# On polynomial invariant rings in modular invariant theory

Manoj Kummini  Chennai Mathematical Institute, Siruseri, Tamilnadu 603103.
India<EMAIL_ADDRESS>and  Mandira Mondal  Chennai Mathematical Institute,
Siruseri, Tamilnadu 603103. India<EMAIL_ADDRESS>
###### Abstract.

Let $\Bbbk$ be a field of characteristic $p>0$, $V$ a finite-dimensional
$\Bbbk$-vector-space, and $G$ a finite $p$-group acting $\Bbbk$-linearly on
$V$. Let $S=\operatorname{Sym}V^{*}$. We show that $S^{G}$ is a polynomial
ring if and only if the dimension of its singular locus is less than
$\operatorname{rk}_{\Bbbk}V^{G}$. Confirming a conjecture of Shank-Wehlau-
Broer, we show that if $S^{G}$ is a direct summand of $S$, then $S^{G}$ is a
polynomial ring, in the following cases:

  1. (a)

$\Bbbk=\mathbb{F}_{p}$ and $\operatorname{rk}_{\Bbbk}V^{G}=4$; or

  2. (b)

$|G|=p^{3}$.

In order to prove the above result, we also show that if
$\operatorname{rk}_{\Bbbk}V^{G}\geq\operatorname{rk}_{\Bbbk}V-2$, then the
Hilbert ideal $\mathfrak{h}_{G,S}$ is a complete intersection.

Both authors were partially supported by an Infosys Foundation fellowship.

MM thanks the National Board of Higher Mathematics.

##  1\. Introduction

Let $\Bbbk$ be a field and $V$ a finite-dimensional $\Bbbk$-vector-space. Let
$G\subseteq\operatorname{GL}(V)$ be finite group. Then the action of $G$ on
$V$ induces an action on $V^{*}$ which extends to a graded $\Bbbk$-algebra
automorphism of $S:=\operatorname{Sym}V^{*}=\Bbbk[x_{1},\ldots,x_{n}]$.
Suppose that the characteristic of $\Bbbk$ is $p>0$ and that $G$ is a finite
$p$-group. Various authors have studied necessary and/or sufficient conditions
for the invariant ring $S^{G}$ to be a polynomial ring; see, e.g. [SW99],
[Bro05], [Bro06], [Bro10], [KM22], [Bra22], and references therein. In this
context, we prove the following theorem.

######  1.1 Theorem.

Let $\Bbbk$ be a field of characteristic $p>0$, $V$ a finite-dimensional
$\Bbbk$-vector space and $G\subseteq\operatorname{GL}(V)$ a finite $p$-group.
Write $S=\operatorname{Sym}V^{*}$ and $R=S^{G}$. Then the following are
equivalent:

  1. (a)

$R$ is a polynomial ring.

  2. (b)

$\dim\operatorname{Sing}(R)<\operatorname{rk}_{\Bbbk}V^{G}$, where
$\operatorname{Sing}(R):=\\{{\mathfrak{p}}\in\operatorname{Spec}R\mid
R_{\mathfrak{p}}\;\text{is not regular}\\}$.

  3. (c)

For some $W\subseteq V^{G}$, $R_{\mathfrak{p}}$ is a regular local ring, where
${\mathfrak{q}}$ is the $S$-ideal generated by the $\Bbbk$-subspace
$(V/W)^{*}$ of $V^{*}$ and ${\mathfrak{p}}={\mathfrak{q}}\cap R$.

A. Braun recently gave a characterization of polynomial invariant rings
[Bra22, Theorem B]. Our theorem above is related to one of the lemmas used in
its proof. We state the relevant part below. (Note the disparity in the
notation: Braun works with $\operatorname{Sym}V$ while we work with
$\operatorname{Sym}V^{*}$.)

######  1.2 Theorem ([Bra22, Lemma 2.4]).

In addition to the notation of Theorem 1.1, assume that $\Bbbk$ is perfect.
Let $m=\operatorname{rk}_{\Bbbk}V-\operatorname{rk}_{\Bbbk}V^{G}$. If $R$
satisfies the Serre condition $(\mathrm{R}_{m})\,$, then $R$ is a polynomial
ring.

(Braun shows, more generally, that for each $m$ and for each $U\subseteq V$
with $\operatorname{rk}_{\Bbbk}U=n-m$, $S^{G_{U}}$ is a polynomial ring where
$G_{U}=\\{g\in G\mid gv=v\;\text{for all}\;v\in U\\}$. When
$m=\operatorname{rk}_{\Bbbk}V-\operatorname{rk}_{\Bbbk}V^{G}$, we can take
$U=V^{G}$, in which case $G_{U}=G$.) Theorem 1.1 gives an elementary proof of
Theorem 1.2, without assuming that $\Bbbk$ is perfect. Moreover, Theorem
1.1(c) reduces the checking to just one subspace, viz., $U=V^{G}$.

If $S^{G}$ is a polynomial ring then $S$ is a free $S^{G}$-module, so $S^{G}$
is a direct summand of $S$. (See Section 4 for (well-known) remarks about
direct summands.) The converse is not in general true; in the non-modular
case, every invariant ring is a direct summand of the respective symmetric
algebra. However, for actions of finite $p$-groups in characteristic $p>0$,
the Shank-Wehlau-Broer conjecture (R. J. Shank and D. L. Wehlau [SW99,
Conjecture 1.1], reformulated by A. Broer [Bro05, Corollary 4, p. 14]) is the
following:

######  1.3 Conjecture.

$S^{G}$ is a polynomial ring if it is a direct summand of $S$ as an
$S^{G}$-module.

Using Theorem 1.1 we may rephrase the Shank-Wehlau-Broer conjecture as
follows:

######  1.4 Conjecture.

If $S^{G}$ is a direct summand of $S$ as an $S^{G}$-module, then its singular
locus has dimension less than $\operatorname{rk}_{\Bbbk}V^{G}$.

In this paper we prove a few results in the direction of settling Conjecture
1.3. The _Hilbert ideal_ $\mathfrak{h}_{G,S}$ is the $S$-ideal generated by
the invariant polynomials of positive degree, i.e., by $(S^{G})_{+}$. If
$S^{G}$ is a direct summand, then every generating set of $\mathfrak{h}_{G,S}$
(as an $S$-ideal) consisting of homogeneous invariant elements generates
$S^{G}$ as a $\Bbbk$-algebra; see [Ben93, Theorem 1.6.3] or [DK15, Theorem
2.2.10] for a proof. Hence if $S^{G}$ is a direct summand and
$\mathfrak{h}_{G,S}$ is a complete intersection, then $S^{G}$ is a polynomial
ring. Conversely, if $S^{G}$ is a polynomial ring then it is a direct summand
of $S$ and $\mathfrak{h}_{G,S}$ is a complete intersection.

We now state our results about Conjecture 1.3 when
$\operatorname{rk}_{\Bbbk}V$ is small. If $\operatorname{rk}_{\Bbbk}V=2$, then
$S^{G}$ is a polynomial ring (Proposition 5.6). In higher dimensions, we prove
the following:

######  1.5 Theorem.

If $\operatorname{rk}V^{G}\geq\operatorname{rk}V-2$, then $\mathfrak{h}_{G,S}$
is a complete intersection.

######  1.6 Corollary.

If $\operatorname{rk}_{\Bbbk}V\leq 3$, then $\mathfrak{h}_{G,S}$ is a complete
intersection. In particular, if $\operatorname{rk}_{\Bbbk}V\leq 3$ and $S^{G}$
is a direct summand of $S$ then it is a polynomial ring.

The second assertion of Corollary 1.6 follows from [Bra22, Theorem A] also.

######  1.7 Theorem.

Assume that $\Bbbk=\mathbb{F}_{p}$ and $\operatorname{rk}_{\Bbbk}V=4$. If
$S^{G}$ is a direct summand of $S$ then it is a polynomial ring.

When $|G|$ is small, we prove the following:

######  1.8 Theorem.

Assume that $|G|=p^{3}$. If $S^{G}$ is a direct summand of $S$, then $S^{G}$
is a polynomial ring.

This paper is organized as follows. Theorems 1.1 and 1.5 are proved in Section
3. The key new result, going into their proof, is that the relative Hilbert
ideal $\mathfrak{h}_{G,S,W}$ (for $W\subseteq V^{G}$) is extended from
$\operatorname{Sym}W^{\perp}$; required definitions are given in Section 2. In
Section 5, we describe a composition series of $G$. Section 6 uses the results
of 5 and of [Bro06] to look at how relative Hilbert ideals and the property of
being polynomial rings change as we move along the composition series; results
of this section are used in Section 7 to prove Theorem 1.7. Theorem 1.8 is
proved in Section 8. Finally, in Section 9, we discuss some examples related
to the results in the earlier sections.

##  2\. Preliminaries

######  2.1 Notation.

Let $\Bbbk$ be a field and $V$ an $n$-dimensional $\Bbbk$-vector-space. Let
$G\subseteq\operatorname{GL}(V)$ be finite group. Let
$S=\operatorname{Sym}(V^{*})$. ∎

Note that $S$ is a polynomial ring with $\dim S=n$ and that $G$ acts on $S$ as
degree-preserving $\Bbbk$-algebra automorphisms.

