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\begin{document}

\title{Polynomials over Symmetric Polynomials\\[5pt] {\large {[expository note based on the literature and emails of Darij
Grinberg and Victor Reiner]}}}
\author{GPT-5.6 Sol, edited by Darij Grinberg}
\date{not quite finished yet, \today}
\maketitle

\begin{abstract}
\textbf{Abstract.} Let $\mathbf{k}$ be a commutative ring, and let the
symmetric group $\mathfrak{S}_{n}$ act on $P=\mathbf{k}[x_{1},x_{2}%
,\ldots,x_{n}]$ by permuting the variables. We prove four classical results.
First, the coinvariant algebra (the quotient of $P$ by the ideal generated by
the symmetric polynomials with constant term $0$) is a free $\mathbf{k}%
$-module of rank $n!$, with the residue classes of the Artin monomials as a
basis. Second, $P$ is a free module of rank $n!$ over the ring
$P^{\mathfrak{S}_{n}}$ of symmetric polynomials, again with the Artin
monomials as a basis. Third, if $n!$ is invertible in $\mathbf{k}$, the
coinvariant algebra is the regular $\mathbf{k}[\mathfrak{S}_{n}]$-module.
Fourth, under the same hypothesis, $P$ is a free left $P^{\mathfrak{S}_{n}%
}[\mathfrak{S}_{n}]$-module of rank $1$. The first result follows from an
elementary normal-form lemma for monic polynomials with pairwise relatively
prime leading monomials. The second is proved by lifting the Artin basis. For
the third, we use orbit harmonics with a strongly discrete point orbit, and
the fourth follows by equivariantly lifting a regular basis of the coinvariant
algebra. \medskip

\textbf{Manifest.} What follows is a series of folklore arguments that are not
easily found in the literature, at least not all in one place. None of these
is new; I have learned some of them from Vic Reiner and Brendon Rhoades. The
text below was produced by GPT-5.6 based on a telegraphic outline. I have then
edited it. -- DG\footnote{This work is in the public domain.}

\end{abstract}

\section{Statement and notation}

Let $n$ be a positive integer. Set $[n]=\{1,2,\ldots,n\}$, and let
$\mathfrak{S}_{n}$ denote the $n$-th \emph{symmetric group}, that is, the
group of all bijections $[n]\rightarrow[n]$. Throughout, $\mathbf{k}$ is a
commutative ring with identity and $1_{\mathbf{k}}\neq0$. (The requirement
$1_{\mathbf{k}} \neq0$ is unnecessary in essence; it just serves to make
certain statements hold literally, e.g., ensuring that each free $\mathbf{k}%
$-module has a unique rank.\footnote{Of course, the case when $1_{\mathbf{k}}
= 0$ is trivial, since $\mathbf{k}$ is a one-element set in this case.}) For
specific claims, we will impose further requirements on $\mathbf{k}$.

Consider the polynomial ring
\[
P=\mathbf{k}[x_{1},x_{2},\ldots,x_{n}].
\]
We give $P$ its usual grading by total degree:\footnote{All gradings are
$\mathbb{N}$-gradings in this note. If $V$ is a graded abelian group, then
$V_{d}$ shall mean the $d$-th graded component of $V$.}
\[
P=\bigoplus_{d\geq0}P_{d},
\]
where $P_{d}$ is the free $\mathbf{k}$-module of homogeneous polynomials of
total degree $d$. The group $\mathfrak{S}_{n}$ acts on $P$ (and on each
$P_{d}$) by permuting the variables:
\[
\sigma(x_{i})=x_{\sigma(i)}\qquad\text{for all }\sigma\in\mathfrak{S}%
_{n}\text{ and }i\in[n].
\]
This is an action by $\mathbf{k}$-algebra automorphisms.

\begin{noncompile}
Equivalently, let $\mathfrak{S}_{n}$ act on $\mathbf{k}^{n}$ by
\[
\sigma\cdot(b_{1},b_{2},\ldots,b_{n})=(b_{\sigma^{-1}(1)},b_{\sigma^{-1}%
(2)},\ldots,b_{\sigma^{-1}(n)}).
\]
Then the action on polynomials is the contragredient action
\[
(\sigma f)(v)=f(\sigma^{-1}\cdot v)\qquad(f\in P,\ v\in\mathbf{k}^{n}).
\]

\end{noncompile}

Let
\[
\Lambda=P^{\mathfrak{S}_{n}}=\left\{  f\in P\mid\sigma(f)=f\text{ for all
}\sigma\in\mathfrak{S}_{n}\right\}
\]
be the subring of symmetric polynomials. For each $i\in\lbrack n]$, let
\[
e_{i}=e_{i}(x_{1},x_{2},\ldots,x_{n})
=\sum_{1\leq j_{1}<j_{2}<\cdots<j_{i}\leq n}
x_{j_{1}}x_{j_{2}}\cdots x_{j_{i}}
\]
denote the $i$-th elementary symmetric polynomial. The fundamental theorem of
symmetric polynomials gives
\begin{equation}
\Lambda=\mathbf{k}[e_{1},e_{2},\ldots,e_{n}], \label{eq.Lam=ke}%
\end{equation}
and says that $e_{1},e_{2},\ldots,e_{n}$ are algebraically independent over
$\mathbf{k}$. Equivalently, the substitution map
\[
\mathbf{k}[t_{1},t_{2},\ldots,t_{n}]\longrightarrow\Lambda,\qquad
t_{i}\longmapsto e_{i}%
\]
(a $\mathbf{k}$-algebra homomorphism from the polynomial ring $\mathbf{k}%
[t_{1},t_{2},\ldots,t_{n}]$), is an isomorphism. The ring $\Lambda$ is a
graded subring of $P$; write
\[
\Lambda_{+}=\bigoplus_{d>0}\Lambda_{d}%
\]
for its ideal of elements with constant term $0$. This ideal $\Lambda_{+}$ is
generated by the elementary symmetric polynomials $e_{1},e_{2},\ldots,e_{n}$:

\begin{proposition}
\label{prop.augideal} We have
\begin{equation}
\Lambda_{+}=e_{1}\Lambda+e_{2}\Lambda+\cdots+e_{n}\Lambda.
\label{eq:augmentation-ideal}%
\end{equation}

\end{proposition}

\begin{proof}
The inclusion ``$\supseteq$'' is clear, since each $e_{i}$ belongs to
$\Lambda_{i}$ and thus to $\Lambda_{+}$.

Conversely, let $a\in\Lambda_{+}$. By \eqref{eq.Lam=ke}, we can write
$a=F(e_{1},e_{2},\ldots,e_{n})$ for some $F\in T:=\mathbf{k}[t_{1}%
,t_{2},\ldots,t_{n}]$. Evaluating at $x_{1}=x_{2}=\cdots=x_{n}=0$ gives
\begin{align*}
F(0,0,\ldots,0)  &  =a(0,0,\ldots,0)\qquad\left(  \text{since $e_{i}%
(0,0,\ldots,0)=0$ for all $i>0$}\right) \\
&  =0\qquad\left(  \text{since $a$ has constant term $0$}\right)  .
\end{align*}
Hence $F\in t_{1}T+t_{2}T+\cdots+t_{n}T$. Substituting $t_{i}=e_{i}$ yields
$a\in e_{1}\Lambda+e_{2}\Lambda+\cdots+e_{n}\Lambda$ (since $F(e_{1}%
,e_{2},\ldots,e_{n})=a$). This proves the \textquotedblleft$\subseteq
$\textquotedblright\ inclusion.
\end{proof}

Extending the ideal $\Lambda_{+}=e_{1}\Lambda+e_{2}\Lambda+\cdots+e_{n}%
\Lambda$ from $\Lambda$ to $P$ gives the ideal
\begin{align}
J  &  :=\Lambda_{+}P=\left(  e_{1}\Lambda+e_{2}\Lambda+\cdots+e_{n}%
\Lambda\right)  P\qquad\left(  \text{by \eqref{eq:augmentation-ideal}}\right)
\nonumber\\
&  =e_{1}P+e_{2}P+\cdots+e_{n}P \label{eq:J-two-descriptions}%
\end{align}
of $P$. The \emph{coinvariant algebra} (of $\mathfrak{S}_{n}$ acting on $P$)
is defined to be the quotient $\mathbf{k}$-algebra%
\[
C:=P/J.
\]


For any $f\in\Lambda$, $g\in P$, and $\sigma\in\mathfrak{S}_{n}$, we have
\begin{align}
\sigma(fg)  &  =\sigma(f)\sigma(g)\qquad\left(  \text{since }\mathfrak{S}%
_{n}\text{ acts by algebra automorphisms}\right) \nonumber\\
&  =f\sigma(g)\qquad\left(  \text{since $f\in\Lambda$ entails }\sigma\left(
f\right)  =f\right)  . \label{eq.sigfg}%
\end{align}
Thus the action of $\mathfrak{S}_{n}$ on $P$ is $\Lambda$-linear, and $P$ is
naturally a left module over the group algebra $\Lambda\lbrack\mathfrak{S}%
_{n}]$.

The action of $\mathfrak{S}_{n}$ on $P$ also preserves the ideal $J$, since
$J$ is generated by symmetric polynomials. Hence the quotient algebra $C=P/J$
inherits an action of $\mathfrak{S}_{n}$ and is a left module over
$\mathbf{k}[\mathfrak{S}_{n}]$.

We will prove the following folklore results mentioned, among other places, in
\cite[Section~1.5]{Haiman}:

\begin{theorem}
\label{thm:main}\ \ 

\begin{enumerate}
\item[\textbf{(a)}] The $\mathbf{k}$-module $C$ is free of rank $n!$. More
explicitly, call the $n!$ monomials
\[
x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{n}^{a_{n}} \qquad\text{with}\qquad0\leq
a_{i}<i\quad\text{for every }i\in[n]
\]
the \emph{Artin monomials}. Their residue classes in $C$ form a $\mathbf{k}%
$-basis of $C$.

\item[\textbf{(b)}] The $\Lambda$-module $P$ is free of rank $n!$; the Artin
monomials form a $\Lambda$-basis of $P$.

\item[\textbf{(c)}] Assume that $n!$ is invertible in $\mathbf{k}$. Then
\footnote{Here and in the following, \textquotedblleft$A$%
-module\textquotedblright\ for a ring $A$ shall always mean \textquotedblleft
left $A$-module\textquotedblright. Thus, a $\mathbf{k}[\mathfrak{S}_{n}%
]$-module is the same thing as a representation of $\mathfrak{S}_{n}$ over
$\mathbf{k}$.}
\[
C\cong\mathbf{k}[\mathfrak{S}_{n}]\qquad\text{as a $\mathbf{k}[\mathfrak{S}%
_{n}]$-module.}%
\]
In other words, the coinvariant algebra $C=P/J$ is isomorphic to the regular
$\mathbf{k}[\mathfrak{S}_{n}]$-module.

