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\begin{document}

\title{Polynomials over Symmetric Polynomials\\[5pt] {\large {[semi-expository note based on the literature and emails of Darij
Grinberg and Victor Reiner]}}}
\author{GPT-5.6 Sol, edited by Darij Grinberg}
\date{\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}\left[  x_{1}%
,x_{2},\ldots,x_{n} \right]  $ by permuting the variables. We prove four
results that are known (at least in the case when $\mathbf{k}$ is
a field) but not easily found in the literature.
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}\left[  \mathfrak{S}_{n}
\right]  $-module.
Fourth, under the same hypothesis, $P$ is a free left
$P^{\mathfrak{S}_{n}}\left[  \mathfrak{S}_{n} \right]  $-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.} The main ideas behind this paper are not new;
I have learned some of them from Vic Reiner and Brendon Rhoades. The main
novelty, to the extent it is one,
is extending the proofs from fields to commutative rings.
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}

Set $\mathbb{N}=\left\{  0,1,2,\ldots\right\}  $. For any $r\in\mathbb{N}$,
let $\left[  r\right]  $ denote the set $\left\{  1,2,\ldots,r\right\}  $.

Let $n$ be a positive integer. Let $\mathfrak{S}_{n}$ denote the $n$-th
\emph{symmetric group}, that is, the group of all bijections $\left[
n\right]  \rightarrow\left[  n\right]  $. 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
\[
P=\mathbf{k}\left[  x_{1},x_{2},\ldots,x_{n} \right]  .
\]
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:
\[
\sigma(x_{i})=x_{\sigma(i)}\qquad\text{for all }\sigma\in\mathfrak{S}%
_{n}\text{ and }i\in\left[  n \right]  .
\]
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\left[  n\right]  $,
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}\left[  e_{1},e_{2},\ldots,e_{n}\right]  , \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
\[
\mathbf{k}\left[  t_{1},t_{2},\ldots,t_{n}\right]  \longrightarrow
\Lambda,\qquad t_{i}\longmapsto e_{i}%
\]
(a $\mathbf{k}$-algebra homomorphism from the polynomial ring $\mathbf{k}%
\left[  t_{1},t_{2},\ldots,t_{n}\right]  $), 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}\left[
t_{1},t_{2},\ldots,t_{n}\right]  $. 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\left[  \mathfrak{S}%
_{n} \right]  $.

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}\left[  \mathfrak{S}_{n} \right]  $.

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
\[
x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{n}^{a_{n}} \qquad\text{with}\qquad0\leq
a_{i}<i\quad\text{for every }i\in\left[  n \right]
\]
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}\left[  \mathfrak{S}%
_{n}\right]  $-module is the same thing as a representation of $\mathfrak{S}%
_{n}$ over $\mathbf{k}$.}
\[
C\cong\mathbf{k}\left[  \mathfrak{S}_{n}\right]  \qquad\text{as a
$\mathbf{k}\left[  \mathfrak{S}_{n}\right]  $-module.}%
\]
In other words, the coinvariant algebra $C=P/J$ is isomorphic to the regular
$\mathbf{k}\left[  \mathfrak{S}_{n}\right]  $-module.

\item[\textbf{(d)}] Assume that $n!$ is invertible in $\mathbf{k}$. Then
\[
P\cong\Lambda\left[  \mathfrak{S}_{n} \right]  \qquad\text{as a $\Lambda
\left[  \mathfrak{S}_{n} \right]  $-module.}%
\]
In other words, $P$ is isomorphic to the regular $\Lambda\left[
\mathfrak{S}_{n} \right]  $-module. Equivalently, there exists an element
$f\in P$ such that the map
\begin{equation}
\Phi:\Lambda\left[  \mathfrak{S}_{n} \right]  \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\left[  \mathfrak{S}_{n} \right]  $-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%
\[
A:=R\left[  z_{1},z_{2},\ldots,z_{N}\right]  ,
\]
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
\[
m_{i}=\operatorname{LM}(g_{i})\qquad\text{(for $i\in\left[  k\right]  $)}%
\]
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\left[  k\right]  $). 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\left[  k \right]  $, 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\left[  k \right]  $ 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\left[  k\right]  $, 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}
\rho(qg_{i})=0. \label{eq:rho-kills-ideal}%
\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\left[  n\right]  $, we have the identity
\begin{equation}
\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)
\label{eq:elementary-complete-tail}%
\end{equation}
in the formal power-series ring $P\left[  \left[  t\right]  \right]  $, 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\left[  \left[  t\right]
\right]  $ 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\left[  n \right]  $, 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\left[  n \right]  \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\left[  n \right]  $. 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}\left[  \left[  q\right]  \right]  .
\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 \emph{graded basis} (i.e., a basis consisting of
homogeneous vectors).

\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 graded $\mathbf{k}$-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}%
=\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
\[
x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{n}^{a_{n}}\qquad\text{(with }0\leq
a_{i}<i\text{ for every }i\in\left[  n \right]  \text{)}%
\]
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}\left[  x_{1},x_{2}%
,\ldots,x_{n} \right]  $, 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}

Other proofs of Theorem~\ref{thm:main} \textbf{(b)} can be found in
\cite[(5.1)]{Macdon91} and in \cite[Proposition~3.3]{facpoly}. (In both
sources, the theorem is only stated for $\mathbf{k}=\mathbb{Z}$, but the
general case can be easily derived from this via base change.)

\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.

We begin with a simple and fundamental lemma about evaluation homomorphisms:

\begin{lemma}
\label{lem:eval-ker}Let $y=(y_{1},y_{2},\ldots,y_{n})\in\mathbf{k}^{n}$ be any
point. Then, the map%
\begin{align*}
\epsilon_{y}:P  & \rightarrow\mathbf{k},\\
f  & \mapsto f\left(  y\right)  =f\left(  y_{1},y_{2},\ldots,y_{n}\right)
\end{align*}
is a $\mathbf{k}$-algebra homomorphism, known as \emph{evaluation at }$y$. Its
kernel is the ideal
\begin{equation}
\mathfrak{m}_{y}:=(x_{1}-y_{1})P+(x_{2}-y_{2})P+\cdots+(x_{n}-y_{n}%
)P\label{eq:lem:eval-ker:my}%
\end{equation}
of $P$.
\end{lemma}

\begin{proof}
The fact that $\epsilon_{y}$ is a homomorphism is completely elementary. To
show that $\mathfrak{m}_{y}$ (as defined in \eqref{eq:lem:eval-ker:my}) is the
kernel of $\epsilon_{y}$, we can argue as follows: Any polynomial $f\in P$ can
be transformed into $\epsilon_{y}\left(  f\right)  =f\left(  y\right)  $ by
replacing $x_{1},x_{2},\ldots,x_{n}$ successively by $y_{1},y_{2},\ldots
,y_{n}$. Each replacement changes the polynomial by a difference that is
divisible by $x_{i}-y_{i}$; thus, the total difference $f-\epsilon_{y}\left(
f\right)  $ belongs to $(x_{1}-y_{1})P+(x_{2}-y_{2})P+\cdots+(x_{n}%
-y_{n})P=\mathfrak{m}_{y}$. If $f\in\operatorname*{Ker}\epsilon_{y}$, then
$\epsilon_{y}\left(  f\right)  =0$, so this means that $f$ itself belongs to
$\mathfrak{m}_{y}$. Thus, $\ker\epsilon_{y}\subseteq\mathfrak{m}_{y}$. The
converse implication $\mathfrak{m}_{y}\subseteq\ker\epsilon_{y}$ is even more
obvious, so $\ker\epsilon_{y}=\mathfrak{m}_{y}$ follows.