######  2.2 Definition ([Bro10, p. 406]).

Let $W\subseteq V^{G}$ be a subspace. The _Hilbert ideal of $G$ in $S$ with
respect to $W$_, denoted $\mathfrak{h}_{G,S,W}$, is the $S$-ideal
$((W^{\perp}S)\cap S^{G})S$, where $W^{\perp}=\ker(V^{*}\longrightarrow
W^{*})$, i.e., the subspace of linear forms on $V$ that vanish on $W$.

Note that $\mathfrak{h}_{G,S}$ is $\mathfrak{h}_{G,S,0}$. We can use the
relative Hilbert ideal to check if the Hilbert ideal is a complete
intersection.

######  2.3 Proposition ([Bro10, Lemma 2.2 and its proof]).

Let $W\subseteq V^{G}$ be a subspace of codimension $s$. Let
${\mathfrak{p}}=W^{\perp}S\cap R$. Then:

  1. (a)

There exist homogeneous $f_{s+1},\ldots,f_{n}\in R$ such that
$R/{\mathfrak{p}}$ is an $(n-s)$-dimensional polynomial ring generated by (the
residue classes of) $f_{s+1},\ldots,f_{n}$.

  2. (b)

$\mathfrak{h}_{G,S}=\mathfrak{h}_{G,S,W}+(f_{s+1},\ldots,f_{n})S$.

  3. (c)

$f_{s+1},\ldots,f_{n}$ form a regular sequence on $S/\mathfrak{h}_{G,S,W}$.

  4. (d)

$\mu(\mathfrak{h}_{G,S})=\mu(\mathfrak{h}_{G,S,W})+(n-s)$, where for a
homogeneous $S$-ideal $J$, $\mu(J)$ denotes the cardinality of a minimal
generating set. In particular, if $\mathfrak{h}_{G,S,W}$ is a complete
intersection, then $\mathfrak{h}_{G,S}$ is a complete intersection.

The first two statements are from [Bro10, Lemma 2.2] and its proof. The
remaining two statements will be proved after Corollary 3.2, whose proof will
only use the first statement of this proposition.

######  2.4 Discussion.

Let $R\subseteq R^{\prime}$ be noetherian $\mathbb{N}$-graded algebras with
$R_{0}=(R^{\prime})_{0}=\Bbbk$. Suppose that $R^{\prime}$ is a polynomial ring
and that $R^{\prime}$ is a free $R$-module. Then $R$ is a polynomial ring. To
see this, let $F_{\bullet}$ be a minimal graded free resolution of $R/R_{+}$
as an $R$-module. Then $F_{\bullet}\otimes_{R}R^{\prime}$ is a graded free
resolution of $R^{\prime}/R_{+}R^{\prime}$ as an $R^{\prime}$-module (since
$R^{\prime}$ is a free $R$-module). It is minimal since
$R_{+}\subseteq(R^{\prime})_{+}$. Since $R^{\prime}$ is a polynomial ring,
$F_{\bullet}\otimes_{R}R^{\prime}$ has finite length, so $F_{\bullet}$ has
finite length. Hence $R$ is a polynomial ring. See, e.g., [Ben93, 6.2.3] for
details. ∎

######  2.5 Notation.

In addition to Notation 2.1 we now assume that $\operatorname{char}\Bbbk=p>0$
and that $G$ a finite $p$-group. ∎

The following lemma is known, but we give a brief sketch of the proof, since
we need the details in later sections.

######  2.6 Lemma.

Then there exists a basis $\\{v_{1},\ldots,v_{n}\\}$ of $V$ such that

$g(v_{i})-v_{i}\in\Bbbk\langle v_{i+1},\ldots,v_{n}\rangle.$

(In other words, the elements of $G$ act on $V$ as lower-triangular unipotent
matrices.)

###### Proof.

It is known that $V^{G}\neq 0$ [CW11, 4.0.1]. Let $W$ be any non-zero subspace
of $V^{G}$. Say $t=\dim W$. $G$ acts on $V/W$; then, by induction on
dimension, there exist $v_{1},\ldots,v_{n-t}\in V$ such that their images in
$V/W$ is basis of $V/W$ with respect to which every element of $g$ (for its
action on $V/W$) is lower triangular and unipotent. Let
$v_{n-t+1},\ldots,v_{n}$ be a basis of $W$. Then every element of $G$ is
lower-triangular with respect to $\\{v_{1},\ldots,v_{n}\\}$ and unipotent. ∎

We recall the definition of terminal variables from [ES18] and locally finite
iterative higher derivations corresponding to them. Fix a basis
$\\{x_{1},\ldots,x_{n}\\}$ of $V^{*}$. Say that $x_{j}$ is a _terminal_
variable if $\Bbbk\langle x_{1},\ldots,x_{j-1},x_{j+1},\ldots,x_{n}\rangle$ is
a $G$-stable subspace of $V^{*}$. Suppose that $x_{j}$ is a terminal variable.
Define $\Bbbk$-linear locally finite iterative higher derivations
$\Delta_{j}^{(l)}:S\longrightarrow S$ as follows. Let $t$ be a new variable
and extend the $G$-action on $S$ to $S[t]$ with $G$ acting trivially on $t$.
Let $\phi_{j}$ be the ring map $S\longrightarrow S$ given by

$x_{i}\mapsto\begin{cases}x_{i}&\text{if}\;i\neq j,\\\
x_{j}+t&\text{if}\;i=j.\end{cases}$

Note that $\phi_{j}$ is $G$-equivariant since $x_{j}$ is a terminal variable.
Define $\Delta_{j}^{(l)}$ by $\phi_{j}(f)=\sum\Delta^{(l)}_{j}(f)t^{l}$. Since
$\phi_{j}$ is $G$-equivariant, we see that $\Delta_{j}^{(l)}(f)\in S^{G}$
whenever $f\in S^{G}$.

##  3\. Proof of Theorems 1.1 and 1.5

We first prove that the relative Hilbert ideal is an extended ideal and some
corollaries. Theorems 1.1 and 1.5 would then follow.

######  3.1 Proposition.

Let $W\subseteq V^{G}$. Write $S^{\prime}=\operatorname{Sym}W^{\perp}$. Then
$\mathfrak{h}_{G,S,W}=(\mathfrak{h}_{G,S,W}\cap S^{\prime})S$.

###### Proof.

Note that $\mathfrak{h}_{G,S,W}\supseteq(\mathfrak{h}_{G,S,W}\cap
S^{\prime})S$. Therefore we need to show that
$\mathfrak{h}_{G,S,W}\subseteq(\mathfrak{h}_{G,S,W}\cap S^{\prime})S$, which
we do by induction on degree.

Write $n-r+1=\operatorname{rk}_{\Bbbk}W$. Construct a basis
$\\{v_{1},\ldots,v_{n}\\}$ following the proof of Lemma 2.6. Then
$W=\Bbbk\langle v_{r},\ldots,v_{n}\rangle$. Let $\\{x_{1},\ldots,x_{n}\\}$ be
the basis of $V^{*}$ dual to $\\{v_{1},\ldots,v_{n}\\}$. Then
$W^{\perp}=\Bbbk\langle x_{1},\ldots,x_{r-1}\rangle$ and
$S^{\prime}=\Bbbk[x_{1},\ldots,x_{r-1}]$.

Let $F\in\mathfrak{h}_{G,S,W}$. Note that for every $r\leq j\leq n$, and for
every $t\geq 0$, $\Delta_{j}^{(t)}F\in\mathfrak{h}_{G,S,W}$. Hence if $\deg F$
is minimum among that of the non-zero homogeneous polynomials in
$\mathfrak{h}_{G,S,W}$, then $F\in S^{\prime}$. Hence we assume that for every
homogeneous $F_{1}\in\mathfrak{h}_{G,S,W}$ with $\deg F_{1}<\deg F$,
$F_{1}\in(\mathfrak{h}_{G,S,W}\cap S^{\prime})S$. Hence
$\Delta_{r}^{(t_{r})}\cdots\Delta_{n}^{(t_{n})}F\in(\mathfrak{h}_{G,S,W}\cap
S^{\prime})S$ for every
${(0,\ldots,0)\neq(t_{r},\cdots,t_{n})\in\mathbb{N}^{n-r+1}}$. Since

$\sum_{(t_{r},\cdots,t_{n})\in\mathbb{N}^{n-r+1}}(-1)^{t_{r}+\cdots+t_{n}}x_{r}^{(t_{r})}\cdots
x_{n}^{(t_{n})}\Delta_{r}^{(t_{r})}\cdots\Delta_{n}^{(t_{n})}F,$

which is the coefficient of $x_{r}^{0}\cdots x_{n}^{0}$ in $F$ when $S$ is
identified with $S^{\prime}[x_{r},\cdots,x_{n}]$ (see [ES18, Lemma 2.2]),
belongs to $\mathfrak{h}_{G,S,W}\cap S^{\prime}$, we conclude that
$F\in(\mathfrak{h}_{G,S,W}\cap S^{\prime})S$. ∎

Note that $G$ acts on $S^{\prime}$; however, it is not true that
$\mathfrak{h}_{G,S,W}\cap S^{\prime}=\mathfrak{h}_{G,S^{\prime}}$. See Example
9.1.