\item[\textbf{(d)}] Assume that $n!$ is invertible in $\mathbf{k}$. Then
\[
P\cong\Lambda\lbrack\mathfrak{S}_{n}]\qquad\text{as a $\Lambda\lbrack
\mathfrak{S}_{n}]$-module.}%
\]
In other words, $P$ is isomorphic to the regular $\Lambda\lbrack
\mathfrak{S}_{n}]$-module. Equivalently, there exists an element $f\in P$ such
that the map
\begin{equation}
\Phi:\Lambda\lbrack\mathfrak{S}_{n}]\longrightarrow P,\qquad\sum_{\sigma
\in\mathfrak{S}_{n}}a_{\sigma}\sigma\longmapsto\sum_{\sigma\in\mathfrak{S}%
_{n}}a_{\sigma}\sigma(f) \label{eq:normal-basis-map}%
\end{equation}
is an isomorphism of $\Lambda\lbrack\mathfrak{S}_{n}]$-modules.
\end{enumerate}
\end{theorem}

We first prove an elementary normal-form lemma and use it to compute the
coinvariant algebra, proving \textbf{(a)}. A lifting argument then proves
\textbf{(b)}. Next, an orbit-harmonics argument proves \textbf{(c)}. Finally,
combining the lifting argument with \textbf{(c)} proves \textbf{(d)}. The
invertibility hypothesis in \textbf{(c)} and \textbf{(d)} cannot simply be
omitted; we give a modular counterexample at the end.

\section{A coprime-leading-monomial lemma}

\label{sec:coprime-leading-monomials}

In this section, we shall prove a general property of certain ideals in
polynomial rings. This property is part of the theory of Gr\"{o}bner bases
(see Remark~\ref{rmk.grobner}), but we will give a self-contained proof avoiding the
general theory, as the property is the only thing we will need.

Let $R$ be a commutative ring with identity. Consider the polynomial ring%
\[
A:=R[z_{1},z_{2},\ldots,z_{N}],
\]
and fix a monomial order $\prec$ on $A$. That is, $\prec$ is a total
well-order on the set of all monomials in $A$, and
\[
u\prec v\quad\Longrightarrow\quad uw\prec vw
\]
for all monomials $u,v,w$. For a nonzero polynomial $f\in A$, its
\emph{leading monomial} $\operatorname{LM}(f)$ is the largest monomial that
occurs in $f$ with a nonzero coefficient. We call $f$ \emph{monic} if the
coefficient of $\operatorname{LM}(f)$ is $1$. We say that two monomials $u$
and $v$ are \emph{relatively prime} if the only monomial that divides them
both is $1$; equivalently, this means that they have no variable in common.
(For instance, the monomials $z_{1}z_{3}^{2}$ and $z_{2}z_{5}$ are relatively
prime, but the monomials $z_{1}z_{3}^{2}$ and $z_{3}z_{5}$ are not.)

\begin{lemma}
[Coprime-leading-monomial lemma]\label{lem:coprime-leading-monomials} Let
$g_{1},g_{2},\ldots,g_{k}\in A$ be monic polynomials whose leading monomials
\[
m_{i}=\operatorname{LM}(g_{i})\qquad\text{(for $i\in[k]$)}
\]
are pairwise relatively prime. A monomial $q$ will be called \emph{reduced} if
it is not divisible by any of $m_{1}, m_{2}, \ldots, m_{k}$ (that is, if
$m_{i} \nmid q$ for each $i \in[k]$). Define an ideal $I$ of $A$ by
\[
I= g_{1} A + g_{2} A + \cdots+ g_{k} A.
\]
Then the residue classes of the reduced monomials form an $R$-module basis of
$A/I$.
\end{lemma}

\begin{proof}
For each $i\in\lbrack k]$, write
\begin{equation}
g_{i}=m_{i}-r_{i}, \label{eq:gi-mi-ri}%
\end{equation}
where every monomial occurring in $r_{i}$ is strictly smaller than $m_{i}$.
(We can write $g_{i}$ in this form, since $g_{i}$ is monic with leading
monomial $m_{i}$.) Let $A_{\operatorname{red}}$ be the free $R$-submodule of
$A$ spanned by the reduced monomials.

We shall construct an $R$-linear map
\[
\rho:A\longrightarrow A_{\operatorname{red}}%
\]
by recursively defining the values $\rho\left(  u\right)  $ for all monomials
$u$; the recursion shall proceed along the well-founded order relation $\prec
$. If a monomial $u$ is reduced, set $\rho(u)=u$. If $u$ is not reduced,
choose an $i\in\lbrack k]$ such that $m_{i}\mid u$ and set
\begin{equation}
\rho(u)=\rho\left(  \frac{u}{m_{i}}r_{i}\right)  . \label{eq:rho-recursion}%
\end{equation}
On the right-hand side, $\rho$ is applied termwise and $R$-linearly. This is
legitimate because every monomial occurring in $(u/m_{i})r_{i}$ is strictly
smaller than $u$: indeed, every monomial of $r_{i}$ is smaller than $m_{i}$,
and a monomial order is compatible with multiplication.

We must check that \eqref{eq:rho-recursion} is independent of the choice of
$i$. Suppose that both $m_{i}$ and $m_{j}$ divide $u$, where $i\neq j$. We
must show that
\begin{align}
\rho\left(  \frac{u}{m_{i}}r_{i}\right)  = \rho\left(  \frac{u}{m_{j}}%
r_{j}\right)  . \label{eq:rho-recursion-fork}%
\end{align}


Since the monomials $m_{i}$ and $m_{j}$ are relatively prime and both divide
$u$, their product must divide $u$ as well; thus
\[
u=qm_{i}m_{j}%
\]
for some monomial $q$. By the recursive induction hypothesis, the definition
of $\rho$ is already independent of all choices on monomials strictly smaller
than $u$. Every monomial $t$ occurring in $r_{i}$ satisfies $t\prec m_{i}$ and
thus (since $\prec$ is a monomial order)
\[
qtm_{j}\prec qm_{i}m_{j}=u.
\]
When evaluating $\rho(qtm_{j})$, we may therefore use the divisor $m_{j}$ in
\eqref{eq:rho-recursion}, and obtain
\[
\rho(qtm_{j})=\rho\left(  \dfrac{qtm_{j}}{m_{j}}r_{j}\right)  =\rho(qtr_{j}).
\]
Multiplying by the coefficient of $t$ and summing over all monomials of
$r_{i}$ gives
\[
\rho(qr_{i}m_{j})=\rho(qr_{i}r_{j}).
\]
The same argument (with the roles of $i$ and $j$ switched) yields
\[
\rho(qr_{j}m_{i})=\rho(qr_{j}r_{i}).
\]
Since $A$ is commutative, this is the same as
\[
\rho(qm_{i}r_{j})=\rho(qr_{i}r_{j}).
\]
Hence, from $u=qm_{i}m_{j}$, we obtain
\[
\rho\left(  \frac{u}{m_{i}}r_{i}\right)  =\rho(qr_{i}m_{j})=\rho(qr_{i}%
r_{j})=\rho(qm_{i}r_{j})=\rho\left(  \frac{u}{m_{j}}r_{j}\right)  ,
\]
which proves \eqref{eq:rho-recursion-fork}. Thus we have shown that the
$R$-linear map $\rho:A\rightarrow A_{\operatorname{red}}$ is well-defined.

We next record two properties of $\rho$. First, for every monomial $q$ and
every $i\in[k]$, the definition allows us to reduce the monomial $qm_{i}$
using $m_{i}$ (that is, apply \eqref{eq:rho-recursion} to $u=qm_{i}$); hence
\begin{equation}
\rho(qm_{i})=\rho(qr_{i}). \label{eq:rhom=rhor}%
\end{equation}
By linearity of $\rho$, this entails $\rho(q(m_{i}-r_{i})) = 0$. In view of
\eqref{eq:gi-mi-ri}, this rewrites as
\begin{equation}
\label{eq:rho-kills-ideal}\rho(qg_{i})=0.
\end{equation}
By $R$-linearity, \eqref{eq:rho-kills-ideal} shows that
\[
\rho(I)=0.
\]


Second, every $f\in A$ satisfies
\begin{equation}
f-\rho(f)\in I. \label{eq:f-minus-rho-in-I}%
\end{equation}
Indeed, by $R$-linearity, it is enough to prove this when $f$ is a monomial;
i.e., it is enough to show that $u-\rho\left(  u\right)  \in I$ for any
monomial $u$. We do so by induction on $u$. If $u$ is reduced, the claim is
clear. Otherwise, choose $i$ with $m_{i}\mid u$ and put $q=u/m_{i}$. Then
$u=qm_{i}$ and therefore $\rho(u)=\rho(qm_{i})=\rho(qr_{i})$ by
\eqref{eq:rhom=rhor}; hence,
\begin{align*}
u-\rho(u)  &  =qm_{i}-\rho(qr_{i})\\
&  =q(m_{i}-r_{i})+\bigl(qr_{i}-\rho(qr_{i})\bigr)\\
&  =qg_{i}+\bigl(qr_{i}-\rho(qr_{i})\bigr)\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{eq:gi-mi-ri}}\right)  .
\end{align*}
The first addend belongs to $I$. Every monomial occurring in $qr_{i}$ is
strictly smaller than $qm_{i}=u$. Applying the induction hypothesis to these
monomials and then using $R$-linearity gives $qr_{i}-\rho(qr_{i})\in I$. Thus
the second addend belongs to $I$ as well. Thus, $u-\rho(u)\in I$. This proves \eqref{eq:f-minus-rho-in-I}.

Equation \eqref{eq:f-minus-rho-in-I} shows that
\[
A=I+A_{\operatorname{red}}.
\]
This sum is direct. Indeed, if $h\in I\cap A_{\operatorname{red}}$, then
$\rho(h)=0$ because $\rho(I)=0$, whereas $\rho(h)=h$ because $\rho$ fixes
every reduced monomial. Thus $h=0$. Consequently,
\[
A=I\oplus A_{\operatorname{red}}
\]
as $R$-modules. Therefore, the canonical projection $A \to A/I$, restricted to
$A_{\operatorname{red}}$, becomes an $R$-module isomorphism. Since the reduced
monomials form a basis of $A_{\operatorname{red}}$, this shows that their
residue classes form a basis of $A/I$.
\end{proof}

\begin{remark}
\label{rmk.grobner}In Gr\"{o}bner-basis language,
Lemma~\ref{lem:coprime-leading-monomials} says that $g_{1},g_{2},\ldots,g_{k}$
form a monic Gr\"{o}bner basis and then applies the standard-monomial basis
theorem. Over a field, this follows from Buchberger's first criterion; see,
for example, \cite[Chapter~1, especially the conclusion after Lemma~1.1.38]%
{deGraaf}. The proof above avoids $S$-polynomials and Buchberger's criterion
altogether. It is the particularly simple commutative case of a diamond-lemma
argument in which the only competing reductions are
\[
qm_{i}m_{j}\longrightarrow qr_{i}m_{j}\longrightarrow qr_{i}r_{j}%
\quad\text{and}\quad qm_{i}m_{j}\longrightarrow qm_{i}r_{j}\longrightarrow
qr_{i}r_{j}.
\]
For a more detailed treatment of monic reductions and Gr\"{o}bner bases over
arbitrary commutative rings, see \cite[Section~3.2]{subdiv}. Compare also
Bergman's diamond lemma \cite[Theorem~1.2]{BergmanDiamond}, which makes
similar statements about noncommutative polynomial rings.
\end{remark}

\section{The Artin basis}

We prove Theorem~\ref{thm:main} \textbf{(a)} using Lemma~\ref{lem:coprime-leading-monomials}. Our proof follows the ideas of
\cite[proof of Theorem 1.2.7]{Sturmfels}. Other proofs can be found in
\cite[(DIFF.1.3)]{LLPT95}, \cite[Chapter IV, \S 6, no.~1, Theorem~1,
part~(c)]{Bourba03}, \cite[Theorem, part (c)]{Gailla21} and (over $\mathbb{Z}$
instead of $\mathbf{k}$) in \cite[Proposition 3.4]{FoGePo97} and
\cite[(5.1)]{Macdon91}.