Alternatively, reduce the claim to the case $y=\left(  0,0,\ldots,0\right)  $
by substituting $x_{i}-y_{i}$ for each $x_{i}$.
\end{proof}

Recall that the polynomial ring $P$ 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
\[
P_{\leq d}:=\bigoplus\limits_{i=0}^{d}P_{i}%
\]
(where an empty direct sum is understood to be $\left\{  0\right\}  $, so that
$P_{\leq d}=0$ for all negative $d$).

Thus, $P_{\leq d}$ consists of the polynomials of degree at most $d$, and we
have%
\[
P_{\leq d}P_{\leq e}\subseteq P_{\leq d+e}\qquad\text{for all }d,e\in
\mathbb{Z}.
\]
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 $P$. The \emph{degree filtration} of $P/I$ is the filtration
$0=F_{-1}\subseteq F_{0}\subseteq F_{1}\subseteq\cdots$ of $P/I$ defined by
setting%
\[
F_{d}:=F_{d}^{(P/I)}:=\frac{P_{\leq d}+I}{I}\qquad\text{for each }%
d\in\mathbb{Z}.
\]
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
\begin{align}
(a+F_{d-1})(b+F_{e-1})  &  =ab+F_{d+e-1}\label{eq:gr-mul}\\
\ \ \ \ \ \ \ \ \ \ \text{for all }a  &  \in F_{d}\text{ and }b\in
F_{e}.\nonumber
\end{align}
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 multiplication rule
\eqref{eq:gr-mul} is independent of the chosen representatives. For example,
if $a^{\prime}\in F_{d-1}$, then $a^{\prime}b\in F_{d+e-1}$, so replacing $a$
by $a+a^{\prime}$ does not change the product $(a+F_{d-1})(b+F_{e-1})$ 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 of these $\varphi_{d}$ 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}\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{eq:gr-mul}}%
\right) \\
&  =\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 $q+I\in F_{d}$, where $q\in P_{\leq d}$. Write the polynomial
$q\in P_{\leq d}$ as%
\[
q=q_{0}+q_{1}+\cdots+q_{d}\qquad\text{(with }q_{i}\in P_{i}\text{)}.
\]
Thus, $q-q_{d}=q_{0}+q_{1}+\cdots+q_{d-1}\in P_{\leq d-1}$, so that the class
of $q+I$ in $F_{d}/F_{d-1}$ is $\varphi(q_{d})$. This shows that the class of
$q+I$ in $F_{d}/F_{d-1}$ belongs to the image of $\varphi$. By linearity, we
thus conclude that $\varphi$ is surjective.

We next compute the kernel of $\varphi$. Let $f=f_{0}+f_{1}+\cdots+f_{d}\in I$
be a nonzero polynomial, with $f_{i}\in P_{i}$ and $f_{d}\neq0$. Thus,
$\tau\left(  f\right)  =f_{d}$. But $f_{0}+f_{1}+\cdots+f_{d}\in I$ entails
\[
f_{d}+I=-(f_{0}+f_{1}+\cdots+f_{d-1})+I\in F_{d-1}\ \ \ \ \ \ \ \ \ \ \text{in
}P/I.
\]
Therefore $\varphi\left(  f_{d}\right)  =0$. That is, $\varphi(\tau(f))=0$
(since $\tau\left(  f\right)  =f_{d}$), so that $\tau\left(  f\right)  \in
\ker\varphi$. Since $\ker\varphi$ is an ideal, it follows that
\begin{equation}
\operatorname{in}_{\deg}(I)\subseteq\ker\varphi. \label{eq:in-in-ker}%
\end{equation}


We shall next prove 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 $\varphi(p)=0$. We must show that $p\in\operatorname{in}_{\deg}%
(I)$. 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 $\tau\left(  p-r\right)  $ is therefore $p$. Hence
\[
p=\tau(p-r)\in\operatorname{in}_{\deg}(I)\ \ \ \ \ \ \ \ \ \ \left(
\text{since }p-r\in I\right)  .
\]
This proves $\ker\varphi\subseteq\operatorname{in}_{\deg}(I)$. Combining this
with \eqref{eq:in-in-ker}, we find
\[
\ker\varphi=\operatorname{in}_{\deg}(I).
\]


So $\varphi:P\rightarrow\operatorname{gr}_{F}(P/I)$ is a surjective graded
$\mathbf{k}$-algebra homomorphism with kernel $\ker\varphi=\operatorname{in}%
_{\deg}(I)$. The first isomorphism theorem thus yields a canonical graded
$\mathbf{k}$-algebra isomorphism \eqref{eq:canonical-associated-graded}. Thus,
for each $d$, we have an isomorphism
\[
\left(  P/\operatorname{in}_{\deg}(I)\right)  _{d}\longrightarrow
F_{d}/F_{d-1}=\frac{(P_{\leq d}+I)/I}{(P_{\leq d-1}+I)/I}\cong\frac{P_{\leq
d}+I}{P_{\leq d-1}+I}%
\]
(by the standard quotient-of-a-quotient isomorphism), 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}

We will use one of the staples of group representation theory (see, e.g.,
\cite[proof of Theorem 4.4.14]{sga}), which we prove here for the sake of completeness:

\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}
0\longrightarrow U\longrightarrow V\overset{\pi}{\longrightarrow} W\longrightarrow0
\label{eq:maschke-short-exact}%
\end{equation}
be a short exact sequence of $\mathbf{k}\left[  G\right]  $-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}\left[  G\right]  $-modules.
In particular, this holds whenever $W$ is projective as a $\mathbf{k}$-module.
\end{lemma}

\begin{proof}
Assume that \eqref{eq:maschke-short-exact} splits as a sequence of
$\mathbf{k}$-modules. Choose a $\mathbf{k}$-linear section $s:W\rightarrow V$
of $\pi$. Thus, $\pi s=\operatorname*{id}$. Define a $\mathbf{k}$-linear map
$\widetilde{s}:W\rightarrow V$ by%
\begin{equation}
\widetilde{s}(w)=\frac{1}{|G|}\sum_{g\in G}g\,s(g^{-1}w)\qquad\text{for every
}w\in W. \label{eq:averaged-section}%
\end{equation}
This map $\widetilde{s}$ is a section of $\pi$, since the $G$-equivariance of
$\pi$ gives
\[
\pi\bigl(\widetilde{s}(w)\bigr)=\frac{1}{|G|}\sum_{g\in G}g\,\underbrace{\pi
\bigl(s(g^{-1}w)\bigr)}_{\substack{=g^{-1}w\\\text{(since }\pi
s=\operatorname*{id}\text{)}}}=\frac{1}{|G|}\sum_{g\in G}\underbrace{g(g^{-1}%
w)}_{=w}=w
\]
and thus $\pi\widetilde{s}=\operatorname*{id}$. It is furthermore
$G$-equivariant. Indeed, for $h\in G$ and $w\in W$, we have%
\begin{align*}
\widetilde{s}(hw)  &  =\frac{1}{|G|}\sum_{g\in G}g\,s(g^{-1}hw)\\
&  =\frac{1}{|G|}\sum_{k\in G}hk\,s(\underbrace{\left(  hk\right)  ^{-1}%
h}_{=k^{-1}}w)\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted
}hk\text{ for }g\right) \\
&  =\frac{1}{|G|}\sum_{k\in G}hk\,s(k^{-1}w)=h\cdot\underbrace{\frac{1}%
{|G|}\sum_{k\in G}k\,s(k^{-1}w)}_{=\widetilde{s}\left(  w\right)  }\\
&  =h\widetilde{s}(w).
\end{align*}
Thus $\widetilde{s}$ is a $\mathbf{k}\left[  G\right]  $-linear section of
$\pi$, and therefore the sequence \eqref{eq:maschke-short-exact} splits as a
sequence of $\mathbf{k}\left[  G\right]  $-modules.

