#  $p$-Groups and the Polynomial Ring of Invariants Question

Amiram Braun

###### Abstract.

Let $G\subset GL(V)$ be a finite $p$-group, where $p=\operatorname{char}F$ and
$\dim_{F}V$ is finite. Let $S(V)$ be the symmetric algebra of $V$, $S(V)^{G}$
the subring of $G$-invariants, and $V^{*}$ the dual space of $V$. The
following presents our solution to the above question.

######  Theorem A.

Suppose $\dim_{F}V=3$. Then $S(V)^{G}$ is a polynomial ring if and only if $G$
is generated by transvections.

######  Theorem B.

Suppose $\dim_{F}V\geq 4$. Then $S(V)^{G}$ is a polynomial ring if and only
if:

  1. (1)

$S(V)^{G_{U}}$ is a polynomial ring for each subspace $U\subset V^{*}$ with
$\dim_{F}U=2$, where $G_{U}=\\{g\in G|g(u)=u$, $\forall u\in U\\}$, and

  2. (2)

$S(V)^{G}$ is Cohen-Macaulay.

Alternatively, $(1)$ can be replaced by the equivalent condition:

  1. (3)

$\dim($ non-smooth locus of $S(V)^{G})$ $\leq 1$.

Dept. of Mathematics, University of Haifa, Haifa, Israel 31905

<EMAIL_ADDRESS>
##  1\. Introduction

Let $V$ be a finite dimensional vector space over a field $F$. Let $G\subset
GL(V)$ be a finite group and $S(V)$ the symmetric algebra of $V$. $G$ acts as
a finite group of automorphisms on $S(V)$. Let $S(V)^{G}=\\{x\in
S(V)|g(x)=x,\forall g\in G\\}$ be the ring of $G$-invariants. The
determination of the polynomial property of $S(V)^{G}$ is an old problem. The
basic result is the following. We refer to [1], [17] for an historical
account.

######  Theorem.

(Shephard-Todd-Chevalley-Serre). Suppose in addition that $(p,|G|)=1$, where
$p=\operatorname{char}F$. Then $S(V)^{G}$ is a polynomial ring if and only if
$G$ is generated by pseudo-reflections.

Serre also proved the following implications, where no restrictions on $p$ are
assumed.

######  Theorem.

[18], [3, V, §6, exercise 8]. Suppose $S(V)^{G}$ is a polynomial ring. Then
$G$ is generated by pseudo-reflections and $S(V)^{G_{U}}$ is a polynomial ring
for each subspace $U\subset V^{*}$, where $G_{U}=\\{g\in G|g(u)=u,\forall u\in
U\\}$.

The second implication was firstly proved by R. Steinberg [19] in case
$F=\mathbb{C}$. This theorem suggests a starting point in answering the
polynomiality question in the modular case, that is if $p$ divides $|G|$. The
following result by Nakajima was the first positive result.

######  Theorem.

[16]. Suppose $G\subset GL(V)$ is a finite $p$-group, where $\dim_{F}V=n$ is
finite, and $F=\mathbb{F}_{p}$ is the prime field. Then $S(V)^{G}$ is a
polynomial ring if and only if:
There is a basis of $V$ $\\{z_{1},...,z_{n}\\}$ with $g(z_{1})=z_{1}$,
$(g-1)(z_{i})\in Fz_{1}+\cdots+Fz_{i-1}$, $i=2,...,n$, for each $g\in G$ such
that:

$|G-\operatorname{orbit}(z_{1})|\cdots|G-\operatorname{orbit}(z_{n})|=|G|.$

Actually the ”if” part always implies the polynomiality of $S(V)^{G}$ with no
restriction on $F$. The necessity of the condition is only granted when
$F=\mathbb{F}_{p}$. As such this theorem has no relation to Serre’s result.

Extending and completing earlier results of Nakajima, Kemper-Malle
established,using classification theorems for finite irreducible groups, the
following result.

######  Theorem.

([12, Main Theorem] and the corrections in [5, p. 119]). Suppose $G\subset
GL(V)$ is a finite group, where $V$ is a finite dimensional irreducible
$G$-module. Then $S(V)^{G}$ is a polynomial ring if and only if:

  1. (1)

$S(V)^{G_{U}}$ is a polynomial ring for each subspace $U\subset V^{*}$ with
$\dim_{F}U\geq 1$, and

  2. (2)

$G$ is generated by pseudo-reflections.

Actually one merely needs assuming $\dim_{F}U=1$ in item (1). However, in
spite of a reduction to an indecomposable $V$, Kemper-Malle refrained from
stating the following conjecture.

######  Conjecture C.

Suppose $G\subset GL(V)$ is a finite group, where $V$ is finite dimensional.
Then $S(V)^{G}$ is a polynomial ring if and only if:
$\indent(C1)$ $S(V)^{G_{U}}$ is a polynomial ring for each subspace $U\subset
V^{*}$ with $\dim_{F}U=1$, and $\indent(C2)$ $G$ is generated by pseudo-
reflections.

Our result, in case of $\dim_{F}V\geq 4$, initiated from the trivial
observation that Conjecture C holds trivially for a $p$-group $G$. Indeed the
action of $G$ on $V^{*}$ can always be simultaneously triangulized, in
particular there exists $f\in V^{*}$, $f\neq 0$ with $g(f)=f$, for each $g\in
G$. Therefore $G_{U}=G$, where $U=Ff$ and Conjecture C trivially holds. This
suggests, in the spirit of Conjecture C, that for a finite $p$-group, there
may exist a weaker assumption on subspaces of $V^{*}$, e.g. on the set
$\\{U|U\subset V^{*}$, $\dim_{F}U=2\\}$. Indeed our main result is of this
nature:

######  Theorem B.

Suppose $G\subset GL(V)$ is a finite $p$-group, where
$p=\operatorname{char}F$, $\dim_{F}V\geq 4$. Then $S(V)^{G}$ is a polynomial
ring if and only if:

  1. (1)

$S(V)^{G_{U}}$ is a polynomial ring, for each subspace $U\subset V^{*}$, with
$\dim_{F}U=2$, where $G_{U}=\\{g\in G|g(u)=u$, $\forall u\in U\\}$, and

  2. (2)

$S(V)^{G}$ is Cohen-Macaulay.

Moreover (1) is equivalent to:
(3) $\dim($non-smooth locus of $S(V)^{G})$ $\leq 1$.

So Theorem B has a singularity theory flavour. Note that the assumption
$\dim_{F}V\geq 4$ is essential, as every finite $p$-subgroup of $GL_{3}(F)$
satisfies conditions B(1) and B(2).

One may be puzzled by the omission of the condition ”$G$ is generated by
pseudo-reflections”, in Theorem B. The simple answer is that conditions B(1)
and B(2) actually imply it. This is a consequence of a result of Kemper [11,
Corollary 3.7].

Another seemingly puzzling question is the appearance of the Cohen-Macaulay
assumption in Theorem B(2), as it neither appears in Kemper-Malle [12] result
nor in Conjecture C. In fact the Cohen-Macaulay property is a consequence of
condition (C1). Indeed condition (C1) implies $\dim$(singular locus of
$S(V)^{G})=0$, and hence $\dim$(non-Cohen-Macaulay locus of $S(V)^{G})=0$.
Therefore by Kemper [10, Thm. 3.1] $S(V)^{G}$ is Cohen-Macaulay.

In case of $\dim_{F}V=3$ we have the following.

######  Theorem A.

Suppose $G\subset GL(V)$ is a finite $p$-group, where $\dim_{F}V=3$ and
$p=\operatorname{char}F$. Then $S(V)^{G}$ is a polynomial ring if and only if
$G$ is generated by transvections.

The proof of Theorem A is computational. By a standard result $G$ has the
following presentations of matrices $G\subset\begin{bmatrix}1&F&F\\\ 0&1&F\\\
0&0&1\end{bmatrix}$, namely $V$ has a basis $\\{v_{1},v_{2},v_{3}\\}$ with
$g(v_{1})=v_{1}$, $(g-1)(v_{2})\in Fv_{1}$, $(g-1)(v_{3})\in Fv_{1}+Fv_{2}$,
for each $g\in G$. The simpler cases $G\subset\begin{bmatrix}1&F&F\\\ 0&1&0\\\
0&0&1\end{bmatrix}$ and $G\subset\begin{bmatrix}1&0&F\\\ 0&1&F\\\
0&0&1\end{bmatrix}$ are handled by earlier results of Nakajima. The remaining
case is more involved and the end result $S(V)^{G}=F[v_{1},a,b]$ where $a\in
F[v_{1},v_{2}]^{G}$, is crucial in the proof of Theorem B.

The following is a sketch of the proof of Theorem B, if $F$ is perfect
(Theorem 2.15). It is done by a sequence of hypersurface sections reducing the
problem, after $\dim_{F}V-3$ steps to a $3$-dimensional non-linear problem.
Theorem A cannot be used directly, but provides us with three elements which
are a system of parameters (s.o.p.). The homogenous lifts in $S(V)^{G}$ of
these three elements together with the homogenous lifts in $S(V)^{G}$ of the
elements involved in the definition of each hypersurface section, form a
homogenous s.o.p. with their degree’s product being equal to $|G|$. This
implies that $S(V)^{G}$ is generated by these $n$ lifts and consequently is a
polynomial ring. The general case requires some further considerations about
the non-smooth locus. A simplified proof in case $dim_{F}V=4$ is presented in
the appendix.

We end with two $\dim_{F}V=4$ examples. In the first one $|G|=p^{3}$, $G$ is
elementary abelian and $S(V)^{G}$ is not a polynomial ring. However $S(V)^{G}$
is Cohen-Macaulay, $G$ is generated by transvections, and $G=G_{U}$, $U\subset
V^{*}$, with $\dim_{F}U=2$. In the second example, we have, for the same group
$G$, that $S(V^{*})^{G}$ is a polynomial ring.

######  Convention.

Matrices represent actions on the right.

######  Warning.

Since $g\in GL(V)$ is a pseudo-reflection on $V$ if and only if $g$ is a
pseudo-reflection on $V^{*}$, we shall omit the reference to either $V$ or
$V^{*}$. Note however that the polynomiality of $S(V)^{G}$ or of
$S(V^{*})^{G}$ are not equivalent properties, e.g. [5, p. 120] or Examples
2.20, 2.21.

##  2\. Theorem B

######  Notation.

Let $\mathfrak{m}$ be a prime ideal in $S(V)$. Recall that
$D_{G}(\mathfrak{m}):=\\{g\in G|g(\mathfrak{m})=\mathfrak{m}\\}$ is the
decomposition subgroup of $\mathfrak{m}$ in $G$, $I_{G}(\mathfrak{m}):=\\{g\in
G|(g-1)(S(V))\subseteq\mathfrak{m}\\}$, the inertia subgroup of $\mathfrak{m}$
in $G$, $W:=\mathfrak{m}\cap V$ and $U:=W^{\perp}=\\{f\in V^{*}|f(W)=0\\}$,

######  Lemma 2.1.

Let $G\subset GL(V)$ be a finite group where $\dim_{F}V$ is finite. With the
above notation we have: $G_{U}=I_{G}(\mathfrak{m})$.

Proof: This follows from the observation that $G_{U}=\\{g\in
G|(g-1)(V)\subseteq W\\}$.$\Box$

For a local commutative Noetherian ring $(A,\mathfrak{m})$, $\hat{A}$ denotes
the completion of $A$ at $\mathfrak{m}$.

######  Theorem 2.2.

([6, Exposé V, Prop. 2.2, p. 111]). With the notations above and setting
$\mathfrak{n}:=\mathfrak{m}\cap S(V)^{I_{G}(\mathfrak{m})}$,
$\mathfrak{p}:=\mathfrak{m}\cap S(V)^{G}$, $\mathfrak{q}:=\mathfrak{m}\cap
S(V)^{D_{G}(\mathfrak{m})}$, we have:

  1. (1)

$\widehat{S(V)_{\mathfrak{p}}^{G}}\cong\widehat{S(V)_{\mathfrak{q}}^{D_{G}(\mathfrak{m})}}$;

  2. (2)

$S(V)_{\mathfrak{n}}^{I_{G}(\mathfrak{m})}$ is $\acute{e}$tale over
$S(V)_{\mathfrak{q}}^{D_{G}(\mathfrak{m})}$.

Consequently by [13, the corollary in p. 179 and Theorem 23.7]
$S(V)^{G}_{\mathfrak{p}}$ is regular if and only if
$S(V)_{\mathfrak{n}}^{I_{G}({\mathfrak{m}})}$ is regular.

######  Remark 2.3.

  1. (1)

There is a proof of Theorem 2.2(1), for geometric points, in [10, Prop. 1.1].
However, it relies on [10, Lemma 1.2] which is false in the stated generality.
This can be easily corrected in our setting, as was also observed by David
Mundelius.

  2. (2)

Theorem 2.2 is proved in [6] for any Noetherian ring $A$ with a finite
subgroup $G\subset Aut(A)$.

We have the following consequences, most of which also appear in [10]. A
further complement will be proved in Lemma 2.19.

######  Lemma 2.4.

Let $G\subset GL(V)$ be a finite group and $\dim_{F}V=n$ is finite. Consider
the following conditions:

  1. (1)

$S(V)^{G}$ satisfies Serre’s $R_{m}$ condition [13, §23, p. 183];

  2. (2)

$S(V)^{G_{U}}$ is a polynomial ring for each subspace $U\subset V^{*}$ with
$\dim_{F}U=n-m$;

  3. (3)

$S(V)^{G_{U}}$ is a polynomial ring for each subspace $U\subset V^{*}$ with
$\dim_{F}U\geq n-m$.

Then we have the following implications: $(2)\Leftrightarrow(3)$,
$(3)\Rightarrow(1)$. Moreover if $F=F^{p}$ then $(1)\Rightarrow(3)$.

Proof: Let $\mathfrak{p}$ be a prime ideal in $S(V)^{G}$ with
$\operatorname{height}(\mathfrak{p})=m$. Let $\mathfrak{m}$ be a prime ideal
in $S(V)$ with $\mathfrak{m}\cap S(V)^{G}=\mathfrak{p}$. By ”going up” we have
$\operatorname{height}(\mathfrak{m})=\operatorname{height}(\mathfrak{p})=m$.
Let $W=V\cap\mathfrak{m}$. Then $WS(V)\subseteq\mathfrak{m}$ is a prime ideal
in $S(V)$ and
$\dim_{F}W=\operatorname{height}(WS(V))\leq\operatorname{height}(\mathfrak{m})=m$.
Therefore $\dim_{F}U=\dim_{F}V-\dim_{F}W\geq n-m$, where $U:=W^{\perp}=\\{f\in
V^{*}|f(W)=0\\}$.

We now show the implication $(3)\Rightarrow(1)$. By assumption $S(V)^{G_{U}}$
is a polynomial ring. Therefore by Lemma 2.1, $S(V)^{I_{G}(\mathfrak{m})}$ is
a polynomial ring and consequently $S(V)_{\mathfrak{n}}^{I_{G}(\mathfrak{m})}$
is regular where $\mathfrak{n}:=\mathfrak{m}\cap I_{G}(\mathfrak{m})$. So by
Theorem 2.2 $S(V)_{\mathfrak{p}}^{G}$ is regular.