We shall use the shorthand notation $x^{a}$ for the monomial $x_{1}^{a_{1}%
}x_{2}^{a_{2}}\cdots x_{n}^{a_{n}}$, where $a=(a_{1},a_{2},\ldots,a_{n}%
)\in\mathbb{N}^{n}$.

For $r\in\mathbb{N}$ and a finite list of variables $y_{1},y_{2},\ldots,y_{m}%
$, let $h_{r}(y_{1},y_{2},\ldots,y_{m})$ denote the complete homogeneous
symmetric polynomial of degree $r$ in $y_{1},y_{2},\ldots,y_{m}$; this is
defined as the sum of all monomials $y_{1}^{a_{1}} y_{2}^{a_{2}} \cdots
y_{m}^{a_{m}}$ with $a_{1} + a_{2} + \cdots+ a_{m} = r$. (In particular,
$h_{0}=1$.) For $1\leq i\leq n$, set
\[
g_{i}=h_{i}(x_{i},x_{i+1},\ldots,x_{n}).
\]


\begin{lemma}
[A generating-function identity]\label{lem:elementary-complete-tail} For each
$i\in[n]$, we have the identity
\begin{equation}
\label{eq:elementary-complete-tail}\left(  \sum_{j=0}^{n}(-1)^{j}e_{j}%
t^{j}\right)  \left(  \sum_{r\geq0}h_{r}(x_{i},x_{i+1},\ldots,x_{n}%
)t^{r}\right)  =\prod_{j=1}^{i-1}(1-x_{j}t)
\end{equation}
in the formal power-series ring $P[[t]]$, where $e_{0}=1$.
\end{lemma}

\begin{proof}
The elementary symmetric polynomials satisfy Vi\`{e}te's identity
\begin{equation}
\sum_{j=0}^{n}(-1)^{j}e_{j}t^{j}=\prod_{j=1}^{n}(1-x_{j}t).
\label{eq:e-generating-function}%
\end{equation}
Indeed, when the product on the right-hand side is expanded, choosing the term
$-x_{j}t$ from exactly $r$ of its factors produces $(-1)^{r}e_{r}t^{r}$. On
the other hand, every power series $1-x_{j}t$ in $P[[t]]$ has constant term
$1$ and is therefore a unit. We have
\begin{equation}
\sum_{r\geq0}h_{r}(x_{i},x_{i+1},\ldots,x_{n})t^{r}=\prod_{j=i}^{n}\frac
{1}{1-x_{j}t}. \label{eq:h-generating-function}%
\end{equation}
To see this, expand each factor on the right-hand side as%
\[
\dfrac{1}{1-x_{j}t}=\left(  1-x_{j}t\right)  ^{-1}=\sum_{a\geq0}x_{j}^{a}t^{a}%
\]
by the geometric-series formula. Thus, the whole product rewrites as follows:%
\[
\prod_{j=i}^{n}\frac{1}{1-x_{j}t}=\prod_{j=i}^{n}\ \ \sum_{a\geq0}x_{j}%
^{a}t^{a}=\sum_{\left(  a_{i},a_{i+1},\ldots,a_{n}\right)  \in\mathbb{N}%
^{n-i+1}}x_{i}^{a_{i}}x_{i+1}^{a_{i+1}}\cdots x_{n}^{a_{n}}t^{a_{i}%
+a_{i+1}+\cdots+a_{n}}.
\]
The coefficient of $t^{r}$ in this is the sum of all monomials $x_{i}^{a_{i}%
}x_{i+1}^{a_{i+1}}\cdots x_{n}^{a_{n}}$ with $a_{i}+a_{i+1}+\cdots+a_{n}=r$.
This is precisely $h_{r}(x_{i},x_{i+1},\ldots,x_{n})$. Thus,
\eqref{eq:h-generating-function} is proved.

Multiplying \eqref{eq:e-generating-function} and
\eqref{eq:h-generating-function}, we obtain%
\[
\left(  \sum_{j=0}^{n}(-1)^{j}e_{j}t^{j}\right)  \left(  \sum_{r\geq0}%
h_{r}(x_{i},x_{i+1},\ldots,x_{n})t^{r}\right)  =\prod_{j=1}^{n}(1-x_{j}%
t)\prod_{j=i}^{n}\frac{1}{1-x_{j}t}=\prod_{j=1}^{i-1}(1-x_{j}t),
\]
since the factors $1-x_{i}t,1-x_{i+1}t,\ldots,1-x_{n}t$ cancel. Thus
\eqref{eq:elementary-complete-tail} is proved.
\end{proof}

\begin{lemma}
[Asymmetric generating set of $J$]\label{lem:J-g} We have the equality
\begin{equation}
g_{1} P + g_{2} P + \cdots+ g_{n} P = J \label{eq:g=e-id}%
\end{equation}
of ideals of $P$.
\end{lemma}

\begin{proof}
For each $i \in[n]$, the right-hand side of
\eqref{eq:elementary-complete-tail} has degree at most $i-1$ in $t$. Comparing
coefficients of $t^{i}$ in \eqref{eq:elementary-complete-tail} therefore
gives
\begin{equation}
\sum_{j=0}^{i}(-1)^{j}e_{j}h_{i-j}(x_{i},x_{i+1},\ldots,x_{n})=0.
\label{eq:gi-ei0}%
\end{equation}
The $j=0$ addend on the left-hand side of this equality is $(-1)^{0}e_{0}%
h_{i}(x_{i},x_{i+1},\ldots,x_{n})=h_{i}(x_{i},x_{i+1},\ldots,x_{n})=g_{i}$;
thus, we can rewrite \eqref{eq:gi-ei0} as
\[
g_{i}+\sum_{j=1}^{i}(-1)^{j}e_{j}h_{i-j}(x_{i},x_{i+1},\ldots,x_{n})=0,
\]
or, equivalently,
\begin{align}
g_{i}  &  =-\sum_{j=1}^{i}(-1)^{j}e_{j}h_{i-j}(x_{i},x_{i+1},\ldots
,x_{n})\nonumber\\
&  = -\sum_{j=1}^{i-1}(-1)^{j}e_{j}h_{i-j}(x_{i},x_{i+1},\ldots,x_{n})
-(-1)^{i}e_{i} \label{eq:gi-ei}%
\end{align}
(since $h_{i-i}(x_{i},x_{i+1},\ldots,x_{n})=1$). Thus every $g_{i}$ belongs to
$J$ (since $e_{1},e_{2},\ldots,e_{i} \in J$). This proves the ``$\subseteq$''
inclusion of \eqref{eq:g=e-id}.

Conversely, \eqref{eq:gi-ei} gives
\begin{equation}
(-1)^{i}e_{i}=-g_{i}-\sum_{j=1}^{i-1}(-1)^{j}e_{j}h_{i-j}(x_{i},x_{i+1}%
,\ldots,x_{n}). \label{eq:ei-gi}%
\end{equation}
We now prove by strong induction on $i$ that%
\[
e_{i}\in g_{1}P+g_{2}P+\cdots+g_{i}P.
\]
Indeed, assume that this claim is known for all smaller indices. Every $e_{j}$
occurring on the right-hand side of \eqref{eq:ei-gi} has $j<i$, so the
induction hypothesis places it in $g_{1}P+g_{2}P+\cdots+g_{j}P$. Hence the
left-hand side $(-1)^{i}e_{i}$, and therefore also $e_{i}$, belongs to
$g_{1}P+g_{2}P+\cdots+g_{i}P$. Thus we conclude that $e_{i}\in g_{1}%
P+g_{2}P+\cdots+g_{n}P$ for each $i\in\left[  n\right]  $. In view of
\eqref{eq:J-two-descriptions}, this entails $J\subseteq g_{1}P+g_{2}%
P+\cdots+g_{n}P$. This proves the \textquotedblleft$\supseteq$%
\textquotedblright\ inclusion in \eqref{eq:g=e-id}. Hence the equality of
ideals \eqref{eq:g=e-id} is proved.
\end{proof}

\begin{proposition}
[Artin basis]\label{prop:artin} The residue classes of the Artin monomials
\begin{equation}
x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{n}^{a_{n}}\qquad\text{with}\qquad0\leq
a_{i}<i\quad\text{for every }i\in\lbrack n] \label{eq:artin-monomials}%
\end{equation}
form a $\mathbf{k}$-basis of $P/J=C$.
\end{proposition}

\begin{proof}
From Lemma~\ref{lem:J-g}, we know that $J=g_{1}P+g_{2}P+\cdots+g_{n}P$. To
obtain a $\mathbf{k}$-module basis of $C=P/J$, we shall apply
Lemma~\ref{lem:coprime-leading-monomials} with $R=\mathbf{k}$, $N=n$, $A=P$,
$k=n$, and $I=J$, using the generators $g_{1},g_{2},\ldots,g_{n}$. We
therefore need a monomial order $\succ$ on $P$ for which the $g_{i}$ are monic
and have pairwise relatively prime leading monomials. There are several valid
choices here. The simplest one is to let $\succ$ be the lexicographic order
$>_{\operatorname{lex}}$ with
\[
x_{1}>x_{2}>\cdots>x_{n}.
\]
Thus, for exponent vectors $a=(a_{1},a_{2},\ldots,a_{n})$ and $b=(b_{1}%
,b_{2},\ldots,b_{n})$, we have $x^{a}\succ x^{b}$ if, at the first index $r$
for which $a_{r}\neq b_{r}$, we have $a_{r}>b_{r}$. Alternatively, we can let
$\succ$ be the degree-lexicographic order $>_{\operatorname{grlex}}$, in which
two monomials of equal degree are compared as in the lexicographic order,
whereas two monomials of distinct degrees are compared by their degrees (that
is, the monomial of larger degree is declared to be larger). Both of these
orders are monomial orders (this is particularly easy to check for
$>_{\operatorname{grlex}}$, since there are only finitely many monomials of a
given degree), and both have the property that each of the polynomials $g_{i}$
is monic with leading monomial $\operatorname{LM}(g_{i})=x_{i}^{i}$ (since
$g_{i}=h_{i}(x_{i},x_{i+1},\ldots,x_{n})=x_{i}^{i}+\text{ smaller monomials}%
$). These leading monomials are pairwise relatively prime.

