_m},(m-1)^{i_{m-1}},\ldots,1^{i_1})
% \]
% be the partition having $i_j$ parts equal to $j$.  Then
% \[
        % \lambda^t=(g_1,g_2,\ldots,g_m),
        % \qquad
        % g_s=i_s+i_{s+1}+\cdots+i_m .
% \]
% In the Laurent polynomial ring obtained by inverting the $b_i$'s, put
% \[
        % x_i=\frac{a_i}{b_i}\qquad(1\leq i\leq m).
% \]
% Then
% \[
        % E_j=(b_1b_2\cdots b_m)e_j(x_1,\ldots,x_m),
% \]
% and therefore, if $N=i_0+i_1+\cdots+i_m$, then
% \[
        % E_{\mathbf i}=(b_1b_2\cdots b_m)^N e_\lambda(x_1,\ldots,x_m).
% \]
% The standard symmetric-functions triangularity says that the coefficient of
% $m_{\lambda^t}$ in $e_\lambda$ is $1$.  Equivalently, the coefficient of each
% monomial with exponent multiset $(g_1,\ldots,g_m)$, in particular
% \[
        % x_{\pi_1}^{g_1}x_{\pi_2}^{g_2}\cdots x_{\pi_m}^{g_m},
% \]
% is $1$ in $e_\lambda$.  After homogenizing back, this says that the coefficient of
% \[
        % \prod_{j=0}^m\left(z_{P_j^\pi}\right)^{i_j}
% \]
% in $E_{\mathbf i}$ is $1$.  Multiplication by the monomial $u_\pi$ is injective on monomials, so the coefficient of
% $M_{\pi,\mathbf i}$ in $u_\pi E_{\mathbf i}$ is also $1$.
% Since the expansion of $E_{\mathbf i}$ is obtained by choosing one positive-coefficient addend from each factor $E_j$, this coefficient-$1$ statement also implies the uniqueness of the prefix choice asserted in Lemma~\ref{lem:unique-top-contribution}.
% \end{remark}

\begin{lemma}[antichain triangularity]\label{lem:antichain-triangularity}
Let $\pi\in S_m$ and $\mathbf i=(i_0,\ldots,i_m)\in\NN^{m+1}$. Then the
polynomial
\[
        F_{\pi,\mathbf i}:=u_\pi E_{\mathbf i}
        =u_\pi E_0^{i_0}E_1^{i_1}\cdots E_m^{i_m}\in A_m
\]
has leading chain monomial, with respect to the dominance--lex order, equal
to
\[
        M_{\pi,\mathbf i}
        =u_\pi\prod_{j=0}^m\left(z_{P_j^\pi}\right)^{i_j}.
\]
The coefficient of this leading chain monomial is $1$.
\end{lemma}

\begin{proof}
Let $h^*$ be the height vector of $M_{\pi,\mathbf i}$, and let $h$ be the
height vector of any monomial $M'$ appearing in $F_{\pi,\mathbf i}$.
Our first goal is to prove that $M_{\pi,\mathbf i} \geq M'$ in the
dominance-lex order.
By Lemma~\ref{lem:dominance-after-u}, the sorted height vector
$h^\downarrow$ is dominated by $(h^*)^\downarrow$.  If this dominance is
strict, then $M_{\pi,\mathbf i} > M'$ follows immediately.

It remains to consider the case $h^\downarrow=(h^*)^\downarrow$.  By
Lemma~\ref{lem:dominance-equality-case}, we then have
$\spr(h)\geq_{\mathrm{lex}}\pi = \spr(h^*)$ (since $\spr(h^*)=\pi$ by
Lemma~\ref{lem:spr-of-M}).
If this inequality is strict, then the tie-breaker in the
dominance--lex order again ensures $M_{\pi,\mathbf i} > M'$.

The remaining case is when $h^\downarrow=(h^*)^\downarrow$ and
$\spr(h) = \spr(h^*)$. But in this case,
Lemma~\ref{lem:dominance-equality-case} gives $h=h^*$,
which forces $M_{\pi,\mathbf i} = M'$: the two monomials have the
same height vector and the same diagonal degree, and these data determine
a diagonal monomial.