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 $\operatorname{in}_{\deg}(I)$ and the degree filtration
$0=F_{-1}\subseteq F_{0}\subseteq F_{1}\subseteq\cdots$ from Lemma
\ref{lem:associated-graded} are $G$-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 (i.e., we have
$F_{D}=P/I$ for some $D\in\mathbb{N}$), and that every $\mathbf{k}$-module
$F_{d}/F_{d-1}$ is projective. Then there is a (generally noncanonical)
isomorphism of ungraded $\mathbf{k}\left[  G\right]  $-modules
\begin{equation}
P/I\cong\operatorname{gr}_{F}(P/I). \label{eq:filtered-split}%
\end{equation}


\item The projectivity hypothesis from part \textbf{(b)} (i.e., the assumption
that every $\mathbf{k}$-module $F_{d}/F_{d-1}$ is projective) holds
automatically if the associated graded algebra $\operatorname*{gr}%
\nolimits_{F}\left(  P/I\right)  $ has a graded $\mathbf{k}$-basis (i.e., a
$\mathbf{k}$-basis consisting of homogeneous elements).
\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. That
the filtration $0=F_{-1}\subseteq F_{0}\subseteq F_{1}\subseteq\cdots$ is
$G$-stable is even more obvious (since $P_{\leq d}$ and $I$ are $G$-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. \medskip

\textbf{(b)} Since the filtration stabilizes, there exists a large enough
$D\in\mathbb{N}$ such that $F_{D}=P/I$. Consider this $D$. Of course, this
entails that $F_{d}=P/I$ for all $d\geq D$; therefore, all successive
quotients $F_{d}/F_{d-1}$ with $d>D$ are zero.

For every $d$, the 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}\left[
G\right]  $-linear splitting. Thus, we can choose a $G$-stable complement
$L_{d}\subseteq F_{d}$ such that
\[
F_{d}=F_{d-1}\oplus L_{d}\qquad\text{and therefore}\qquad L_{d}\cong
F_{d}/F_{d-1}%
\]
as $\mathbf{k}\left[  G\right]  $-modules. Choose such $L_{d}$ for each
$d\geq0$; note that all $d>D$ will satisfy $L_{d}=0$ (since $F_{d}/F_{d-1}$ is zero).

Now, successively applying the equalities $F_{d}=F_{d-1}\oplus L_{d}$, we see
that each $k\geq0$ satisfies%
\begin{align*}
F_{k}  &  =\underbrace{F_{k-1}}_{=F_{k-2}\oplus L_{k-1}}\oplus\,L_{k}%
=\underbrace{F_{k-2}}_{=F_{k-3}\oplus L_{k-2}}\oplus\,L_{k-1}\oplus L_{k}\\
&  =F_{k-3}\oplus L_{k-2}\oplus L_{k-1}\oplus L_{k}=\cdots=\underbrace{F_{-1}%
}_{=0}\oplus\,L_{0}\oplus L_{1}\oplus\cdots\oplus L_{k}\\
&  =L_{0}\oplus L_{1}\oplus\cdots\oplus L_{k}=\bigoplus_{d=0}^{k}L_{d}.
\end{align*}
Applying this to $k=D$, we obtain an (ungraded) $\mathbf{k}\left[  G\right]
$-module isomorphism
\[
F_{D}=\bigoplus_{d=0}^{D}\underbrace{L_{d}}_{\cong F_{d}/F_{d-1}}%
\cong\bigoplus_{d=0}^{D}F_{d}/F_{d-1}=\operatorname{gr}_{F}(P/I).
\]
(since all successive quotients $F_{d}/F_{d-1}$ with $d>D$ are zero). In view
of $F_{D}=P/I$, this yields precisely \eqref{eq:filtered-split}. \medskip

\textbf{(c)} If $\operatorname*{gr}\nolimits_{F}\left(  P/I\right)  $ has a
graded $\mathbf{k}$-basis, then each of its graded components $F_{d}/F_{d-1}$
is a free $\mathbf{k}$-module (just take the appropriate subset of the basis
of $\operatorname*{gr}\nolimits_{F}\left(  P/I\right)  $), and therefore is projective.
\end{proof}

We shall now construct specific ideals $I$ to apply the above results to. In
the rest of this section, we shall really use the assumption that
$1_{\mathbf{k}}\neq0$. Of course, Theorem~\ref{thm:main} \textbf{(c)} is
trivial if $1_{\mathbf{k}}=0$ (since any two modules over a trivial ring are isomorphic).

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 $P$: that is,
\[
(\sigma f)(\sigma\cdot v)=f(v)\ \ \ \ \ \ \ \ \ \ \text{for any }\sigma
\in\mathfrak{S}_{n}\text{, }f\in P\text{ and }v\in\mathbf{k}^{n}%
\]
(where polynomials in $P$ are evaluated at points in $\mathbf{k}^{n}$ in the
obvious fashion).

\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}

\begin{example}
\label{exa:discrete}Assume that $n!$ is invertible in $\mathbf{k}$. For any
integer $u$, let $u_{\mathbf{k}}$ denote the image of $u$ under the canonical
ring homomorphism $\mathbb{Z}\rightarrow\mathbf{k}$ (that is, the element
$u\cdot1_{\mathbf{k}}$ of $\mathbf{k}$). Then, the pairwise differences of the
$n$ elements $1_{\mathbf{k}},2_{\mathbf{k}},\ldots,n_{\mathbf{k}}$ are units
of $\mathbf{k}$. Indeed, if $i<j$ are any two integers in $\left[  n\right]
$, then $j-i\in\left[  n-1\right]  \subseteq\left[  n\right]  $ and therefore
$j-i\mid n!$, so that $j-i$ is invertible in $\mathbf{k}$ (since $n!$ is
invertible in $\mathbf{k}$); in other words, $j_{\mathbf{k}}-i_{\mathbf{k}}$
is a unit of $\mathbf{k}$.

Thus, the finite subset $\left\{  1_{\mathbf{k}},2_{\mathbf{k}},\ldots
,n_{\mathbf{k}}\right\}  ^{n}\subseteq\mathbf{k}^{n}$ (which consists of all
points in $\mathbf{k}^{n}$ whose all coordinates belong to $\left\{
1_{\mathbf{k}},2_{\mathbf{k}},\ldots,n_{\mathbf{k}}\right\}  $) is strongly
discrete: Any two distinct points in this set differ in at least one
coordinate, and thus (by the preceding paragraph) the difference of these
respective coordinates is a unit of $\mathbf{k}$.
\end{example}

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 define the ideal%
\[
I(Y):=\{p\in P\mid p(y)=0\text{ for every }y\in Y\}
\]
of $P$ (called the \emph{vanishing ideal} of $Y$). Then:

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


\item[\textbf{(b)}] If $Y$ is $\mathfrak{S}_{n}$-stable, then the isomorphism
\eqref{eq:finite-point-evaluation} is $\mathfrak{S}_{n}$-equivariant, where
$\mathfrak{S}_{n}$ acts on $\mathbf{k}^{Y}$ by the rule%
\[
(\sigma\varphi)(y)=\varphi(\sigma^{-1}\cdot y)\ \ \ \ \ \ \ \ \ \ \text{for
all }\sigma\in\mathfrak{S}_{n}\text{, }\varphi\in\mathbf{k}^{Y}\text{ and
}y\in Y.
\]
Thus it is an isomorphism of $\mathbf{k}$-algebras and of $\mathbf{k}\left[
\mathfrak{S}_{n}\right]  $-modules.
\end{enumerate}
\end{lemma}

\begin{proof}
For each point $y=(y_{1},y_{2},\ldots,y_{n})\in Y$, define the ideal
\[
\mathfrak{m}_{y}=(x_{1}-y_{1})P+(x_{2}-y_{2})P+\cdots+(x_{n}-y_{n})P
\]
of $P$. Evaluation at $y$ is a surjective $\mathbf{k}$-algebra homomorphism
$P\rightarrow\mathbf{k}$ with kernel $\mathfrak{m}_{y}$ (by Lemma
\ref{lem:eval-ker}). Hence, by the first isomorphism theorem, we have a
$\mathbf{k}$-algebra isomorphism%
\begin{equation}
P/\mathfrak{m}_{y}\overset{\cong}{\longrightarrow}\mathbf{k},\label{eq:P/my}%
\end{equation}
which sends each $p+\mathfrak{m}_{y}$ to $p\left(  y\right)  $.