Hence Lemma~\ref{lem:coprime-leading-monomials} shows that the residue classes
of the reduced monomials (i.e., of the monomials not divisible by any of
$x_{1}^{1},x_{2}^{2},\ldots,x_{n}^{n}$) form a $\mathbf{k}$-basis of $P/J$. A
monomial $x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{n}^{a_{n}}$ is reduced
%not divisible by any $x_i^i$
if and only if $0\leq a_{i}<i$ for every $i\in\lbrack n]$. These are exactly
the Artin monomials in \eqref{eq:artin-monomials}. Thus their residue classes
form a basis of $P/J$.

This proves Proposition~\ref{prop:artin} and therefore Theorem~\ref{thm:main}
\textbf{(a)}.
\end{proof}

If $V=\bigoplus_{d\geq0}V_{d}$ is a graded $\mathbf{k}$-module such that every
$V_{d}$ is finite free, its \emph{Hilbert series} is the formal power series
\begin{equation}
\operatorname{Hilb}(V;q):=\sum_{d\geq0}(\operatorname{rank}_{\mathbf{k}}%
V_{d})q^{d}\in\mathbb{Z}[[q]]. \label{eq:hilbert-definition}%
\end{equation}
We will use Hilbert series only for graded $\mathbf{k}$-modules with finite
free homogeneous components. Note that $P$, $C$ and $\Lambda$ are such graded
$\mathbf{k}$-modules; indeed, the basis of $C$ constructed in Proposition
\ref{prop:artin} is a graded basis (i.e., each of its basis vectors is homogeneous).

\begin{corollary}
\label{cor:hilb-C} We have
\[
\operatorname{Hilb}(C;q)=\prod_{i=1}^{n}(1+q+\cdots+q^{i-1}) \qquad\text{ and
} \qquad\operatorname{rank}_{\mathbf{k}}C=n!.
\]

\end{corollary}

\begin{proof}
By Proposition~\ref{prop:artin}, the degree-$d$ component of $C$ has as a
basis the residue classes of the Artin monomials $x_{1}^{a_{1}}x_{2}^{a_{2}%
}\cdots x_{n}^{a_{n}}$ satisfying $0\leq a_{i}<i$ and $a_{1}+a_{2}%
+\cdots+a_{n}=d$. Hence its rank $\operatorname{rank}_{\mathbf{k}} C_{d}$ is
the number of exponent tuples $\left(  a_{1},a_{2},\ldots,a_{n}\right)  $
satisfying $0\leq a_{i}<i$ for all $i$ as well as $a_{1}+a_{2}+\cdots+a_{n}%
=d$. Therefore
\begin{align*}
\operatorname{Hilb}(C;q)  &  =\sum_{d\geq0} \sum_{\substack{(a_{1}%
,a_{2},\ldots,a_{n});\\0\leq a_{i}<i\text{ for each }i;\\a_{1}+a_{2}%
+\cdots+a_{n}=d}} q^{d}\\
&  =\sum_{\substack{(a_{1},a_{2},\ldots,a_{n});\\0\leq a_{i}<i\text{ for each
}i}} q^{a_{1}+a_{2}+\cdots+a_{n}}\\
&  =\sum_{0\leq a_{1}<1}\sum_{0\leq a_{2}<2}\cdots\sum_{0\leq a_{n}<n}%
q^{a_{1}+a_{2}+\cdots+a_{n}}\\
&  =\prod_{i=1}^{n}\left(  \sum_{a=0}^{i-1}q^{a}\right) \\
&  =\prod_{i=1}^{n}(1+q+\cdots+q^{i-1}).
\end{align*}
Evaluating this polynomial at $q=1$ counts all Artin monomials and gives
$\operatorname{rank}_{\mathbf{k}}C=1\cdot2\cdot\cdots\cdot n=n!$.
\end{proof}

\section{Lifting the coinvariant algebra}

We next prove Theorem~\ref{thm:main} \textbf{(b)}. Over a field, one can start
with an arbitrary graded vector-space complement to $J$ in $P$. Over a general
coefficient ring such complements need not exist, so we formulate the lifting
argument for a chosen graded set of representatives. The Artin monomials
provide such representatives.

We shall use the following elementary fact.

\begin{lemma}
\label{lem:finite-free-surjection} Let $R$ be a commutative ring. Any
surjective $R$-linear map between two finite free $R$-modules of the same rank
is an isomorphism.
\end{lemma}

\begin{proof}
Let $f$ be a surjective $R$-linear map between two free $R$-modules of the
same rank. We must prove that $f$ is an isomorphism.

After choosing bases, the map $f$ is represented by a square matrix $A$. Since
its target is free, the surjection $f$ has an $R$-linear right inverse $g$,
represented by a matrix $B$. Thus $AB=I$. Taking determinants gives
$\det(A)\det(B)=1$, so $\det(A)$ is a unit. The adjugate formula
$A\operatorname*{adj}\left(  A\right)  =\operatorname*{adj}\left(  A\right)
A=\det\left(  A\right)  \cdot I$ then shows that $A$ is invertible. Hence, the
corresponding linear map $f$ is an isomorphism.
\end{proof}

\begin{proposition}
\label{prop:multiplication} Let
\[
H=\bigoplus_{d\geq0}H_{d}%
\]
be a graded $\mathbf{k}$-submodule of $P$ such that each $H_{d}$ is finite
free and the quotient map $P\rightarrow C=P/J$ restricts to a graded
$\mathbf{k}$-module isomorphism
\begin{equation}
H\xrightarrow{\ \sim\ }C. \label{eq:H-C}%
\end{equation}
Then:

\begin{enumerate}
\item[\textbf{(a)}] Multiplication induces an isomorphism of graded $\Lambda
$-modules
\[
m:\Lambda\otimes_{\mathbf{k}}H\xrightarrow{\ \sim\ }P,\qquad a\otimes
h\longmapsto ah.
\]
Here the tensor product has its usual grading:
\[
(\Lambda\otimes_{\mathbf{k}}H)_{d}=\bigoplus_{r+s=d}\Lambda_{r}\otimes
_{\mathbf{k}}H_{s}.
\]


\item[\textbf{(b)}] If $H$ is $\mathfrak{S}_{n}$-stable, then $m$ is also
$\mathfrak{S}_{n}$-equivariant, where $\mathfrak{S}_{n}$ acts on the second
tensor factor.
\end{enumerate}
\end{proposition}

\begin{proof}
The map $m$ is plainly graded and $\Lambda$-linear. We first prove that it is
surjective, or equivalently that $P=\Lambda H$, by strong induction on degree.
Let $p\in P_{d}$ be homogeneous. By \eqref{eq:H-C}, the quotient map
$H\overset{\sim}{\rightarrow}C$ is an isomorphism; since it is graded, it thus
restricts to an isomorphism $H_{d}\overset{\sim}{\rightarrow}C_{d}$. Thus, the
residue class of $p$ in $C_{d}$ lies in the image of this restricted
isomorphism; that is, there is an $h\in H_{d}$ such that $p-h\in J_{d}$.
Since
\[
J=e_{1}P+e_{2}P+\cdots+e_{n}P,
\]
we can thus write
\[
p-h=\sum_{i=1}^{n}e_{i}q_{i},
\]
where each $q_{i}\in P$. Moreover, by projecting both sides of this equality
onto the $d$-th graded component of $P$, we obtain%
\begin{equation}
p-h=\sum_{i=1}^{n}e_{i}\widetilde{q}_{i},\label{eq:p-h2}%
\end{equation}
where $\widetilde{q}_{i}$ is the $\left(  d-i\right)  $-th graded component of
$q_{i}$ (since each $e_{i}$ is homogeneous of degree $i$). As usual,
$\widetilde{q}_{i}=0$ when $i>d$. By induction, each $\widetilde{q}_{i}$ lies
in $\Lambda H$ (since $\deg\widetilde{q}_{i}=d-i<d$). Since $e_{i}\in\Lambda$,
the equality \eqref{eq:p-h2} thus yields $p-h\in\underbrace{\Lambda\Lambda
}_{=\Lambda}H=\Lambda H$, and therefore $p\in\Lambda H$ as well (since $h\in
H_{d}\subseteq H\subseteq\Lambda H$). So we have proved the surjectivity of
$m$.

It remains to prove the injectivity of $m$. We do this degree by degree. By
\eqref{eq.Lam=ke}, the monomials
\[
e_{1}^{a_{1}}e_{2}^{a_{2}}\cdots e_{n}^{a_{n}}\qquad\text{(with }a_{1}%
,a_{2},\ldots,a_{n}\in\mathbb{N}\text{)}%
\]
form a homogeneous $\mathbf{k}$-basis of $\Lambda$, and they are homogeneous
of degree $1a_{1}+2a_{2}+\cdots+na_{n}$. Hence
\begin{align}
\operatorname{Hilb}(\Lambda;q)  & =\sum_{d\geq0}\sum_{\substack{a_{1}%
,a_{2},\ldots,a_{n}\in\mathbb{N};\\1a_{1}+2a_{2}+\cdots+na_{n}=d}%
}q^{d}\nonumber\\
& =\sum_{a_{1},a_{2},\ldots,a_{n}\in\mathbb{N}}q^{1a_{1}+2a_{2}+\cdots+na_{n}%
}\nonumber\\
& =\prod_{i=1}^{n}\sum_{a\in\mathbb{N}}q^{ia}=\prod_{i=1}^{n}\frac
{1}{1-q^{i}}\label{eq:HilbL}%
\end{align}
(since $\sum_{a\in\mathbb{N}}q^{ia}=\dfrac{1}{1-q^{i}}$ for each $i>0$).
Likewise, the ordinary monomial basis of $P$ gives%
\begin{equation}
\operatorname{Hilb}(P;q)=\frac{1}{(1-q)^{n}}.\label{eq:HilbP}%
\end{equation}
Furthermore, \eqref{eq:H-C} and Corollary~\ref{cor:hilb-C} give
\begin{align}
\operatorname{Hilb}(H;q)  & =\operatorname{Hilb}(C;q)=\prod_{i=1}%
^{n}\underbrace{(1+q+\cdots+q^{i-1})}_{=\dfrac{1-q^{i}}{1-q}}\nonumber\\
& =\frac{\prod_{i=1}^{n}(1-q^{i})}{(1-q)^{n}}.\label{eq:HilbH}%
\end{align}
If $V$ and $W$ are two graded $\mathbf{k}$-modules whose homogeneous
components are finite free, then their Hilbert series multiply under tensor
products, i.e., we have%
\[
\operatorname*{Hilb}\left(  V\otimes_{\mathbf{k}}W;q\right)
=\operatorname*{Hilb}\left(  V;q\right)  \cdot\operatorname*{Hilb}\left(
W;q\right)  ,
\]
since each $d\geq0$ satisfies
\[
\operatorname{rank}_{\mathbf{k}}(V\otimes_{\mathbf{k}}W)_{d}=\sum
_{r+s=d}(\operatorname{rank}_{\mathbf{k}}V_{r})(\operatorname{rank}%
_{\mathbf{k}}W_{s}).
\]
Consequently,
\begin{align*}
\operatorname{Hilb}(\Lambda\otimes_{\mathbf{k}}H;q)  & =\operatorname{Hilb}%
(\Lambda;q)\cdot\operatorname{Hilb}(H;q)\\
& =\left(  \prod_{i=1}^{n}\frac{1}{1-q^{i}}\right)  \cdot\frac{\prod_{i=1}%
^{n}(1-q^{i})}{(1-q)^{n}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{eq:HilbL} and \eqref{eq:HilbH}}\right)  \\
& =\frac{1}{(1-q)^{n}}=\operatorname{Hilb}(P;q)\ \ \ \ \ \ \ \ \ \ \left(
\text{by \eqref{eq:HilbP}}\right)  .
\end{align*}
In other words, for each $d\geq0$, we have $\operatorname*{rank}%
\nolimits_{\mathbf{k}}\left(  \Lambda\otimes_{\mathbf{k}}H\right)
_{d}=\operatorname*{rank}\nolimits_{\mathbf{k}}P_{d}$.