So we have shown that the monomial $M_{\pi,\mathbf i}$ is the
leading chain monomial of $F_{\pi,\mathbf i}$. It remains to
show that its coefficient is $1$.
But this follows from Lemma~\ref{lem:unique-top-contribution}.
\end{proof}

\begin{remark}
\label{rem:triangularity-not-groebner}
Lemma~\ref{lem:antichain-triangularity} is not a Gr\"obner-basis statement about the elements $E_0,E_1,\ldots,E_m$. The dominance--lex order is a partial order on the diagonal monomial basis, not a multiplicative monomial order on $B_m$.
% In particular, the leading choice in $E_j$ depends on the descent monomial $u_\pi$ multiplying it: in $u_\pi E_j$, the leading contribution comes from the prefix $P_j^\pi$. This dependence on $\pi$ is exactly the $P$-partition phenomenon.
\end{remark}

\subsection{\label{subsec:split-pf}Proof of Theorem~\ref{thm:splitting}}

We now recall once again the three nested rings we are working with:
\begin{align*}
& \ C_m = \ZZ[E_0, E_1, \ldots, E_m] \\
\subseteq & \ A_m = \left(\text{diagonal subring of $B_m$}\right)
= \bigoplus_{n \geq 0} (B_m)_{n\delta} \\
\subseteq & \ B_m = \ZZ[a_1, b_1, a_2, b_2, \ldots, a_m, b_m].
\end{align*}

\begin{proposition}\label{prop:A-free-over-C}
The $C_m$-module $A_m$ is free with basis
\[
        \{u_\pi\mid \pi\in S_m\}.
\]
In particular, since $u_{12\cdots m}=1$, the inclusion $C_m\hookrightarrow A_m$ has a $C_m$-linear retraction.
\end{proposition}

\begin{proof}
Since $C_m$ is the polynomial ring $\ZZ[E_0,\ldots,E_m]$, it has a basis consisting of all monomials $E_{\mathbf i} = E_0^{i_0}\cdots E_m^{i_m}$ with $\mathbf i = (i_0, i_1, \ldots, i_m) \in \NN^{m+1}$. The products
\[
        F_{\pi,\mathbf i}=u_\pi E_0^{i_0}\cdots E_m^{i_m}
        \qquad(\pi\in S_m,\ \mathbf i\in\NN^{m+1})
\]
are precisely the elements obtained by multiplying the proposed basis elements $u_\pi$ by this monomial basis of $C_m$.

By Lemma~\ref{lem:antichain-triangularity}, each $F_{\pi,\mathbf i}$ has leading chain monomial $M_{\pi,\mathbf i}$ with coefficient $1$. By Lemma~\ref{lem:ppartition-bijection}, the monomials $M_{\pi,\mathbf i}$ are pairwise distinct and, in fact, run over the entire chain-monomial basis of $A_m$.

Therefore, in each diagonal degree, the transition matrix from the family
\[
        \{F_{\pi,\mathbf i}\}
\]
to the chain-monomial basis of $A_m$ is unitriangular after choosing any linear extension of the partial order used in Lemma~\ref{lem:antichain-triangularity}. Hence the $F_{\pi,\mathbf i}$ form a $\ZZ$-basis of $A_m$.

Equivalently,
\[
        A_m=\bigoplus_{\pi\in S_m} C_m u_\pi
\]
as a $C_m$-module. Since the identity permutation $12\cdots m$ has no descents, $u_{12\cdots m}=1$, so the summand for the identity permutation is exactly $C_m$. Projection onto this summand is a $C_m$-linear retraction $A_m\to C_m$.
\end{proof}

\begin{remark}[What is free after base change]
\label{rem:what-is-free}
The freeness part of Proposition~\ref{prop:A-free-over-C} is stronger than the retraction part: it says that the diagonal subring $A_m$ is a free $C_m$-module of rank $m!$, with a basis containing $1$.
Consequently, for every $C_m$-algebra $R$, the base change
\[
        R\otimes_{C_m}A_m
\]
is a free $R$-module of rank $m!$, with basis $1\otimes u_\pi$ for $\pi\in S_m$.

The analogous freeness statement for the whole ring $B_m$ is false for $m\geq2$, even if one allows an infinite basis. In fact, $B_m$ is not flat over $C_m$ for $m\geq2$. See Corollary~\ref{cor:not-flat} below for a proof of this.

Thus the freeness used in this paper is precisely the freeness of the diagonal summand $A_m$, not of all of $B_m$. What we use for $B_m$ is only that $A_m$ is a $C_m$-direct summand of $B_m$ via the multidegree projection. Hence, after any base change $C_m\to R$, the $R$-algebra
\[
        R\otimes_{C_m}B_m
\]
contains the finite free $R$-module $R\otimes_{C_m}A_m$ as an $R$-module direct summand, and in particular contains the copy of $R=R\otimes_{C_m}C_m$ as a direct summand.
This is what we will apply below.
\end{remark}