Moreover, by the definition of $I\left(  Y\right)  $, we have
\begin{equation}
I(Y)=\bigcap_{y\in Y}\mathfrak{m}_{y}. \label{eq:IY=cut}%
\end{equation}


The ideals $\mathfrak{m}_{y}$ for $y\in Y$ are pairwise comaximal. Indeed, if
$y\neq z$, then strong discreteness of $Y$ gives an $i$ for which
$u:=y_{i}-z_{i}$ is a unit, and thus we have
\[
1=u^{-1}\left(  y_{i}-z_{i}\right)  =\underbrace{-u^{-1}(x_{i}-y_{i})}%
_{\in\mathfrak{m}_{y}}+\underbrace{u^{-1}(x_{i}-z_{i})}_{\in\mathfrak{m}_{z}%
}\in\mathfrak{m}_{y}+\mathfrak{m}_{z},
\]
which shows that $\mathfrak{m}_{y}$ and $\mathfrak{m}_{z}$ are comaximal. The
Chinese remainder theorem thus gives $P/\bigcap_{y\in Y}\mathfrak{m}_{y}%
\cong\prod_{y\in Y}P/\mathfrak{m}_{y}$ as $\mathbf{k}$-algebras (and even as
$P$-algebras). In view of \eqref{eq:IY=cut}, this rewrites as%
\[
P/I(Y)\cong\prod_{y\in Y}\underbrace{P/\mathfrak{m}_{y}}_{\substack{\cong%
\mathbf{k}\\\text{(by \eqref{eq:P/my})}}}\cong\prod_{y\in Y}\mathbf{k}%
=\mathbf{k}^{Y}\ \ \ \ \ \ \ \ \ \ \text{as }\mathbf{k}\text{-algebras.}%
\]
This is precisely the evaluation map in \eqref{eq:finite-point-evaluation}.
Thus, \textbf{(a)} is proved. \medskip

\textbf{(b)} Assume that $Y$ is $\mathfrak{S}_{n}$-stable. Let $\sigma
\in\mathfrak{S}_{n}$ and $p\in P$. Then, each $y\in Y$ satisfies%
\begin{align*}
\bigl(\sigma\,\operatorname{ev}_{Y}(p+I(Y))\bigr)(y)  &
=\bigl(\operatorname{ev}_{Y}(p+I(Y))\bigr)(\sigma^{-1}\cdot y)\\
&  =p(\sigma^{-1}\cdot y)=(\sigma p)(y)=\bigl(\operatorname{ev}_{Y}(\sigma
p+I(Y))\bigr)(y).
\end{align*}
In other words, $\sigma\,\operatorname{ev}_{Y}(p+I(Y))=\operatorname{ev}%
_{Y}(\sigma p+I(Y))$. This shows that the map $\operatorname{ev}_{Y}$ is
$\mathfrak{S}_{n}$-equivariant (since $\sigma p+I\left(  Y\right)
=\sigma\left(  p+I\left(  Y\right)  \right)  $). Thus, \textbf{(b)} is proved.
\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}\left[  \mathfrak{S}_{n} \right]  $-module, to the regular
module $\mathbf{k}\left[  \mathfrak{S}_{n} \right]  $.
\end{proposition}

\begin{proof}
Let $1_{\mathbf{k}},2_{\mathbf{k}},\ldots,n_{\mathbf{k}}$ be as in Example
\ref{exa:discrete}. Define a point
\[
v:=(1_{\mathbf{k}},2_{\mathbf{k}},\ldots,n_{\mathbf{k}})\in\mathbf{k}%
^{n}\qquad\text{and its }\mathfrak{S}_{n}\text{-orbit}\qquad X:=\mathfrak{S}%
_{n}\cdot v.
\]
The elements of $X$ are the $n!$ points in $\mathbf{k}^{n}$ obtained from $v$
by permuting its coordinates. Recall from Example \ref{exa:discrete} that the
pairwise differences of the $n$ elements $1_{\mathbf{k}},2_{\mathbf{k}}%
,\ldots,n_{\mathbf{k}}$ are units of $\mathbf{k}$; in particular, these $n$
elements $1_{\mathbf{k}},2_{\mathbf{k}},\ldots,n_{\mathbf{k}}$ are distinct
(since $1_{\mathbf{k}}\neq0$ entails that all units of $\mathbf{k}$ are
nonzero). Hence, permuting the coordinates of the vector $v$ leads to $n!$
distinct vectors. In other words, $X$ is an $n!$-element set, and is an
$\mathfrak{S}_{n}$-torsor (i.e., isomorphic to the regular $\mathfrak{S}_{n}%
$-set $\mathfrak{S}_{n}$ as an $\mathfrak{S}_{n}$-set). Moreover, $X$ is
strongly discrete (since the pairwise differences of the $n$ elements
$1_{\mathbf{k}},2_{\mathbf{k}},\ldots,n_{\mathbf{k}}$ are units of
$\mathbf{k}$). Lemma~\ref{lem:finite-point-coordinate-ring} \textbf{(a)}
(applied to $Y=X$) therefore gives a $\mathbf{k}$-algebra isomorphism%
\begin{equation}
P/I(X)\cong\mathbf{k}^{X}. \label{eq:coordinate-algebra}%
\end{equation}
Moreover, Lemma~\ref{lem:finite-point-coordinate-ring} \textbf{(b)} (applied
to $Y=X$) shows that this isomorphism is $\mathfrak{S}_{n}$-equivariant (since
$X$ is $\mathfrak{S}_{n}$-stable). Thus, it is an isomorphism of
$\mathbf{k}\left[  \mathfrak{S}_{n}\right]  $-modules.

For $x\in X$, let $\delta_{x}\in\mathbf{k}^{X}$ be its indicator function
(sending $x$ to $1$ and sending all other elements of $X$ to $0$). The
functions $(\delta_{x})_{x\in X}$ form a $\mathbf{k}$-basis of $\mathbf{k}%
^{X}$, and satisfy $\sigma\delta_{x}=\delta_{\sigma\cdot x}$ for all $x\in X$
and $\sigma\in\mathfrak{S}_{n}$. Since $X$ is an $\mathfrak{S}_{n}$-torsor,
the map%
\begin{equation}
\mathbf{k}\left[  \mathfrak{S}_{n} \right]  \longrightarrow\mathbf{k}%
^{X},\qquad\sigma\longmapsto\delta_{\sigma\cdot v}
\label{eq:coordinate-regular}%
\end{equation}
is an isomorphism of $\mathbf{k}\left[  \mathfrak{S}_{n} \right]  $-modules.