Thus, in every degree $d$, the map $m:\left(  \Lambda\otimes_{\mathbf{k}%
}H\right)  _{d}\rightarrow P_{d}$ is a surjection between finite free
$\mathbf{k}$-modules of the same rank. Therefore,
Lemma~\ref{lem:finite-free-surjection} shows that each homogeneous component
of $m$ is an isomorphism. Altogether, $m$ is thus an isomorphism of graded
$\mathbf{k}$-modules. This proves part \textbf{(a)}. \medskip

\textbf{(b)} If $H$ is $\mathfrak{S}_{n}$-stable, then $m$ is equivariant
because of \eqref{eq.sigfg}.
\end{proof}

\begin{corollary}
\label{cor:freeness-all-char} The polynomial ring $P$ is a free $\Lambda
$-module of rank $n!$. More explicitly, the Artin monomials
\[
x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{n}^{a_{n}}\qquad\text{(with }0\leq
a_{i}<i\text{ for every }i\in\lbrack n]\text{)}%
\]
form a $\Lambda$-basis of $P$.
\end{corollary}

\begin{proof}
Let $H$ be the free $\mathbf{k}$-submodule of $P$ spanned by the Artin
monomials. This is a graded $\mathbf{k}$-submodule of $P$, since each Artin
monomial is homogeneous. The Artin monomials form a $\mathbf{k}$-basis of $H$.
Meanwhile, their residue classes in $P/J=C$ form a $\mathbf{k}$-basis of $C$,
by Proposition~\ref{prop:artin}.

Restricting the quotient map $P\rightarrow P/J=C$ to $H$, we obtain a graded
$\mathbf{k}$-linear map $\phi:H\rightarrow C$. This map $\phi$ sends the Artin
monomials in $H$ to their residue classes in $C$. Since the former form a
$\mathbf{k}$-basis of $H$ while the latter form a $\mathbf{k}$-basis of $C$,
we thus conclude that $\phi$ is a $\mathbf{k}$-module isomorphism. Hence,
Proposition~\ref{prop:multiplication} \textbf{(a)} shows that multiplication
induces an isomorphism of graded $\Lambda$-modules
\[
m:\Lambda\otimes_{\mathbf{k}}H\xrightarrow{\ \sim\ }P,\qquad a\otimes
h\longmapsto ah.
\]
But the Artin monomials $x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{n}^{a_{n}}$ form
a $\mathbf{k}$-basis of $H$. Hence, the tensors $1\otimes_{\mathbf{k}}%
x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{n}^{a_{n}}$ form a $\Lambda$-basis of
$\Lambda\otimes_{\mathbf{k}}H$ (since base change respects bases). Therefore,
the images of these tensors under $m$ form a $\Lambda$-basis of $P$ (since $m$
is a $\Lambda$-module isomorphism). But these images are again the Artin
monomials $x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{n}^{a_{n}}$. So we have shown
that the Artin monomials form a $\Lambda$-basis of $P$. This proves Corollary~\ref{cor:freeness-all-char} and, with it, Theorem~\ref{thm:main} \textbf{(b)}.
\end{proof}

\begin{remark}
[Harmonic representatives]\label{rem:harmonics} When $\mathbf{k}$ is a field
of characteristic $0$, one customary choice of $H$ is the space of harmonic
polynomials; compare Haiman's discussion of harmonics and the one-set case
\cite[Sections~1.3 and~1.5]{Haiman}. On $\mathbb{Q}[x_{1},x_{2},\ldots,x_{n}%
]$, consider the \emph{apolar pairing}, i.e., the $\mathbb{Q}$-bilinear form
given by%
\[
\langle p,q\rangle=\bigl(p(\partial_{x_{1}},\partial_{x_{2}},\ldots
,\partial_{x_{n}})q\bigr)(0).
\]
The monomials are orthogonal with respect to this pairing, and satisfy%
\[
\langle x^{a},x^{a}\rangle=a_{1}!a_{2}!\cdots a_{n}!.
\]
Thus, after extension to $\mathbb{R}$, this pairing is positive definite, and
$J^{\perp}$ is a graded $\mathfrak{S}_{n}$-stable complement to $J$. After
base change to any characteristic-$0$ field, one may therefore take
$H=J^{\perp}$ in Proposition~\ref{prop:multiplication}. This realization is
not needed for the ring-valued theorem.
\end{remark}

Another proof of Theorem~\ref{thm:main} \textbf{(b)} can be found in
\cite[Proposition~3.3]{facpoly}.

\section{Orbit harmonics and the regular representation}

We now prove Theorem~\ref{thm:main} \textbf{(c)}. The construction is the
orbit-harmonics, or point-orbit, method used by Oh and Rhoades
\cite[Section~1]{OhRhoades}; their Section~3.1 applies it to the coinvariant
algebra of a reflection group. We first record the associated-graded facts in
the form needed over a commutative coefficient ring.

Recall that $P$ is graded. If
\[
f=f_{0}+f_{1}+\cdots+f_{d}\qquad\text{(with }f_{i}\in P_{i}\text{ and}%
\ f_{d}\neq0\text{)}%
\]
is a nonzero polynomial in $P$, then its \emph{leading homogeneous part} is
defined to be
\[
\tau(f):=f_{d}.
\]
For an arbitrary ideal $I\subseteq P$, define the graded ideal
\begin{equation}
\operatorname{in}_{\deg}(I)=\sum_{\substack{f\in I;\\f\neq0}}\tau
(f)P.\label{eq:degree-initial-ideal}%
\end{equation}
Thus $\operatorname{in}_{\deg}(I)$ is generated by leading homogeneous parts
(not by leading monomials!). Reiner and Rhoades call this ideal
$\operatorname{gr}I$; compare \cite[Proposition~2.1]{ReinerRhoades}.

For any $d\in\mathbb{Z}$, define the $\mathbf{k}$-submodule
\[
P_{\leq d}:=
\begin{cases}
\displaystyle\bigoplus_{i=0}^{d}P_{i}, & \text{if }d\geq0,\\[6pt]
0, & \text{if }d<0.
\end{cases}
\]
Thus, $P_{\leq d}$ consists of the polynomials of degree at most $d$ (when
$d\geq0$), and
\[
P_{\leq d}P_{\leq e}\subseteq P_{\leq d+e}
\qquad\text{for all }d,e\in\mathbb{Z}.
\]
Hence $P$ is a filtered $\mathbf{k}$-algebra, with filtration
$0=P_{\leq-1}\subseteq P_{\leq0}\subseteq P_{\leq1}\subseteq\cdots$.

\begin{lemma}
[The associated graded quotient]\label{lem:associated-graded} Let $I$ be any
ideal of $P$. Give $P/I$ the filtration $0=F_{-1}\subseteq F_{0}\subseteq
F_{1}\subseteq\cdots$, where
\[
F_{d}:=F_{d}(P/I):=\frac{P_{\leq d}+I}{I}
\qquad\text{for each }d\in\mathbb{Z}.
\]
Thus $F_{d}$ is the image of $P_{\leq d}$ under the quotient map
$P\rightarrow P/I$. This is a multiplicative filtration: $F_{d}F_{e}\subseteq
F_{d+e}$.

The \emph{associated graded algebra} of $P/I$ is the graded
$\mathbf{k}$-algebra
\[
\operatorname{gr}_{F}(P/I):=\bigoplus_{d\geq0}F_{d}/F_{d-1}.
\]
Its multiplication is given explicitly by
\[
(a+F_{d-1})(b+F_{e-1})=ab+F_{d+e-1}
\qquad(a\in F_{d},\ b\in F_{e}).
\]
There is a canonical isomorphism of graded $\mathbf{k}$-algebras
\begin{equation}
P/\operatorname{in}_{\deg}(I)\xrightarrow{\ \sim\ }
\operatorname{gr}_{F}(P/I).\label{eq:canonical-associated-graded}
\end{equation}
More explicitly, if $p\in P_{d}$ is homogeneous, then this isomorphism sends
\[
p+\operatorname{in}_{\deg}(I)
\quad\longmapsto\quad
(p+I)+F_{d-1}\in F_{d}/F_{d-1}.
\]
Equivalently, it identifies the degree-$d$ component of the left-hand side
with
\begin{equation}
\left(P/\operatorname{in}_{\deg}(I)\right)_{d}
\cong \frac{P_{\leq d}+I}{P_{\leq d-1}+I}.
\label{eq:associated-graded-components}
\end{equation}
\end{lemma}

\begin{proof}
The inclusion $P_{\leq d}P_{\leq e}\subseteq P_{\leq d+e}$ implies
$F_{d}F_{e}\subseteq F_{d+e}$. It also shows that the displayed
multiplication is independent of the chosen representatives. For example, if
$a'\in F_{d-1}$, then $a'b\in F_{d+e-1}$, so replacing $a$ by $a+a'$ does
not change the product in $F_{d+e}/F_{d+e-1}$; the same argument applies to
$b$. Thus $\operatorname{gr}_{F}(P/I)$ is indeed a graded algebra.

For each $d\geq0$, define a $\mathbf{k}$-linear map
\[
\varphi_{d}:P_{d}\longrightarrow F_{d}/F_{d-1},
\qquad
p\longmapsto(p+I)+F_{d-1}.
\]
Taking the direct sum over all $d$ gives a graded $\mathbf{k}$-linear map
\[
\varphi:P=\bigoplus_{d\geq0}P_{d}
\longrightarrow\operatorname{gr}_{F}(P/I).
\]
This map is an algebra homomorphism. Indeed, if $p\in P_{d}$ and $q\in P_{e}$
are homogeneous, then
\begin{align*}
\varphi(p)\varphi(q)
&=\bigl((p+I)+F_{d-1}\bigr)\bigl((q+I)+F_{e-1}\bigr)\\
&=(pq+I)+F_{d+e-1}\\
&=\varphi(pq).
\end{align*}
It also sends $1$ to $1$, so it is a homomorphism of graded
$\mathbf{k}$-algebras.