######  3.2 Corollary.

$S/\mathfrak{h}_{G,S,W}$ is Cohen-Macaulay and
$\operatorname{Ass}_{S}(S/\mathfrak{h}_{G,S,W})=\\{W^{\perp}S\\}$.

###### Proof.

Adopt the notation from the proof of Proposition 3.1. Hence, by Proposition
3.1, $x_{r},\ldots,x_{n}$ is a regular sequence on $S/\mathfrak{h}_{G,S,W}$.
Note that $\dim S/\mathfrak{h}_{G,S,W}=\operatorname{rk}_{\Bbbk}W=n-r+1$, so
$S/\mathfrak{h}_{G,S,W}$ is Cohen-Macaulay.

Since $S$ is a polynomial ring over $S^{\prime}$, it follows that
$\operatorname{ht}(\mathfrak{h}_{G,S,W}\cap
S^{\prime})=\operatorname{ht}\mathfrak{h}_{G,S,W}=\operatorname{ht}W^{\perp}S=r-1=\dim
S^{\prime}$. Hence $S^{\prime}/(\mathfrak{h}_{G,S,W}\cap S^{\prime})$ has a
finite filtration with successive quotients isomorphic to
$S^{\prime}/(x_{1},\ldots,x_{r-1})S^{\prime}$. Since $S$ is a flat
$S^{\prime}$-algebra, $S/\mathfrak{h}_{G,S,W}$ has a filtration with
successive quotients isomorphic to $S/(x_{1},\ldots,x_{r-1})S$. Then
$\operatorname{Ass}_{S}(S/\mathfrak{h}_{G,S,W})=\\{(x_{1},\ldots,x_{r-1})S\\}=\\{W^{\perp}S\\}$.
∎

######  3.3 Corollary.

$S/\mathfrak{h}_{G,S,W}$ is a free $R/{\mathfrak{p}}$-module.

###### Proof.

$R/{\mathfrak{p}}$ is a polynomial ring (Proposition 2.3(a)). By Corollary
3.2, $S/\mathfrak{h}_{G,S,W}$ is a finitely generated graded Cohen-Macaulay
$R/{\mathfrak{p}}$-module with $\dim S/\mathfrak{h}_{G,S,W}=\dim
R/{\mathfrak{p}}$. Now use the Auslander-Buchsbaum formula. ∎

###### Proof of Proposition 2.3.

(a) and (b) are from [Bro10, Lemma 2.2] and its proof.

(c): By Corollary 3.3, $f_{1},\ldots,f_{s}$, which is an
$R/{\mathfrak{p}}$-regular sequence is regular on $S/\mathfrak{h}_{G,S,W}$.

(d): Note that for every homogeneous $S$-ideal $J$ and homogeneous $a\in S$
regular on $S/J$, $\mu(J+(a))=\mu(J)+1$. ∎

###### Proof of Theorem 1.1.

(a) $\implies$ (b): Immediate.

(b) $\implies$ (c): Write ${\mathfrak{p}}=(V^{G})^{\perp}S\cap R$. Then $\dim
R/{\mathfrak{p}}=\operatorname{rk}_{\Bbbk}V^{G}$, so $R_{\mathfrak{p}}$ is
regular.

(c) $\implies$ (a): Write $\kappa({\mathfrak{p}})$ for the residue field at
${\mathfrak{p}}$ and $K$ for the fraction field of $R$. We compute
$S\otimes_{R}\kappa({\mathfrak{p}})$ in two different ways:

$S\otimes_{R}\kappa({\mathfrak{p}})=(S/{\mathfrak{p}}S)\otimes_{R/{\mathfrak{p}}}\kappa({\mathfrak{p}})=(R\smallsetminus{\mathfrak{p}})^{-1}S\otimes_{R_{\mathfrak{p}}}\kappa({\mathfrak{p}}).$

By Corollary 3.3, $S/\mathfrak{h}_{G,S,W}$ is is a free
$R/{\mathfrak{p}}$-module. Therefore

$\operatorname{rk}_{\kappa({\mathfrak{p}})}(S\otimes_{R}\kappa({\mathfrak{p}}))=\operatorname{rk}_{\Bbbk}(S\otimes_{R}\Bbbk).$

On the other hand, since $R_{\mathfrak{p}}$ is a regular local ring,
$(R\smallsetminus{\mathfrak{p}})^{-1}S$ is a free $R_{\mathfrak{p}}$-module.
Then

$\operatorname{rk}_{\kappa({\mathfrak{p}})}(S\otimes_{R}\kappa({\mathfrak{p}}))=\operatorname{rk}_{K}(S\otimes_{R}K).$

Therefore
$\operatorname{rk}_{K}(S\otimes_{R}K)=\operatorname{rk}_{\Bbbk}(S\otimes_{R}\Bbbk)$.
Using NAK, we see that $S$ is a free $R$-module. Therefore, by Discussion 2.4,
$R$ is a polynomial ring. ∎

######  3.4 Proposition.

If $\operatorname{rk}V^{G}\geq\operatorname{rk}V-2$, then
$\mathfrak{h}_{G,S,V^{G}}$ is a complete intersection.

###### Proof.

We will apply Proposition 3.1 with $W=V^{G}$. First, construct a basis
$\\{v_{1},\ldots,v_{n}\\}$ of $V$ following the proof of Lemma 2.6 with
$W=V^{G}$. Let $\\{x_{1},\ldots,x_{n}\\}$ be the basis of $V^{*}$ dual to
$\\{v_{1},\ldots,v_{n}\\}$.

Then $v_{i}\in V^{G}$ for all $i\geq 3$. Therefore $W^{\perp}=\Bbbk\langle
x_{1}\rangle$ or $W^{\perp}=\Bbbk\langle x_{1},x_{2}\rangle$. Hence
$S^{\prime}=\Bbbk[x_{1}]$ or $S^{\prime}=\Bbbk[x_{1},x_{2}]$. In both cases
$x_{1}\in S^{G}$. Therefore $x_{1}\in\mathfrak{h}_{G,S,V^{G}}\cap S^{\prime}$.
Hence $\mathfrak{h}_{G,S,V^{G}}\cap S^{\prime}$ is a complete intersection. By
Proposition 3.1, $\mathfrak{h}_{G,S,V^{G}}$ is a complete intersection. ∎

###### Proof of Theorem 1.5.

Apply Propositions 3.4 and 2.3(d). ∎

##  4\. Direct summands

In this section, we collect some (well-known) remarks about direct summands of
rings. Recall that a subring $R$ of a ring $S$ is a _direct summand_ if there
exists an $R$-linear map $\psi:S\longrightarrow R$ such that $\psi\circ
i=\mathrm{id}_{R}$, where $i:R\longrightarrow S$ is the inclusion map.

######  4.1 Proposition.

Let $R\subseteq S$ be integral domains such that there exists $s_{0}\in S$ the
$R$-submodule $Rs_{0}\subseteq S$ is a free $R$-module and is a direct summand
of $S$ as an $R$-module. Then $R$ is a direct summand of $S$. In particular,
if $R$ is an $\mathbb{N}$-graded polynomial ring and $S$ is an
$\mathbb{N}$-graded Cohen-Macaulay domain that is finitely generated
$R$-module, then $R$ is a direct summand of $S$.

###### Proof.

Let $h:S\longrightarrow S$ be the $R$-linear map $s\mapsto ss_{0}$,
$\pi:S\longrightarrow Rs_{0}$ be the $R$-linear projection map and
$j:Rs_{0}\longrightarrow R$ the $R$-linear map $rs_{0}\mapsto r$. Define
$\psi=j\circ\pi\circ h$. Then $\psi\circ i=\mathrm{id}_{R}$; this proves the
first assertion. To prove the second assertion, note, by the Auslander-
Buchsbaum formula, that $S$ is a free $R$-module. ∎

######  4.2 Proposition.

Let $R\subseteq A\subseteq S$ be rings. If $R$ is a direct summand of $S$,
then it is a direct summand of $A$.

###### Proof.

There exists an $R$-linear $\psi:S\longrightarrow R$ such that $\psi(r)=r$ for
each $r\in R$. Take $\psi|_{A}:A\longrightarrow R$. ∎

##  5\. A composition series of $G$

As mentioned in the introduction, we are interested in the following
situation: $\operatorname{char}\Bbbk=p>0$ and $G$ is a $p$-group generated by
elements that act as transvections (i.e. non-diagonalizable pseudoreflections)
on $V$.

######  5.1 Notation.

$\\{v_{1},\ldots,v_{n}\\}$ will denote a basis of $V$ as in Lemma 2.6. We
write $\\{x_{1},\ldots,x_{n}\\}$ for the dual basis of $V^{*}$ dual to
$\\{v_{1},\ldots,v_{n}\\}$. For $g\in G$, define

$\beta_{g}:=\max\\{j\mid x_{j}\;\text{appears in}\;gx_{i}-x_{i}\;\text{for
some}\;i\\}.$

Let $\mathcal{P}$ denote the set of pseudoreflections of $G$. Define
$\beta_{G}:=\max\\{\beta_{g}\mid g\in\mathcal{P}\\}$. ∎

######  5.2 Remark.

$\beta_{g}=\max\\{j\mid gv_{j}\neq v_{j}\\}$. Note that $\beta_{g}$ depends on
the choice of the _ordered basis_ $\\{v_{1},\ldots,v_{n}\\}$. Note also that
$v_{i}\in V^{G}$ for each $\beta_{G}+1\leq i\leq n$.