\begin{remark}[Hilbert-series consistency]\label{rem:hilbert-series-consistency}
Taking Hilbert series in the free decomposition of Proposition~\ref{prop:A-free-over-C} gives again the generating-function identity of Remark~\ref{rem:eulerian-from-bijection}. Thus the algebraic basis theorem and the labelled-antichain $P$-partition count are two explanations of the same numerical identity for Eulerian numbers.
\end{remark}

\begin{remark}[Garsia--Stanton context]\label{rem:garsia-stanton-context}
The monomials $u_\pi$ are variants of the Garsia--Stanton descent monomials (\cite[(7.21)]{GarsiaStanton1984}, \cite[(6.7)]{Garsia1980}, \cite[(2)]{Allen1994}).  The idea goes back to Garsia's and Stanton's work on group actions on Stanley--Reisner rings and invariant rings \cite{Garsia1980}, \cite{GarsiaStanton1984}.  In the classical coinvariant algebra of type $A$, the corresponding monomials form what is now usually called the Garsia--Stanton descent basis; see, for example, Adin--Brenti--Roichman \cite{AdinBrentiRoichman2005}.  The proof above can be viewed as the labelled-antichain, diagonal-monomial shadow of this construction.  A close precursor to our proof, but situated in the ordinary polynomial ring $\ZZ[x_1, x_2, \ldots, x_m]$ instead of $A_m$, appears in Allen's \cite{Allen1994}.
\end{remark}

\begin{remark}[Cohen--Macaulay and quotient-ring context]\label{rem:cm-context}
Another standard route is to identify $A_m$ with the homogeneous coordinate ring of the Segre embedding of $(\mathbb P^1)^m$, or equivalently with the Hibi ring of the Boolean lattice. This ring is Cohen--Macaulay; the elements $E_0,\ldots,E_m$ form a homogeneous system of parameters; hence they form a regular sequence. This gives the Hilbert series of $A_m/(E_0,\ldots,E_m)A_m$, after which one can lift a basis of the quotient. The present proof avoids that quotient and proves the free $C_m$-basis directly by triangularity. For background on Stanley--Reisner rings and the Cohen--Macaulay viewpoint, see \cite{StanleyCCA}.
\end{remark}

\begin{proof}[Proof of Theorem~\ref{thm:splitting}]
By Proposition~\ref{prop:A-free-over-C}, the inclusion $C_m\hookrightarrow A_m$ has a $C_m$-linear retraction $\rho : A_m \to C_m$.
The decomposition of $B_m$ into multihomogeneous components gives a direct sum decomposition into $C_m$-submodules according to the cosets of $\ZZ\delta$ in $\ZZ^m$, because multiplication by any $E_r$ shifts multidegree by $\delta$.
The summand corresponding to the coset of $0$ is $A_m$. Hence the multihomogeneous projection
\[
        B_m\to A_m
\]
is $C_m$-linear (even $A_m$-linear) and restricts to the identity on $A_m$. Composing this projection with the retraction $\rho : A_m\to C_m$ gives a $C_m$-linear retraction
\[
        B_m\to A_m\to C_m.
\]
This proves the theorem.
\end{proof}

\subsection{Non-freeness of the polynomial ring over the diagonal subring}

This subsection is meant to establish the non-freeness (and non-flatness) of $B_m$ as a $C_m$-module, which was claimed in Remark~\ref{rem:what-is-free}.
Other than this, it is inconsequential.

\begin{proposition}[A root syzygy among the coefficients]\label{prop:root-syzygy}
For each $i\in[m]$, we have the identity
\[
        \sum_{j=0}^m (-1)^j a_i^{m-j} b_i^j E_j=0
        \qquad \text{in the polynomial ring $B_m$.}
\]
\end{proposition}

\begin{proof}
Fix $i\in[m]$. Recall from Proposition \ref{prop:hg-vieta}
that
\[
        \sum_{j=0}^m E_jX^j
        =
        \prod_{r=1}^m (a_rX+b_r)
        \qquad \text{in the univariate polynomial ring $B_m[X]$.}
\]
Homogenizing this equality (by substituting $X/Y$ for $X$ and
multiplying the result by $Y^m$), we obtain
\[
        \sum_{j=0}^m E_jX^jY^{m-j}
        =
        \prod_{r=1}^m (a_rX+b_rY)
        \qquad \text{in the bivariate polynomial ring $B_m[X,Y]$.}
\]
Now, substitute $X=-b_i$ and $Y=a_i$ here. The right hand side becomes
\[
        \prod_{r=1}^m (a_r(-b_i)+b_ra_i),
\]
which is $0$, since its $i$-th factor is $a_i(-b_i)+b_ia_i=0$. Thus the left hand side is $0$ as well; that is,
\[
        \sum_{j=0}^m E_j(-b_i)^j a_i^{m-j}=0.
\]
In other words,
\[
        \sum_{j=0}^m (-1)^j a_i^{m-j} b_i^j E_j=0.
\]
This proves the proposition.
\end{proof}