The map $\varphi$ is surjective. Indeed, let an element of $F_{d}/F_{d-1}$ be
represented by $q+I$, where $q\in P_{\leq d}$. Write
\[
q=q_{0}+q_{1}+\cdots+q_{d}
\qquad(q_{i}\in P_{i}).
\]
Since $q-q_{d}\in P_{\leq d-1}$, the class of $q+I$ in $F_{d}/F_{d-1}$ is
$\varphi(q_{d})$.

We next compute the kernel. Let $0\neq f=f_{0}+f_{1}+\cdots+f_{d}\in I$, with
$f_{i}\in P_{i}$ and $f_{d}\neq0$. In $P/I$ we have
\[
f_{d}+I=-(f_{0}+f_{1}+\cdots+f_{d-1})+I\in F_{d-1}.
\]
Therefore $\varphi(\tau(f))=\varphi(f_{d})=0$. Since $\ker\varphi$ is an ideal,
it follows that
\[
\operatorname{in}_{\deg}(I)\subseteq\ker\varphi.
\]

For the reverse inclusion, first note that $\varphi$ is graded. Thus, if
$p=\sum_{d\geq0}p_{d}\in\ker\varphi$, where $p_{d}\in P_{d}$, then the direct
sum decomposition of the target gives $\varphi(p_{d})=0$ for every $d$.
Hence it is enough to consider a homogeneous element $p\in P_{d}$ satisfying
$\varphi(p)=0$. If $p=0$, then there is nothing to prove. Otherwise,
$\varphi(p)=0$ means that $p+I\in F_{d-1}$. By the definition of $F_{d-1}$,
there exists $r\in P_{\leq d-1}$ such that
\[
p-r\in I.
\]
The polynomial $p-r$ is nonzero, because $p$ is a nonzero homogeneous
polynomial of degree $d$, whereas $r$ has degree at most $d-1$. Its leading
homogeneous part is therefore $p$. Hence
\[
p=\tau(p-r)\in\operatorname{in}_{\deg}(I).
\]
This proves $\ker\varphi\subseteq\operatorname{in}_{\deg}(I)$, and thus
\[
\ker\varphi=\operatorname{in}_{\deg}(I).
\]
The first isomorphism theorem now yields
\eqref{eq:canonical-associated-graded}. For each $d$, the standard
quotient-of-a-quotient isomorphism gives
\[
F_{d}/F_{d-1}
=\frac{(P_{\leq d}+I)/I}{(P_{\leq d-1}+I)/I}
\cong\frac{P_{\leq d}+I}{P_{\leq d-1}+I},
\]
which proves the componentwise description
\eqref{eq:associated-graded-components}. No choices were made in the
construction of $\varphi$, which is why the isomorphism is canonical.
\end{proof}

\begin{lemma}
[Maschke splitting lemma]\label{lem:maschke-splitting} Let $G$ be a finite
group, and assume that $|G|$ is invertible in $\mathbf{k}$. Let
\begin{equation}
\label{eq:maschke-short-exact}0\longrightarrow U\longrightarrow
V\xrightarrow{\pi}W\longrightarrow0
\end{equation}
be a short exact sequence of $\mathbf{k}[G]$-modules. If
\eqref{eq:maschke-short-exact} splits as a sequence of $\mathbf{k}$-modules,
then it also splits as a sequence of $\mathbf{k}[G]$-modules. In particular,
this holds whenever $W$ is projective as a $\mathbf{k}$-module.
\end{lemma}

\begin{proof}
Choose a $\mathbf{k}$-linear section $s:W\to V$ of $\pi$. Define
\begin{equation}
\label{eq:averaged-section}\widetilde{s}(w) =\frac{1}{|G|}\sum_{g\in
G}g\,s(g^{-1}w) \qquad\text{for every }w\in W.
\end{equation}
This is a $\mathbf{k}$-linear map. It is a section of $\pi$, since the
$G$-equivariance of $\pi$ gives
\[
\pi\bigl(\widetilde{s}(w)\bigr)
=\frac{1}{|G|}\sum_{g\in G}g\,\pi\bigl(s(g^{-1}w)\bigr)
=\frac{1}{|G|}\sum_{g\in G}g(g^{-1}w) =w.
\]
It is also $G$-equivariant. Indeed, for $h\in G$, substitute $g=hk$ in the sum
to obtain
\[
\widetilde{s}(hw) =\frac{1}{|G|}\sum_{k\in G}hk\,s(k^{-1}w) =h\widetilde{s}%
(w).
\]
Thus $\widetilde{s}$ is a $\mathbf{k}[G]$-linear section of $\pi$. The final
assertion follows because a surjection onto a projective $\mathbf{k}$-module
has a $\mathbf{k}$-linear section. This is the usual averaging argument behind
Maschke's theorem; compare \cite[Theorem~4.4.14]{sga}.
\end{proof}

\begin{lemma}
[Equivariance and splitting]\label{lem:equivariant-splitting} Let a group $G$
act on $P$ by degree-preserving $\mathbf{k}$-algebra automorphisms, and let
$I\subseteq P$ be a $G$-stable ideal.

\begin{enumerate}
[label=\textbf{(\alph*)}]

\item The ideal $\operatorname{in}_{\deg}(I)$ is $G$-stable, and the canonical
isomorphism \eqref{eq:canonical-associated-graded} is $G$-equivariant.

\item Assume, in addition, that $G$ is finite, that $|G|$ is invertible in
$\mathbf{k}$, that the filtration of $P/I$ stabilizes, and that every
$\mathbf{k}$-module
\[
F_{d}(P/I)/F_{d-1}(P/I)
\]
is projective. Then there is a generally noncanonical isomorphism of ungraded
$\mathbf{k}[G]$-modules
\begin{equation}
\label{eq:filtered-split}P/I\cong\operatorname{gr}_{F}(P/I).
\end{equation}
In particular, the projectivity hypothesis holds if the associated graded
algebra has a homogeneous $\mathbf{k}$-basis.
\end{enumerate}
\end{lemma}

\begin{proof}
\textbf{(a)} For $g\in G$ and $0\neq f\in I$, degree preservation gives
$\tau(gf)=g\tau(f)$. Hence $\operatorname{in}_{\deg}(I)$ is $G$-stable. The
map $\varphi$ in the proof of Lemma~\ref{lem:associated-graded} is defined
only from the grading, the quotient map, and the induced filtration, all of
which are $G$-equivariant. Thus \eqref{eq:canonical-associated-graded} is $G$-equivariant.

\textbf{(b)} Write $F_{d}=F_{d}(P/I)$. Since the filtration stabilizes, only
finitely many of its successive quotients are nonzero. For every $d$,
projectivity of $F_{d}/F_{d-1}$ gives a $\mathbf{k}$-linear splitting of the
short exact sequence
\[
0\longrightarrow F_{d-1}\longrightarrow F_{d} \longrightarrow F_{d}%
/F_{d-1}\longrightarrow0.
\]
Lemma~\ref{lem:maschke-splitting} upgrades this to a $\mathbf{k}[G]$-linear
splitting. Choose a $G$-stable complement $L_{d}\subseteq F_{d}$ such that
\[
F_{d}=F_{d-1}\oplus L_{d}
\qquad\text{and}\qquad
L_{d}\cong F_{d}/F_{d-1}
\]
as $\mathbf{k}[G]$-modules. If $D$ is large enough that $F_{D}=P/I$, then
induction on $d$ gives
\[
P/I=F_{D}=\bigoplus_{d=0}^{D}L_{d}
\cong\bigoplus_{d=0}^{D}F_{d}/F_{d-1}
=\operatorname{gr}_{F}(P/I),
\]
which is \eqref{eq:filtered-split}.
\end{proof}

The symmetric group $\mathfrak{S}_{n}$ acts on $\mathbf{k}^{n}$ by
\[
\sigma\cdot(b_{1},b_{2},\ldots,b_{n}) =(b_{\sigma^{-1}(1)},b_{\sigma^{-1}%
(2)},\ldots,b_{\sigma^{-1}(n)}).
\]
This action is compatible with the action on $P$:
\[
(\sigma f)(\sigma\cdot v)=f(v).
\]


\begin{definition}
\label{def:strongly-discrete} A finite subset $Y\subseteq\mathbf{k}^{n}$ will
be called \emph{strongly discrete} if, for any two distinct points
\[
y=(y_{1},y_{2},\ldots,y_{n}) \quad\text{and}\quad z=(z_{1},z_{2},\ldots
,z_{n})
\]
of $Y$, at least one coordinate difference $y_{i}-z_{i}$ is a unit of
$\mathbf{k}$.
\end{definition}

If $Y$ is a finite set, let $\mathbf{k}^{Y}$ denote the $\mathbf{k}$-algebra
of all functions $Y\to\mathbf{k}$, with pointwise addition and multiplication.

\begin{lemma}
\label{lem:finite-point-coordinate-ring} Let $Y$ be a finite strongly discrete
subset of $\mathbf{k}^{n}$, and set
\[
I(Y)=\{p\in P\mid p(y)=0\text{ for every }y\in Y\}.
\]
Then:

\begin{enumerate}
\item[(a)] Evaluation induces a canonical $\mathbf{k}$-algebra isomorphism
\begin{equation}
\label{eq:finite-point-evaluation}\operatorname{ev}_{Y}%
:P/I(Y)\xrightarrow{\ \sim\ }\mathbf{k}^{Y}, \qquad p+I(Y)\longmapsto
(y\longmapsto p(y)).
\end{equation}


\item[(b)] If $Y$ is $\mathfrak{S}_{n}$-stable, then
\eqref{eq:finite-point-evaluation} is $\mathfrak{S}_{n}$-equivariant, where
\[
(\sigma\varphi)(y)=\varphi(\sigma^{-1}\cdot y).
\]
Thus it is an isomorphism of $\mathfrak{S}_{n}$-algebras and, in particular,
of $\mathbf{k}[\mathfrak{S}_{n}]$-modules.
\end{enumerate}
\end{lemma}

\begin{proof}
For $y=(y_{1},y_{2},\ldots,y_{n})\in Y$, set
\[
\mathfrak{m}_{y} =(x_{1}-y_{1})P+(x_{2}-y_{2})P+\cdots+(x_{n}-y_{n})P.
\]
Evaluation at $y$ is a surjective $\mathbf{k}$-algebra homomorphism
$P\to\mathbf{k}$ with kernel $\mathfrak{m}_{y}$. Indeed, $p-p(y)$ lies in
$\mathfrak{m}_{y}$ for every $p\in P$, as follows by replacing $x_{1}%
,x_{2},\ldots,x_{n}$ successively by $y_{1},y_{2},\ldots,y_{n}$ and factoring
each successive difference by $x_{i}-y_{i}$. Hence
\[
P/\mathfrak{m}_{y}\cong\mathbf{k}.
\]
Moreover,
\[
I(Y)=\bigcap_{y\in Y}\mathfrak{m}_{y}.
\]


The ideals $\mathfrak{m}_{y}$ are pairwise comaximal. Indeed, if $y\neq z$,
strong discreteness gives an $i$ for which $u:=y_{i}-z_{i}$ is a unit, and
\[
1=u^{-1}(x_{i}-z_{i})-u^{-1}(x_{i}-y_{i}) \in\mathfrak{m}_{z}+\mathfrak{m}%
_{y}.
\]
The Chinese remainder theorem now gives
\[
P/I(Y) \cong\prod_{y\in Y}P/\mathfrak{m}_{y} \cong\prod_{y\in Y}\mathbf{k}
=\mathbf{k}^{Y}.
\]
This is precisely the evaluation map in \eqref{eq:finite-point-evaluation}.