######  5.3 Lemma.

Let $g,h\in\mathcal{P}$.

  1. (a)

$ghg^{-1}\in\mathcal{P}$ and $\beta_{ghg^{-1}}=\beta_{h}$.

  2. (b)

If $\beta_{g}=\beta_{h}$ then $gh=hg$.

###### Proof.

(a): Let $W\subseteq V$ be the hyperplane fixed by $h$. Then $ghg^{-1}$ fixes
the hyperplane $gW:=\\{gv\mid v\in W\\}$. Hence $ghg^{-1}\in\mathcal{P}$. Let
$l\in S_{1}$ be a polynomial defining $W$. Then $gW$ is defined by $gl$. Hence

$\beta_{h}=\max\\{i\mid x_{i}\;\text{is a term of}\;l\\}=\max\\{i\mid
x_{i}\;\text{is a term of}\;gl\\}=\beta_{ghg^{-1}}.$

(b): Let $m=\beta_{g}=\beta_{h}$. Let $1\leq i\leq n$. Then
$gx_{i}-x_{i}\in\Bbbk\langle x_{1},\ldots,x_{m}\rangle$. Similarly
$hx_{i}-x_{i}\in\Bbbk\langle x_{1},\ldots,x_{m}\rangle$. Hence
$g(hx_{i}-x_{i})=hx_{i}-x_{i}$ and $h(gx_{i}-x_{i})=gx_{i}-x_{i}$. Therefore

$g(h(x_{i}))=g(x_{i}+(hx_{i}-x_{i}))=x_{i}+(gx_{i}-x_{i})+(hx_{i}-x_{i})=h(g(x_{i})).\qed$

Using the above lemma, we see that there is a composition series of $G$
consisting of subgroups generated by transvections.

######  5.4 Proposition.

$G$ has a composition series $0=G_{0}\subsetneq
G_{1}\subsetneq\cdots\subsetneq G_{k}=G$ such that for each $1\leq l\leq k$,
$G_{l}$ is a transvection group and $G_{l}/G_{l-1}$ is isomorphic to
$\mathbb{Z}/p\mathbb{Z}$ and is generated by the residue class of a
transvection.

###### Proof.

For $1\leq l\leq\beta_{G}$, define $G_{l}$ to be the group generated by
$\\{g\in\mathcal{P}\mid\beta_{g}\leq l\\}$. By Lemma 5.3 $G_{l}$ is normal in
$G_{l+1}$ and $G_{l+1}/G_{l}$ is an abelian group generated by the residue
classes of $\\{g\in\mathcal{P}\mid\beta_{g}=l+1\\}$. Refine this filtration,
if necessary. ∎

Various authors (see, e.g., [Bra22, Section 1]) have used inertia subgroups of
prime ideals generated by linear subspaces of $V^{*}$ (the space of
homogeneous linear forms in $S$) to study homological properties of $S^{G}$.
In Example 9.2, we show that our filtration is not necessarily given by such
inertia subgroups. Moreover, the reason for looking at the above composition
series is the hope that we could prove Conjecture 1.3 by induction on the
order of the group. When $G$ is a $p$-group and $\operatorname{char}\Bbbk=p$,
if the invariant ring is a direct summand, then $G$ is generated by
transvections [Bro05, Theorem 2(ii)]. Therefore we would like the subgroups in
the composition series to be generated by transvections. Example 9.2 shows
that inertia subgroups need not generated by transvections, in general.
Moreover, When $G$ is a $p$-group and $\operatorname{char}\Bbbk=p$, there
exist $U$ of small dimensions (since $V^{G}\neq 0$) such that $G_{U}=G$, so
checking whether $S^{G_{U}}$ is a polynomial ring for all $U$ is not,
logically, easier than checking whether $S^{G}$ is a polynomial ring.

We end this section with some observations. A partial converse to the next
proposition will be proved in Section 6 (see Proposition 6.7).

######  5.5 Proposition.

$0=G_{0}\subsetneq G_{1}\subsetneq\cdots\subsetneq G_{k}=G$ be the filtration
of Proposition 5.4. Assume that $S^{G_{l}}$ is a direct summand of $S$ for
each $l$. Then $S^{G_{l}}$ is a polynomial ring for each $l$.

###### Proof.

Since $S=S^{G_{0}}$ is a polynomial ring, we may assume, by induction, that
$S^{G_{l-1}}$ is a polynomial ring. Note that $S^{G_{l}}$ is a direct summand
of $S^{G_{l-1}}$. Write $A=S^{G_{l-1}}$ and $R=S^{G_{l}}$. Then there exists a
homogeneous $a\in A$ such that $A=R[a]$; see [Bro06, Theorem 4, p. 588].
Moreover, there exist $r_{1},r_{2}\in R$ such that $R[a]\simeq
R[T]/(T^{p}-r_{1}T-r_{2})$. Hence $A$ is a free $R$-module. Hence $R$ is a
polynomial ring, by Discussion 2.4. ∎

######  5.6 Proposition.

If $\operatorname{rk}_{\Bbbk}V=2$, then $S^{G}=\Bbbk[x_{1},\prod_{G}x_{2}]$.

###### Proof.

Write $R^{\prime}=\Bbbk[x_{1},\prod_{G}x_{2}]$. Since $R^{\prime}\subseteq
S^{G}$, and $S$ is finite over $R^{\prime}$, we see that

$|G|=[\operatorname{Frac}(S):\operatorname{Frac}(S^{G})]\leq[\operatorname{Frac}(S):\operatorname{Frac}(R^{\prime})]=\deg\left(\prod_{G}x_{2}\right)\leq|G|,$

so $\operatorname{Frac}(S^{G})=\operatorname{Frac}(R^{\prime})$. Since
$R^{\prime}$ is integrally closed, $R^{\prime}=S^{G}$. ∎

##  6\. Relative Hilbert ideals at consecutive stages in the composition
series

######  6.1 Notation.

Let $0=G_{0}\subsetneq G_{1}\subsetneq\cdots\subsetneq G_{k}=G$ be the
filtration from Proposition 5.4. Write $G^{\prime}=G_{k-1}$. We denote by
$\sigma$ a transvection in $G\smallsetminus G^{\prime}$. Let $R=S^{G}$ and
$A=S^{G^{\prime}}$. Let $<$ denote the degree-lexicographic monomial order on
$S$ with $x_{1}<x_{2}<\cdots<x_{n}$.

######  6.2 Lemma.

Suppose that $R$ is a direct summand of $A$. Let $d_{0}=\min\\{d\mid A_{d}\neq
R_{d}\\}$.

  1. (a)

$\operatorname{rk}_{\Bbbk}A_{d_{0}}/R_{d_{0}}=1$.

  2. (b)

$A=R[a]$ for all $a\in A_{d_{0}}\smallsetminus R_{d_{0}}$.

###### Proof.

(a): By [Bro06, Theorem 4, p. 588], there exists a homogeneous $a_{0}\in
A\smallsetminus R$ such that $A=R[a_{0}]$. It is immediate that $\deg
a_{0}=d_{0}$. Suppose there exists $a\in A_{d_{0}}\smallsetminus R_{d_{0}}$
that is $\Bbbk$-linearly independent of $a_{0}$ in $A_{d_{0}}/R_{d_{0}}$. Then
$a\not\in R[a_{0}]$ since the degrees of $a$ and $a_{0}$ are the same. This
proves the assertion.

(b): Suppose that $a\in A_{d_{0}}\smallsetminus R_{d_{0}}$. Let $a_{0}$ be as
in (a). Then, by (a), $a+\lambda a_{0}\in R_{d_{0}}$ for some
$\lambda\in\Bbbk^{\times}$. Hence $a_{0}\in R[a]$, so $A=R[a]$. ∎

######  6.3 Lemma.

We may assume $\sigma(x_{i})=x_{i}$ for all $i\neq n$.

###### Proof.

Let $i_{\sigma}=\text{max}\\{i\mid\sigma x_{i}\neq x_{i}\\}$. Note that
$i_{\sigma}>\beta_{\sigma}$. For $i>\beta_{\sigma}$, define $\lambda_{i}$ by
$\sigma x_{i}-x_{i}=\lambda_{i}(\sigma x_{i_{\sigma}}-x_{i_{\sigma}})$. We
define a change of coordinates as follows:

$x^{\prime}_{i}:=\begin{cases}x_{i},&\text{if}\;i\leq\beta_{\sigma},\\\
x_{i}-\lambda_{i}x_{i_{\sigma}},&\text{if}\;i>\beta_{\sigma}\;\text{and}\;i\not\in\\{i_{\sigma},n\\},\\\
x_{n},&\text{if}\;i=i_{\sigma},\\\ x_{i_{\sigma}},&\text{if}\;i=n.\\\
\end{cases}$

Let $\tau\in G$ be a transvection and $1\leq i\leq n$. Then

$\max\\{j\mid x_{j}\;\text{appears in}\;\tau x_{i}-x_{i}\\}=\max\\{j\mid
x_{j}\;\text{appears in}\;\tau x^{\prime}_{i}-x^{\prime}_{i}\\}.$

Hence the action of $G$ on $\Bbbk\langle
x^{\prime}_{1},\ldots,x^{\prime}_{n}\rangle$ is ‘upper-triangular’ and, for
all $g\in G$, $\beta_{g}$ does not change after this change of coordinates. ∎

######  6.4 Lemma.

Suppose that $R$ is a direct summand of $A$. Then there exists a minimal
algebra generating set $\\{x_{1},a_{2},\ldots,a_{m}\\}$ of $A$ with $m\geq n$
such that

  1. (a)

there exists a unique $2\leq i_{0}\leq m$ such that $a_{i_{0}}\not\in R$.