\begin{corollary} \label{cor:not-flat}
Assume that $m \geq 2$.
Then, the $C_m$-module $B_m$ is not flat.
\end{corollary}

\begin{proof}
Taking $i=1$ in Proposition~\ref{prop:root-syzygy}, we find
\[
        a_1^mE_0-a_1^{m-1}b_1E_1+a_1^{m-2}b_1^2E_2-
        \cdots+(-1)^m b_1^mE_m=0.
\]
Hence
\[
        b_1^mE_m\in (E_0,E_1,\ldots,E_{m-1})B_m.
\]
However,
\[
        b_1^m\notin (E_0,E_1,\ldots,E_{m-1})B_m.
\]
Indeed, specialize $b_2=b_3=\cdots=b_m=0$. Then
\[
        \prod_{r=1}^m(a_rX+b_r)
        \mapsto
        (a_1X+b_1)a_2X\cdots a_mX,
\]
so the ideal $(E_0,E_1,\ldots,E_{m-1})B_m$ maps into the principal ideal
\[
        (b_1a_2a_3\cdots a_m),
\]
which does not contain $b_1^m$ when $m\geq2$.
Thus $E_m$ is a zero-divisor modulo $(E_0,E_1,\ldots,E_{m-1})B_m$. Hence
\[
        E_0,E_1,\ldots,E_m
\]
is not a regular sequence in $B_m$.
But $E_0,E_1,\ldots,E_m$ is a regular sequence in the polynomial ring $C_m=\ZZ[E_0,E_1,\ldots,E_m]$. Since flat base change preserves regular sequences, $B_m$ cannot be flat over $C_m$.
\end{proof}

In particular, $B_m$ is not free over $C_m$ for $m\geq2$. The case $m=1$ is exceptional and trivial: then $C_1=B_1$.

\section{\label{sec:split}Application to splitting arbitrary polynomials}

We now deduce some consequences for univariate polynomials.

\begin{theorem}\label{thm:main}
Let $R$ be a commutative ring, and let $m$ be a positive integer.
Let $P = p_0 + p_1 X + \cdots + p_m X^m \in R[X]$ (with $p_0, p_1, \ldots, p_m \in R$) be a univariate polynomial of degree $\leq m$.
Then there exists a commutative ring $S$ and an injective ring homomorphism $R\hookrightarrow S$ such that $P$ factors in $S[X]$ as a product of $m$ polynomials of degree at most $1$.
\end{theorem}

\begin{proof}
% If $P=0$, choose $m=1$. If $P\neq0$, choose an integer
% \[
        % m\geq \max\{1,\deg P\}.
% \]
% Write
% \[
        % P=p_0+p_1X+\cdots+p_mX^m,
% \]
% where $p_i=0$ for $i>\deg P$.
%
Let $A_m$, $B_m$, $C_m$, and $E_0,E_1,\ldots,E_m$ be as in Section~\ref{sec:ABC}. By Lemma~\ref{lem:alg-independent}, $C_m=\ZZ[E_0,E_1,\ldots,E_m]$ is a polynomial ring over $\ZZ$ on the generators $E_0,E_1,\ldots,E_m$. Hence there is a unique ring homomorphism
\[
        \varphi:C_m\to R,
        \qquad
        E_i\longmapsto p_i.
\]
Define
\[
        S=R\otimes_{C_m}B_m,
\]
where $R$ is regarded as a $C_m$-algebra via $\varphi$.

The inclusion $C_m\hookrightarrow B_m$ is split as a $C_m$-module map by Theorem~\ref{thm:splitting}. Tensoring this split injection with $R$ over $C_m$ yields a split injection of $R$-modules
\begin{align}
        R=R\otimes_{C_m}C_m
        \hookrightarrow
        R\otimes_{C_m}B_m=S
\label{pf:thm:main:RtoS}
\end{align}
(since split injections remain split injections under base change).
In particular, the natural ring homomorphism $R\to S$ is injective. We identify $R$ with its image in $S$,
equating each $r \in R$ with $r \otimes_{C_m} 1 \in S$.

More precisely, Proposition~\ref{prop:A-free-over-C} yields that
\[
        R\otimes_{C_m}A_m
\]
is a free $R$-module of rank $m!$, with basis $1\otimes u_\pi$ for $\pi\in S_m$, and the copy of $R$ generated by $1\otimes u_{12\cdots m}=1$ is a direct summand of this free module.  The splitting algebra $S=R\otimes_{C_m}B_m$ contains this module as an $R$-module direct summand.  Thus the construction gives not merely injectivity of $R\to S$, but a split injection of underlying $R$-modules.