For equivariance, let $\sigma\in\mathfrak{S}_{n}$, $p\in P$, and $y\in Y$.
Then
\[
\bigl(\sigma\,\operatorname{ev}_{Y}(p+I(Y))\bigr)(y) =p(\sigma^{-1}\cdot y)
=(\sigma p)(y) =\operatorname{ev}_{Y}(\sigma p+I(Y))(y).
\]
This proves \textbf{(b)}.
\end{proof}

\begin{proposition}
\label{prop:coinvariant-regular} Assume that $n!$ is invertible in
$\mathbf{k}$. Then the coinvariant algebra $C=P/J$ is isomorphic, as an
ungraded $\mathbf{k}[\mathfrak{S}_{n}]$-module, to the regular module
$\mathbf{k}[\mathfrak{S}_{n}]$.
\end{proposition}

\begin{proof}
Put
\[
v=(1_{\mathbf{k}},2\cdot1_{\mathbf{k}},\ldots,n\cdot1_{\mathbf{k}}%
)\in\mathbf{k}^{n} \qquad\text{and}\qquad X=\mathfrak{S}_{n}\cdot v.
\]
If $r,s\in[n]$ are distinct, then $(r-s)\cdot1_{\mathbf{k}}$ is a unit.
Indeed, up to sign, the positive integer $|r-s|$ divides $n!$; since the image
of $n!$ in $\mathbf{k}$ is a unit, so is the image of $|r-s|$. It follows that
the coordinates of $v$ are pairwise distinct, that $X$ is an $\mathfrak{S}%
_{n}$-torsor, and that $X$ is strongly discrete.
Lemma~\ref{lem:finite-point-coordinate-ring} therefore gives
\begin{equation}
\label{eq:coordinate-algebra}P/I(X)\cong\mathbf{k}^{X}%
\end{equation}
as $\mathfrak{S}_{n}$-algebras.

For $x\in X$, let $\delta_{x}\in\mathbf{k}^{X}$ be its indicator function. The
functions $(\delta_{x})_{x\in X}$ form a $\mathbf{k}$-basis, and $\sigma
\delta_{x}=\delta_{\sigma\cdot x}$. Since $X$ is a torsor,
\begin{equation}
\label{eq:coordinate-regular}\mathbf{k}[\mathfrak{S}_{n}]\longrightarrow
\mathbf{k}^{X}, \qquad\sigma\longmapsto\delta_{\sigma\cdot v},
\end{equation}
is an isomorphism of $\mathbf{k}[\mathfrak{S}_{n}]$-modules.

Set
\[
c_{i}=e_{i}(v)\in\mathbf{k} \qquad\text{and}\qquad K=(e_{1}-c_{1}%
)P+(e_{2}-c_{2})P+\cdots+(e_{n}-c_{n})P.
\]
Every generator of $K$ vanishes on $X$, so $K\subseteq I(X)$.

We next identify $P/K$ explicitly. Evaluation at $v$ restricts to a surjective
homomorphism
\[
\epsilon_{v}:\Lambda\longrightarrow\mathbf{k}, \qquad f\longmapsto f(v),
\]
whose kernel is
\[
(e_{1}-c_{1})\Lambda+(e_{2}-c_{2})\Lambda+\cdots+(e_{n}-c_{n})\Lambda.
\]
Indeed, under the isomorphism
\[
\mathbf{k}[t_{1},t_{2},\ldots,t_{n}]\xrightarrow{\sim}\Lambda, \qquad
t_{i}\longmapsto e_{i},
\]
this map is evaluation at $(c_{1},c_{2},\ldots,c_{n})$, whose kernel is
$(t_{1}-c_{1})\mathbf{k}[t_{1},t_{2},\ldots,t_{n}]+\cdots+ (t_{n}%
-c_{n})\mathbf{k}[t_{1},t_{2},\ldots,t_{n}]$.
Corollary~\ref{cor:freeness-all-char} gives the graded direct-sum
decomposition
\[
P=\bigoplus_{m\in\mathcal{A}}\Lambda m,
\]
where $\mathcal{A}$ is the set of Artin monomials. Reducing this decomposition
modulo $K=(\ker\epsilon_{v})P$ shows that the residue classes of the Artin
monomials form a $\mathbf{k}$-basis of $P/K$.

The inclusion $K\subseteq I(X)$ gives a surjection
\[
P/K\twoheadrightarrow P/I(X)\cong\mathbf{k}^{X}.
\]
The source has the $n!$ Artin monomials as a basis, while the target has the
$n!$ indicator functions as a basis. Hence
Lemma~\ref{lem:finite-free-surjection} shows that this surjection is an
isomorphism. Thus
\begin{equation}
\label{eq:K=IX}K=I(X).
\end{equation}


We now determine the degree filtration on $P/I(X)=P/K$. For every $d\geq0$,
\begin{equation}
\label{eq:filtered-artin-basis}F_{d}(P/K) =\bigoplus_{\substack{m\in
\mathcal{A};\\\deg m\leq d}}\mathbf{k}\,\overline{m},
\end{equation}
where $\overline{m}$ denotes the residue class of $m$ modulo $K$. The
inclusion ``$\supseteq$'' is immediate. For the reverse inclusion, let $p\in
P_{\leq d}$. Write it uniquely as
\[
p=\sum_{m\in\mathcal{A}}\lambda_{m} m \qquad(\lambda_{m}\in\Lambda).
\]
Since the decomposition $P=\bigoplus_{m}\Lambda m$ is graded, every
homogeneous component of $\lambda_{m}$ has degree at most $d-\deg m$. In
particular, $\lambda_{m}=0$ when $\deg m>d$. Modulo $K$, each $\lambda_{m}$
becomes the scalar $\epsilon_{v}(\lambda_{m})$, so the class of $p$ lies in
the right-hand side of \eqref{eq:filtered-artin-basis}. Directness follows
from the fact that all Artin classes form a basis of $P/K$.

Consequently, $\operatorname{gr}_{F}(P/I(X))$ has the homogeneous $\mathbf{k}%
$-basis
\[
m^{\ast}\qquad(m\in\mathcal{A}),
\]
where $m^{\ast}$ is the class of $\overline m$ in filtration degree $\deg m$.

Since the leading homogeneous part of $e_{i}-c_{i}$ is $e_{i}$, we have
\[
J=e_{1}P+e_{2}P+\cdots+e_{n}P \subseteq\operatorname{in}_{\deg}(I(X)).
\]
Thus Lemma~\ref{lem:associated-graded} gives a canonical surjective graded
homomorphism
\begin{equation}
\label{eq:C-to-gr}C=P/J\twoheadrightarrow P/\operatorname{in}_{\deg}(I(X))
\xrightarrow{\ \sim\ }\operatorname{gr}_{F}(P/I(X)).
\end{equation}
This map sends the residue class of each Artin monomial $m$ to $m^{\ast}$.
Proposition~\ref{prop:artin} says that the former classes are a basis of $C$,
and the preceding paragraph says that the latter classes are a basis of the
associated graded algebra. Hence \eqref{eq:C-to-gr} is an isomorphism. In
particular,
\begin{equation}
\label{eq:J=degree-initial}J=\operatorname{in}_{\deg}(I(X)).
\end{equation}


The filtration in \eqref{eq:filtered-artin-basis} stabilizes, and each of its
successive quotients is free, with basis given by the Artin monomials of the
corresponding degree. Lemma~\ref{lem:equivariant-splitting} therefore gives an
isomorphism of ungraded $\mathbf{k}[\mathfrak{S}_{n}]$-modules
\[
P/I(X)\cong\operatorname{gr}_{F}(P/I(X))\cong C.
\]
Combining this with \eqref{eq:coordinate-algebra} and
\eqref{eq:coordinate-regular} yields
\[
C\cong\mathbf{k}[\mathfrak{S}_{n}].
\]
This proves Theorem~\ref{thm:main} \textbf{(c)}. It also recovers the
classical statement recorded by Haiman that the coinvariant algebra affords
the regular representation \cite[Section~1.5]{Haiman}.
\end{proof}

\begin{remark}
The preceding proof uses the orbit-harmonics argument of Oh and Rhoades at two
distinct points: the passage from the point quotient to its associated graded
algebra, and the use of a free orbit whose function module is regular. The
proof only concerns the coinvariant algebra as an \emph{ungraded}
$\mathfrak{S}_{n}$-module. Special choices involving roots of unity produce an
additional cyclic symmetry that encodes the grading, but no such refinement is
needed here.
\end{remark}

\begin{remark}
The part of the construction preceding the equivariant splitting requires less
than the invertibility of $n!$. Namely, suppose that $a_{1},a_{2},\ldots
,a_{n}\in\mathbf{k}$ have pairwise invertible differences, and let
\[
v=(a_{1},a_{2},\ldots,a_{n}), \qquad X=\mathfrak{S}_{n}\cdot v.
\]
Then $X$ is a strongly discrete $\mathfrak{S}_{n}$-torsor, and the same
argument gives the canonical graded isomorphism
\[
C\cong\operatorname{gr}_{F}(P/I(X)).
\]
The invertibility of $n!$ is used only to average the splittings that identify
the filtered module $P/I(X)$ with its associated graded module equivariantly.
\end{remark}

\section{Proof of the equivariant normal-basis statement}

\begin{proof}
[Proof of Theorem~\ref{thm:main} \textbf{(d)}]Assume that $n!$ is invertible
in $\mathbf{k}$. For each $d$, the quotient map
\[
P_{d}\twoheadrightarrow C_{d}
\]
has a $\mathbf{k}$-linear section, because $C_{d}$ is free by
Proposition~\ref{prop:artin}. Applying Lemma~\ref{lem:maschke-splitting} to
the short exact sequence
\[
0\longrightarrow J_{d}\longrightarrow P_{d}\longrightarrow C_{d}
\longrightarrow0
\]
gives an $\mathfrak{S}_{n}$-equivariant section. Let $H_{d}$ be its image and
set
\[
H=\bigoplus_{d\geq0}H_{d}.
\]
Then $H$ is a graded $\mathfrak{S}_{n}$-stable $\mathbf{k}$-submodule of $P$,
and the quotient map restricts to an isomorphism
\[
H\xrightarrow{\ \sim\ }C
\]
of graded $\mathbf{k}[\mathfrak{S}_{n}]$-modules.