Set
\begin{align*}
c_{i}  &  :=e_{i}(v)\in\mathbf{k}\qquad\text{for all }i\in\left[  n\right]
,\ \ \ \ \ \ \ \ \ \ \text{and}\\
K  &  :=(e_{1}-c_{1})P+(e_{2}-c_{2})P+\cdots+(e_{n}-c_{n})P.
\end{align*}
The ideal $K$ satisfies
\[
K\subseteq I\left(  X\right)  .
\]
(Indeed, for each $i\in\left[  n\right]  $, we have $e_{i}\left(  x\right)
=e_{i}\left(  v\right)  $ for each $x\in X$, because the polynomial $e_{i}$ is
symmetric (and $x$ is obtained from $v$ by permuting the coordinates). That
is, $e_{i}\left(  x\right)  =e_{i}\left(  v\right)  =c_{i}$ for each $x\in X$;
in other words, the polynomial $e_{i}-c_{i}$ vanishes on $x$. Since this holds
for each $x\in X$, we thus obtain $e_{i}-c_{i}\in I\left(  X\right)  $. Since
this holds for each $i\in\left[  n\right]  $, this shows that all the $n$
generators of the ideal $K$ belong to $I\left(  X\right)  $, and thus
$K\subseteq I(X)$ follows.)

We next identify $P/K$ explicitly. Evaluation at $v$ restricts to a surjective
$\mathbf{k}$-algebra homomorphism
\[
\epsilon_{v}:\Lambda\longrightarrow\mathbf{k},\qquad f\longmapsto f(v),
\]
whose kernel is
\[
\ker\epsilon_{v}=(e_{1}-c_{1})\Lambda+(e_{2}-c_{2})\Lambda+\cdots+(e_{n}%
-c_{n})\Lambda.
\]
(Indeed, if we identify $\Lambda$ with the polynomial ring $\mathbf{k}\left[
t_{1},t_{2},\ldots,t_{n}\right]  $ using the isomorphism
\[
\mathbf{k}\left[  t_{1},t_{2},\ldots,t_{n}\right]  \xrightarrow{\sim}\Lambda
,\qquad t_{i}\longmapsto e_{i},
\]
then this map is simply evaluation at $(c_{1},c_{2},\ldots,c_{n})$, and its
kernel is therefore the ideal $(t_{1}-c_{1})\mathbf{k}\left[  t_{1}%
,t_{2},\ldots,t_{n}\right]  +(t_{2}-c_{2})\mathbf{k}\left[  t_{1},t_{2}%
,\ldots,t_{n}\right]  +\cdots+(t_{n}-c_{n})\mathbf{k}\left[  t_{1}%
,t_{2},\ldots,t_{n}\right]  $, by Lemma \ref{lem:eval-ker}.) Thus,%
\begin{align*}
\left(  \ker\epsilon_{v}\right)  P &  =\left(  (e_{1}-c_{1})\Lambda
+(e_{2}-c_{2})\Lambda+\cdots+(e_{n}-c_{n})\Lambda\right)  P\\
&  =(e_{1}-c_{1})P+(e_{2}-c_{2})P+\cdots+(e_{n}-c_{n})P\\
&  =K.
\end{align*}


Let $\mathcal{A}$ be the set of all $n!$ Artin monomials.
Corollary~\ref{cor:freeness-all-char} shows that $\mathcal{A}$ is a $\Lambda
$-basis of $P$. This gives the graded direct-sum decomposition
\begin{equation}
P=\bigoplus_{m\in\mathcal{A}}\Lambda m. \label{eq:P-Artin-sum}%
\end{equation}
Since $K=(\ker\epsilon_{v})P$, this decomposition also gives
\begin{equation}
K=\bigoplus_{m\in\mathcal{A}}(\ker\epsilon_{v})m. \label{eq:K-Artin-sum}%
\end{equation}
Indeed, multiplying the decomposition \eqref{eq:P-Artin-sum} by the ideal
$\ker\epsilon_{v}$ yields
\[
(\ker\epsilon_{v})P=(\ker\epsilon_{v})\left(  \bigoplus_{m\in\mathcal{A}%
}\Lambda m\right)  =\sum_{m\in\mathcal{A}}\underbrace{(\ker\epsilon
_{v})\Lambda}_{=\ker\epsilon_{v}}m=\sum_{m\in\mathcal{A}}(\ker\epsilon_{v})m,
\]
and this sum is direct because it is contained in the direct sum
$\bigoplus_{m\in\mathcal{A}}\Lambda m$.

Now, let $H$ be the $\mathbf{k}$-submodule of $P$ spanned by the Artin
monomials. Thus, $\mathcal{A}$ is a basis of $H$. Moreover, $\Lambda
=\mathbf{k}\oplus\ker\epsilon_{v}$ (where $\mathbf{k}$ denotes the span of
$1\in\Lambda$), since $\epsilon_{v}:\Lambda\rightarrow\mathbf{k}$ is a
$\mathbf{k}$-algebra homomorphism that sends $1\in\Lambda$ to $1\in\mathbf{k}$
(so that each $f\in\Lambda$ satisfies $f-\epsilon_{v}\left(  f\right)
\cdot1\in\ker\epsilon_{v}$). Hence, each $m\in\mathcal{A}$ satisfies
\[
\Lambda m=\left(  \mathbf{k}\oplus\ker\epsilon_{v}\right)  m=\mathbf{k}%
m\oplus\left(  \ker\epsilon_{v}\right)  m.
\]
Indeed, multiplication by the monomial $m$ is injective on $P$ (it merely
shifts the exponent vector of every monomial), so it preserves the directness
of the decomposition $\Lambda=\mathbf{k}\oplus\ker\epsilon_{v}$. Therefore,
\eqref{eq:P-Artin-sum} becomes%
\[
P=\bigoplus_{m\in\mathcal{A}}\underbrace{\Lambda m}_{=\mathbf{k}m\oplus\left(
\ker\epsilon_{v}\right)  m}=\bigoplus_{m\in\mathcal{A}}\left(  \mathbf{k}%
m\oplus\left(  \ker\epsilon_{v}\right)  m\right)  =\underbrace{\bigoplus
_{m\in\mathcal{A}}\mathbf{k}m}_{=\operatorname*{span}\mathcal{A}=H}%
\oplus\underbrace{\bigoplus_{m\in\mathcal{A}}\left(  \ker\epsilon_{v}\right)
m}_{\substack{=K\\\text{(by \eqref{eq:K-Artin-sum})}}}=H\oplus K.
\]
Consequently, the canonical projection $P\rightarrow P/K$ becomes an
isomorphism when restricted to $H$. Thus the residue classes of the Artin
monomials form a $\mathbf{k}$-basis of $P/K$ (since the Artin monomials
themselves form a $\mathbf{k}$-basis of $H$). In other words, the family
$\left(  \overline{m}\right)  _{m\in\mathcal{A}}$ is a basis of $P/K$, where
$\overline{m}$ denotes the residue class $m+K$ of a given $m\in\mathcal{A}$ in
$P/K$.

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


We now consider the degree filtration of $P/I(X)=P/K$, that is, the filtration
$0=F_{-1}^{\left(  P/K\right)  }\subseteq F_{0}^{\left(  P/K\right)
}\subseteq F_{1}^{\left(  P/K\right)  }\subseteq\cdots$ defined by%
\[
F_{d}^{(P/K)}:=\frac{P_{\leq d}+K}{K}\qquad\text{for every }d\geq-1.
\]
This is the filtration $0=F_{-1}\subseteq F_{0}\subseteq F_{1}\subseteq\cdots$
from Lemma~\ref{lem:associated-graded}, but constructed for $P/K$ instead of
$P/I$. Thus, $\operatorname{gr}_{F}(P/K)=\bigoplus_{d\geq0}F_{d}^{\left(
P/K\right)  }/F_{d-1}^{\left(  P/K\right)  }$. Note that $P_{\leq-1}=0$.