The canonical ring morphism $\xi : B_m \to R \otimes_{C_m} B_m = S$ (sending each $u \in B_m$ to $1 \otimes u \in S$) sends each $E_j \in B_m$ to $1 \otimes_{C_m} E_j = \varphi\left(E_j\right) \otimes_{C_m} 1 = p_j \otimes_{C_m} 1 = p_j \in S$.
Let $\alpha_i,\beta_i\in S$ be the images of $a_i,b_i\in B_m$, respectively, under this morphism $\xi$. In $B_m[X]$ we have the universal identity
\[
        \prod_{i=1}^m(a_iX+b_i)=\sum_{j=0}^m E_jX^j
\]
(a restatement of \eqref{prop:hg-vieta:1}).
After tensoring with $R$ (that is, applying $\xi$), this becomes
\[
        \prod_{i=1}^m(\alpha_iX+\beta_i)
        =
        \sum_{j=0}^m p_jX^j
        =P
\]
in $S[X]$. Hence $P$ splits in $S[X]$ as a product of $m$ degree-$\leq1$ polynomials.
\end{proof}

\begin{proposition}\label{prop:explicit-S}
The ring $S$ constructed in the above proof of Theorem~\ref{thm:main} can be written explicitly as
\[
        S\cong
        R[a_1,b_1,\ldots,a_m,b_m]
        \Big/
        (E_0-p_0,E_1-p_1,\ldots,E_m-p_m),
\]
where now the same formula for $E_r$ is interpreted inside the polynomial ring
$R[a_1,b_1,\ldots,a_m,b_m]$:
\[
        E_r=\sum_{\substack{I\subseteq[m];\\ |I|=r}}
        \prod_{i\in I}a_i\prod_{i\notin I}b_i.
\]
\end{proposition}

\begin{proof}
Let $a_{1}^{\prime},b_{1}^{\prime},a_{2}^{\prime},b_{2}^{\prime},\ldots
,a_{m}^{\prime},b_{m}^{\prime}$ be $2m$ further indeterminates, and let%
\[
E_{r}^{\prime}=\sum_{\substack{I\subseteq [m];\\|I|=r}}\prod_{i\in
I}a_{i}^{\prime}\prod_{i\notin I}b_{i}^{\prime}%
\]
over the ring $\ZZ$. Then, we claim that
\begin{equation}
B_{m}\cong C_{m}\left[  a_{1}^{\prime},b_{1}^{\prime},\ldots,a_{m}^{\prime
},b_{m}^{\prime}\right]  \Big/(E_{0}^{\prime}-E_{0},E_{1}^{\prime}%
-E_{1},\ldots,E_{m}^{\prime}-E_{m})\label{pf:prop:explicit-S:1}%
\end{equation}
as $C_m$-algebras.
Indeed, Lemma \ref{lem:alg-independent} shows that $C_{m}$ is the polynomial
ring $\ZZ[E_{0},E_{1},\ldots,E_{m}]$; thus,%
\begin{align*}
& C_{m}\left[  a_{1}^{\prime},b_{1}^{\prime},\ldots,a_{m}^{\prime}%
,b_{m}^{\prime}\right]  \Big/(E_{0}^{\prime}-E_{0},E_{1}^{\prime}-E_{1}%
,\ldots,E_{m}^{\prime}-E_{m})\\
& \cong\left(  \ZZ[E_{0},E_{1},\ldots,E_{m}]\right)  \left[
a_{1}^{\prime},b_{1}^{\prime},\ldots,a_{m}^{\prime},b_{m}^{\prime}\right]
\Big/(E_{0}^{\prime}-E_{0},E_{1}^{\prime}-E_{1},\ldots,E_{m}^{\prime}%
-E_{m})\\
& \cong\left(  \ZZ\left[  a_{1}^{\prime},b_{1}^{\prime},\ldots
,a_{m}^{\prime},b_{m}^{\prime}\right]  \right)  \left[  E_{0},E_{1}%
,\ldots,E_{m}\right]  \Big/(E_{0}^{\prime}-E_{0},E_{1}^{\prime}-E_{1}%
,\ldots,E_{m}^{\prime}-E_{m})\\
& \cong\ZZ\left[  a_{1}^{\prime},b_{1}^{\prime},\ldots,a_{m}^{\prime
},b_{m}^{\prime}\right] ,
\end{align*}
since any ring $P$ and any elements $u_{0},u_{1},\ldots,u_{m}\in P$ satisfy
the $P$-algebra isomorphism
\[
P\left[  E_{0},E_{1},\ldots,E_{m}\right]  \Big/\left(  u_{0}-E_{0}%
,u_{1}-E_{1},\ldots,u_{m}-E_{m}\right)  \cong P
\]
(because adjoining a variable
to a ring $P$ and then setting this variable equal to an element of $P$ gives
back $P$). Hence,%
\begin{align*}
& C_{m}\left[  a_{1}^{\prime},b_{1}^{\prime},\ldots,a_{m}^{\prime}%
,b_{m}^{\prime}\right]  \Big/(E_{0}^{\prime}-E_{0},E_{1}^{\prime}-E_{1}%
,\ldots,E_{m}^{\prime}-E_{m})\\
& \cong\ZZ\left[  a_{1}^{\prime},b_{1}^{\prime},\ldots,a_{m}^{\prime
},b_{m}^{\prime}\right]  \cong\ZZ\left[  a_{1}^{\prime},b_{1}^{\prime
},\ldots,a_{m}^{\prime},b_{m}^{\prime}\right]  =B_{m},
\end{align*}
which proves \eqref{pf:prop:explicit-S:1}.