  2. (b)

$x_{1}<\operatorname{in}_{>}(a_{2})<\cdots<\operatorname{in}_{>}(a_{m})$;

  3. (c)

there exists a unique $2\leq i_{1}\leq m$ such that
$\operatorname{in}_{>}(a_{i_{1}})=x_{n}^{\deg a_{i_{1}}}$.

Moreover, $\deg a_{i_{1}}$ is a power of $p$.

###### Proof.

By Lemma 6.2 $A=\Bbbk[x_{1},a_{2},\ldots,a_{m}]$ for some
$a_{2},\ldots,a_{m}\in A$ with $m\geq n+1$, such that all but one of them is
in $R$. >From this we choose a minimal generating set which we continue to
call $x_{1},a_{2},\ldots,a_{m}$, $m\geq n$, after relabelling the elements, if
necessary. Note that all but one of the elements of the minimal generating set
is in $R$. This proves (a).

Now suppose that $a,a^{\prime}\in A$ are part of a $\Bbbk$-algebra generating
set $X$ of $A$ with
$\operatorname{in}_{>}(a)=\operatorname{in}_{>}(a^{\prime})$ that satisfies
(a). If $a\not\in R$ or $a^{\prime}\not\in R$, assume, after relabelling if
necessary, that $a\in R$ and $a^{\prime}\not\in R$. Note that
$(X\cup\\{a^{\prime}+\lambda a\\})\smallsetminus\\{a^{\prime}\\}$ generates
$A$ for all $\lambda\in\Bbbk^{\times}$. For a suitable
$\lambda\in\Bbbk^{\times}$, $\operatorname{in}_{>}(a^{\prime}+\lambda
a)<\operatorname{in}_{>}(a)$. Therefore, without loss of generality we can
assume
$\operatorname{in}_{>}x_{1}<\operatorname{in}_{>}a_{2}<\cdots<\operatorname{in}_{>}a_{m}$.
This proves (b).

To prove (c), let

$i_{1}:=\min\\{i\mid\operatorname{in}_{>}(a_{i})=x_{n}^{\deg a_{i}}\\}.$

(This set is non-empty since $S$ is integral over $A$.) Now, for all $r\in A$,
if $\deg_{x_{n}}(r)=p^{e}m$ with $p\nmid m$, then
$\deg_{x_{n}}(\Delta_{x_{n}}^{(m-1)p^{e}}r)=p^{e}$. Hence, if
$\deg_{x_{n}}(a_{i_{1}})$ is not a power of $p$, we can find an element $r\in
A$ such that $\deg r<\deg a_{i_{1}}$ and $\operatorname{in}_{>}(r)=x_{n}^{\deg
r}$. Hence there must exist $j<i_{1}$ such that $a_{j}\in A$ and
$\operatorname{in}_{>}(a_{j})=x_{n}^{\deg a_{j}}$, contradicting the
minimality of $i_{1}$. Therefore write $\deg a_{i_{1}}=p^{e_{1}}$.

Now suppose that $j>i_{1}$ and $\operatorname{in}_{>}(a_{j})=x_{n}^{\deg
a_{j}}$. If $p^{e_{1}}\nmid\deg a_{j}$, then applying $\Delta_{x_{n}}^{(t)}$
to it for a suitable $t$, we obtain $a\in A$ with
$\operatorname{in}_{>}(a)=x_{n}^{\deg a}$ and $\deg a<\deg(a_{i_{1}})$, which
implies that there exists $i<i_{1}$ with
$\operatorname{in}_{>}(a_{i})=x_{n}^{\deg a_{i}}$, contradicting the choice of
$i_{1}$. Hence $p^{e_{1}}\mid\deg a_{j}$. We now consider the different cases.

Case 1: $i_{0}=i_{1}$. Then $a_{j}\in R$. Assume, for now, that
$p^{e_{1}+1}\nmid\deg a_{j}$. Applying $\Delta_{x_{n}}^{(t)}$ to it for a
suitable $t$, we obtain $a\in R$ with $\operatorname{in}_{>}(a)=x_{n}^{\deg
a}$ and $\deg a=\deg(a_{i_{1}})$. Hence $a$ belongs to the subalgebra of $A$
generated by $\\{a_{i}\mid 1\leq i\leq m,\deg a_{i}\leq\deg(a_{i_{1}})\\}$.
However, since $i_{0}=i_{1}$ and $\operatorname{in}_{>}(a)=x_{n}^{\deg
a_{i_{1}}}$ it follows that $a=\lambda a_{i_{1}}+a^{\prime}$ for some
$\lambda\in\Bbbk^{\times}$ and $a^{\prime}\in R$, contradicting the fact that
$a\in R$. Hence $p^{e_{1}+1}\mid\deg a_{j}$. Then we may replace

$a_{j}-\lambda\left(\prod_{\sigma}a_{i_{1}}\right)^{\frac{\deg a_{j}}{p\deg
a_{i_{1}}}}$

for a suitable $\lambda\in\Bbbk^{\times}$. This may violate (b), but the steps
in its proof may be repeated to ensure that (b) is satisfied.

Case 1: $i_{0}\neq i_{1}$. Then $a_{i_{1}}\in R$. We may replace $a_{j}$ by

$a_{j}-\lambda a_{i_{1}}^{\frac{\deg a_{j}}{\deg a_{i_{1}}}}$

for a suitable $\lambda\in\Bbbk^{\times}$. Again, steps in the proof of (b)
may be repeated to ensure that (b) is satisfied. This completes the proof of
the lemma. ∎

######  6.5 Lemma.

Adopt the notation from Lemma 6.4. Assume that $\beta_{\sigma}>x_{1}$. Then
$a_{2}\in\Bbbk[x_{1},\ldots,x_{n-1}]^{G}$.

###### Proof.

Since $A\cap\Bbbk[x_{1},\ldots,x_{n-1}]\subseteq R$, it suffices to show that
$a_{2}\in\Bbbk[x_{1},\ldots,x_{n-1}]$. By way of contradiction assume that
${x_{n}}$ divides some terms of $a_{2}$. Then in every such term, the exponent
of ${x_{n}}$ must be a power of $p$; for otherwise, by taking
$\Delta_{x_{n}}^{(t)}a_{2}$ for a suitable $t$, we get a (non-zero) element of
$R\smallsetminus\Bbbk[x_{1}]$, contradicting the minimality of
$\operatorname{in}_{>}(a_{2})$. Using the same argument, we see that $a_{2}$
must be of the form

$a_{2,0}+\sum_{i\geq 0}\lambda_{i}x_{1}^{\deg a_{2}-p^{i}}{x_{n}}^{p^{i}}$

with $a_{2,0}\in\Bbbk[x_{1},\ldots,x_{n-1}]$ and $\lambda_{i}\in\Bbbk$. Hence

$(\sigma-1)a_{2}=(\sigma-1)a_{2,0}+\sum_{i\geq 0}\lambda_{i}x_{1}^{\deg
a_{2}-p^{i}}(\sigma-1){x_{n}}^{p^{i}}.$

By Lemma 6.3, $(\sigma-1)a_{2,0}=0$. Note that
$(\sigma-1)x_{n}\not\in\Bbbk\langle x_{1}\rangle$. Therefore, unless
$\lambda_{i}=0$ for all $i\geq 0$, $(\sigma-1)a_{2}$ would be a (non-zero)
element of $A\smallsetminus\Bbbk[x_{1}]$ with
$\operatorname{in}_{>}((\sigma-1)a_{2})<\operatorname{in}_{>}(a_{2})$,
contradicting the choice of $a_{2}$. Hence $\lambda_{i}=0$ for all $i\geq 0$.
∎

######  6.6 Lemma.

Adopt the notation of Lemma 6.4. Suppose that $R$ is a direct summand of $A$.
Let $W=\Bbbk\langle v_{n}\rangle$. If $i_{0}=i_{1}$, then

$\mathfrak{h}_{G^{\prime},S,W}=\mathfrak{h}_{G,S,W}=(x_{1},a_{2},\ldots,a_{i_{0}-1},a_{i_{0}+1},\ldots,a_{m})S.$

###### Proof.

Write $\Lambda=\\{x_{1},a_{2},\ldots,a_{i_{0}-1},a_{i_{0}+1},\ldots,a_{m}\\}$.
Let $a\in W^{\perp}S\cap A$ be a homogeneous element. Write
$a=\Phi(x_{1},a_{2},\ldots,a_{m})$ where $\Phi$ is a polynomial in $m$
variables $\xi_{1},\ldots,\xi_{m}$. Since $\Lambda\subseteq W^{\perp}S$, it
follows that no term of $\Phi(\xi_{1},\ldots,\xi_{m})$ is a power of
$\xi_{i_{0}}$. Hence $a\in\Lambda S$, and, therefore,
$\mathfrak{h}_{G^{\prime},S,W}\subseteq\Lambda S$. Moreover, $\Lambda\subseteq
W^{\perp}S\cap R$, so we get

$\mathfrak{h}_{G^{\prime},S,W}\subseteq\Lambda
S\subseteq\mathfrak{h}_{G,S,W}\subseteq\mathfrak{h}_{G^{\prime},S,W},$

proving the lemma. ∎

As an application, we prove a partial converse to Proposition 5.5.