We show that $(2)\Leftrightarrow(3)$. It is clear that $(3)\Rightarrow(2)$.
Assume $(2)$. Let $U^{\prime}\subseteq V^{*}$ be a subspace with
$\dim_{F}U^{\prime}\geq n-m$. Let $U\subseteq U^{\prime}$ be a subspace with
$\dim_{F}U=n-m$. Therefore by assumption $S(V)^{H}$ is a polynomial ring,
where $H:=G_{U}$. Consequently by Serre [2, V, $\S$6, Ex.8]
$S(V)^{H_{U^{\prime}}}$ is a polynomial ring. Let $g\in G_{U^{\prime}}$. Then
$g\in G_{U}=H$. Therefore $G_{U^{\prime}}=H_{U^{\prime}}$. Hence
$S(V)^{G_{U^{\prime}}}$ is a polynomial ring, as needed.

Assume now that $F=F^{p}$ and $(1)$ holds. Let $U\subseteq V^{*}$ be a
subspace with $\dim_{F}U\geq n-m$. Set $W:=U^{\perp}=\\{v\in V|f(v)=0,\forall
f\in U\\}$. Then $\dim_{F}W=\dim_{F}V-\dim_{F}U\leq n-(n-m)=m$. Hence $WS(V)$
is a prime ideal with $\operatorname{height}(WS(V))=\dim_{F}W\leq m$. Set
$\mathfrak{p}=WS(V)\cap S(V)^{G}$. Then by ”going down”
$\operatorname{height}(\mathfrak{p})\leq m$. Consequently by $(1)$
$S(V))_{\mathfrak{p}}^{G}$ is regular. Therefore by [10, Prop. 1.9, Example
1.10 (where $F=F^{p}$ is mistakenly missing)],
$S(V)^{I_{G}(WS(V))}=S(V)^{G_{U}}$ is a polynomial ring.$\Box$

We shall need the following observation.

######  Lemma 2.5.

Let $G\subset GL(V)$ be a finite $p$-group where $n:=\dim_{F}V$ is finite and
$p=\operatorname{char}F$. Suppose:

  1. (1)

$S(V)^{G}$ is Cohen-Macaulay;

  2. (2)

$S(V)^{G_{U}}$ is a polynomial ring for each subspace $U\subset V^{*}$ with
$\dim_{F}U=n-2$.

Then $G$ is generated by transvections.

Proof: By $(1)$ and [11, Cor. 3.7] we have that $G$ is generated by bi-
reflections on $V^{*}$, $G=<\sigma_{1},...,\sigma_{r}>$. Since
$\ker(\sigma_{i}-1):=U_{i}$, $U_{i}\subset V^{*}$ with $\dim_{F}U_{i}=n-2$,
$i=1,...,r$, it follows by $(2)$ that $\sigma_{i}\in G_{U_{i}}$ where
$S(V)^{G_{U_{i}}}$ is a polynomial ring. Therefore $G_{U_{i}}$ is generated by
transvections for $i=1,...,r$, implying that the combined set of transvection
generators of $G_{U_{i}}$, $i=1,...,r$, will generate $G$.$\Box$

The following is easy to verify.

######  Lemma 2.6.

Let $\sigma$ be a transvection on $V$ (equivalently on $V^{*}$) and $v\in
V^{\sigma}$. Then $\bar{\sigma}$, the induced map on $\bar{V}=V/Fv$ is a
transvection (or the identity) on $\bar{V}$ (equivalently on $\bar{V}^{*}$).

Proof: Clearly $\widebar{\ker(\sigma-1)}\subseteq\ker(\bar{\sigma}-1)$, so the
result follows since
$\dim_{F}\widebar{\ker(\sigma-1)}=\dim_{F}\widebar{V}-1$.$\Box$

The next result is a slight extension of Nakajima’s [15, Prop. 3.5], [17, Thm.
6.2.7] where the restriction $F=\mathbb{F}_{q}$ appears, but it is not used.

Let $A=\oplus_{n\geq 0}A_{n}$, $A_{0}=F$, be a graded ring. We denote by
$A_{+}:=\oplus_{n>0}A_{n}$, the irrelevant maximal ideal of $A$.

######  Proposition 2.7.

Let $\\{g_{1},...,g_{k}\\}\subset GL(V)$, where $\dim_{F}V$ is finite and
$\operatorname{char}F>0$. Let $\\{v_{1},...,v_{r}\\}$ be a $F$-linearly
independent subset of $V$ and $x\in S(V)-F[v_{1},...,v_{r}]$, a homogenous
element with $\deg x=p^{e}$. Suppose $(g_{j}-1)(x)\in F[v_{1},...,v_{r}]_{+}$,
$g_{j}(v_{i})=v_{i}$, $j=1,...,k$, $i=1,...,r$. Then:

  1. (1)

$<g_{1},...,g_{k}>$ induces an elementary abelian finite $p$-group of graded
automorphisms on the polynomial ring $F[v_{1},...,v_{r},x]$.

  2. (2)

$F[v_{1},...,v_{r},x]^{<g_{1},...,g_{k}>}=F[v_{1},...,v_{r},y]$ is a
polynomial ring, $\deg y=p^{t}=\\\ p^{e}\cdot|<g_{1},...,g_{k}>|$, and $y$ is
a monic $p$-polynomial in $x$ with coefficients in $F[v_{1},...,v_{r}]_{+}$.

Proof: Item (1) clearly holds. We shall establish item (2) by induction on
$k$. Set $y_{1}:=x^{p}-[(g_{1}-1)(x)]^{p-1}x$ so $\deg y_{1}=p^{e+1}$, since
$(g_{1}-1)(x)\in F[v_{1},...,v_{r}]_{+}$, it follows that
$(g_{1}-1)(y_{1})=0$, as well as
$(g_{j}-1)(y_{1})=[(g_{j}-1)(x)]^{p}-[(g_{1}-1)(x)]^{p-1}(g_{j}-1)(x)\in
F[v_{1},...,v_{r}]_{+}$. We claim that
$F[v_{1},...,v_{r},x]^{<g_{1}>}=F[v_{1},...,v_{r},y_{1}]$.

By Galois theory and since
$Q(F[v_{1},...,v_{r},x])^{<g_{1}>}=Q(F[v_{1},...,v_{r},x]^{<g_{1}>})$, we have
$[Q(F[v_{1},...,v_{r},x]):Q(F[v_{1},..,v_{r},x]^{<g_{1}>})]=p$. Since
$\\{1,x,...,x^{p-1}\\}$ generate $F[v_{1},...,v_{r},x]$ as a
$F[v_{1},...,v_{r},y_{1}]$-module, it follows that $\\{1,x,...,x^{p-1}\\}$ is
a free basis of $F[v_{1},...,v_{r},x]$ over $F[v_{1},...,v_{r},y_{1}]$.

Let $z\in F[v_{1},...,v_{r},x]^{<g_{1}>}$. Then
$z=c_{1}+c_{2}x+\cdots+c_{p-1}x^{p-1}$, $c_{i}\in F[v_{1},...,v_{r},y_{1}]$,
$i=1,,,.p-1$. Let $s=\max\\{i|c_{i}\neq 0\\}$. Suppose $s\geq 2$.

Recall that if $\phi$ is an automorphism of a commutative ring $A$ and
$\delta:=\phi-1$, then for $a\in A$ and $i\geq 1$ we have:

$\delta(a^{i})=ia^{i-1}\delta(a)+\sum_{j=0}^{i-2}\binom{i}{j}a^{j}\delta(a)^{i-j}.$

Taking $\phi=g_{1}$, we get that $0=\delta(z)=sc_{s}x^{s-1}\delta(x)+b$, where
$b\in\sum_{j=0}^{s-2}F[v_{1},...,v_{r},y_{1}]x^{j}$. Since $\delta(x)\neq 0$
(otherwise $g_{1}=1$), it follows from the freeness of $\\{1,x,...,x^{p-1}\\}$
and $\delta(x)\in F[v_{1},...,v_{r}]_{+}$ that $c_{s}=0$. Hence $z\in
F[v_{1},...,v_{r},y_{1}]$ and the case $k=1$ is established. Observe that
$K.\dim F[v_{1},...,v_{r},y_{1}]=K.\dim F[v_{1},...,v_{r},x]^{<g_{1}>}=K.\dim
F[v_{1},...,v_{r},x]=r+1$, implying that $F[v_{1},...,v_{r},y_{1}]$ is a
polynomial ring in $r+1$ variables.

Since $(g_{j}-1)(y_{1})\in F[v_{1},...,v_{r}]_{+}$, $j=2,...,k$, as seen
above, the general case follows now by induction, having by (1) the
$<g_{2},...g_{k}>$-stability of $F[v_{1},...,v_{r},v_{x}]^{<g_{1}>}$ and the
following:

$F[v_{1},...,v_{r},x]^{<g_{1},...,g_{k}>}=(F[v_{1},...,v_{r},x]^{<g_{1}>})^{<g_{2},...,g_{k}>}=F[v_{1},...,v_{r},y_{1}]^{<g_{2},...,g_{k}>}.\Box$

######  Remark 2.8.

[15, Prop. 3.5] is obtained by taking $\\{v_{1},...,v_{n-1},v_{n}\\}$ a basis
of $V$, $x=v_{n}$ and $r=n-1$.

The next result extends to a non-linear setting Nakajima’s [15, Prop. 3.4,
where $F=\mathbb{F}_{q}$ is unnecessarily assumed]. The proof is markedly
different.

######  Proposition 2.9.

Let $\\{g_{1},...,g_{k}\\}\subset GL(V)$, where $\dim_{F}V=n$ and
$\operatorname{char}F=p>0$. Let $\\{v,m_{2},...,m_{n}\\}\subset S(V)$ satisfy:

  1. (1)

$\\{v,m_{2},...,m_{n}\\}$ are algebraically independent over $F$;

  2. (2)

$v\in V$, $g_{j}(v)=v$, $j=1,...,k$;

  3. (3)

$m_{i}$ is homogenous, $\deg m_{i}=p^{e_{i}}$ and $(g_{j}-1)(m_{i})\in
Fv^{p^{e_{i}}}$, $j=1,...,k$, $i=2,...,n$. Then:
(a) $<g_{1},...,g_{k}>$ induces an elementary abelian finite $p$-group of
graded automorphisms on the polynomial ring $F[v,m_{2},...,m_{n}]$ and,
(b) $F[v,m_{2},...,m_{n}]^{<g_{1},...g_{k}>}=F[v,y_{2},...,y_{n}]$, is a
polynomial ring where $y_{i}$ is homogenous, $i=2,...,n$.

Moreover suppose:

  1. (4)

$F=F^{p}$ and,

  2. (5)

$m_{i}-v_{i}^{p^{e_{i}}}\in vS(V)$ where $v_{i}\in V$, $i=2,...,n$.
Then:
(c) $y_{i}-u_{i}^{p^{f_{i}}}\in vS(V)$, where $u_{i}\in V$, $\deg
y_{i}=p^{f_{i}}$, $i=2,...,n$

Proof: $<g_{1},...g_{k}>$ extends to a group of graded automorphisms on $S(V)$
in the obvious way. We have by (3) that
$(g_{j}-1)(m_{i})=a_{ji}v^{p^{e_{i}}}$, $a_{ji}\in F$, $i=2,...,n$,
$j=1,...,k$. This implies that
$(g_{j}-1)^{2}(m_{i})=(g_{j}-1)(a_{ji}v^{p^{e_{i}}})=0$, and
$(g_{j}-1)^{p}(m_{i})=0$. Hence $g_{j|F[v,m_{1},...,m_{n}]}^{p}=1$ for
$j=1,...,k$. Also
$(g_{j}g_{l})(m_{i})=g_{j}(m_{i}+a_{li}v^{p^{e_{i}}})=m_{i}+a_{ji}v^{p^{e_{i}}}+a_{li}v^{p^{e_{i}}}=(g_{l}g_{j})(m_{i}).$
Hence $<g_{1},...,g_{k}>$ induces an elementary abelian finite $p$-group of
graded automorphism on $F[v,m_{1},...,m_{n}]$, and (a) is verified.

Let $m_{r}$ be such that $g_{1}(m_{r})\neq m_{r}$. Equivalently $a_{1r}\neq
0$. We have by Proposition 2.7 that $F[v,m_{r}]^{<g_{1}>}=F[v,y]$, where
$y:=m_{r}^{p}-(a_{1r}v^{p^{e_{r}}})^{p-1}m_{r}$.

We shall now handle the case $k=1$. We may assume that $e_{2}\leq
e_{3}\leq\cdots\leq e_{n}$. Suppose $r$ is the $1^{st}$ index with
$(g_{1}-1)(m_{r})\neq 0$, that is $a_{1r}\neq 0$. For $j\neq r$,
$j\in\\{2,...,n\\}$ set
$m^{\prime}_{j}:=m_{j}-\frac{a_{1j}}{a_{1r}}v^{p^{e_{j}-e_{r}}}m_{r}\in
F[v,m_{2},...,m_{r}]$. Clearly $m_{j}^{\prime}=m_{j}$ if $j<r$. Hence
$(g_{1}-1)(m_{j}^{\prime})=0$, $\forall j\neq r$. Clearly
$F[v,m_{2},...,m_{r}]=F[v,m_{r},m_{j}^{\prime},j\neq r]$. Consequently
$\\{v,m_{r},m_{2}^{\prime},...,m_{r-1}^{\prime},m_{r+1}^{\prime},...,m_{n}\\}$
are algebraically independent over $F$.
Therefore $F[v,m_{2},...,m_{n}]^{<g_{1}>}=F[v,m_{r},m_{j}^{\prime},j\neq
r]^{<g_{1}>}=F[v,m_{r}]^{<g_{1}>}[m_{j}^{\prime},j\neq
r]=F[v,y,m_{2}^{\prime},...,m^{\prime}_{r-1},m^{\prime}_{r+1},...,m^{\prime}_{n}]$.
Since $n=K.\dim F[v,m_{2},...,m_{n}]=K.\dim F[v,m_{2},...,m_{n}]^{<g_{1}>}$,
it follows that the latter is a polynomial ring, and the case $k=1$ is
established.

We shall now settle the general case by induction on $k$. We have by (a) that
$F[v,m_{2},...,m_{n}]^{<g_{1}>}$ is $<g_{2},...,g_{k}>$-stable, so it follows
that:

$\begin{split}F[v,m_{2},...,m_{n}]^{<g_{1},...,g_{k}>}&=(F[v,m_{2},...,m_{n}]^{<g_{1}>})^{<g_{2},...,g_{k}>}\\\
&=F[v,y,m_{2}^{\prime},....,m^{\prime}_{r-1},m^{\prime}_{r+1},...,m_{n}]^{<g_{2},...g_{k}>}.\end{split}$

It is also evident from
$m_{j}^{\prime}=m_{j}-\frac{a_{1j}}{a_{1r}}v^{p^{e_{j}-e_{r}}}m_{r}$, that
$\\{g_{2},...,g_{k}\\}$ satisfy $(1)$, $(2)$ ,$(3)$ on
$\\{v,y,m_{2}^{\prime},...,m^{\prime}_{r-1},m^{\prime}_{r+1},...,m^{\prime}_{n}\\}$,
so by induction $F[v,y,m_{j}^{\prime},j\neq r]^{<g_{2},...,g_{n}>}$ is a
polynomial ring.