For every $d\geq0$, we claim that%
\begin{equation}
\text{the }\mathbf{k}\text{-module }F_{d}^{(P/K)}\text{ has basis }\left(
\overline{m}\right)  _{m\in\mathcal{A};\ \deg m\leq d},
\label{eq:filtered-artin-basis}%
\end{equation}
where $\overline{m}$ denotes the residue class $m+K$ of $m\in P$ modulo $K$.

(\textit{Proof:} The residue classes $\overline{m}$ for $m\in\mathcal{A}$
satisfying $\deg m\leq d$ clearly belong to $F_{d}^{(P/K)}$, since the
corresponding $m$'s belong to $P_{\leq d}$. Moreover, the family $\left(
\overline{m}\right)  _{m\in\mathcal{A};\ \deg m\leq d}$ is a subfamily of the
basis $\left(  \overline{m}\right)  _{m\in\mathcal{A}}$ of $P/K$, and thus is
$\mathbf{k}$-linearly independent. It remains to show that this family spans
$F_{d}^{(P/K)}$. To prove this, we let $\overline{p}=p+K$ be an arbitrary
element of $F_{d}^{(P/K)}$, with $p\in P_{\leq d}$. We must show that
$\overline{p}$ is a $\mathbf{k}$-linear combination of the family $\left(
\overline{m}\right)  _{m\in\mathcal{A};\ \deg m\leq d}$.

By decomposing $p\in P_{\leq d}$ into its homogeneous components, it is enough
to treat the case in which $p$ is homogeneous of some degree at most $d$;
replacing $d$ by this degree, we may assume that $p$ is homogeneous of degree
$d$. We can write $p$ uniquely as
\begin{equation}
p=\sum_{m\in\mathcal{A}}\lambda_{m}m\qquad\text{with }\lambda_{m}\in
\Lambda\label{eq:filtered-artin-basis:pf1}%
\end{equation}
(since $\mathcal{A}$ is a basis of the $\Lambda$-module $P$). By projecting
this equality onto the $d$-th graded component of $P$, we obtain%
\begin{equation}
p=\sum_{\substack{m\in\mathcal{A};\\\deg m\leq d}}\widetilde{\lambda}_{m}m,
\label{eq:filtered-artin-basis:pf2}%
\end{equation}
where $\widetilde{\lambda}_{m}$ is the $\left(  d-\deg m\right)  $-th
homogeneous component of $\lambda_{m}$ (since each $m\in\mathcal{A}$ is
homogeneous, and since $p$ is homogeneous of degree $d$). Projecting this
equality onto $P/K$ yields%
\[
\overline{p}=\sum_{\substack{m\in\mathcal{A};\\\deg m\leq d}%
}\underbrace{\overline{\widetilde{\lambda}_{m}}}_{\substack{=\epsilon
_{v}\left(  \widetilde{\lambda}_{m}\right)  \\\text{(since }\widetilde{\lambda
}_{m}-\epsilon_{v}\left(  \widetilde{\lambda}_{m}\right)  \in\ker\epsilon
_{v}\subseteq\left(  \ker\epsilon_{v}\right)  P=K\\\text{and thus
}\widetilde{\lambda}_{m}\equiv\epsilon_{v}\left(  \widetilde{\lambda}%
_{m}\right)  \operatorname{mod}K\text{)}}}\overline{m}=\sum_{\substack{m\in
\mathcal{A};\\\deg m\leq d}}\epsilon_{v}\left(  \widetilde{\lambda}%
_{m}\right)  \overline{m},
\]
which is clearly a $\mathbf{k}$-linear combination of the family $\left(
\overline{m}\right)  _{m\in\mathcal{A};\ \deg m\leq d}$, just as we desired to
show. Thus, \eqref{eq:filtered-artin-basis} is proved.)

Consequently, for every $d\geq0$,%
\begin{equation}
\text{the }\mathbf{k}\text{-module }F_{d}^{(P/K)}/F_{d-1}^{(P/K)}\text{ has
basis }\left(  m^{\ast}\right)  _{m\in\mathcal{A};\ \deg m=d},
\label{eq:filtered-artin-basis2}%
\end{equation}
where $m^{\ast}$ is the residue class of $\overline{m}\in F_{d}^{(P/K)}$ in
$F_{d}^{(P/K)}/F_{d-1}^{(P/K)}$.

(\textit{Proof:} The $\mathbf{k}$-module $F_{d}^{(P/K)}$ has basis $\left(
\overline{m}\right)  _{m\in\mathcal{A};\ \deg m\leq d}$ (by
\eqref{eq:filtered-artin-basis}), whereas the $\mathbf{k}$-module
$F_{d-1}^{(P/K)}$ has basis $\left(  \overline{m}\right)  _{m\in
\mathcal{A};\ \deg m\leq d-1}$ (again by \eqref{eq:filtered-artin-basis}). The
latter basis is clearly a subfamily of the former basis. Thus, a basis of the
quotient $\mathbf{k}$-module $F_{d}^{(P/K)}/F_{d-1}^{(P/K)}$ can be obtained
by listing the (residue classes of the) elements of the former basis that
don't belong to the latter basis. This means listing the $m^{\ast}$ for all
$m\in\mathcal{A}$ that satisfy $\deg m\leq d$ but don't satisfy $\deg m\leq
d-1$ -- that is, that satisfy $\deg m=d$. Thus,
\eqref{eq:filtered-artin-basis2} is proved.)

Note that if $d$ is larger than the degrees of all Artin monomials
$m\in\mathcal{A}$, then the basis $\left(  m^{\ast}\right)  _{m\in
\mathcal{A};\ \deg m=d}$ in \eqref{eq:filtered-artin-basis2} is empty, so that
\eqref{eq:filtered-artin-basis2} shows that $F_{d}^{(P/K)}/F_{d-1}^{(P/K)}=0$,
that is, $F_{d}^{(P/K)}=F_{d-1}^{(P/K)}$. This shows that the filtration
$0=F_{-1}^{\left(  P/K\right)  }\subseteq F_{0}^{\left(  P/K\right)
}\subseteq F_{1}^{\left(  P/K\right)  }\subseteq\cdots$ of $P/K$ stabilizes.