Now,%
\begin{align*}
S  & =R\otimes_{C_{m}}B_{m}\\
& \cong R\otimes_{C_{m}}\left(  C_{m}\left[  a_{1}^{\prime},b_{1}^{\prime
},\ldots,a_{m}^{\prime},b_{m}^{\prime}\right]  \Big/(E_{0}^{\prime}%
-E_{0},E_{1}^{\prime}-E_{1},\ldots,E_{m}^{\prime}-E_{m})\right)  \\
& \qquad\qquad\left(  \text{by \eqref{pf:prop:explicit-S:1}}\right)  \\
& \cong\left(  R\otimes_{C_{m}}C_{m}\left[  a_{1}^{\prime},b_{1}^{\prime
},\ldots,a_{m}^{\prime},b_{m}^{\prime}\right]  \right)  \Big/(E_{0}^{\prime
}-E_{0},E_{1}^{\prime}-E_{1},\ldots,E_{m}^{\prime}-E_{m})\\
& \qquad\qquad\left(  \text{since quotienting by an ideal commutes with base
change}\right)  \\
& \cong R\left[  a_{1}^{\prime},b_{1}^{\prime},\ldots,a_{m}^{\prime}%
,b_{m}^{\prime}\right]  \Big/(E_{0}^{\prime}-p_{0},E_{1}^{\prime}-p_{1}%
,\ldots,E_{m}^{\prime}-p_{m})\\
& \qquad\qquad\left(
\begin{array}
[c]{c}%
\text{since }R\otimes_{C_{m}}C_{m}\left[  a_{1}^{\prime},b_{1}^{\prime}%
,\ldots,a_{m}^{\prime},b_{m}^{\prime}\right]  \cong R\left[  a_{1}^{\prime
},b_{1}^{\prime},\ldots,a_{m}^{\prime},b_{m}^{\prime}\right]  \text{ as
}R\text{-algebras,}\\
\text{and the isomorphism sends each }E_{j}\text{ to }p_{j}%
\end{array}
\right)  \\
& \cong R[a_{1},b_{1},\ldots,a_{m},b_{m}]\Big/(E_{0}-p_{0},E_{1}-p_{1}%
,\ldots,E_{m}-p_{m}).
\end{align*}
This proves the proposition.
% Let
% \[
        % T=\ZZ[t_0,t_1,\ldots,t_m].
% \]
% By Lemma~\ref{lem:alg-independent}, the map
% \[
        % T\longrightarrow C_m,\qquad t_j\longmapsto E_j
% \]
% is an isomorphism. We identify $T$ with $C_m$ by this isomorphism. Under this identification, the $C_m$-algebra structure on $R$ used in the proof of Theorem~\ref{thm:main} is the $T$-algebra structure determined by
% \[
        % t_j\longmapsto p_j \qquad (0\leq j\leq m).
% \]
% Thus
% \[
        % S=R\otimes_{C_m}B_m
        % \cong
        % R\otimes_T \ZZ[a_1,b_1,\ldots,a_m,b_m],
% \]
% where $T$ acts on the second factor by $t_j\mapsto E_j$.
%
% We claim that this tensor product is naturally isomorphic to
% \[
        % R[a_1,b_1,\ldots,a_m,b_m]
        % /(E_0-p_0,E_1-p_1,\ldots,E_m-p_m).
% \]
% Indeed, the canonical $R$-algebra map
% \[
        % R[a_1,b_1,\ldots,a_m,b_m]
        % \longrightarrow
        % R\otimes_T \ZZ[a_1,b_1,\ldots,a_m,b_m]
% \]
% that sends each $r\in R$ to $r\otimes 1$ and sends the variables $a_i,b_i$ to
% $1\otimes a_i,1\otimes b_i$ sends each $E_j-p_j$ to $0$, since
% \[
        % 1\otimes E_j=t_j\otimes 1=p_j\otimes 1.
% \]
% Hence it factors through the displayed quotient.
%
% Conversely, the quotient has the universal property of the tensor product. More explicitly, let $Q$ be an $R$-algebra. To give an $R$-algebra homomorphism
% \[
        % R[a_1,b_1,\ldots,a_m,b_m]
        % /(E_0-p_0,E_1-p_1,\ldots,E_m-p_m)
        % \longrightarrow Q
% \]
% is the same as to choose elements $x_i,y_i\in Q$ such that
% \[
        % E_j(x_1,y_1,\ldots,x_m,y_m)=p_j
        % \qquad(0\leq j\leq m).
% \]
% This is the same as to give a ring homomorphism
% \[
        % \psi:\ZZ[a_1,b_1,\ldots,a_m,b_m]\longrightarrow Q,
        % \qquad a_i\longmapsto x_i,
        % \quad b_i\longmapsto y_i,
% \]
% that is compatible with the two induced $T$-algebra structures on $Q$. Let us spell
% out what this means. The polynomial $P=\sum_{j=0}^m p_jX^j$ makes $R$ into a
% $T$-algebra by
% \[
        % t_j\longmapsto p_j \qquad(0\leq j\leq m).
% \]
% Since $Q$ is an $R$-algebra, with structure map say $\alpha:R\to Q$, this also
% makes $Q$ into a $T$-algebra via the composite
% \[
        % T\longrightarrow R\xrightarrow{\alpha} Q,
        % \qquad
        % t_j\longmapsto p_j\longmapsto \alpha(p_j).
% \]
% On the other hand, the map $T\to \ZZ[a_1,b_1,\ldots,a_m,b_m]$ sends
% $t_j$ to $E_j$, so the map $\psi$ makes $Q$ into a $T$-algebra with
% \[
        % t_j\longmapsto \psi(E_j)=E_j(x_1,y_1,\ldots,x_m,y_m).
% \]
% Thus the compatibility condition is precisely
% \[
        % E_j(x_1,y_1,\ldots,x_m,y_m)=\alpha(p_j)
        % \qquad(0\leq j\leq m),
% \]
% which, as usual, we write simply as $E_j(x_1,y_1,\ldots,x_m,y_m)=p_j$ inside
% $Q$. This is precisely the universal property of
% \[
        % R\otimes_T\ZZ[a_1,b_1,\ldots,a_m,b_m].
% \]
% Therefore the two $R$-algebras are naturally isomorphic. This proves the proposition.
\end{proof}