We now need to verify item $(c)$. For this we need to show the existence of
$\\{u,u_{j},j\neq r\\}\subset V$ with $y-u^{\deg y}\in vS(V)$,
$m^{\prime}_{j}-u_{j}^{\deg m_{j}^{\prime}}\in vS(V)$, $j\neq r$. Now if
$e_{j}>e_{r}$ take $u_{j}:=v_{j}$ and
$m_{j}^{\prime}-u_{j}^{p^{e_{j}}}=(m_{j}-v_{j}^{p^{e_{j}}})-\frac{a_{1j}}{a_{1r}}v^{p^{e_{j}-e_{r}}}m_{r}\in
vS(V)$.

If $e_{j}=e_{r}$, then $m_{j}^{\prime}=m_{j}-\frac{a_{1j}}{a_{1r}}m_{r}$ is of
degree $p^{e_{r}}$, so set
$u_{j}:=v_{j}-(\frac{a_{1j}}{a_{1r}})^{\frac{1}{p^{e^{r}}}}v_{r}$. Then
$m_{j}^{\prime}-u_{j}^{p^{e_{r}}}=(m_{j}-v_{j}^{p^{e_{r}}})-\frac{a_{1j}}{a_{1r}}(m_{r}-v_{r}^{p^{e_{r}}})\in
vS(V)$. Now $u:=v-a_{1r}^{\frac{p-1}{p^{e_{r}+1}}}v\in V$ satisfies:
$y-u^{p^{e_{r}+1}}=[m_{r}^{p}-(a_{1r}v^{p^{e_{r}}})^{p-1}m_{r}]-[v^{p^{e_{r}+1}}-a_{1r}^{p-1}v^{p^{e_{r}+1}}]=(m_{r}-v^{p^{e_{r}}})^{p}-(a_{1r}v^{p^{e_{r}}})^{p-1}(m_{r}-v^{p^{e_{r}}})\in
vS(V)$. This settles $(c)$ for $k=1$. The rest follows by induction as
above.$\Box$

The next result is also an extension of Nakajima’s [15, Prop. 3.4].

######  Corollary 2.10.

Let $<g_{1},...,g_{k}>\subset GL(V)$, where $\dim_{F}V=n$,
$\operatorname{char}F=p>0$ and $F=F^{p}$. Let $\\{v,v_{2},...,v_{n}\\}$ be a
basis of $V$ and assume that $g_{j}(v)=v$, $(g_{j}-1)(v_{i})\in Fv$,
$j=1,...,k$, $i=2,...,n$. Let $0\leq e_{1}\leq e_{2}\leq\cdots\leq e_{k}$ be
an increasing sequence of integers. Then:

  1. (1)

$<g_{1},...,g_{k}>\subseteq
Aut_{\operatorname{gr}}(F[v^{p^{e_{1}}},v_{2}^{p^{e_{2}}},...,v_{n}^{p^{e_{n}}}])\cap
Aut_{\operatorname{gr}}(F[v,v_{2}^{p^{e_{2}-e_{1}}},...,v_{n}^{p^{e_{n}-e_{1}}}])$;

  2. (2)

$F[v^{p^{e_{1}}},v_{2}^{p^{e_{2}}},...,v_{n}^{p^{e_{n}}}]^{<g_{1},...,g_{k}>}=F[v^{p^{e_{1}}},z_{2},...,z_{n}]$,
and is a polynomial ring in $n$-variables, where $z_{i}$ is homogenous of
degree $p^{h_{i}}$, $i=2,...,n$;

  3. (3)

$z_{i}-u_{i}^{p^{h_{i}}}\in v^{p^{e_{1}}}S(V)$, $i=2,...,n$ and
$\\{v,u_{2},...,u_{n}\\}$ is a basis of $V$;

  4. (4)

$e_{1}\leq h_{2}\leq\cdots\leq h_{n}$.

Proof: $<g_{1},...,g_{k}>\subset AutS(V)$ and the above assumptions grant that
$F[v,v_{2}^{p^{e_{2}-e_{1}}},...,v_{n}^{p^{e_{n}-e_{1}}}]$ and
$F[v^{p^{e_{1}}},v_{2}^{p^{e_{2}}},...,v_{n}^{p^{e_{n}}}]$ are
$<g_{1},...,g_{k}>$-stable. Hence $(1)$ holds.

We next observe that the map $\phi(f)=f^{p^{e_{1}}}$,
$\phi:F[v,v_{2}^{p^{e_{2}-e_{1}}},...,v_{n}^{p^{e_{n}-e_{1}}}]\rightarrow
F[v^{p^{e_{1}}},v_{2}^{p^{e_{2}}},...,v_{n}^{p^{e_{n}}}]$, is a ring
isomorphism (but not an algebra homomorphism since $F=F^{p}$ is being used).
It is also easily checked that $g_{i}\phi=\phi g_{i}$, $i=1,...,k$.
Consequently the restriction of $\phi$ induces the following isomorphism:

(2.1)
$\phi:F[v,v_{2}^{p^{e_{2}-e_{1}}},...,v_{n}^{p^{e_{n}-e_{1}}}]^{<g_{1},...,g_{k}>}\cong
F[v^{p^{e_{1}}},v_{2}^{p^{e_{2}}},...,v_{n}^{p^{e_{n}}}]^{<g_{1},...,g_{k}>}.$

Now by Proposition 2.9, taking $m_{i}=v_{i}^{p^{e_{i}-e_{1}}}$, $i=2,...,n$,
we have that
$F[v,v_{2}^{p^{e_{2}-e_{1}}},...,v_{n}^{p^{e_{n}-e_{1}}}]^{<g_{1},...,g_{k}>}=F[v,y_{2},...,y_{n}]$
is a polynomial ring, where $y_{i}$ is homogenous of degree $p^{f_{i}}$, and
$y_{i}-u_{i}^{p^{f_{i}}}\in vS(V)$, $i=2,...,n$. Consequently by applying
$\phi$ and using isomorphism (2.1) we have that
$F[v^{p^{e_{1}}},v_{2}^{p^{e_{2}}},...,v_{n}^{p^{e_{n}}}]^{<g_{1},...,g_{k}>}=F[v^{p^{e_{1}}},y_{2}^{p^{e_{1}}},...,y_{n}^{p^{e_{1}}}]$,
so (2), (4) and part of (3) follow by setting $z_{i}:=y_{i}^{p^{e_{1}}}$,
$z_{i}-u_{i}^{p^{f_{i}+e_{1}}}\in v^{p^{e_{1}}}S(V)$ and $h_{i}:=f_{i}+e_{1}$,
$i=2,...,n$.

We shall now establish the basis property of $\\{v,u_{2},...,u_{n}\\}$.
$v^{p^{e_{1}}}$ is a prime element in
$F[v^{p^{e_{1}}},v_{2}^{p^{e_{2}}},...,v_{n}^{p^{e_{n}}}]$. Since
$S(V)=F[v,v_{2},...,v_{n}]$ is finite over the former, it follows that
$vS(V)\cap
F[v^{p^{e_{1}}},v_{2}^{p^{e_{2}}},...,v_{n}^{p^{e_{n}}}]=v^{p^{e_{1}}}F[v^{p^{e_{1}}},v_{2}^{p^{e_{2}}},...,v_{n}^{p^{e_{n}}}]$.
Since
$F[v^{p^{e_{1}}},v_{2}^{p^{e_{2}}},...,v_{n}^{p^{e_{n}}}]^{<g_{1},...,g_{k}>}=F[v^{p^{e_{1}}},z_{2},...,z_{n}]$
it follows that
$v^{p^{e_{1}}}F[v^{p^{e_{1}}},v_{2}^{p^{e_{2}}},...,v_{n}^{p^{e_{n}}}]\cap
F[v^{p^{e_{1}}},z_{2},...,z_{n}]=v^{p^{e_{1}}}F[v^{p^{e_{1}}},z_{2},...,z_{n}]$.
Consequently $vS(V)\cap
F[v^{p^{e_{1}}},z_{2},...,z_{n}]=v^{p^{e_{1}}}F[v^{p^{e_{1}}},z_{2},...,z_{n}]:=(v^{p^{e_{1}}})$.
Note that $F[v^{p^{e_{1}}},z_{2},...,z_{n}]/(v^{p^{e_{1}}})$ is a polynomial
ring in $n-1$ variables. Therefore:

$\begin{split}S(\bar{V})=S(V)/vS(V)&\supset
F[v^{p^{e_{1}}},z_{2},...,z_{n}]/(v^{p^{e_{1}}})=\\\
&=F[\bar{z_{2}},...,\bar{z_{n}}]=F[\bar{u_{2}}^{p^{h_{2}}},...,\bar{u_{n}}^{p^{h_{2}}}],\end{split}$

where $\bar{V}:=V/Fv$ and $\bar{z_{i}}$ (respectively $\bar{u_{i}}$) is the
image of $z_{i}$ (respectively $u_{i}$) in $S(\bar{V})$ (respectively
$\bar{V}$), $i=2,...,n$. If $\\{\bar{u}_{2},...,\bar{u}_{n}\\}$ is linearly
dependent over $F$, it violates the polynomial ring property of
$F[\bar{z_{2}},...,\bar{z_{n}}]$. It is now also clear that
$\\{v,u_{2},...,u_{n}\\}$ is a basis of $V$ over $F$.$\Box$

It is a well-known classical result that for a a finite $p$-group $G\subset
GL(V)$, with finite $\dim_{F}V$ and $\operatorname{char}F=p>0$, that there
exists a basis of $V$, putting $G$ simultaneously into a triangular form. The
next result is an extension of this in the non-linear setting. The generators
of the given polynomial ring are changed in a way reflecting the triangulating
action of $G$ on $V$.

######  Proposition 2.11.

Let $G\subset GL(V)$ be a finite $p$-group, where $n:=\dim_{F}V$ is finite,
$p=\operatorname{char}F>0$ and $F=F^{p}$. Let
$A=F[v^{p^{e_{1}}},x_{2},...,x_{n}]$ be a graded polynomial subring of $S(V)$
(in $n$-variables) with $G\subset Aut_{\operatorname{gr}}(A)$. Assume:

  1. (1)

$v\in A^{G}\cap V$;

  2. (2)

$x_{i}$ is homogenous, $\deg x_{i}=p^{e_{i}}$, $x_{i}-v_{i}^{p^{e_{i}}}\in
v^{p^{e_{1}}}S(V)$, where $v_{i}\in V$, $i=2,...,n$, and
$\\{v,v_{2},...,v_{n}\\}$ is a basis of $V$;

  3. (3)

$e_{1}\leq e_{2}\leq\cdots\leq e_{n}$.

Then there exist homogenous elements $y_{2},...,y_{n}\in A$, $\deg
y_{i}=p^{e_{i}}$, $y_{i}-u_{i}^{p^{e_{i}}}\in v^{p^{e_{1}}}S(V)$, where
$u_{i}\in V$, $i=2,...,n$, having the following properties:

  1. (4)

$(g-1)(y_{2})\in F[v^{p^{e_{1}}}]_{+}$, $(g-1)(y_{i})\in
F[v^{p^{e_{1}}},y_{2},...,y_{i-1}]_{+}$, $i>2$, for each $g\in G$;

  2. (5)

$A=F[v^{p^{e_{1}}},y_{2},...,y_{n}]$;

  3. (6)

$\\{v,u_{2},...,u_{n}\\}$ is a triangulating basis for the action of $G$ on
$V$;

Proof:
$\\{x_{2},...,x_{n}\\}=\cup_{r=1}^{l}\\{x_{i_{r-1}+1},...,x_{i_{r}}\\}$, is a
union of $l$ disjoint blocks, where $\deg x_{i}=\deg x_{i_{r}}=p^{e_{i_{r}}}$,
for $i_{r-1}+1\leq i\leq i_{r}$. Set $x_{i_{0}}=v^{p^{e_{1}}}$. Suppose $(4)$
holds for $i\leq i_{r-1}:=s$ as well as
$F[v^{p^{e_{1}}},x_{2},...,x_{s}]=F[v^{p^{e_{1}}},y_{2},...,y_{s}]$. We shall
now establish (4) for $i\leq t:=i_{r}$, as well as
$F[v^{p^{e_{1}}},x_{2},...,x_{t}]=F[v^{p^{e_{1}}},y_{2},...,y_{t}]$. We
therefore consider the variables $\\{x_{s+1},...,x_{t}\\}$, $\deg
x_{i}=p^{e}$, $e=e_{i}$, $i_{r-1}+1\leq i\leq i_{r}$. Since $g\in G$ preserves
degrees we have:

(2.2) $\begin{split}(g-1)(x_{i})&=\sum_{j=s+1}^{t}b_{ij}(g)x_{j}+x,\\\ &\text{
where }b_{ij}(g)\in F,\text{ }x\in F[v^{p^{e_{1}}},x_{2},...,x_{s}]_{+},\text{
for each }g\in G.\end{split}$

Similarly $(v^{p^{e_{1}}},x_{2},...,x_{s})$, the prime graded ideal generated
by $\\{v^{p^{e_{1}}},x_{2},...,x_{s}\\}$ in
$F[v^{p^{e_{1}}},x_{2},...,x_{t}]$, is $G$-stable. Consequently $G$ acts by
graded automorphisms on
$F[v^{p^{e_{1}}},x_{2},...,x_{t}]/(v^{p^{e_{l}}},x_{2},...,x_{s})=F[\bar{x}_{s+1},...,\bar{x}_{t}]$,
inducing by equality (2.2) a linear action on
$F\bar{x}_{s+1}+\cdots+F\bar{x}_{t}:=W$, a $(t-s)$-dimensional $F$-subspace of
the polynomial ring $F[\bar{x}_{s+1},...,\bar{x}_{t}]$. It implies, since $G$
is a $p$-group, the existence of a $W$-basis $\\{z_{s+1},...,z_{t}\\}$ which
triangulates the action of $G$ on $W$. Thus:

(2.3) $(g-1)(z_{s+1})=0,\text{
}(g-1)(z_{i})=\sum_{j=s+1}^{i-1}a_{ij}(g)z_{j},\text{ }a_{ij}(g)\in F,\text{
}s+2\leq i\leq t.$

Since $\\{\bar{x}_{s+1},...,\bar{x}_{t}\\}$ is a basis of $W$ it follows that
$z_{i}=\sum_{j=s+1}^{t}\lambda_{ij}\bar{x}_{j}$, $\lambda_{ij}\in F$,
$i=s+1,...,t$ and $\lambda:=(\lambda_{ij})$ is an invertible matrix. Set
$y_{i}:=\sum_{j=s+1}^{t}\lambda_{ij}x_{j}$, $i=s+1,...,t$. Let
$\lambda^{-1}=(\mu_{ij})$. Then $x_{i}=\sum_{j=s+1}^{t}\mu_{ij}y_{j}$,
$i=s+1,...,t$. Therefore
$F[v^{p^{e_{1}}},x_{2},...,x_{t}]=F[v^{p^{e_{1}}},x_{2},...,x_{s}][x_{s+1},...,x_{t}]=F[v^{p^{e_{1}}},y_{2},...,y_{s}][y_{s+1},....,y_{t}]=F[v^{p^{e_{1}}},y_{2},...,y_{t}]$.
Set $u_{i}:=\sum_{j=s+1}^{t}\lambda_{ij}^{\frac{1}{p^{e}}}v_{j}$,
$i=s+1,...,t$. Then
$y_{i}-u_{i}^{p^{e}}=\sum_{j=s+1}^{t}\lambda_{ij}x_{j}-\sum_{j=s+1}^{t}\lambda_{ij}v_{j}^{p^{e}}=\sum_{j=s+1}^{t}\lambda_{ij}(x_{j}-v_{j}^{p^{e}})\in
v^{p^{e_{1}}}S(V)$, $i=2,...,t$.