######  6.7 Proposition.

Adopt the notation of Lemma 6.4. Suppose that $i_{0}=i_{1}$ and that $R$ is a
polynomial ring. Then $A$ is a polynomial ring.

###### Proof.

Note that $R$ is a polynomial ring if and only if $S$ is a free $R$-module.
The ‘only if’ direction follows from the Auslander-Buchsbaum formula; See
Discussion 2.4 for the ‘if’ direction. $R$ is a polynomial ring if and only if
$\operatorname{rk}_{\Bbbk}S/\mathfrak{h}_{G,S}=|G|$. Similarly $A$ is a
polynomial ring if and only if
$\operatorname{rk}_{\Bbbk}S/\mathfrak{h}_{G^{\prime},S}=|G^{\prime}|$. Let
$W=\Bbbk\langle v_{n}\rangle$.

Step 1. $\min\\{\deg r\mid r\in R\smallsetminus W^{\perp}S\\}=p\deg
a_{i_{0}}$. Proof: Let $r\in R\smallsetminus W^{\perp}S$ be of the smallest
degree. Then $\operatorname{in}_{>}(r)=x_{n}^{\deg r}$. By considering
$\Delta_{n}^{(t)}r,t\geq 1$, we see that $\deg r$ is a power of $p$. By the
definition of $i_{1}$, $\deg r\geq\deg a_{i_{1}}$. Since $i_{0}=i_{1}$, $r\neq
a_{i_{1}}$. Hence $\deg r>\deg a_{i_{1}}$. Since $\deg a_{i_{1}}$ is a power
of $p$, it follows that $\deg r\geq p\deg a_{i_{1}}=p\deg a_{i_{0}}$. On the
other hand $\prod_{\sigma}a_{i_{0}}\in R\smallsetminus W^{\perp}S$, so $\deg
r\leq p\deg a_{i_{0}}$.

Step 2. Let $r_{0}\in R\smallsetminus W^{\perp}S$ be such that $\deg
r_{0}=p\deg a_{i_{0}}$. Since
$\operatorname{Ass}_{S}(S/\mathfrak{h}_{G,S,W})=W^{\perp}S$ (Corollary 3.2)
$a_{i_{0}}$ and $r_{0}$ are non-zero-divisors on $S/\mathfrak{h}_{G,S,W}$.

Step 3. Note that $R/(W^{\perp}S\cap R)$ is a $1$-dimensional polynomial ring
(Proposition 2.3), $R/(W^{\perp}S\cap R)\simeq\Bbbk[r_{0}]$. Similarly
$A/(W^{\perp}S\cap A)\simeq\Bbbk[a_{i_{0}}]$. Hence,
$\mathfrak{h}_{G,S}=\mathfrak{h}_{G,S,W}+(r_{0})$ and
$\mathfrak{h}_{G^{\prime},S}=\mathfrak{h}_{G^{\prime},S,W}+(a_{i_{0}})$.

Step 4. For a homogeneous $S$-ideal $J$, write $\deg(S/J)$ for the _degree_
(or _multiplicity_) of $S/J$. Then, by Lemma 6.6,

$\displaystyle|G|=\operatorname{rk}_{\Bbbk}S/\mathfrak{h}_{G,S}$
$\displaystyle=\deg r_{0}\deg(S/\mathfrak{h}_{G,S,W})$
$\displaystyle=p\deg(a_{i_{0}})\deg(S/\mathfrak{h}_{G^{\prime},S,W})=p\operatorname{rk}_{\Bbbk}S/\mathfrak{h}_{G^{\prime},S}\geq
p|G^{\prime}|=|G|,$

so $\operatorname{rk}_{\Bbbk}S/\mathfrak{h}_{G^{\prime},S}=|G^{\prime}|$. In
other words,

$\operatorname{rk}_{\Bbbk}S\otimes_{A}A/A_{+}=\operatorname{rk}_{\operatorname{Frac}(A)}S\otimes_{A}\operatorname{Frac}(A)$

so, by NAK, $S$ is a free $A$-module. By Discussion 2.4, $A$ is a polynomial
ring. ∎

##  7\. Hilbert ideals in dimension $4$

In the following we assume $\operatorname{rk}_{\Bbbk}V=4$ and we adopt the
notation of Lemma 6.4. Let $W=\Bbbk\langle v_{4}\rangle\subseteq V^{G}$.

######  7.1 Lemma.

Suppose that $R$ is a direct summand of $S$ and that $\beta_{\sigma}>1$. If
$a_{3}\in R$ and $\mathfrak{h}_{G^{\prime},S,W}$ is a complete intersection,
then

$\mathfrak{h}_{G^{\prime},S,W}=\mathfrak{h}_{G,S,W}=(x_{1},a_{2},a_{3})S.$

###### Proof.

It suffices to show that

(7.2) $\mathfrak{h}_{G^{\prime},S,W}=(x_{1},a_{2},a_{3})S,$

for, then, it would follow that

$\mathfrak{h}_{G^{\prime},S,W}=(x_{1},a_{2},a_{3})S\subseteq\mathfrak{h}_{G,S,W}\subseteq\mathfrak{h}_{G^{\prime},S,W},$

proving the proposition. (Note that $a_{2}\in R$ by Lemma 6.5.)

To prove (7.2), in turn, it suffices to show that $x_{1},a_{2},a_{3}$ are part
of a minimal generating set of $\mathfrak{h}_{G^{\prime},S,W}$, i.e., none of
them belong to $S_{+}\mathfrak{h}_{G^{\prime},S,W}$. Note that
$\operatorname{ht}\mathfrak{h}_{G^{\prime},S,W}=3$. Since
$\mathfrak{h}_{G^{\prime},S,W}$ is a complete intersection, (7.2) would then
follow.

It is immediate that $x_{1}\notin S_{+}\mathfrak{h}_{G^{\prime},S,W}$. Suppose
that $a_{2}\in S_{+}\mathfrak{h}_{G^{\prime},S,W}$. Since $a_{2}\not\in
x_{1}S$, for $a_{2}$ to belong to $S_{+}\mathfrak{h}_{G^{\prime},S,W}$, it is
necessary that there is a homogeneous element $b$ of $S^{G^{\prime}}$ such
that $b\not\in\Bbbk[x_{1}]$ and $\deg b<\deg a_{2}$. This contradicts the
choice of $a_{2}$. Hence $a_{2}\not\in S_{+}\mathfrak{h}_{G^{\prime},S,W}$.

We now show that $a_{3}\notin(x_{1},a_{2})S$. Since $R$ is a direct summand of
$S$ and $a_{3}\in R$, if $a_{3}\in(x_{1},a_{2})S$ then
$a_{3}\in(x_{1},a_{2})R$. This implies $a_{3}\in(x_{1},a_{2})A$. Write
$a_{3}=a^{\prime}x_{1}+a^{\prime\prime}a_{2}$ with
$a^{\prime},a^{\prime\prime}\in A$. By degree considerations, we have that
$a^{\prime},a^{\prime\prime}\in\Bbbk[x_{1},a_{2}]$ which is a contradiction of
the fact that $\\{x_{1},a_{2},a_{3}\\}$ is a part of minimal generating set of
$A$ as a $\Bbbk$-algebra. Hence $a_{3}\notin(x_{1},a_{2})S$.

Now suppose that $a_{3}\in S_{+}\mathfrak{h}_{G^{\prime},S,W}$. Since
$a_{3}\notin(x_{1},a_{2})S$, it is necessary that there is a homogeneous
element $b$ of $S^{G^{\prime}}$ such that $b\not\in(x_{1},a_{2})S$ and $\deg
b<\deg a_{3}$. Note that $b\not\in\Bbbk[x_{1},a_{2}]$. This contradicts the
choice of $a_{3}$. Hence $a_{3}\not\in S_{+}\mathfrak{h}_{G^{\prime},S,W}$. ∎

######  7.3 Lemma.

Suppose $\Bbbk=\mathbb{F}_{p}$. Then $\beta_{G^{\prime}}\leq 2$. In particular
$\mathfrak{h}_{G^{\prime},S,W}$ is a complete intersection.