Note that Proposition~\ref{prop:explicit-S} gives what is probably the simplest construction of $S$.
The point of Theorem~\ref{thm:splitting} is that the natural map $R\to S$ is injective for every specialization $E_i\mapsto p_i$.

\begin{theorem}[Universal property of the splitting algebra]\label{thm:universal-property}
Let $R$, $m$, $P$ and $p_0,\ldots,p_m$ be as in Theorem~\ref{thm:main}. Let
\[
        S\cong
        R[a_1,b_1,\ldots,a_m,b_m]
        \Big/
        (E_0-p_0,E_1-p_1,\ldots,E_m-p_m)
\]
be the $R$-algebra of Proposition~\ref{prop:explicit-S}, and let $\alpha_i,\beta_i\in S$ denote the images of $a_i,b_i$ in this algebra $S$.

For every $R$-algebra $T$, the map
\[
        \operatorname{Hom}_{R\text{-alg}}(S,T)
        \longrightarrow
        \left\{(c_1,d_1,\ldots,c_m,d_m)\in T^{2m}\ \middle|\
        P=\prod_{i=1}^m(c_iX+d_i)\text{ in }T[X]\right\}
\]
that sends an $R$-algebra homomorphism $f:S\to T$ to
\[
        (f(\alpha_1),f(\beta_1),\ldots,f(\alpha_m),f(\beta_m))
\]
is a bijection.
Equivalently, $S$ is universal for ordered factorizations of $P$ into exactly $m$ linear-or-constant factors.
\end{theorem}