We now verify $(4)$,$(5)$ for $i\leq t$. Since $\bar{y}_{i}=z_{i}$, for
$i=s+1,...,t$, it follows from the equalities (2.3) that each member in
$B:=\\{(g-1)(y_{s+1}),(g-1)(y_{i})-\sum_{j=s+1}^{i-1}a_{ij}(g)y_{j}$, $s+2\leq
i\leq t$, $g\in G\\}$ is homogenous of degree $p^{e}$ (unless it is $0$), and
is in $(v^{p^{e_{1}}},x_{2},...,x_{s})$. This shows that each member in $B$,
considered as sum of monomials in $F[v^{p^{e_{1}}},x_{2},...,x_{t}]$, will not
have $x_{i}$, with $i\geq s+1$ in any of its monomials because $\deg
x_{i}=p^{e}$, for $i\geq s+1$ and each such monomial will contain at least one
of $\\{v^{p^{e_{1}}},x_{2},...,x_{s}\\}$. Therefore each member of $B$ is in
$F[v^{p^{e_{1}}},x_{2},...,x_{s}]_{+}=F[v^{p^{e_{1}}},y_{2},...,y_{s}]_{+}$.
This shows that $(4)$ holds for $i_{r-1}+1\leq i\leq i_{r}$, as well.
Iterating this by induction will establish $(4)$, $(5)$.

We aim now at verifying $(6)$. We shall firstly show that
$\\{v,u_{2},...,u_{n}\\}$ is a basis of $V$. $v^{p^{e_{1}}}$ is a prime
element in $A$, and hence $\mathfrak{q}=v^{p^{e_{1}}}A$ is a height one prime
ideal in $A$. Since $v^{p^{e_{1}}}S(V)\subseteq vS(V)$ it follows that
$vS(V)\cap A=\mathfrak{p}$ is a prime ideal in $A$ with
$\mathfrak{p}\supseteq\mathfrak{q}$. Since
$A/\mathfrak{p}=F[\bar{v}_{2}^{p^{e_{2}}},...,\bar{v}_{n}^{p^{e_{n}}}]$
(regarded as a subring of $S(V)/vS(V)$) it follows since
$\\{\bar{v}_{2},...,\bar{v}_{n}\\}$, is a basis of $V/Fv$, that $K.\dim
A/\mathfrak{p}=n-1$. Therefore $K.\dim A=n$ implies that
$\operatorname{height}\mathfrak{p}=1$ and $\mathfrak{q}=\mathfrak{p}$. Hence
$A/\mathfrak{p}=A/\mathfrak{q}=F[\bar{u}_{2}^{p^{e_{2}}},...,\bar{u}_{n}^{p^{e_{n}}}]$
is a polynomial ring in $(n-1)$ variables, implying that
$\\{\bar{u}_{2},...,\bar{u}_{n}\\}$ are linearly independent over $F$ and
therefore a basis of $V/Fv$. Consequently $\\{v,u_{2},...,u_{n}\\}$ is a basis
of $V$.

We finally establish the triangulating property of $\\{v,u_{2},...,u_{n}\\}$.
Recall that $y_{i}-u_{i}^{p^{e_{i}}}\in v^{p^{e_{1}}}S(V)$, $i=2,...,n$. By
$(4)$ $(g-1)(y_{2})\in F[v]_{+}$, for each $g\in G$. Hence
$(g-1)(y_{2})=\alpha v^{p^{e_{2}}}$, $\alpha\in F$. Since
$y_{2}-u_{2}^{p^{e_{2}}}\in v^{p^{e_{1}}}S(V)$ it follows that
$(g-1)(u_{2})^{p^{e_{2}}}\in vS(V)$, a prime ideal in $S(V)$. Hence
$(g-1)(u_{2})\in vS(V)\cap V=Fv$, for each $g\in G$.

For $i>2$ we have by $(4)$ that $(g-1)(y_{i})\in F[v,y_{2},...,y_{i-1}]_{+}$
Hence $(g-1)(u_{i})^{p^{e_{i}}}\in(v,u_{2},...,u_{i-1})$ (meaning now the
ideal in $S(V)$ generated by $\\{v,u_{2},...,u_{i-1}\\}$). Since the latter is
a prime ideal then $(g-1)(u_{i})\in(v,u_{2},...,u_{i-1})\cap
V=Fv+Fu_{2}+\cdots+Fu_{i-1}$, for each $g\in F$.$\Box$

We shall need the following old result of Nagata.

######  Proposition 2.12.

[14, p. 143]. Let $R$ be a commutative Noetherian regular local ring and
$\mathfrak{m}$ its unique maximal ideal. Let $\mathfrak{p}$ be a prime ideal
in $R$. Then $\mathfrak{p}^{(e)}\subseteq\mathfrak{m}^{e}$, for every
$e\in\mathbb{N}$, where
$\mathfrak{p}^{(e)}:=\mathfrak{p}_{\mathfrak{p}}^{e}\cap R$ is the $e$-th
symbolic power of $\mathfrak{p}$.

The next result provides an hypersurface section reduction which will be used
repeatedly in Theorem 2.15.

######  Proposition 2.13.

Let $A=\oplus_{n\geq 0}A_{n}$ be an $\mathbb{N}$-graded polynomial ring over a
field $F=A_{0}$ and $G\subset Aut_{\operatorname{gr}}(A)$ a finite subgroup of
graded automorphisms. Let $x\in A^{G}$ be a homogenous element of $A$. Set
$H:=\ker(G\rightarrow G|_{A/xA})$, $\mathfrak{m}:=A_{+}$, the irrelevant
maximal ideal of $A$ and $\mathfrak{n}:=A_{+}^{G}$. Suppose:

  1. (1)

$x\in\mathfrak{m}-\mathfrak{m}^{2}$;

  2. (2)

$G/H$ acts faithfully on $A^{H}/xA^{H}$;

  3. (3)

$A_{\mathfrak{n}}^{G}$ satisfies Serre’s $S_{m+1}$ and $R_{m}$ conditions with
$m\geq 2$.

Then $A^{G}/(x)=(A^{H}/xA^{H})^{G/H}$, and $[A^{G}/(x)]_{\mathfrak{n}/(x)}$
satisfies $S_{m}$ and $R_{m-1}$.

Proof: By $(1)$ and [7, Theorem 161]
$A_{\mathfrak{m}}/xA_{\mathfrak{m}}=[A/xA]_{\bar{\mathfrak{m}}}$ is a regular
local ring, where $\bar{\mathfrak{m}}=\mathfrak{m}/(x)$. Hence it is a domain
implying the same for $A/xA$, so $x$ is a prime element in $A$. Since $xA\cap
A^{H}=xA^{H}$ and $xA\cap A^{G}=xA^{G}:=(x)$, it follows that $x$ is also a
prime element in $A^{H}$ and $A^{G}$.

We next show that $[A^{G}/(x)]_{\bar{\mathfrak{n}}}$ satisfies $R_{m-1}$ where
$\bar{\mathfrak{n}}=\mathfrak{n}/(x)$. Let $\mathfrak{p}\subset\mathfrak{n}$
be a prime ideal in $A^{G}$ with $x\in\mathfrak{p}$,
$\operatorname{height}(\mathfrak{p})=m$. Hence $A^{G}_{\mathfrak{p}}$ is
regular. Let $\\{\mathfrak{p}_{1},...,\mathfrak{p}_{k}\\}$ be the set of prime
ideals in $A$ lying over $\mathfrak{p}$. Since $\mathfrak{m}$ is the unique
maximal ideal in $A$ lying over $\mathfrak{n}$, it follows by ”going up” that
$\mathfrak{p}_{i}\subset\mathfrak{m}$, $i=1,...,k$. We have:
$\mathfrak{p}_{i}^{(2)}=\mathfrak{p}_{i}^{2}A_{\mathfrak{p}_{i}}\cap
A=\mathfrak{p}_{i\mathfrak{m}}^{2}(A_{\mathfrak{m}})_{\mathfrak{p}_{im}}\cap
A_{\mathfrak{m}}\cap A=\mathfrak{p}_{i\mathfrak{m}}^{(2)}\cap
A\subset\mathfrak{m}_{\mathfrak{m}}^{2}\cap A=\mathfrak{m}^{2}$, where the
inclusion is due to Proposition 2.12 and the last equality is by the
maximality of $\mathfrak{m}$. Consequently by $(1)$
$x\notin\mathfrak{p}_{i}^{(2)}$. Since
$\mathfrak{p}^{(2)}=\mathfrak{p}^{2}A_{\mathfrak{p}}^{G}\cap
A^{G}\subset\mathfrak{p}_{i}^{2}A_{\mathfrak{p}_{i}}\cap
A=\mathfrak{p}_{i}^{(2)}$, it follows that $x\notin\mathfrak{p}^{(2)}$. Hence
$x\in\mathfrak{p}_{\mathfrak{p}}-\mathfrak{p}_{\mathfrak{p}}^{2}$ and
therefore, by the regularity at $A_{\mathfrak{p}}^{G}$ and by [7, Thm. 161],
we have that
$A_{\mathfrak{p}}^{G}/xA_{\mathfrak{p}}^{G}=[A^{G}/(x)]_{\bar{\mathfrak{p}}}$
is regular where $\bar{\mathfrak{p}}=\mathfrak{p}/(x)$. Moreover
$K.\dim[A^{G}/(x)]_{\bar{\mathfrak{p}}}=m-1$. So
$[A^{G}/(x)]_{\bar{\mathfrak{n}}}$ satisfies $R_{m-1}$.

Similarly suppose $\bar{\mathfrak{p}}=\mathfrak{p}/(x)$ is a prime ideal in
$A^{G}/(x)$ with $\bar{\mathfrak{p}}\subset\bar{\mathfrak{n}}$. Since
$A_{\mathfrak{n}}^{G}$ satisfies $S_{m+1}$ it follows that
$\operatorname{depth}\mathfrak{p}_{\mathfrak{p}}=\operatorname{depth}(\mathfrak{p}_{\mathfrak{n}})_{\mathfrak{p}_{\mathfrak{n}}}\geq
min\\{m+1,\operatorname{height}\mathfrak{p}_{\mathfrak{n}}\\}=min\\{m+1,\operatorname{height}\mathfrak{p}\\}$.
Since $(x)\subset\mathfrak{p}$, $x$ can be completed to a regular sequence
inside $\mathfrak{p}_{\mathfrak{p}}$, implying that
$\operatorname{depth}\bar{\mathfrak{p}}_{\bar{\mathfrak{p}}}=\operatorname{depth}\mathfrak{p}_{\mathfrak{p}}-1$.
Since
$\operatorname{height}\bar{\mathfrak{p}}=\operatorname{height}\mathfrak{p}-1$,
the inequality
$\operatorname{depth}\bar{\mathfrak{p}}_{\bar{\mathfrak{p}}}\geq
min\\{m,\operatorname{height}\bar{\mathfrak{p}}\\}$ clearly holds and the
$S_{m}$ property of $(A^{G}/(x))_{\bar{\mathfrak{n}}}$ is established.

Consequently, since $m\geq 2$, $[A^{G}/(x)]_{\bar{\mathfrak{n}}}$ satisfies
$S_{2}$ and $R_{1}$, implying that it is normal.

Now by $(2)$ and Galois theory we have that
$[Q(A^{H}/xA^{H}):Q(A^{H}/xA^{H})^{G/H}]=|G/H|$. Since $G$ is finite,
$Q(A^{H}/xA^{H})^{G/H}=Q((A^{H}/xA^{H})^{G/H})$ and we therefore have:

(2.4) $[Q(A^{H}/xA^{H}):Q((A^{H}/xA^{H})^{G/H}]=|G/H|.$

Similarly $(A^{H})^{G/H}=A^{G}$, $(Q(A^{H}))^{G/H}=Q(A^{G})$ and
$[Q(A)^{H}:Q(A^{G})]=|G/H|$. Since $A_{(x)}^{G}$ is a DVR it follows that
$A_{(x)}^{H}$ is free of rank $|G/H|$ over $A_{(x)}^{G}$. Consequently:

$[A^{H}_{(x)}/xA^{H}_{(x)}:A^{G}_{(x)}/(x)_{(x)}]=|G/H|,\text{implying: }$
(2.5) $[Q(A^{H}/xA^{H}):Q(A^{G}/(x))]=|G/H|.$

Since $A^{G}/(x)\subseteq(A^{H}/xA^{H})^{G/H}$, it follows from equalities
(2.4) and (2.5) that $Q(A^{G}/(x))=Q((A^{H}/xA^{H})^{G/H})$. Now since
$[A^{G}/(x)]_{\bar{\mathfrak{n}}}$ is normal and $(A^{H}/xA^{H})^{G/H}$ is
finite over $A^{G}/(x)$, it follows that:

(2.6)
$[A^{G}/(x)]_{\bar{\mathfrak{n}}}=[(A^{H}/xA^{H})^{G/H}]_{\bar{\mathfrak{n}}}.$

Finally since $A^{G}/(x)\subseteq(A^{H}/xA^{H})^{G/H}$ is a finite extension
and both are $\mathbb{N}$-graded with $F$ as the $0$-component, it follows
from equality (2.6) that there exists $y=\alpha+z$, $\alpha\in F-\\{0\\}$,
$z\in\bar{\mathfrak{n}}$, with $y[A^{H}/xA^{H}]\subseteq A^{G}/(x)$. Hence
$[A^{H}/xA^{H}]=\bar{\mathfrak{n}}[A^{H}/xA^{H}]+[A^{G}/(x)]$. So by the
graded version of Nakayama’s lemma (e.g. [4, 1.5.24])
$(A^{H}/xA^{H})^{G/H}=A^{G}/(x)$.$\Box$

The following is a useful criterion on a given system of $n$($=\dim_{F}V$)
homogenous elements of $S(V)^{G}$.

######  Proposition 2.14.

[5, Thm. 3.9.4]. Let $G\subset GL(V)$ is a finite group, $\dim_{F}V=n$ and
$S(V)=F[x_{1},...,x_{n}]$. Let $f_{1},...,f_{n}\in S(V)^{G}$ be homogenous
elements. Then the following statements are equivalent:

  1. (1)

$S(V)^{G}=F[f_{1},...,f_{n}]$ (in particular $S(V)^{G}$ is a polynomial ring);

  2. (2)

$\\{f_{1},....,f_{n}\\}$ are algebraically independent over $F$ and
$\prod_{i=1}^{n}\deg(f_{i})=|G|$;

  3. (3)

The Jacobian determinant $\det(\frac{\partial f_{i}}{\partial x_{j}})$ is non-
zero and $\prod_{i=1}^{n}\deg(f_{i})=|G|$;

  4. (4)

$\\{f_{1},...,f_{n}\\}$ is a system of parameters (s.o.p.) and
$\prod_{i=1}^{n}\deg(f_{i})=|G|$.

Proof: The equivalence of items $(1)$, $(2)$ and $(3)$ are due to [9]. The
equivalence of items (1) and (4) is standard and follows from (e.g. [5, Prop.
3.11.2]).

The following is the proof of Theorem B in case $F$ is perfect.

######  Theorem 2.15.

Let $G$ be a finite $p$-group, where $n:=\dim_{F}V\geq 4$,
$\operatorname{char}F=p$, and $F=F^{p}$. Then $S(V)^{G}$ is a polynomial ring
if and only if:

  1. (1)

$S(V)^{G_{U}}$ is a polynomial ring, for each subspace $U\subset V^{*}$ with
$\dim_{F}U=2$, and

  2. (2)

$S(V)^{G}$ is Cohen-Macaulay.