###### Proof.

If $\beta_{\sigma}=2$, then by the construction of the filtration
$\beta_{\tau}\leq 2$ for all pseudoreflection $\tau\in G$. Therefore assume
$\beta_{\sigma}=3$. By way of contradiction, if $x_{3}$ appears in
$(\tau-1)x_{4}$, then for some $k=0,\ldots,p-1$, we have
$\beta_{\sigma^{k}\tau}\leq 2$. By our choice of filtration,
$\sigma^{k}\tau\in G^{\prime}$, which is a contradiction. Hence the claim
follows. The second statement follows from Proposition 3.4 since
$\beta_{G}\leq 2$ implies $\operatorname{rk}_{\Bbbk}V^{G}\geq
2=\operatorname{rk}_{\Bbbk}V-2$. ∎

######  7.4 Lemma.

Adopt the notation from Lemma 6.4. Assume that $\beta_{G^{\prime}}\leq 2$,
$i_{0}=3$ (i.e., $a_{3}\not\in R$), and $\beta_{\sigma}=3$. Then

  1. (a)

$\operatorname{in}_{>}((\sigma-1)a_{3})=x_{3}^{\deg a_{3}}$;

  2. (b)

No term of $a_{3}$ is divisible by $x_{3}x_{4}$, i.e., coefficients of
$x_{4}^{t}$ in $a_{3}$ are in $\Bbbk[x_{1},x_{2}]$ for all $t>0$.

  3. (c)

$\operatorname{in}_{>}(a_{3})=x_{4}^{\deg a_{3}}$, i.e., $i_{1}=3$.

###### Proof.

(a): By [Bro06, Theorem 4(i)], $((\sigma-1)a_{3})^{p-1}$ generates the
Dedekind different ideal $\mathcal{D}_{A/R}$. Fix generators $\Delta_{S/R}$
and $\Delta_{S/A}$ for the Dedekind differents $\mathcal{D}_{S/R}$ and
$\mathcal{D}_{S/A}$ respectively. Then

$\Delta_{S/R}=\Delta_{S/A}\cdot((\sigma-1)a_{3})^{p-1}$

up to multiplication by an element of $\Bbbk^{\times}$.

For each transvection $\tau\in G$, fix a linear form $l_{\tau}$ that defines
its reflecting hyperplane; $l_{\tau}$ is defined uniquely up to multiplication
by elements of $\Bbbk^{\times}$. Hence $\operatorname{in}_{>}l_{\tau}=x_{3}$
if and only if $\tau\in G\smallsetminus G^{\prime}$. Moreover, the irreducible
factors of $\Delta_{S/R}$ (respectively, $\Delta_{S/A}$) are precisely the
$l_{\tau}$ with $\tau$ a transvection in $G$ (respectively, $G^{\prime}$); see
[Ben93, Proposition 3.10.3] or [Bro05, Proposition 9, p. 17]. Therefore the
irreducible factors of $(\sigma-1)a_{3}$ are precisely the $l_{\tau}$ with
$\tau$ a transvection in $G\smallsetminus G^{\prime}$. This proves (a).

(b): Since $\beta_{G^{\prime}}\leq 2$, $x_{3}$ is a terminal variable for
$G^{\prime}$. Hence, for all $t>0$, $\Delta_{x_{3}}^{(t)}(a_{3})$ belongs to
$A$ and, hence, to $\Bbbk[x_{1},a_{2}]$, by degree considerations. If
$x_{3}x_{4}$ divides some term of $a_{3}$, then there exists $t>0$ such that
$x_{4}$ divides a term of $\Delta_{x_{3}}^{(t)}(a)$. This is a contradiction
since $\\{x,a_{2}\\}\subseteq\Bbbk[x_{1},x_{2},x_{3}]$.

(c): For all $a,b,c\in\mathbb{N}$ with $c\neq 0$, If
$\operatorname{in}_{>}(a_{3})=x_{1}^{a}x_{2}^{b}x_{4}^{c}$, then
$\operatorname{in}_{>}((\sigma-1)a_{3})=x_{1}^{a}x_{2}^{b}x_{3}^{c}$. Moreover
for all $a,b,c\in\mathbb{N}$, $(\sigma-1)(x_{1}^{a}x_{2}^{b}x_{3}^{c})=0$.
Hence the assertion follows from (a) and (b). ∎

###### Proof of Theorem 1.7.

We show that $\mathfrak{h}_{G,S}$ is a complete intersection.

Case I: $\beta_{G}\leq 2$. Then $\mathfrak{h}_{G,S,V^{G}}$ is a complete
intersection (Proposition 3.4). This case does not use the hypothesis that
$\Bbbk=\mathbb{F}_{p}$.

Case II: $\beta_{G}=\beta_{\sigma}=3$. Let $W=\Bbbk\langle v_{4}\rangle$.
Apply Lemma 7.3 to see that $\mathfrak{h}_{G^{\prime},S,W}$ is a complete
intersection. By Lemma 6.5, $i_{0}\geq 3$. If $i_{0}=3$, then Lemmas 7.3, 7.4
and 6.6 imply that $\mathfrak{h}_{G,S,W}$ is a complete intersection. If
$i_{0}>3$, then Lemma 7.1 implies that $\mathfrak{h}_{G,S,W}$ is a complete
intersection.

Finally use Proposition 2.3(d) to see that $\mathfrak{h}_{G,S}$ is a complete
intersection. ∎

##  8\. Proof of Theorem 1.8

The following lemma might be known, but we give a proof for the sake of
completeness.

######  8.1 Lemma.

Let $G$ be a transvection group of order $p^{2}$. Then $S^{G}$ is a polynomial
ring.

###### Proof.

Every group of order $p^{2}$ is abelian. Let $\sigma,\tau\in G$ be
transvections that generate $G$. Fix equations $l_{\sigma}$ and $l_{\tau}$
defining the reflecting hyperplanes of $\sigma$ and $\tau$ respectively. Note
that $\sigma\tau\sigma^{-1}$ is a transvection, with reflecting hyperplane
defined by $\sigma l_{\tau}$; similarly, $\tau\sigma\tau^{-1}$ is a
transvection, with reflecting hyperplane defined by $\tau l_{\sigma}$. Since
$\sigma\tau=\tau\sigma$, we see that $\sigma l_{\tau}=l_{\tau}$; in any case,
$\tau l_{\tau}=l_{\tau}$. Therefore $l_{\tau}\in S^{G}$. Similarly
$l_{\sigma}\in S^{G}$.

If $l_{\tau}$ and $l_{\sigma}$ are linearly dependent over $\Bbbk$, then $G$
has exactly one reflecting hyperplane. Therefore using [LS87, Theorem 5,
Appendix] we see that $S^{G}$ is a polynomial ring.

Otherwise, we could assume that $l_{\sigma}=x_{1}$ and $l_{\tau}=x_{2}$. After
permuting the variables $x_{3},\ldots,x_{n}$ and replacing $x_{1}$ by a
suitable scalar multiple of it, we may assume that $\sigma x_{3}=x_{3}+x_{1}$.
For all $i\geq 4$, after replacing $x_{i}$ by $x_{i}+\lambda x_{3}$ for a
suitable $\lambda\in\Bbbk$, we may assume that $\sigma x_{i}=x_{i}$.
Similarly, if there exists $i\geq 4$ such that $\tau x_{i}\neq x_{i}$, we may
assume that $\tau x_{4}\neq x_{4}$ and that that $\tau x_{i}=x_{i}$ for all
$i\geq 5$. Hence
$S^{G}=\Bbbk[x_{1},\cdots,x_{4}]^{G}\otimes_{\Bbbk}\Bbbk[x_{5},\ldots,x_{n}]$.
Therefore we may assume that $n=4$. Further replacing $x_{3}$ with
$x_{3}+\lambda x_{4}$ for a suitable $\lambda$, we may assume that $\tau
x_{3}=x_{3}$. Hence

$S^{G}=\Bbbk[x_{1},x_{2},\prod_{i=0}^{p-1}\sigma^{i}x_{3},\prod_{i=0}^{p-1}\tau^{i}x_{4}]$

which is a polynomial ring. ∎

######  8.2 Proposition.

Assume that $|G|=p^{3}$. If $S^{G}$ is a direct summand of $S$, then $S^{G}$
is a polynomial ring.

###### Proof.

Note that $G$ is a transvection group. Let $0=G_{0}\subsetneq G_{1}\subsetneq
G_{2}\subsetneq G_{3}=G$ be the filtration from Proposition 5.4. For $l=1,2$,
$S^{G_{l}}$ is a direct summand of $S$ by [LS87, Theorem 5, Appendix] and
Lemma 8.1, respectively. Now apply Proposition 5.5. ∎

##  9\. Examples

The first example shows that
$\mathfrak{h}_{G,S,W}\neq\mathfrak{h}_{G,\operatorname{Sym}W^{\perp}}S$, in
general. This is related to Proposition 3.1.