\begin{proof}
Let $T$ be an $R$-algebra. Since $S$ is presented as a quotient of the polynomial $R$-algebra
\[
        R[a_1,b_1,\ldots,a_m,b_m],
\]
an $R$-algebra homomorphism $f:S\to T$ is the same as a choice of images
\[
        c_i=f(\alpha_i),\qquad d_i=f(\beta_i)\qquad(1\leq i\leq m)
\]
for the variables $a_i,b_i$ satisfying the defining relations of the quotient. These relations are
\[
        E_j(c_1,d_1,\ldots,c_m,d_m)=p_j
        \qquad(0\leq j\leq m),
\]
where
\[
        E_j(c_1,d_1,\ldots,c_m,d_m)
        =
        \sum_{\substack{I\subseteq[m];\\ |I|=j}}
        \prod_{i\in I}c_i\prod_{i\notin I}d_i .
\]
But these are exactly the coefficient identities obtained by expanding
\[
        \prod_{i=1}^m(c_iX+d_i)
        =
        \sum_{j=0}^m
        E_j(c_1,d_1,\ldots,c_m,d_m)X^j .
\]
Thus the defining relations hold if and only if
\[
        \prod_{i=1}^m(c_iX+d_i)=\sum_{j=0}^m p_jX^j=P
        \qquad\text{in }T[X].
\]
This proves that the displayed map is well-defined and surjective.

It remains only to note uniqueness. Once the tuple $(c_1,d_1,\ldots,c_m,d_m)$ is specified, there is a unique $R$-algebra map
\[
        R[a_1,b_1,\ldots,a_m,b_m]\to T
\]
sending $a_i\mapsto c_i$ and $b_i\mapsto d_i$. If the tuple gives a factorization of $P$, then the relations $E_j-p_j$ vanish under this map, so it factors uniquely through $S$. Hence two homomorphisms $S\to T$ with the same tuple are equal. This proves injectivity of the displayed map and completes the proof.
\end{proof}

A factorization with fewer than $m$ factors can be regarded as one with exactly $m$ factors by adjoining extra factors $0X+1$.

\begin{thebibliography}{9}

\bibitem{AdinBrentiRoichman2005}
\href{https://doi.org/10.1090/S0002-9947-04-03494-4}{R. M. Adin, F. Brenti, and Y. Roichman,
\textit{Descent representations and multivariate statistics},
Transactions of the American Mathematical Society \textbf{357} (2005), no.~8, 3051--3082.}

\bibitem{Allen1994}
\href{https://doi.org/10.1023/A:1022481303750}{E. E. Allen,
\textit{The Descent Monomials and a Basis for the Diagonally Symmetric Polynomials},
Journal of Algebraic Combinatorics \textbf{3} (1994), 5--16.}

\bibitem{Bosch2018}
\href{https://doi.org/10.1007/978-3-319-95177-5}{Siegfried Bosch,
\textit{Algebra: From the Viewpoint of Galois Theory},
Springer 2018.}

\bibitem{CoxLittleOShea2025}
\href{https://doi.org/10.1007/978-3-031-91841-4}{David A. Cox, John Little, Donal O'Shea,
\textit{Ideals, Varieties, and Algorithms},
Undergraduate Texts in Mathematics,
5th edition, Springer 2025.}

\bibitem{Garsia1980}
\href{https://doi.org/10.1016/0001-8708(80)90006-7}{Adriano M. Garsia,
\textit{Combinatorial Methods in the Theory of Cohen--Macaulay Rings},
Advances in Mathematics \textbf{38} (1980), 229--266.}

\bibitem{GarsiaStanton1984}
\href{https://doi.org/10.1016/0001-8708(84)90005-7}{A. M. Garsia and D. Stanton,
\textit{Group actions on Stanley--Reisner rings and invariants of permutation groups},
Advances in Mathematics \textbf{51} (1984), no.~2, 107--201.}

\bibitem{Hibi1987}
\href{https://doi.org/10.2969/aspm/01110093}{T. Hibi,
\textit{Distributive lattices, affine semigroup rings and algebras with straightening laws},
in \textit{Commutative Algebra and Combinatorics}, Advanced Studies in Pure Mathematics \textbf{11}, North-Holland, Amsterdam, 1987, pp.~93--109.}

\bibitem{StanleyEC1}
\href{https://doi.org/10.1017/CBO9781139058520}{R. P. Stanley,
\textit{Enumerative Combinatorics, Volume 1}, second edition,
Cambridge Studies in Advanced Mathematics \textbf{49}, Cambridge University Press, 2012.}

\bibitem{StanleyCCA}
\href{https://doi.org/10.1007/b139094}{R. P. Stanley,
\textit{Combinatorics and Commutative Algebra}, second edition,
Progress in Mathematics \textbf{41}, Birkh\"auser, Boston, 1996.}

\end{thebibliography}

\end{document}