Proof: By condition $(1)$ and Lemma 2.4 $S(V)^{G}$ satisfies Serre’s $R_{n-2}$
condition. Hence by Lemma 2.5 and assumption $(2)$ $G$ is generated by
transvections. $G$ is $p$-group so by a classical result [8, Theorem C, p.100]
there is a $G$-triangulating basis $\\{v_{1},y_{2}^{(1)},...,y_{n}^{(1)}\\}$
of $V$, for the action of $G$ on $V$, with $g(v_{1})=v_{1}$, and
$(g-1)(y_{j}^{(1)})\in Fv_{1}+Fy_{2}^{(1)}+\cdots+Fy_{j-1}^{(1)}$, $j>2$,
$(g-1)(y_{2}^{(1)})\in Fv_{1}$, $\forall g\in G$.

Set $A_{1}=S(V))$, $x_{1}=v_{1}$, $G_{1}=G$, $V_{1}=V$. For $i=1$ the next
(I),(II),(III),(IV),(V) hold by the previous paragraph. Given by induction the
existence of a quadruple $(A_{i},x_{i},V_{i},G_{i})$, $i\geq 2$, with the
following properties:

  1. (I)

$V_{i}:=V/Fv_{1}+\cdots+Fv_{i-1}$, $\dim_{F}V_{i}=n-i+1$,
$\\{v_{1},...,v_{i-1}\\}\subseteq V$;

  2. (II)

$G_{i}:=G/G_{(Fv_{1}+\cdots+Fv_{i-1})^{\perp}}$, $G_{i}$ acts faithfully on
$V_{i}$, where this action is induced from the action of $G$ on $V$;

  3. (III)

$A_{i}:=F[x_{i},y_{i+1}^{(i)},...,y_{n}^{(i)}]\subset S(V_{i})$ is a
polynomial ring in $(n-i+1)$ homogenous elements,
$x_{i}=\bar{v}_{i}^{p^{e_{i}}}$, $y_{j}^{(i)}=\bar{u}_{j}^{p^{f_{j}}}$,
$j=i+1,...,n$, where $\\{\bar{v}_{i},\bar{u}_{i+1},...,\bar{u}_{n}\\}$ is a
triangulating basis for the action of $G_{i}$ on $V_{i}$ and $e_{i}\leq
f_{i+1}\leq\cdots\leq f_{n}$;

  4. (IV)

$(g-1)(x_{i})=0$, $(g-1)(y_{i+1}^{(i)})\in F[x_{i}]_{+}$,
$(g-1)(y_{j}^{(i)})\in F[x_{i},y_{i+1}^{(i)},...,y_{j-1}^{(i)}]_{+}$ for each
$g\in G_{i}$ and $j=i+2,...,n$;

  5. (V)

$A_{i-1}^{G_{i-1}}/(x_{i-1})=A_{i}^{G_{i}}$, $[A_{i}^{G_{i}}]_{\mathfrak{n}}$
is Cohen-Macaulay and satisfies Serre’s $R_{n-i-1}$ condition, where
$\mathfrak{n}:=(A_{i}^{G_{i}})_{+}$.

We shall now proceed to construct the $(i+1)$-th quadruple with the above
properties.

Step 1: $G$ acts in a triangular way on $Fv_{1}+\cdots+Fv_{i}$, and
consequently $(v_{1},...,v_{i})$ is a
$G$-stable prime ideal in $S(V)$.

We have, from properties (III) and (IV), which are valid by induction for
$j\leq i$, that $(g-1)(\bar{v}_{j})^{p^{e_{j}}}=(g-1)(x_{j})=0$, $\forall g\in
G_{j}$. Therefore, using $S(V_{j})=S(V)/(v_{1},...,v_{j-1})$,
$(g-1)(v_{j})^{p^{e_{j}}}\in(v_{1},...,v_{j-1})$, $\forall g\in G$. So since
the latter is a prime ideal in $S(V)$, we get by degree considerations that
$(g-1)(v_{j})\in(v_{1},...,v_{j-1})\cap V=Fv_{1}+\cdots+Fv_{j-1}$, $\forall
g\in G$, and $2\leq j\leq i$. Consequently $(v_{1},...,v_{i})$ is a $G$-stable
ideal in $S(V)$.

Step 2: $x_{i}A_{i}=\bar{v}_{i}S(V_{i})\cap A_{i}$, and consequently
$A_{i}/x_{i}A_{i}\subset S(V_{i+1})$, where $V_{i+1}:=$
$V/Fv_{1}+\cdots+Fv_{i-1}+Fv_{i}$.

By the polynomial property in (III) we have that $\mathfrak{q}:=x_{i}A_{i}$ is
a prime ideal in $A_{i}$ and $\operatorname{height}(\mathfrak{q})=1$. Since
$\bar{v}_{i}S(V_{i})$ is a (height one) prime ideal in $S(V_{i})$, it follows
that $\mathfrak{p}:=\bar{v}_{i}S(V_{i})\cap A_{i}$ is a prime ideal in $A_{i}$
with $\mathfrak{p}\supseteq\mathfrak{q}$. Since
$\\{\bar{v}_{i},\bar{u}_{i+1},...,\bar{u}_{n}\\}$ is a basis of
$V_{i}=V/Fv_{1}+\cdots+Fv_{i-1}$, it follows that the images
$\\{\tilde{u}_{i+1},...,\tilde{u}_{n}\\}$ form a basis of $V_{i+1}$, so they
are generators of
$S(V_{i+1})=S(V_{i})/\bar{v}_{i}S(V_{i})=S(V)/(v_{1},...,v_{i})$. Therefore
$A_{i}/\mathfrak{p}=F[\bar{y}_{i+1}^{(i)},...,\bar{y}_{n}^{(i)}]=F[\tilde{u}_{i+1}^{p^{f_{i+1}}},...,\tilde{u}_{n}^{p^{f_{n}}}]$
is a polynomial ring in $(n-i)$ variables, implying since $K.\dim
A_{i}=(n-i+1)$, that $\operatorname{height}(\mathfrak{p})=1$ and hence
$\mathfrak{p}=\mathfrak{q}$.

Step 3:
$H_{i}=G_{(Fv_{1}+\cdots+Fv_{i})^{\perp}}/G_{(Fv_{1}+\cdots+Fv_{i-1})^{\perp}}$,
where $H_{i}:=\\{g\in G_{i}|(g-1)(A_{i})\subseteq x_{i}A_{i}\\}$.

Let $g\in H_{i}$, then $(g-1)(y_{j}^{(i)})\in x_{i}A_{i}$, $j=i+1,...,n$.
Hence $(g-1)(y_{j}^{(i)})\in
x_{i}S(V_{i})\subseteq\bar{v}_{i}S(V_{i})=(v_{1},...,v_{i})/(v_{1},...,v_{i-1})$.
Write $g=g^{\prime}G_{(Fv_{1}+\cdots Fv_{i-1})^{\perp}}$, $g^{\prime}\in G$.
It follows that $[(g^{\prime}-1)(u_{j})]^{p^{f_{j}}}\in(v_{1},...,v_{i})$.
Since the latter ia a prime ideal in $S(V)$, we get that
$(g^{\prime}-1)(u_{j})\in(v_{1},...,v_{i})\cap V=Fv_{1}+\cdots+Fv_{i}$,
$j=i+1,...,n$. Also $0=(g-1)(x_{i})=[(g-1)(\bar{v}_{i})]^{p^{e_{i}}}$,
implying that $(g-1)(\bar{v}_{i})=0$, hence $(g^{\prime}-1)(v_{i})\in
Fv_{1}+\cdots+Fv_{i-1}$. Consequently $(g^{\prime}-1)(V)\subseteq
Fv_{1}+\cdots+Fv_{i}$. Equivalently $g^{\prime}\in
G_{(Fv_{1}+\cdots+Fv_{i})^{\perp}}$ and hence $g\in
G_{(Fv_{1}+\cdots+Fv_{i})^{\perp}}/G_{(Fv_{1}+\cdots+Fv_{i-1})^{\perp}}$, and
$H_{i}\subseteq
G_{(Fv_{1}+\cdots+Fv_{i})^{\perp}}/G_{(Fv_{1}+\cdots+Fv_{i-1})^{\perp}}$. The
reversed inclusion is clear.

Step 4: $A_{i}^{H_{i}}$ is a polynomial ring.

Recall that $A_{i}=F[x_{i},y_{i+1}^{(i)},...,y_{n}^{(i)}]\subset S(V_{i})$ is
a polynomial ring in $(n-i+1)$ homogenous elements, and
$x_{i}=\bar{v}_{i}^{p^{e_{i}}}$, $y_{j}^{(i)}=\bar{u}_{j}^{p^{f_{j}}}$,
$j=i+1,...,n$, $e_{i}\leq f_{i+1}\leq\cdots\leq f_{n}$. We have from the
previous paragraph that $(g^{\prime}-1)(u_{j})\in Fv_{1}+...+Fv_{i}$.
Consequently $(g-1)(\bar{u}_{j})\in F\bar{v}_{i}$ and
$g(\bar{v}_{i})=\bar{v}_{i}$, $\forall g\in H_{i}$, $j=i+1,...,n$. It follows
from Corollary 2.10 that $A_{i}^{H_{i}}=F[x_{i},z_{i+1},...,z_{n}]$ is a
polynomial ring in $(n-i+1)$ homogenous elements, $\deg z_{j}=p^{h_{j}}$,
$z_{j}-\bar{w}_{j}^{p^{h_{j}}}\in\bar{v}_{i}^{p^{e_{i}}}S(V_{i})$, for
$j=i+1,...,n$, where $e_{i}\leq h_{i+1}\leq\cdots\leq h_{n}$ and
$\\{\bar{v}_{i},\bar{w}_{i+1},...,\bar{w}_{n}\\}$ is a basis of $V_{i}$.

Step 5: $G_{i}/H_{i}$ acts faithfully on $A_{i}^{H_{i}}/x_{i}A_{i}^{H_{i}}$.

Indeed if $g\in G_{i}$, with $(g-1)(A_{i}^{H_{i}})\subseteq
x_{i}A_{i}^{H_{i}}=\bar{v}_{i}^{p^{e_{i}}}A_{i}^{H_{i}}\subseteq\bar{v}_{i}^{p^{e_{i}}}S(V_{i})$,
then
$(g-1)(\bar{w}_{j})^{p^{h_{j}}}=(g-1)(\bar{w}_{j}^{p^{h_{j}}})\in\bar{v}_{i}S(V_{i})$.
Since the latter is a prime in $S(V_{i})$, it follows that
$(g-1)(\bar{w}_{j})\in\bar{v}_{i}S(V_{i})\cap V_{i}=F\bar{v}_{i}$, for
$j=i+1,...,n$. Since
$V_{i}=\operatorname{span}_{F}\\{\bar{v}_{i},\bar{w}_{i+1},...,\bar{w}_{n}\\}$
it follows that $(g-1)(V_{i})\subseteq F\bar{v}_{i}$. Equivalently if
$g=g^{\prime}G_{(Fv_{1}+\cdots+Fv_{i-1})^{\perp}}$ then
$(g^{\prime}-1)(V)\subseteq Fv_{1}+\cdots+Fv_{i}$, and hence $g^{\prime}\in
G_{(Fv_{1}+\cdots+Fv_{i})^{\perp}}$. Consequently $g\in H_{i}$ as claimed.

Step 6: Changing the generators of $A_{i}^{H_{i}}$.

We shall now apply Proposition 2.11 with $G_{i}$ acting on $A_{i}^{H_{i}}$.
Then $A_{i}^{H_{i}}=F[x_{i},m_{i+1},...,m_{n}]$,
$m_{j}-\bar{t}_{j}^{p^{h_{j}}}\in\bar{v}_{i}^{p^{e_{i}}}S(V_{i})$ for
$j=i+1,...,n$, $\\{\bar{v}_{i},\bar{t}_{i+1},...,\bar{t}_{n}\\}$ is a
triangulating basis for the action of $G_{i}$ on $V_{i}$,
$\deg(m_{j})=p^{h_{j}}$, and $m_{j}$ is a homogenous element in $S(V_{i})$,
$j=i+1,...,n$. Moreover $(g-1)(m_{i+1})\in
F[\bar{v}_{i}^{p^{e_{i}}}]_{+}=F[x_{i}]_{+}$, $(g-1)(m_{j})\in
F[x_{i},m_{i+1},...,m_{j-1}]_{+}$, for $j>i+1$, $g\in G_{i}$.

Step 7: $(A_{i+1},x_{i+1},V_{i+1},G_{i+1})$ and its properties.

Set $A_{i+1}:=A_{i}^{H_{i}}/x_{i}A_{i}^{H_{i}}$,
$V_{i+1}:=V/Fv_{1}+\cdots+Fv_{i}$ (as in Step 2),
$G_{i+1}:=G_{i}/H_{i}=G/G_{(Fv_{1}+\cdots Fv_{i})^{\perp}}$ and
$x_{i+1}:=\bar{m}_{i+1}$, the image of $m_{i+1}$ in $A_{i+1}\subset
S(V_{i+1})$, where the last inclusion is by Step 2.

We next verify property $(II)$ for $(A_{i+1},x_{i+1},V_{i+1},G_{i+1})$. By
Step 1, the action of $G$ on $V_{i+1}$ is well defined and linear, so inducing
a linear action of $G_{i+1}=G/G_{(Fv_{1}+\cdots Fv_{i})^{\perp}}$ on
$V_{i+1}$. We now show that this action is faithful. Say $(g-1)(V_{i+1})=0$,
for some $g\in G_{i+1}=G/H_{i}$. Then $(g-1)(S(V_{i+1}))=0$. Hence
$(g-1)(A_{i}^{H_{i}}/x_{i}A_{i}^{H_{i}})=0$ and $g=1$ follows from Step 5.

Set $e_{i+1}:=h_{i+1}$, $v_{i+1}:=t_{i+1}$, $y_{j}^{(i+1)}:=\bar{m}_{j}$, the
image of $m_{j}$ in $A_{i+1}$, $j=i+2,...,n$. Since
$A_{i}^{H_{i}}=F[x_{i},m_{i+1},...,m_{n}]$ is a polynomial ring, it follows
that $A_{i+1}=A_{i}^{H_{i}}/x_{i}A_{i}^{H_{i}}$ is a polynomial ring in
$(n-i)$ variables. This implies that
$A_{i+1}=F[x_{i+1},y_{i+2}^{(i+1)},...,y_{n}^{i+1}]$,
$x_{i+1}=\bar{v}_{i+1}^{p^{e_{i+1}}}$,
$y_{j}^{(i+1)}=\bar{t}_{j}^{p^{h_{j}}}$, $j=i+2,...,n$ (by abuse of notation
we still denote by $\bar{t}_{j}$ the image of $t_{j}$ in $V_{i+1})$. It now
follows from Step 6 that $(g-1)(x_{i+1})=0$,
$(g-1)(y_{i+2}^{(i+1)})=(g-1)(\bar{m}_{i+2})\in
F[\bar{m}_{i+1}]_{+}=F[x_{i+1}]_{+}$, $(g-1)(y_{j}^{(i+1)})\in
F[x_{i+1},y_{i+2}^{(i+1)},...,j_{j-1}^{(i+1)}]_{+}$, $j>i+2$, $\forall g\in
G_{i+1}$. It also follows, as in the proof of Proposition 2.11(6), that
$\\{\bar{v}_{i+1},\bar{t}_{i+2},...,\bar{t}_{n}\\}$ is a triangulating basis
for the action of $G_{i+1}$ in $V_{i+1}$.