Taking the direct sum of \eqref{eq:filtered-artin-basis2} over all $d\geq0$,
we conclude that the associated graded algebra
\[
\operatorname{gr}_{F}(P/K)=\bigoplus_{d\geq0}F_{d}^{\left(  P/K\right)
}/F_{d-1}^{\left(  P/K\right)  }%
\]
has the graded $\mathbf{k}$-basis
\begin{equation}
\left(  m^{\ast}\right)  _{m\in\mathcal{A}}. \label{eq:m*bas}%
\end{equation}


For each $i\in\left[  n\right]  $, we have $\tau\left(  e_{i}-c_{i}\right)
=e_{i}$ (since $e_{i}\in P_{i}$ while $c_{i}\in\mathbf{k}=P_{0}$), so that
$e_{i}=\tau\left(  e_{i}-c_{i}\right)  \in\operatorname{in}_{\deg}(I\left(
X\right)  )$ (since $e_{i}-c_{i}\in K=I\left(  X\right)  $). Therefore, by
\eqref{eq:J-two-descriptions}, we obtain%
\[
J\subseteq\operatorname{in}_{\deg}(I(X)).
\]
Thus we obtain a canonical surjective graded homomorphism
\begin{equation}
C=P/J\twoheadrightarrow P/\operatorname{in}_{\deg}%
(I(X))\xrightarrow{\ \sim\ }\operatorname{gr}_{F}(P/I(X)) \label{eq:C-to-gr}%
\end{equation}
by Lemma~\ref{lem:associated-graded} (applied to $I=I\left(  X\right)  $);
moreover, this homomorphism is $\mathfrak{S}_{n}$-equivariant by Lemma
\ref{lem:equivariant-splitting} \textbf{(a)} (since the ideal $I\left(
X\right)  $ is $\mathfrak{S}_{n}$-stable). This homomorphism sends the residue
class $m+J$ of each Artin monomial $m$ to $m^{\ast}\in\operatorname{gr}%
_{F}(P/I(X))$. Proposition~\ref{prop:artin} says that the former classes are a
$\mathbf{k}$-basis of $C$, whereas our observation \eqref{eq:m*bas} says that
the latter classes are a $\mathbf{k}$-basis of $\operatorname{gr}%
_{F}(P/K)=\operatorname{gr}_{F}(P/I(X))$ (since $K=I\left(  X\right)  $).
Hence, \eqref{eq:C-to-gr} is an isomorphism (since a $\mathbf{k}$-linear map
sending a basis to a basis must be an isomorphism). In particular, the first
arrow $P/J\twoheadrightarrow P/\operatorname{in}_{\deg}(I(X))$ in
\eqref{eq:C-to-gr} must be injective, so that
\begin{equation}
J=\operatorname{in}_{\deg}(I(X)). \label{eq:J=degree-initial}%
\end{equation}
Also, \eqref{eq:C-to-gr} is an $\mathfrak{S}_{n}$-equivariant $\mathbf{k}%
$-module isomorphism, hence a $\mathbf{k}\left[  \mathfrak{S}_{n}\right]
$-module isomorphism.

The filtration $0=F_{-1}^{\left(  P/K\right)  }\subseteq F_{0}^{\left(
P/K\right)  }\subseteq F_{1}^{\left(  P/K\right)  }\subseteq\cdots$ stabilizes
(as we saw above). Moreover, the basis \eqref{eq:m*bas} is graded, so
Lemma~\ref{lem:equivariant-splitting} \textbf{(c)} (applied to $I=I(X)=K$)
shows that every successive quotient $F_{d}^{(P/K)}/F_{d-1}^{(P/K)}$ is
projective. Hence Lemma~\ref{lem:equivariant-splitting} \textbf{(b)} (applied
to $I=I(X)=K$) gives an isomorphism of ungraded $\mathbf{k}\left[
\mathfrak{S}_{n} \right]  $-modules
\begin{equation}
P/I(X)\cong\operatorname{gr}_{F}(P/I(X)) \label{eq:PIXgriso}%
\end{equation}
(since $\left\vert \mathfrak{S}_{n}\right\vert =n!$ is invertible in
$\mathbf{k}$). Now, we have the following chain of isomorphisms of ungraded
$\mathbf{k}\left[  \mathfrak{S}_{n} \right]  $-modules:%
\begin{align*}
C  &  \cong\operatorname{gr}_{F}(P/I(X))\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{eq:C-to-gr}}\right) \\
&  \cong P/I(X)\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{eq:PIXgriso}}%
\right) \\
&  \cong\mathbf{k}^{X}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{eq:coordinate-algebra}}\right) \\
&  \cong\mathbf{k}\left[  \mathfrak{S}_{n} \right]
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{eq:coordinate-regular}}\right)  .
\end{align*}
This proves Proposition \ref{prop:coinvariant-regular}, i.e.,
Theorem~\ref{thm:main} \textbf{(c)}.
\end{proof}

Thus, we have recovered the classical statement recorded by Haiman that the
coinvariant algebra affords the regular representation \cite[Section~1.5]%
{Haiman}. Over a field of characteristic $0$, this result is due to
Chevalley \cite[last sentence of Theorem (B)]{Cheval55} (in the
more general setting of a Coxeter group).

\begin{remark}
The preceding proof follows the orbit-harmonics strategy of Oh and Rhoades
\cite[Section~1]{OhRhoades} at two distinct points: it passes from the
coordinate ring of the finite orbit $X$ to its associated graded algebra, and
it uses the fact that the function module of a free orbit is the regular
module. The proof concerns the coinvariant algebra only as an \emph{ungraded}
$\mathfrak{S}_{n}$-module. In graded refinements of orbit harmonics, special
point sets involving roots of unity carry an additional cyclic symmetry whose
eigenvalues record the grading; 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
\[
v=(a_{1},a_{2},\ldots,a_{n})\in\mathbf{k}^{n},\qquad X=\mathfrak{S}_{n}\cdot
v.
\]
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
(as we did in the proof of Lemma~\ref{lem:maschke-splitting}).
\end{remark}

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

We are now ready to prove the last part of Theorem \ref{thm:main}: part
\textbf{(d)}, about the $\Lambda\left[  \mathfrak{S}_{n}\right]  $-module
structure of $P$.

\begin{proof}
[Proof of Theorem~\ref{thm:main} \textbf{(d)}]Assume that $n!$ is invertible
in $\mathbf{k}$. For each $d\geq0$, the $\mathbf{k}$-module $C_{d}$ is free
(since Proposition~\ref{prop:artin} shows that $C$ has a graded basis), and
thus the quotient map
\[
P_{d}\twoheadrightarrow C_{d}%
\]
has a $\mathbf{k}$-linear section. That is, the short exact sequence
\[
0\longrightarrow J_{d}\longrightarrow P_{d}\longrightarrow C_{d}%
\longrightarrow0
\]
of $\mathbf{k}\left[  \mathfrak{S}_{n}\right]  $-modules (where $J_{d}%
\rightarrow P_{d}$ is the inclusion and $P_{d}\rightarrow C_{d}$ the
projection) splits as a sequence of $\mathbf{k}$-modules. Since $\left\vert
\mathfrak{S}_{n}\right\vert =n!$ is invertible in $\mathbf{k}$, this sequence
thus also splits as a sequence of $\mathbf{k}\left[  \mathfrak{S}_{n}\right]
$-modules by Lemma~\ref{lem:maschke-splitting}. That is, the quotient map
$P_{d}\twoheadrightarrow C_{d}$ has a $\mathbf{k}\left[  \mathfrak{S}%
_{n}\right]  $-linear section. Let $H_{d}\subseteq P_{d}$ be the image of this
section, and set
\[
H=\bigoplus_{d\geq0}H_{d}.
\]
Then $H$ is a graded $\mathbf{k}\left[  \mathfrak{S}_{n}\right]  $-submodule
of $P$, and the quotient map $P\rightarrow C$ restricts to an isomorphism
\[
H\xrightarrow{\ \sim\ }C
\]
of graded $\mathbf{k}\left[  \mathfrak{S}_{n}\right]  $-modules.