######  9.1 Example.

Let $\Bbbk=\mathbb{F}_{9}$ and choose $a\in\Bbbk\setminus\mathbb{F}_{3}$.
Consider the action of $G=\langle\tau,\sigma\rangle$ on a $3$-dimensional
vector space $V=\Bbbk\langle v_{1},v_{2},v_{3}\rangle$ such that the action on
$V^{*}$ with dual basis $\\{x,y,z\\}$ is given by
$\tau(x)=x,\tau(y)=y+ax,\tau(z)=z$ and
$\sigma(x)=x,\sigma(y)=y+x,\sigma(z)=z+x$. Write $S=\Bbbk[x,y,z]$ and
$W=\Bbbk\langle v_{3}\rangle\subset V^{G}$. We get
$S^{G}=\Bbbk[x,(a-1)y^{3}-ayx^{2}+zx^{2},z^{3}-x^{2}z]$, hence
$\mathfrak{h}_{G,S,W}=(x,y^{3})$. Let
$S^{\prime}:=\operatorname{Sym}W^{\perp}=\Bbbk[x,y]$. Then $(S^{\prime})^{G}$
is generated by $x$ and the $G$-orbit product of $y$, which is a polynomial of
degree $9$. Hence
$\mathfrak{h}_{G,S^{\prime}}=(x,y^{9})\neq\mathfrak{h}_{G,S,W}\cap
S^{\prime}$.

######  9.2 Example.

Let $\Bbbk$ be a field of characteristic $3$ and $V$ is a $\Bbbk$-vector space
of dimension $5$. Let $G$ be a $3$-group generated by $\langle
g_{1},g_{2},g_{3}\rangle$ with actions on $V^{*}=\Bbbk\langle
x_{1},\ldots,x_{5}\rangle$ given as follows:

$\displaystyle g_{1}(x_{i})=x_{i}\;\text{for}\;$ $\displaystyle i\neq
4\;\text{and}\;g_{1}(x_{4})=x_{4}+x_{3}+x_{2};$ $\displaystyle
g_{2}(x_{i})=x_{i}\;\text{for}\;$ $\displaystyle i\neq
5\;\text{and}\;g_{2}(x_{5})=x_{5}+x_{3}-x_{2}+x_{1};$ $\displaystyle
g_{3}(x_{i})=x_{i}\;\text{for}\;$ $\displaystyle i\neq
4,5\;\text{and}\;g_{3}(x_{4})-x_{4}=g_{3}(x_{5})-x_{5}=-x_{3}-x_{2}-x_{1}.$

For a prime ideal ${\mathfrak{q}}$ of $S$, write $I_{\mathfrak{q}}$ for its
inertia subgroup, $\\{g\in G\mid(g-1)s\in{\mathfrak{q}};\text{for all}\;s\in
S\\}$. We use this representation to show that the filtration in Proposition
5.4 need not be same as the filtration obtained by refining the filtration by
the inertia groups of the ideals $(x_{1},\ldots,x_{i})$, i.e., the group
generated by $\\{g\in G\mid\beta_{g}\leq i\\}$. We can check that
$I_{(x_{1},x_{2})}$ is a cyclic group of order $3$ generated by
$g_{1}g_{2}g_{3}$ which is not a pseudoreflection. Hence no refinement of the
filtration $I_{(x_{1},x_{2})}\subsetneq I_{(x_{1},x_{2},x_{3})}=G$ will give
us a filtration of the type described in Proposition 5.4.

######  9.3 Example.

Let $G$ be a Nakajima group with Nakajima basis $\\{x_{1},\ldots,x_{n}\\}$ of
$V^{*}$. We shall construct a filtration of the form appearing in Proposition
5.4 (with respect to the basis $\\{x_{1},\ldots,x_{n}\\}$), such that each
$G_{i}$ in the filtration is a Nakajima group with respect to the same basis.
We have

$G=P_{n}P_{n-1}\cdots P_{1}:=\\{g_{n}g_{n-1}\cdots g_{1}\mid g_{k}\in
P_{k}\\}$

where $P_{k}=\\{g\in G\mid gx_{j}=x_{j}\text{ for all }j\neq k\\}$.

For any subgroup $H$ of $G$, let us define $P_{k}(H)=P_{k}\cap H$. Then $H$ is
a Nakajima group with same Nakajima basis as $G$, if
$H=P_{n}(H)P_{n-1}(H)\cdots P_{1}(H)$. First we show that each
$G_{l}=\\{g\in\mathcal{P}\mid\beta_{g}\leq l\\}$ is a Nakajima group. Let
$g\in G_{l}$ and write $g=g_{n}g_{n-1}\cdots g_{1}$ with $g_{k}\in
P_{k}$.Since $g(x_{k})=g_{k}(x_{k})$, we must have $\beta_{g_{k}}\leq l$,
i.e., $g_{k}\in P_{k}(G_{l})$. Hence $G_{l}$ is a Nakajima group.

Now we describe a filtration as in Proposition 5.4 such that each $G_{i}$ in
the filtration is a Nakajima group. Note that classes of the elements of the
set $\bigcup_{k=l+1}^{n}P_{k}(G_{l+1})\setminus P_{k}(G_{l})$ generate the
group $G_{l+1}/G_{l}$. Let $l+1<s\leq n$ be the smallest integer such that
$P_{s}(G_{l+1})\setminus P_{s}(G_{l})\neq\varnothing$. Set $G_{l,1}=\langle
G_{l},g\rangle$ for some $g\in P_{s}(G)$ with $\beta_{g}=l+1$. Inductively,
define a filtration

$G_{l}<G_{l,1}<\cdots<G_{l,l_{0}}<G_{l+1}$

such that if $G_{l,i}\cap\big{(}P_{k}(G_{l+1})\setminus
P_{k}(G_{l})\big{)}\neq\varnothing$, then $P_{k-1}(G_{l+1})\setminus
P_{k-1}(G_{l})\subset G_{l,i}$ for all $k>l+1$ and that each quotients in the
above filtration is a group of order $p$.

We claim that each $G_{l,i}$ in the above filtration is a Nakajima group. For
$g\in G_{l,i}$, write $g=g_{n}g_{n-1}\cdots g_{1}$ with $g_{k}\in P_{k}$. We
can also write $g=h_{1}\tau_{1}\cdots h_{r}\tau_{r}$ where $h_{j}\in G_{l}$
and
$\tau_{j}\in\big{\langle}G_{l,i}\cap\big{\\{}\bigcup_{k=l+1}^{k_{0}}P_{k}(G_{l+1})\setminus
P_{k}(G_{l})\big{\\}}\big{\rangle}$. Suppose $l+1\leq k_{0}\leq n$ is the
largest integer such that $G_{l,i}\cap\big{(}P_{k_{0}}(G_{l+1})\setminus
P_{k_{0}}(G_{l})\big{)}\neq\varnothing$. If $k_{0}<k_{1}\leq n$, then we have
$g(x_{k_{1}})=h_{1}h_{2}\cdots h_{r}(x_{k_{1}})$. Hence $\beta_{g_{k_{1}}}\leq
l$ which implies $g_{k}\in P_{k}(G_{l})$ for all $k>k_{0}$. Since
$\beta_{g_{k}}\leq l+1$, by definition of $G_{l,i}$, we have $g_{k}\in
P_{k}(G_{l,i})$ for all $k<k_{0}$. This forces $g_{k_{0}}\in G_{l,i}$. Hence
each $G_{l,i}$ is a Nakajima group.

## References

  * [Ben93] D. J. Benson.  Polynomial invariants of finite groups, volume 190 of London Mathematical Society Lecture Note Series.  Cambridge University Press, Cambridge, 1993.
  * [Bra22] A. Braun.  $p$-groups and the polynomial ring of invariants question.  Adv. Math., 397:Paper No. 108120, 29, 2022.
  * [Bro05] A. Broer.  The direct summand property in modular invariant theory.  Transform. Groups, 10(1):5–27, 2005.
  * [Bro06] A. Broer.  Hypersurfaces in modular invariant theory.  J. Algebra, 306(2):576–590, 2006.
  * [Bro10] A. Broer.  Invariant theory of abelian transvection groups.  Canad. Math. Bull., 53(3):404–411, 2010.
  * [CW11] H. E. A. E. Campbell and D. L. Wehlau.  Modular invariant theory, volume 139 of Encyclopaedia of Mathematical Sciences.  Springer-Verlag, Berlin, 2011.  Invariant Theory and Algebraic Transformation Groups, 8.
  * [DK15] H. Derksen and G. Kemper.  Computational invariant theory, volume 130 of Encyclopaedia of Mathematical Sciences.  Springer, Heidelberg, enlarged edition, 2015.  With two appendices by Vladimir L. Popov, and an addendum by Norbert A’Campo and Popov, Invariant Theory and Algebraic Transformation Groups, VIII.
  * [ES18] J. Elmer and M. Sezer.  Locally finite derivations and modular coinvariants.  Q. J. Math., 69(3):1053–1062, 2018.  arXiv:1605.06363.
  * [KM22] M. Kummini and M. Mondal.  On Hilbert ideals for a class of $p$-groups in characteristic $p$.  Proc. Amer. Math. Soc., 150(1):145–151, 2022.  arXiv:2105.10527 [math.AC].
  * [LS87] P. S. Landweber and R. E. Stong.  The depth of rings of invariants over finite fields.  In Number theory (New York, 1984–1985), volume 1240 of Lecture Notes in Math., pages 259–274. Springer, Berlin, 1987.
  * [SW99] R. J. Shank and D. L. Wehlau.  The transfer in modular invariant theory.  J. Pure Appl. Algebra, 142(1):63–77, 1999.