This settles properties $(I)$, $(II)$, $(III)$ and $(IV)$ of
$(A_{i+1},x_{i+1},V_{i+1},G_{i+1})$.

Now we verify property $(V)$ for the quadruple
$(A_{i+1},x_{i+1},V_{i+1},G_{i+1})$.

Set $\mathfrak{n}:=[A_{i}^{G_{i}}]_{+}=$ the irrelevant maximal ideal of
$A_{i}^{G_{i}}$, $\mathfrak{m}:=[A_{i}]_{+}=$ the irrelevant maximal ideal of
$A_{i}$. Since $x_{i}\in\mathfrak{m}-\mathfrak{m}^{2}$ (by the fact that it is
a minimal generator of $A_{i}$), it is also a prime element of the polynomial
ring $A_{i}$ as well as being a $G_{i}$-invariant. By Step 5 we have that
$G_{i}/H_{i}$ acts faithfully on $A_{i}^{H_{i}}/x_{i}A_{i}^{H_{i}}$. Also by
property ($V)$ for $(A_{i},x_{i},V_{i},G_{i})$,
$[A_{i}^{G_{i}}]_{\mathfrak{n}}$ satisfies Serre’s $R_{n-i-1}$ condition , as
well as being Cohen-Macaulay. This implies by Proposition 2.13 that:
$A_{i}^{G_{i}}/(x_{i})=(A_{i}^{H_{i}}/x_{i}A_{i}^{H_{i}})^{G_{i}/H_{i}}=A_{i+1}^{G_{i+1}}$,
and
$[A_{i}^{G_{i}}/(x_{i})]_{\mathfrak{n}/(x_{i})}=[A_{i+1}^{G_{i+1}}]_{\mathfrak{n}/(x_{i})}$
satisfies Serre’s $R_{n-i-2}$ condition. It is also a Cohen-Macaulay ring, so
property $(V)$ is verified for the quadruple
$(A_{i+1},x_{i+1},V_{i+1},G_{i+1})$.

This settles the inductive step.

Step 8: The chain of hypersurface sections of $S(V)^{G}$.

Due to the requirement of $m\geq 2$ in Serre’s condition $R_{m}$, in
Proposition 2.13, this procedure can be carried only for $i\leq n-3$.

Consequently we have the following chain of equalities:

(2.7) $S(V)^{G}/(v)=A_{1}^{G_{1}}/(x_{1})=A_{2}^{G_{2}},\text{
}A_{2}^{G_{2}}/(x_{2})=A_{3}^{G_{3}},...,A_{n-3}^{G{n-3}}/(x_{n-3})=A_{n-2}^{G_{n-2}}.$

Observe that
$A_{n-2}=F[\bar{w}_{n-2}^{p^{f}},\bar{w}_{n-1}^{p^{g}},\bar{w}_{n}^{p^{h}}]\subset
S(V_{n-2})$, $V_{n-2}=V/Fv_{1}+\cdots+v_{n-3}$, $G_{n-2}=G/G_{(Fv_{1}+\cdots
Fv_{n-3})^{\perp}}$, $f\leq g\leq h$ and $\\{\bar{w}_{n-2}$,
$\bar{w}_{n-1},\bar{w}_{n}\\}$ is a triangulating basis for the faithful,
linear action of $G_{n-2}$ on
$V_{n-2}=F\bar{w}_{n-2}+F\bar{w}_{n-1}+F\bar{w}_{n}$.

Step 9: The chain of hypersurface sections of $S(V)^{T}$, where
$T:=G_{(Fv_{1}+\cdots+Fv_{n-3})^{\perp}}$.

Let $U:=(Fv_{1}+...+Fv_{n-3})^{\perp}$. $\dim_{F}U=n-(n-3)=3$ and therefore by
assumption $(1)$ and Lemma 2.4, $S(V)^{G_{U}}$ is a polynomial ring. For each
$1\leq i\leq n-3$, we have $G_{(Fv_{1}+\cdots+Fv_{i})^{\perp}}=\\{g\in
G|(g-1)(V)\subseteq Fv_{1}+\cdots+Fv_{i}\\}\subseteq\\{g\in
G|(g-1)(V)\subseteq Fv_{1}+\cdots+Fv_{n-3}\\}=T$. Consequently
$G_{(Fv_{1}+\cdots+Fv_{i})^{\perp}}=T_{(Fv_{1}+\cdots+Fv_{i})^{\perp}}$, and
therefore for $i\geq 2$
$H_{i}=G_{(Fv_{1}+\cdots+Fv_{i})^{\perp}}/G_{(Fv_{1}+\cdots+Fv_{i-1})^{\perp}}=T_{(Fv_{1}+\cdots+Fv_{i})^{\perp}}/T_{(Fv_{1}+\cdots+Fv_{i-1})^{\perp}}$.
Also for $i=1$, we have $H_{1}=G_{(Fv_{1})^{\perp}}=T_{(Fv_{1})^{\perp}}$.

Set $T_{i}:=T/T_{(Fv_{1}+\cdots+Fv_{i-1})^{\perp}}$, $i=2,...,n-2$. Then since
$S(V)^{T}$ is a polynomial ring we can similarly apply Proposition 2.13 with
the same $A_{i}$, $H_{i}$, $x_{i}$, for each $i$, and get, as in Step 8, the
following chain of equalities.

(2.8) $S(V)^{T}/(v_{1})=S(V)^{T}/(x_{1})=A_{2}^{T_{2}},\text{
}A_{2}^{T_{2}}/(x_{2})=A_{3}^{T_{3}},...,A_{n-3}^{T_{n-3}}=A_{n-2}^{T_{n-2}}.$

But $T_{n-2}=T/T_{(Fv_{1}+\cdots+Fv_{n-3})^{\perp}}=1$, implying that
$A_{n-2}^{T_{n-2}}=A_{n-2}=F[\bar{w}_{n-2}^{p^{f}},\bar{w}_{n-1}^{p^{g}},\bar{w}_{n}^{p^{h}}]\subseteq
S(V_{n-2})$. Also $f\leq g\leq h$ and
$\\{\bar{w}_{n-2},\bar{w}_{n-1},\bar{w}_{n}\\}$ is a triangulating basis for
the action of $G_{n-2}$ on
$V_{n-2}=F\bar{w}_{n-2}+F\bar{w}_{n-1}+F\bar{w}_{n}$.

Let
$\\{x^{\prime}_{1},...,x^{\prime}_{n-3},x^{\prime}_{n-2},x^{\prime}_{n-1},x^{\prime}_{n}\\}$
be homogenous preimages of
$\\{x_{1},...,x_{n-3},\bar{w}_{n-2}^{p^{f}},\bar{w}_{n-1}^{p^{g}},\bar{w}_{n}^{p^{h}}\\}$
in $S(V)^{T}$. They exist by equalities (2.8) and they generate $S(V)^{T}$.
Consequently by Proposition 2.14:

(2.9) $|T|=\prod_{i=1}^{n}(\deg x^{\prime}_{i})=\prod_{i=1}^{n-3}(\deg
x_{i})p^{f}p^{g}p^{h}.$

Step 10: The conclusion of the proof.

We may clearly assume that $G\neq T$.
$G_{n-2}=G/G_{(Fv_{1}+\cdots+Fv_{n-3})^{\perp}}=G/T$ acts linearly on
$V_{n-2}=F\bar{w}_{n-2}+F\bar{w}_{n-1}+R\bar{w}_{n}$, and is generated by
transvections (since $G$ is generated by transvections on $V$ by Lemma 2.5,
and by Lemma 2.6, this property is inherited by $G_{i}$ on $V_{i}$, for each
$i$). Moreover $\\{\bar{w}_{n-2},\bar{w}_{n-1},\bar{w}_{n}\\}$ is a
triangulating basis for the action of $G_{n-2}$ on $V_{n-2}$. Therefore (as
e.g. in Proposition 3.2) $G_{n-2}$ is represented, with respect to this basis,
by matrices in $\begin{bmatrix}1&F&F\\\ 0&1&F\\\ 0&0&1\end{bmatrix}$. We shall
split the discussion into three cases.

Case 1: $G_{n-2}\subseteq\begin{bmatrix}1&0&F\\\ 0&1&F\\\ 0&0&1\end{bmatrix}$.
In this case we have that $G_{n-2}=H_{n-2}$, implying by Corollary 2.10 that:

$A_{n-2}^{G_{n-2}}=F[\bar{w}_{n-2}^{p^{f}},\bar{w}_{n-1}^{p^{g}},\bar{w}_{n}^{p^{h}}]^{H_{n-2}}=F[x_{n-2},x_{n-1},x_{n}].$

Therefore if
$\\{x^{\prime}_{1},...,x^{\prime}_{n-3},x^{\prime}_{n-2},x^{\prime}_{n-1},x^{\prime}_{n}\\}$
are homogenous preimages in $S(V)^{G}$ of
$\\{x_{1},...,x_{n-3},x_{n-2},x_{n-1},x_{n}\\}$ which exist by the chain of
equalities (2.7), then they generate $S(V)^{G}$ and consequently $S(V)^{G}$ is
a polynomial ring.

Case 2: $G_{n-2}\subset\begin{bmatrix}1&F&F\\\ 0&1&0\\\ 0&0&1\end{bmatrix}$.
Hence $g(\bar{w}_{n-2})=\bar{w}_{n-2}$, $g(\bar{w}_{n-1}=\bar{w}_{n-1}$ for
each $g\in G_{n-2}$. Consequently by Proposition 2.7
$F[\bar{w}_{n-2},\bar{w}_{n-1},\bar{w}_{n}]^{G_{n-2}}=F[\bar{w}_{n-2},\bar{w}_{n-1},a]$
is a polynomial ring where $\deg a=|G_{n-2}|=|G/T|$. Therefore using $f\leq
g\leq h$, $a^{p^{h}}\in
F[\bar{w}_{n-2}^{p^{f}},\bar{w}_{n-1}^{p^{g}},\bar{w}_{n}^{p^{h}}]^{G_{n-2}}=A_{n-2}^{G_{n-2}}$.
Clearly $\\{\bar{w}_{n-2}^{p^{f}},\bar{w}_{n-1}^{p^{g}},a^{p^{h}}\\}$ is a
system of parameters of $A_{n-2}^{G_{n-2}}$. By the chain of equalities (2.7)
we can pick in $S(V)^{G}$ homogenous preimages
$\\{x^{\prime}_{1},...,x^{\prime}_{n-3},x^{\prime}_{n-2},x^{\prime}_{n-1},x^{\prime}_{n}\\}$
of
$\\{x_{1},...,x_{n-3},\bar{w}_{n-2}^{p^{f}},\bar{w}_{n-1}^{p^{g}},a^{p^{h}}\\}$.
These $n$ preimages form a system of homogenous parameters in $S(V)^{G}$, and
by equality (2.9):

$\begin{split}\prod_{i=1}^{n}(\deg x^{\prime}_{i})&=\prod_{i=1}^{n-3}(\deg
x^{\prime}_{i})p^{f}p^{g}\deg a^{p^{h}}\\\ &=\prod_{i=1}^{n-3}(\deg
x_{i})p^{f}p^{g}(\deg a)p^{h}=|T|(\deg a)=|T||G/T|=|G|.\end{split}$

Consequently by Proposition 2.14
$S(V)^{G}=F[x^{\prime}_{1},...,x^{\prime}_{n-3},x^{\prime}_{n-2},x^{\prime}_{n-1},x^{\prime}_{n}]$
is a polynomial ring.

Case 3: $G_{n-2}\not\subset\begin{bmatrix}1&0&F\\\ 0&1&F\\\
0&0&1\end{bmatrix}$, $\begin{bmatrix}1&F&F\\\ 0&1&0\\\ 0&0&1\end{bmatrix}$.
Therefore by Theorem 3.3,
$F[\bar{w}_{n-2},\bar{w}_{n-1},\bar{w}_{n}]^{G_{n-2}}=F[\bar{w}_{n-2},a,b]$,
where $a\in F[\bar{w}_{n-2},\bar{w}_{n-1}]^{G_{n-2}}$, and hence $(\deg
a)(\deg b)=|G_{n-2}|$. Since $f\leq g\leq h$, then $\bar{w}_{n-2}^{p^{f}}$,
$a^{p^{g}}\in F[\bar{w}_{n-2}^{p^{f}},\bar{w}_{n-1}^{p^{g}}]^{G_{n-2}}$,
$b^{p^{h}}\in
F[\bar{w}_{n-2}^{p^{f}},\bar{w}_{n-1}^{p^{g}},\bar{w}_{n}^{p^{h}}]^{G_{n-2}}$.
In particular $\\{\bar{w}_{n-2}^{p^{f}},a^{p^{g}},b^{p^{h}}\\}$ is a system of
parameters (s.o.p.) of
$A_{n-2}^{G_{n-2}}=F[\bar{w}_{n-2}^{p^{f}},\bar{w}_{n-1}^{p^{g}},\bar{w}_{n}^{p^{h}}]^{G_{n-2}}$.
Therefore by the chain of equalities (2.7) there are $n$ homogenous preimages
in $S(V)^{G}$ of
$\\{x_{1},...,x_{n-3},\bar{w}_{n-2}^{p^{f}},a^{p^{g}},b^{p^{h}}\\}$,
$\\{x^{\prime}_{1},...,x^{\prime}_{n-3},x^{\prime}_{n-2},x^{\prime}_{n-1},x^{\prime}_{n}\\}$.
These $n$ preimages form a homogenous s.o.p. of $S(V)^{G}$ and by (2.9):

$\begin{split}\prod_{i=1}^{n}(\deg x^{\prime}_{i})&=\prod_{i=1}^{n-3}(\deg
x_{i})(\deg x^{\prime}_{n-2})(\deg x^{\prime}_{n-1})(\deg x^{\prime}_{n})\\\
&=\prod_{i=1}^{n-3}(\deg x_{i})p^{f}p^{g}(\deg a)p^{h}(\deg b)=|T|(\deg
a)(\deg b)\\\ &=|T||G/T|=|G|.\end{split}$

Therefore by Proposition 2.14
$S(V)^{G}=F[x^{\prime}_{1},...,x^{\prime}_{n-3},x^{\prime}_{n-2},x^{\prime}_{n-1},x^{\prime}_{n}]$
is a polynomial ring.$\Box$

######  Definition 2.16.

Let $A$ be an affine domain over a field $F$ and
$\mathfrak{p}\in\operatorname{spec}A$. We say that $A_{\mathfrak{p}}$ is
smooth if $A_{\mathfrak{p}}$ is geometrically regular, that is if
$A_{\mathfrak{p}}\otimes_{F}K$ is regular for each finite field extension $K$
of $F$.

######  Remark 2.17.

Note the following:

  1. (1)

By [13, Theorem 28.7] this is equivalent to $A_{\mathfrak{\mathfrak{p}}}$
being $\mathfrak{p}_{\mathfrak{p}}$-smooth over $F$;

  2. (2)

[3, X, $\S 6$, Definition 1, p. 75] uses the terminology ”absolutely regular”.
It is slightly differently defined, but is an equivalent notion.