By Proposition~\ref{prop:coinvariant-regular}, choose a $\mathbf{k}%
[\mathfrak{S}_{n}]$-module isomorphism
\[
\theta:\mathbf{k}[\mathfrak{S}_{n}]\xrightarrow{\ \sim\ }H,
\]
and put $f=\theta(1)$. For every $\sigma\in\mathfrak{S}_{n}$,
\[
\theta(\sigma) =\theta(\sigma\cdot1) =\sigma\cdot\theta(1) =\sigma(f).
\]
Thus the family
\[
\{\sigma(f)\mid\sigma\in\mathfrak{S}_{n}\}
\]
is a $\mathbf{k}$-basis of $H$. Proposition~\ref{prop:multiplication} shows
that it is a $\Lambda$-basis of $P$.

Under the canonical identification
\[
\Lambda[\mathfrak{S}_{n}] \cong\Lambda\otimes_{\mathbf{k}}\mathbf{k}%
[\mathfrak{S}_{n}],
\]
the map $\Phi$ from \eqref{eq:normal-basis-map} is the composite
\[
\Lambda\otimes_{\mathbf{k}}\mathbf{k}[\mathfrak{S}_{n}]
\xrightarrow{\ \id_\Lambda\otimes\theta\ } \Lambda\otimes_{\mathbf{k}}H
\xrightarrow{\ m\ }P.
\]
Both arrows are $\Lambda[\mathfrak{S}_{n}]$-linear isomorphisms. Hence so is
$\Phi$.
\end{proof}

\begin{remark}
[The isomorphism is not graded]The element $f$ cannot be chosen homogeneous
when $n\geq2$. Indeed, if $\deg f=d$, then all $\sigma(f)$ have degree $d$,
and the resulting isomorphism would imply
\[
\operatorname{Hilb}(P;q)=q^{d} n!\,\operatorname{Hilb}(\Lambda;q),
\]
which is false. What is graded is the decomposition
\[
P\cong\Lambda\otimes_{\mathbf{k}}H,
\]
where $H$ has the nontrivial coinvariant grading. Only after forgetting this
grading is $H$ the regular module.
\end{remark}

\begin{example}
Let $n=2$ and assume that $2$ is invertible in $\mathbf{k}$. Then
\[
\Lambda=\mathbf{k}[x_{1}+x_{2},x_{1}x_{2}] \qquad\text{and}\qquad
H=\operatorname{span}_{\mathbf{k}}\{1,x_{1}-x_{2}\}.
\]
For the transposition $s=(1\ 2)$, the element
\[
f=1+(x_{1}-x_{2})
\]
has orbit
\[
f=1+(x_{1}-x_{2}), \qquad s(f)=1-(x_{1}-x_{2}),
\]
which is a $\mathbf{k}$-basis of $H$ and therefore a $\Lambda$-basis of $P$.
\end{example}

\section{A modular counterexample}

The invertibility hypothesis in Theorem~\ref{thm:main} \textbf{(c)} and
\textbf{(d)} cannot simply be omitted. Take $n=2$ and $\mathbf{k}%
=\mathbb{F}_{2}$. Writing $s=(1\ 2)$, we have
\[
C =\frac{\mathbf{k}[x_{1},x_{2}]} {(x_{1}+x_{2})P+x_{1}x_{2}P} \cong%
\frac{\mathbf{k}[t]}{t^{2}\mathbf{k}[t]}.
\]
The relation $x_{1}+x_{2}=0$ becomes $x_{1}=x_{2}$, so $s$ acts trivially on
all of $C$. On the other hand, $s$ does not act trivially on the regular
module $\mathbf{k}[\mathfrak{S}_{2}]$: it interchanges the basis vectors $1$
and $s$. Hence, $C$ is not isomorphic to $\mathbf{k}[\mathfrak{S}_{2}]$ as
$\mathbf{k}[\mathfrak{S}_{2}]$-modules, and Theorem~\ref{thm:main}
\textbf{(c)} fails.

Theorem~\ref{thm:main} \textbf{(d)} fails as well. Indeed, tensoring a
hypothetical $\Lambda[\mathfrak{S}_{2}]$-module isomorphism $P\cong%
\Lambda[\mathfrak{S}_{2}]$ over $\Lambda$ with $\Lambda/\Lambda_{+}%
\cong\mathbf{k}$ --- or, equivalently, reducing modulo $\Lambda_{+}%
P=e_{1}P+e_{2}P$ --- would produce the impossible isomorphism $C\cong%
\mathbf{k}[\mathfrak{S}_{2}]$.

\section*{Notes on the origins of the proofs}

The two field-valued ingredients underlying Theorem~\ref{thm:main} are stated
in Haiman's treatment of the ordinary coinvariant algebra \cite[Section~1.5]%
{Haiman}: the coinvariant algebra affords the regular representation, and a
homogeneous space of representatives freely generates the polynomial ring over
the invariant subring. The proofs above formulate both statements over a
commutative coefficient ring, under the hypotheses stated in
Theorem~\ref{thm:main}.

The degeneration from a finite point locus to the quotient by leading
homogeneous parts follows the orbit-harmonics framework of Oh and Rhoades
\cite[Section~1]{OhRhoades}; their Section~3.1 explains the reflection-group
coinvariant case via a regular orbit. Our orbit
\[
(1_{\mathbf{k}},2\cdot1_{\mathbf{k}},\ldots,n\cdot1_{\mathbf{k}})
\]
is chosen so that distinct orbit points are strongly discrete. The additional
cyclic action used in graded refinements of orbit harmonics is not needed
here. The associated-graded construction is also recorded by Reiner and
Rhoades \cite[Propositions~2.1 and~2.7]{ReinerRhoades}.

Our proof of Theorem~\ref{thm:main} \textbf{(a)} imitates \cite[proof of
Theorem~1.2.7]{Sturmfels}, but replaces the additional variables $y_{1}%
,y_{2},\ldots,y_{n}$ by $0,0,\ldots,0$.

\begin{thebibliography}{99}                                                                                               %


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\bibitem {Artin71}%
\href{https://projecteuclid.org/ebooks/notre-dame-mathematical-lectures/Galois-Theory/toc/ndml/1175197041}{Emil
Artin, \textit{Galois theory}, lectures delivered at the University of Notre
Dame, edited and supplemented by Arthur N. Milgram, Notre Dame Mathematical
Lectures \textbf{2}, University of Notre Dame Press, 2nd edition, 6th printing
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\bibitem {Bourba03}\href{https://doi.org/10.1007/978-3-642-61698-3}{Nicolas
Bourbaki, \textit{Algebra II: Chapters 4--7}, Springer 2003}.

\bibitem {deGraaf}Willem~de Graaf, \emph{Computational Algebra}, lecture
notes; see Chapter~1, especially Sections~1.1.6--1.1.7 and
Theorem~1.1.33.\newline%
\url{https://degraaf.maths.unitn.it/algnotes/compalg.pdf}. See
\url{https://www.cip.ifi.lmu.de/~grinberg/algebra/compalg-errata-v2.pdf} for
unofficial errata.

\bibitem {FoGePo97}\href{https://doi.org/10.1090/S0894-0347-97-00237-3}{Sergey
Fomin, Sergei Gelfand, Alexander Postnikov, \textit{Quantum Schubert
polynomials}, Journal of the American Mathematical Society \textbf{10}, number
3 (1997), pp. 565--596}.

\bibitem {Gailla21}Pierre-Yves Gaillard, Math StackExchange answer \#4261642,
``Basis for $\mathbb{Z}[x_{1},x_{2},\ldots,x_{n}]$ over $\mathbb{Z}%
[e_{1},e_{2},\ldots,e_{n}]$.''\newline\url{https://math.stackexchange.com/q/4261642}.

\bibitem {Garsia02}Adriano M. Garsia, \textit{Pebbles and Expansions in the
Polynomial Ring}, 21 July 2002.\newline\url{http://www.math.ucsd.edu/~garsia/somepapers/fnewpebbles.pdf}

\bibitem {facpoly}GPT-5.5 and Darij Grinberg, \textit{Splitting a Polynomial
into Linear Factors after an Injective Ring Extension}, 19 July 2026. \newline\url{https://www.cip.ifi.lmu.de/~grinberg/algebra/factorpoly-gpt.pdf}

\bibitem {subdiv}Darij Grinberg, \textit{$t$-unique reductions for
M\'{e}sz\'aros's subdivision algebra}, updated version of 16 June
2016.\newline\url{https://www.cip.ifi.lmu.de/~grinberg/algebra/subdiv-v7.pdf}

\bibitem {sga}Darij Grinberg, \textit{An introduction to the symmetric group
algebra}, arXiv:2507.20706v1. \newline\url{https://arxiv.org/abs/2507.20706v1}

\bibitem {Haiman}Mark D.~Haiman, \emph{Conjectures on the quotient ring by
diagonal invariants}, J. Algebraic Combin. \textbf{3} (1994), no.~1, 17--76;
see Section~1.5 for the classical one-set-of-variables case, \url{https://math.berkeley.edu/~mhaiman/ftp/diagonal/diagonal.pdf}.

\bibitem {LLPT95}D. Laksov, A. Lascoux, P. Pragacz, and A. Thorup, \textit{The
LLPT Notes}, edited by A. Thorup, 1995--2018,\newline\url{http://web.math.ku.dk/noter/filer/sympol.pdf}.

\bibitem {Macdon91}Ian~G. Macdonald, \emph{Notes on Schubert polynomials},
Laboratoire de combinatoire et d'informatique math\'ematique, Universit\'e du
Qu\'ebec \`a Montr\'eal, 1991.
\url{https://lacim.uqam.ca/les-parutions/LACIM-Publications-Volume-06.pdf}
\newline See
\url{https://www.cip.ifi.lmu.de/~grinberg/algebra/mcd-schub-errata.pdf} for errata.

\bibitem {OhRhoades}Jaeseong~Oh and Brendon~Rhoades, \emph{Cyclic sieving and
orbit harmonics}, Math. Z. \textbf{300} (2022), 639--660,
\href{https://doi.org/10.1007/s00209-021-02800-z}{doi:10.1007/s00209-021-02800-z}%
.

\bibitem {ReinerRhoades}Victor~Reiner and Brendon~Rhoades, \emph{Harmonics and
graded Ehrhart theory}, preprint, arXiv:2407.06511v3 (2024); see
Propositions~2.1 and~2.7, \url{https://arxiv.org/abs/2407.06511v3}.

\bibitem {Sturmfels}%
\href{https://www.math.ens.psl.eu/~benoist/refs/Sturmfels.pdf}{Bernd
Sturmfels, \emph{Algorithms in Invariant Theory}, 2nd edition, Springer 2008.}
See
\url{https://www.cip.ifi.lmu.de/~grinberg/algebra/sturmfels-ait-errata.pdf}
for unofficial errata.
\end{thebibliography}


\end{document}