Forgetting the gradings in this isomorphism gives an isomorphism
$H\xrightarrow{\sim}C$ of ungraded $\mathbf{k}\left[  \mathfrak{S}_{n}\right]
$-modules. Proposition~\ref{prop:coinvariant-regular} gives an isomorphism
$\mathbf{k}\left[  \mathfrak{S}_{n}\right]  \xrightarrow{\sim}C$ of such
modules. Composing the latter isomorphism with the inverse of
$H\xrightarrow{\sim}C$, we obtain an isomorphism $\mathbf{k}\left[
\mathfrak{S}_{n}\right]  \xrightarrow{\sim}H$. So choose a $\mathbf{k}\left[
\mathfrak{S}_{n}\right]  $-module isomorphism
\[
\theta:\mathbf{k}\left[  \mathfrak{S}_{n}\right]  \xrightarrow{\ \sim\ }H,
\]
and put $f:=\theta(1)$. For every $\sigma\in\mathfrak{S}_{n}$, we have
$\sigma=\sigma\cdot1$ and thus%
\begin{align}
\theta(\sigma)  &  =\theta(\sigma\cdot1)=\sigma\cdot\theta
(1)\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\theta\text{ is }\mathfrak{S}%
_{n}\text{-equivariant}\right) \nonumber\\
&  =\sigma(f)\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\theta(1)=f\right)  .
\label{eq:thetasig=}%
\end{align}
Thus the family $\left(  \sigma\left(  f\right)  \right)  _{\sigma
\in\mathfrak{S}_{n}}$ is the image of the basis $\left(  \sigma\right)
_{\sigma\in\mathfrak{S}_{n}}$ of $\mathbf{k}\left[  \mathfrak{S}_{n}\right]  $
under the isomorphism $\theta$. Hence, the former family $\left(
\sigma\left(  f\right)  \right)  _{\sigma\in\mathfrak{S}_{n}}$ is a
$\mathbf{k}$-basis of $H$ (since a $\mathbf{k}$-module isomorphism sends bases
to bases). Consequently, the family $\left(  1\otimes\sigma(f)\right)
_{\sigma\in\mathfrak{S}_{n}}$ is a $\Lambda$-basis of $\Lambda\otimes
_{\mathbf{k}}H$.

But Proposition~\ref{prop:multiplication} \textbf{(a)} says that the
multiplication map
\[
m:\Lambda\otimes_{\mathbf{k}}H\xrightarrow{\ \sim\ }P,\qquad a\otimes
h\longmapsto ah
\]
is an isomorphism of graded $\Lambda$-modules. This isomorphism sends each
$1\otimes\sigma(f)$ to $\sigma(f)$. Hence the family $\left(  \sigma
(f)\right)  _{\sigma\in\mathfrak{S}_{n}}$ is a $\Lambda$-basis of $P$ (since
it is the image of the $\Lambda$-basis $\left(  1\otimes\sigma(f)\right)
_{\sigma\in\mathfrak{S}_{n}}$ of $\Lambda\otimes_{\mathbf{k}}H$ under this
$\Lambda$-module isomorphism $m$).

Note that the $\Lambda$-module isomorphism $m$ is also $\mathfrak{S}_{n}%
$-equivariant by Proposition~\ref{prop:multiplication} \textbf{(b)}, and thus
is a $\Lambda\left[  \mathfrak{S}_{n}\right]  $-module isomorphism. Also,
$\operatorname*{id}\nolimits_{\Lambda}\otimes\theta:\Lambda\otimes
_{\mathbf{k}}\mathbf{k}\left[  \mathfrak{S}_{n}\right]  \rightarrow
\Lambda\otimes_{\mathbf{k}}H$ is a $\Lambda\left[  \mathfrak{S}_{n}\right]
$-module isomorphism (since $\theta:\mathbf{k}\left[  \mathfrak{S}_{n}\right]
\rightarrow H$ is a $\mathbf{k}\left[  \mathfrak{S}_{n}\right]  $-module
isomorphism), and there is a canonical $\Lambda\left[  \mathfrak{S}%
_{n}\right]  $-module isomorphism%
\[
\kappa:\Lambda\otimes_{\mathbf{k}}\mathbf{k}\left[  \mathfrak{S}_{n}\right]
\rightarrow\Lambda\left[  \mathfrak{S}_{n}\right]
,\ \ \ \ \ \ \ \ \ \ a\otimes\sigma\mapsto a\sigma.
\]


Composing the $\Lambda\left[  \mathfrak{S}_{n}\right]  $-module isomorphisms%
\[
\Lambda\left[  \mathfrak{S}_{n}\right]  \xrightarrow{\ \kappa^{-1}\ }\Lambda
\otimes_{\mathbf{k}}\mathbf{k}\left[  \mathfrak{S}_{n}\right]
\xrightarrow{\ \id_\Lambda\otimes\theta\ }\Lambda\otimes_{\mathbf{k}%
}H\xrightarrow{\ m\ }P,
\]
we thus obtain a $\Lambda\left[  \mathfrak{S}_{n}\right]  $-module isomorphism
from $\Lambda\left[  \mathfrak{S}_{n}\right]  $ to $P$, which shows that
$P\cong\Lambda\left[  \mathfrak{S}_{n}\right]  $ as $\Lambda\left[
\mathfrak{S}_{n}\right]  $-modules. Moreover, this composed isomorphism is
easily seen to send each $a\sigma$ (with $a\in\Lambda$ and $\sigma
\in\mathfrak{S}_{n}$) to $a\theta\left(  \sigma\right)  =a\sigma\left(
f\right)  $ (by \eqref{eq:thetasig=}), and thus must send each $\sum
_{\sigma\in\mathfrak{S}_{n}}a_{\sigma}\sigma$ (with $a_{\sigma}\in\Lambda$) to
$\sum_{\sigma\in\mathfrak{S}_{n}}a_{\sigma}\sigma\left(  f\right)  $. So it is
precisely the map $\Phi$ from \eqref{eq:normal-basis-map}. Thus the proof of
Theorem \ref{thm:main} \textbf{(d)} is complete.
\end{proof}

\begin{remark}
[The isomorphism is not graded]The element $f$ in the above proof 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
\[
\Lambda=\mathbf{k}\left[  x_{1}+x_{2},x_{1}x_{2} \right]  \qquad
\text{and}\qquad H=\operatorname{span}_{\mathbf{k}}\{1,x_{1}-x_{2}\}.
\]
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. For example, take $n=2$ and
$\mathbf{k}=\mathbb{F}_{2}$. Writing $s=(1\ 2)$, we have
\[
C=\frac{\mathbf{k}\left[  x_{1},x_{2}\right]  }{(x_{1}+x_{2})P+x_{1}x_{2}%
P}\cong\frac{\mathbf{k}\left[  t\right]  }{t^{2}\mathbf{k}\left[  t\right]
}.
\]
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}\left[  \mathfrak{S}_{2}\right]  $: it interchanges the
basis vectors $1$ and $s$. Hence, $C$ is not isomorphic to $\mathbf{k}\left[
\mathfrak{S}_{2}\right]  $ as $\mathbf{k}\left[  \mathfrak{S}_{2}\right]
$-modules, and Theorem~\ref{thm:main} \textbf{(c)} fails.

Theorem~\ref{thm:main} \textbf{(d)} fails as well in this generality. Indeed,
tensoring a hypothetical $\Lambda\left[  \mathfrak{S}_{2}\right]  $-module
isomorphism $P\cong\Lambda\left[  \mathfrak{S}_{2}\right]  $ 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}\left[  \mathfrak{S}_{2}\right]  $.

\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
\[
(1_{\mathbf{k}},2_{\mathbf{k}},\ldots,n_{\mathbf{k}})
\]
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}. Significant changes
were necessary to make the proofs of \textbf{(c)} and \textbf{(d)} apply not
only to fields of \textquotedblleft good\textquotedblright\ characteristic,
but to arbitrary commutative rings $\mathbf{k}$ in which $n!$ is invertible,
as this level of generality rendered several linear-algebraic shortcuts inaccessible.

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 $0,0,\ldots,0$.

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\end{document}