######  Lemma 2.18.

Let $A$ be an affine domain over a field $F$. Then:

  1. (1)

$A_{\mathfrak{p}}$ is geometrically regular if and only if
$\hat{A}_{\mathfrak{p}}$ is geometrically regular;

  2. (2)

Let $A\subset B$ be a finite extension, and
$\mathfrak{p}\in\operatorname{spec}A$, $\mathfrak{n}\in\operatorname{spec}B$
with $\mathfrak{n}\cap A=\mathfrak{p}$. Suppose $B_{\mathfrak{n}}$ is étale
over $A_{\mathfrak{p}}$. Then $A_{\mathfrak{p}}$ is geometrically regular if
and only if $B_{\mathfrak{p}}$ is geometrically regular.

Proof: $(1)$ follows from [13, $\S$28, Example 1(ii), p.215].

$(2)$ Let $K\supset F$ be a finite radical field extension of $F$.
$A_{\mathfrak{p}}\otimes_{F}K\subset B_{\mathfrak{n}}\otimes_{F}K$ is a
faithfully flat extension [13, p. 46, (2)]. Both rings are local, hence, by
using [13, Thm 23.7(i)], the regularity of $A_{\mathfrak{p}}\otimes_{F}K$
follows from the given regularity of $B_{\mathfrak{p}}\otimes_{F}K$. This
shows by Remark 2.17(2) that $A_{\mathfrak{p}}$ is geometrically regular.

To show the opposite direction, suppose that $A_{\mathfrak{p}}$ is
geometrically regular. Recall that
$\mathfrak{p}_{\mathfrak{p}}B_{\mathfrak{n}}=\mathfrak{n}_{\mathfrak{n}}$ and
$B_{\mathfrak{n}}/\mathfrak{n}_{\mathfrak{n}}$ is a finite separable extension
of $A_{\mathfrak{p}}/\mathfrak{p}_{\mathfrak{p}}$. Consider the sequence
$F\subset A_{\mathfrak{p}}\subset B_{\mathfrak{n}}$. By [13, Thm. 28.1
”transitivity”] it suffices to show that $B_{\mathfrak{n}}$ is
$\mathfrak{n}_{\mathfrak{n}}$-smooth over $A_{\mathfrak{p}}$. For this purpose
we apply [13, Thm. 28.10] observing that
$B_{\mathfrak{n}}\otimes_{A_{\mathfrak{p}}}A_{\mathfrak{p}}/\mathfrak{p}_{\mathfrak{p}}=B_{\mathfrak{n}}/\mathfrak{p}_{\mathfrak{p}}B_{\mathfrak{n}}=B_{\mathfrak{n}}/\mathfrak{n}_{\mathfrak{n}}$
is $0$-smooth over $A_{\mathfrak{p}}/\mathfrak{p}_{\mathfrak{p}}$, being a
finite separable field extension of
$A_{\mathfrak{p}}/\mathfrak{p}_{\mathfrak{p}}$.$\Box$

The next result is a complement to Lemma 2.4.

######  Lemma 2.19.

Let $G\subset GL(V)$ be a finite group and $\dim_{F}V=n$. Let $n\geq m\geq 1$
be an integer. Consider the following properties:

  1. (3)

$S(V)^{G_{U}}$ is a polynomial ring for each subspace $U\subset V^{*}$ with
$dim_{F}U\geq n-m$;

  2. (4)

$S(V)^{G}_{\mathfrak{p}}$ is geometrically regular for each
$\mathfrak{p}\in\operatorname{spec}S(V)^{G}$ with
$\operatorname{height}(\mathfrak{p})=m$.

Then we have the following implications: $(3)\Leftrightarrow(4)$.

Proof: Firstly we consider $(3)\Rightarrow(4)$. Let
$\mathfrak{\mathfrak{p}}\in\operatorname{spec}S(V)^{G}$,
$\mathfrak{m}\in\operatorname{spec}S(V)$ with $\mathfrak{m}\cap
S(V)^{G}=\mathfrak{p}$. Set $\mathfrak{n}:=\mathfrak{m}\cap
S(V)^{I_{G}(\mathfrak{m})}$, $\mathfrak{q}:=\mathfrak{m}\cap
S(V)^{D_{G}(\mathfrak{m})}$. Set $W:=\mathfrak{m}\cap V$ and $U:=W^{\perp}$.
Since $WS(V)\subseteq\mathfrak{m}$ is a prime ideal, we have that
$\dim_{F}W=\operatorname{height}(WS(V))\leq\operatorname{height}(\mathfrak{m})=\operatorname{height}(\mathfrak{p})=m$.
Consequently $\dim_{F}U=n-\dim_{F}W\geq n-m$. Therefore by condition (3)
$S(V)^{G_{U}}$ is a polynomial ring, implying that $S(V)^{G_{U}}$ is
geometrically regular (every scalar field extension of a polynomial ring is a
polynomial ring). Since $G_{U}=I_{G}(\mathfrak{m})$ we have that
$S(V)_{\mathfrak{n}}^{I_{G}(\mathfrak{m})}$ is geometrically regular, implying
by Lemma 2.18(2) and Theorem 2.2(2) that
$S(V)_{\mathfrak{q}}^{D_{G}(\mathfrak{m})}$ is geometrically regular. This
implies by Lemma 2.18(1) and Theorem 2.2(1) that $S(V)_{\mathfrak{p}}^{G}$ is
geometrically regular.

We now show the direction $(4)\Rightarrow(3)$. By Lemma 2.4 we may only
consider $U\subset V^{*}$ with $\dim_{F}U=n-m$. We have to show that
$S(V)^{G_{U}}$ is a polynomial ring. Set $W:=U^{\perp}$, $\mathfrak{m}=WS(V)$.
Then $\mathfrak{m}$ is a prime ideal in $S(V)$ and
$m=\dim_{F}W=\operatorname{height}(\mathfrak{m})$. We also set
$\mathfrak{n}:=S(V)^{I_{G}(\mathfrak{m})}\cap\mathfrak{m}$,
$\mathfrak{p}:=S(V)^{G}\cap\mathfrak{m}$,
$\mathfrak{q}:=S(V)^{D_{G}(\mathfrak{m})}\cap\mathfrak{m}$. We have
$\operatorname{height}(\mathfrak{p})=\operatorname{height}(\mathfrak{\mathfrak{m}})$
by ”going down”, and $I_{G}(\mathfrak{m})=G_{U}=\\{g\in G|(g-1)(V)\subseteq
W\\}$. By assumption $S(V)^{G}_{\mathfrak{p}}$ is geometrically regular and
consequently by Theorem 2.2(1)
$\widehat{S(V)^{G}_{\mathfrak{p}}}\cong\widehat{S(V)^{D_{G}(\mathfrak{m})}_{\mathfrak{q}}}$
is geometrically regular. Consequently
$S(V)^{D_{G}(\mathfrak{m})}_{\mathfrak{q}}$ is geometrically regular and by
Lemma 2.18(2) so is
$S(V)^{I_{G}(\mathfrak{m})}_{\mathfrak{n}}=S(V)^{G_{U}}_{\mathfrak{n}}$.
Consequently by [3, X,§6, Cor. 4, p. 77]
$S(V)^{G_{U}}_{\mathfrak{n}}\otimes_{F}\bar{F}$ is regular, where $\bar{F}$ is
the algebraic closure of $F$.

Let $V_{\bar{F}}:=V\otimes_{F}\bar{F}$, $W_{\bar{F}}=W\otimes_{F}{\bar{F}}$,
$\mathfrak{m}_{\bar{F}}=\mathfrak{m}\otimes_{F}\bar{F}$. Since
$S(V_{\bar{F}})$ is integral over $S(V)^{I_{G}(\mathfrak{m})}$, there exists
$\mathfrak{b}\in\operatorname{spec}S(V_{\bar{F}})$ such that $\mathfrak{b}\cap
S(V)^{I_{G}(\mathfrak{m})}=\mathfrak{n}$ and
$\mathfrak{b}\supseteq\mathfrak{m}_{\bar{F}}$. Set
$\mathfrak{a}=\mathfrak{b}\cap S(V_{\bar{F}})^{I_{G}(\mathfrak{m})}$,
$H:=I_{G}(\mathfrak{m})=G_{U}$.

Since $(g-1)(V_{\bar{F}})\subseteq
W_{\bar{F}}=V_{\bar{F}}\cap\mathfrak{m}_{\bar{F}}\subseteq
V_{\bar{F}}\cap\mathfrak{b}$, $\forall g\in H$, we have
$(g-1)(S(V_{\bar{F}}))\subseteq\mathfrak{b}$, $\forall g\in H$. Consequently
$H=I_{H}(\mathfrak{b})=D_{H}(\mathfrak{b})$ and
$S(V_{\bar{F}})^{H}_{\mathfrak{a}}$ is regular being the localization of
$S(V_{\bar{F}})^{H}_{\mathfrak{n}}=S(V)^{H}_{\mathfrak{n}}\otimes_{F}\bar{F}$.
Since $\bar{F}$ is perfect, we can conclude by [10, Prop. 1.9, Example. 1.10]
that
$S(V_{\bar{F}})^{H}=S(V_{\bar{F}})^{G_{U}}=S(V)^{G_{U}}\otimes_{F}\bar{F}$ is
a polynomial ring, and therefore so is $S(V)^{G_{U}}$.$\Box$

The proof of Theorem B:

Note that by Lemma 2.19 and Lemma 2.4, $\dim$(non-smooth locus of
$S(V)^{G}$)$\leq 1$ is equivalent to: $S(V)^{G_{U}}$ is a polynomial ring for
each $U\subset V^{*}$ with $\dim_{F}U=2$. We have that $S(V)^{G}$ is Cohen-
Macaulay and $S(V)^{G}_{\mathfrak{p}}$ is geometrically regular for each prime
$\mathfrak{p}$ with $\operatorname{height}(\mathfrak{p})=n-2$. Consequently by
[3, X, $\S 6$, Cor. 4, p.77] $S(V)^{G}_{\mathfrak{p}}\otimes_{F}\bar{F}$ is
regular, where $\bar{F}$ is the algebraic closure of $F$. Set
$V_{\bar{F}}=V\otimes_{F}\bar{F}$, then
$S(V_{\bar{F}})_{\mathfrak{p}}^{G}=S(V)^{G}_{\mathfrak{p}}\otimes_{F}\bar{F}$
is regular.

Let $\mathfrak{q}\in\operatorname{spec}S(V_{\bar{F}})^{G}$ with
$\operatorname{height}(\mathfrak{q})=n-2$. Since $S(V_{\bar{F}})^{G}$ is
integral over $S(V)^{G}$, it follows that $\mathfrak{q}\cap S(V)^{G}$ is a
height $(n-2)$ prime ideal in $S(V)^{G}$. Consequently by the previous
paragraph, $S(V_{\bar{F}})^{G}_{\mathfrak{q}\cap
S(V)^{G}}=S(V)^{G}_{\mathfrak{q}\cap S(V)^{G}}\otimes_{F}\bar{F}$ is regular.
Therefore $S(V_{\bar{F}})^{G}_{\mathfrak{q}}$, being a localization of the
latter ring, is also regular.

To sum up, $S(V_{\bar{F}})^{G}$ satisfies Serre’s $R_{n-2}$ condition,
$\bar{F}$ is perfect and
$S(V_{\bar{F}})^{G}=S(V)^{G}\otimes_{F}\bar{F}$ is Cohen-Macaulay. So by Lemma
2.4 all the requirements of Theorem 2.15 are in place and $S(V_{\bar{F}})^{G}$
is therefore a polynomial ring, implying that $S(V)^{G}$ is a polynomial
ring.$\Box$

In the next example we exhibit a $p$-group $T\subset GL(V)$, $\dim_{F}V=4$,
with the additional properties $|T|=p^{3}$, $T$ is elementary abelian and is
generated by transvections. However, $S(V)^{T}$ is not a polynomial ring,
$S(V)^{T}$ is Gorenstein and $T=T_{U}$, for some subspace $U\subset V^{*}$
with $\dim U=2$.

######  Example 2.20.

$T=<\sigma,\tau_{1},\tau_{2}>\subset GL_{4}(F)$ and $F\supset\mathbb{F}_{p}$
where:

$\sigma=\begin{bmatrix}1&0&1&0\\\ 0&1&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{bmatrix},\text{ }\tau_{1}=\begin{bmatrix}1&0&0&1\\\ 0&1&0&1\\\
0&0&1&0\\\ 0&0&0&1\end{bmatrix},\text{ }\tau_{2}=\begin{bmatrix}1&0&0&0\\\
0&1&0&\xi\\\ 0&0&1&0\\\ 0&0&0&1\end{bmatrix},\text{ }\xi\in F-\mathbb{F}_{p}.$

The matrices are given with respect to the basis $\\{v,w_{3},w_{2},w_{1}\\}$.
We read the action of the matrices along the rows, so e.g.
$\sigma(v)=v+w_{2}$, $\tau_{1}(v)=v+w_{1}$, $\tau_{1}(w_{3})=w_{3}+w_{1}$,
$\tau_{2}(w_{3})=w_{3}+\xi w_{1}$, etc.

Set $W:=Fw_{3}+Fw_{2}+Fw_{1}$. We have: $V^{\sigma}=Fw_{3}+Fw_{2}+Fw_{1}$,
$V^{\tau_{1}}=F(v-w_{3})+Fw_{2}+Fw_{1}$, $V^{\tau_{2}}=Fv+Fw_{2}+Fw_{1}$. So
$V^{T}=Fw_{2}+Fw_{1}$. Clearly $|<\sigma,\tau_{2}>|=p^{2}$ and
$S(V)^{<\sigma,\tau_{2}>}=F[w_{1},w_{2},w^{p}_{3}-(\xi
w_{1})^{p-1}w_{3},v^{p}-w_{2}^{p-1}v]$.

Also $S(W)^{T}=S(W)^{<\tau_{1},\tau_{2}>}$, so by Proposition 2.7
$S(W)^{T}=F[w_{1},w_{2},a]$, with $\deg a=|<\tau_{1},\tau_{2}>|=p^{2}$.

Suppose $S(V)^{T}$ is a polynomial ring. So $S(V)^{T}=F[w_{1},w_{2},x,y]$, and
consequently $(\deg x)(\deg y)=|T|=p^{3}$ (this is so since
$V^{T}=Fw_{2}+Fw_{1}$). Hence by Proposition 2.14, $\deg x=p$, $\deg y=p^{2}$.
At least one of $\\{x,y\\}$ must be in $S(V)^{T}-S(W)$, otherwise
$\\{w_{1},w_{2},x,y\\}$ are algebraically dependent over $F$.

Suppose $x\in S(V)^{T}-S(W)$, then $S(V)^{T}\subset S(V)^{<\sigma,\tau_{2}>}$
and $x\in S(V)^{<\sigma,\tau_{2}>}$. Therefore since $\deg x=p$, we have
$x=\beta(w_{1},w_{2})+\lambda[w_{3}^{p}-(\xi
w_{1})^{p-1}w_{3}]+\mu[v^{p}-w_{2}^{p-1}v]$, $\mu\neq 0$, $\lambda,\mu\in F$,
$\beta(w_{1},w_{2})\in F[w_{1},w_{2}]$, $\deg\beta(w_{1},w_{2})=p$. Hence:

$\begin{split}0&=(\tau_{1}-1)(x)\\\ &=0+\lambda[(\tau_{1}-1)(w_{3})^{p}-(\xi
w_{1})^{p-1}(\tau_{1}-1)(w_{3})]+\mu[(\tau_{1}-1)(v)^{p}-(w_{2})^{p-1}(\tau_{1}-1)(v)]\\\
&=\lambda[w_{1}^{p}-(\xi
w_{1})^{p-1}w_{1}]+\mu[w_{1}^{p}-w_{2}^{p-1}w_{1}]=\lambda(1-\xi^{p-1})w_{1}^{p}+\mu[w_{1}^{p}-w_{2}^{p-1}w_{1}],\end{split}$

a contradiction, since $\mu\neq 0$ and $\\{w_{1},w_{2}\\}$ are algebraically
independent.

Therefore $x\in S(W)^{T}$. By the argument above $S(W)^{T}=F[w_{1},w_{2},a]$,
where $\deg a=p^{2}$. Since $\deg x=p$ we have $x\in F[w_{1},w_{2}]$, which is
another contradiction since $\\{w_{1},w_{2},x,y\\}$ is a minimal generating
set of $S(V)^{T}$, so $x$ must be algebraically independent from
$\\{w_{1},w_{2}\\}$.

Finally we show that $S(V)^{T}$ is a Gorenstein ring. Let
$\mathfrak{m}:=(w_{1},w_{2})S(V),\mathfrak{n}:=\mathfrak{m}\cap
S(V)^{I_{T}(\mathfrak{m})}$, $\mathfrak{p}:=\mathfrak{m}\cap S(V)^{T}$. We
have that $V\cap\mathfrak{m}=Fw_{2}+Fw_{1}$ and consequently
$T_{U}=I_{T}(\mathfrak{m})$, where $U:=(Fw_{2}+Fw_{1})^{\perp}\subset V^{*}$.
Since $S(V)$ is integral over $S(V)^{T_{U}}$, it follows that
$\operatorname{height}(\mathfrak{n})=\operatorname{height}(\mathfrak{m})=2$.
Moreover since $S(V)^{T}$ is normal, it follows that $S(V)^{T}_{\mathfrak{p}}$
is Cohen-Macaulay, implying by [10, Theorem 1.9] that $S(V)^{T_{U}}$ is Cohen-
Macaulay. However by inspection $T_{U}=\\{g\in T|(g-1)(V)\subseteq
Fw_{2}+Fw_{1}\\}=T$, and hence $S(V)^{T}$ is Cohen-Macaulay. Since $T$ is
generated by pseudo-reflections, it is a U.F.D, implying by the Cohen-Macaulay
property, that it is Gorenstein.

The nature of $S(V^{*})^{T}$ is surprisingly different.

######  Example 2.21.

Let $T\subset GL(V)$ be as in Example 2.20. Then $S(V^{*})^{T}$ is a
polynomial ring, and $T=T_{Fw_{1}+Fw_{2}}$.

Proof: Recall that the matrices of $\\{\sigma,\tau_{1},\tau_{2}\\}$ in $GL(V)$
where given in Example 2.20 with respect to the ordered basis
$\\{v,w_{3},w_{2},w_{1}\\}$.

Let $\\{g,f_{3},f_{2},f_{1}\\}\subset V^{*}$ be the dual basis of
$\\{w_{1},w_{2},w_{3},v\\}$, that is:
$f_{1}(v)=1$, $f_{1}(w_{3})=f_{1}(w_{2})=f_{1}(w_{1})=0$; $f_{2}(w_{3})=1$,
$f_{2}(v)=f(w_{2})=f_{2}(w_{1})=0$; $f_{3}(w_{2})=1$,
$f_{3}(v)=f_{3}(w_{3})=f_{3}(w_{1})=0$; $g(w_{1})=1$,
$g(v)=g(w_{3})=g(w_{2})=0$.

With respect to this ordered basis the matrices of $\sigma$, $\tau_{1}$,
$\tau_{2}$ are given by:

$\sigma=\begin{bmatrix}1&0&0&0\\\ 0&1&0&-1\\\ 0&0&1&0\\\
0&0&0&1\end{bmatrix},\text{ }\tau_{1}=\begin{bmatrix}1&0&-1&-1\\\ 0&1&0&0\\\
0&0&1&0\\\ 0&0&0&1\end{bmatrix},\text{ }\tau_{2}=\begin{bmatrix}1&0&-\xi&0\\\
0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&1\end{bmatrix}.$

Set $a_{1}:=g^{p}-(\tau_{1}-1)(g)^{p-1}g=g^{p}-(-f_{2}-f_{1})^{p-1}g$. Clearly
$\tau_{1}(a_{1})=a$. Set
$a=a_{1}^{p}-(\tau_{2}-1)(a_{1})^{p-1}a_{1}=a_{1}^{p}-[(-\xi
f_{2})^{p}-(-f_{2}-f_{1})^{p-1}(-\xi f_{2})]^{p-1}a_{1}$. It is easily seen to
be a $p$-polynomial of degree $p^{2}$ in $g$, and $(\sigma-1)(a)=0$ holds,
since $(\sigma-1)(g)=(\sigma-1)(f_{2})=(\sigma-1)(f_{1})=0$. Also
$\tau_{1}(a)=a$, $\tau_{2}(a)=a$.

Therefore $\\{f_{1},f_{2},f_{3},a\\}$ are homogenous elements in
$S(V^{*})^{<\tau_{1},\tau_{2}>}$, as well as algebraically independent over
$F$ and $(\deg f_{1})(\deg f_{2})(\deg f_{3})(\deg a)=\deg
a=|<\tau_{1},\tau_{2}>|=p^{2}$. Hence
$S(V^{*})^{<\tau_{1},\tau_{2}>}=S(U_{1})[a]=F[f_{1},f_{2},f_{3},a]$, where
$U_{1}:=Ff_{1}+Ff_{2}+Ff_{3}$. Consequently
$S(V^{*})^{<\tau_{1},\tau_{2},\sigma>}=(S(V^{*})^{<\tau_{1},\tau_{2}>})^{<\sigma>}=(S(U_{1})[a])^{<\sigma>}=S(U_{1})^{<\sigma>}[a]=F[f_{1},f_{2},f_{3}^{p}-(-f_{1})^{p-1}f_{3},a]$,
which is a polynomial ring.$\Box$

##  3\. Theorem A

Recall that we adhere to the convention of acting with matrices on the right,
so the action is read along rows.

######  Proposition 3.1.

Let $dim_{F}V=3$, $p=charF>0$ and $T\subset GL(V)$ a finite transvection
group, $T=<\sigma,\tau_{1},...,\tau_{m},\epsilon_{1},...,\epsilon_{n}>$, where
with respect to a fixed basis $\\{v,w_{2},w_{1}\\}$, we have:

  1. (1)

$\tau_{i}:=\begin{bmatrix}1&0&\tau_{13}^{(i)}\\\ 0&1&\tau_{23}^{(i)}\\\
0&0&1\end{bmatrix}$, $\tau_{23}^{(i)}\neq 0$, $i=1,...,m$,

  2. (2)

$\sigma:=\begin{bmatrix}1&\sigma_{12}&\sigma_{13}\\\ 0&1&0\\\
0&0&1\end{bmatrix}$, $\sigma_{12}\neq 0$,

  3. (3)

$\epsilon_{i}:=\begin{bmatrix}1&0&\epsilon_{13}^{(i)}\\\ 0&1&0\\\
0&0&1\end{bmatrix}$, $\epsilon_{13}^{(i)}\neq 0$, $i=1,...,n$.

Suppose further:

  1. (4)

$(H,H)\subseteq<\epsilon_{1},...,\epsilon_{n}>$, where
$H:=<\sigma,\tau_{1},...,\tau_{m}>$,

  2. (5)

$\\{\epsilon_{1},...,\epsilon_{n}\\}$ is a minimal generating set of
$<\epsilon_{1},...,\epsilon_{n}>$, equivalently
$\\{\epsilon_{13}^{(i)}|i=1,...,n\\}$ is $\mathbb{F}_{p}$-linearly
independent.

  3. (6)

$\\{\tau_{1|W},...,\tau_{m|W}\\}$ is a minimal generating set for the subgroup
$<\tau_{1|W},....,\tau_{m|W}>$, equivalently $\\{\tau_{23}^{(i)}|i=1,...,m\\}$
are $\mathbb{F}_{p}$-linearly independent.

Then $S(V)^{T}$ is a polynomial ring.

Proof: Set $W:=Fw_{1}+Fw_{2}$. The plan of the proof is as follows. We aim at
showing that $S(V)^{T}=F[w_{1},y,X]$, where $S(W)^{T}=F[w_{1},y]$, $\deg
y=|T_{|W}|$. This crucially uses the fact that $m\leq n$, where
$p^{m}=|T_{|W}|$ and $p^{n}=|<\epsilon_{1},...,\epsilon_{n}>|$. Also
$S(V)^{<\sigma,\epsilon_{1},...,\epsilon_{n}>}=F[w_{1},w_{2},a]$, $\deg
a=p^{n+1}$ and $X$ is obtained from the element $a$ by a sequence of $m$
adjustments. Since $(\deg w_{1})(\deg y)(\deg X)=1\cdot p^{m}\cdot
p^{n+1}=|T|$, the result will follow from Proposition 2.14.

Now
$\sigma\tau_{i}\sigma^{-1}\tau_{i}^{-1}=\begin{bmatrix}1&0&\sigma_{12}\tau^{(i)}_{23}\\\
0&1&0\\\ 0&0&1\end{bmatrix}$, $i=1,...,m$, and $\sigma_{12}\neq 0$, so by (6)
$\\{\sigma_{12}\tau_{23}^{(i)}|i=1,...,m\\}$ are $\mathbb{F}_{p}$-linearly
independent, implying by (4), (5) that $m\leq n$.

Set $z_{0}:=v$, $z_{i+1}:=z_{i}^{p}-(\epsilon_{i+1}-1)(z_{i})^{p-1}z_{i}$,
$i=0,...,n-1$. Then by Proposition 2.7 $z:=z_{n}$ is of degree $p^{n}$ and
$S(V)^{<\epsilon_{1},...,\epsilon_{n}>}=S(W)[z]=F[w_{2},w_{1},z]$. Set
$a:=z^{p}-(\sigma-1)(z)^{p-1}z$. Since $\deg(z)=p^{n}$ we have
$\deg(a)=p^{n+1}$.

Similarly let $y_{i+1}:=y_{i}^{p}-(\tau_{i+1}-1)(y_{i})^{p-1}y_{i}$,
$i=0,...,m-1$, $y_{0}=w_{2}$. Clearly $\deg(y_{i})=p^{i}$. Then by Proposition
2.7 $S(W)^{<\tau_{1},...,\tau_{i}>}=F[w_{1},y_{i}]$, $i=1,...,m$. Set
$y:=y_{m}$, so $\deg(y)=p^{m}$.

Moreover:

  1. (7)

$z_{i}=v^{p^{i}}+\sum_{j=0}^{i-1}\eta_{j}v^{p^{j}}w_{1}^{p^{i}-p^{j}}$,
$\eta_{j}\in F$, $i=0,...,n$;

  2. (8)

$(\epsilon_{j}-1)(z_{i})=\mu_{j,i}w_{1}^{p^{i}}$, $\mu_{j,i}\in F_{i}$,
$j=1,...,n$, $i=0,...,n$;

  3. (9)

$(\tau_{l}-1)(y_{j})=\theta_{l,j}w_{1}^{p^{j}}$, $\theta_{l,j}\in F$,
$l=1,...,m$, $j=0,...,m$;

  4. (10)

$(\tau_{l}-1)(z_{i})=\alpha_{l,i}w_{1}^{p^{i}}$, $\alpha_{l,i}\in F$,
$l=1,...,m$, $i=0,...,n$.

We shall next show that $(\tau_{l}-1)((\sigma-1)(z))=0$, $l=1,...,m$.

Indeed
$(\tau_{l}\sigma)(z)=(\tau_{l}\sigma\tau_{l}^{-1}\sigma^{-1})(\sigma\tau_{l})(z)=(\sigma\tau_{l})((\tau_{l}\sigma\tau_{l}^{-1}\sigma^{-1})(z))=(\sigma\tau_{l})(z)$,
where the second equality follows since
$\tau_{l}\sigma\tau_{l}^{-1}\sigma^{-1}$ is in
$<\epsilon_{1},...,\epsilon_{n}>\subset Z(T)$ and the last equality holds
since $\epsilon_{i}(z)=z$ for each $i$. Hence
$(\tau_{l}-1)((\sigma-1)(z))=(\sigma-1)((\tau_{l}-1)(z))$. Thus by (10),
$(\sigma-1)((\tau_{l}-1)(z))=(\sigma-1)(\alpha_{l,n}w_{1}^{p^{n}})=0$, for
$l=1,...,m$. Consequently $(\tau_{l}-1)((\sigma-1)(z))=0$, $l=1,...,m$, as
claimed.

Since $(\sigma-1)(z)\in F[w_{1},w_{2}]=S(W)$, it follows that
$(\sigma-1)(z)\in S(W)^{<\tau_{1},...,\tau_{m}>}=F[w_{1},y]$.

Assume, by induction, the existence of $b_{i}$ of the following form:
$b_{i}=\xi_{i}w_{1}^{p^{n+1}-p^{i-1}}y_{i-1}-\eta_{i}[(\sigma-1)(z)]^{p-1}w_{1}^{p^{n}-p^{i-1}}y_{i-1}$,
$\xi_{i},\eta_{i}\in F$, and satisfying:

  1. (11)

$\tau_{i}(a-b_{1}-\cdots-b_{i})=a-b_{1}-\cdots-b_{i}$.

Since $(\sigma-1)(z)\in S(W)^{<\tau_{1},...,\tau_{m}>}=F[w_{1},y_{m}]$, it
follows that $b_{i}\in F[w_{1},y_{i-1}]=S(W)^{<\tau_{1},...,\tau_{i-1}>}$,
$i=2,...,m$.

We shall next construct $b_{i+1}\in F[w_{1},y_{i}]$, in the above form and
satisfying:
$\tau_{i+1}(a-b_{1}-\cdots-b_{i+1})=a-b_{1}-\cdots-b_{i+1}$. We have by (9):

$\begin{split}(\tau_{i+1}-1)(b_{j})&=\xi_{j}w_{1}^{p^{n+1}-p^{j-1}}(\tau_{i+1}-1)(y_{j-1})-\eta_{j}[(\sigma-1)(z)]^{p-1}w_{1}^{p^{n}-p^{j-1}}(\tau_{i+1}-1)(y_{j-1}))\\\
&=\xi_{j}\theta_{i+1,j-1}w_{1}^{p^{n+1}}-\eta_{j}\theta_{i+1,j-1}[(\sigma-1)(z)]^{p-1}w_{1}^{p^{n}},\text{
}j=1,....,i.\end{split}$

Therefore by using (10):

  1. (12)

$(\tau_{i+1}-1)(a-b_{1}-\cdots-b_{i})$
$=(\tau_{i+1}-1)(a)-\sum_{j=1}^{i}(\tau_{i+1}-1)(b_{j})$
$=[(\tau_{i+1}-1)(z)^{p}-(\sigma-1)(z)^{p-1}(\tau_{i+1}-1)(z)]$
$\indent-\sum_{j=1}^{i}\\{\xi_{j}\theta_{i+1,j-1}w_{1}^{p^{n+1}}-\eta_{j}\theta_{i+1,j-1}[(\sigma-1)(z)]^{p-1}w_{1}^{p^{n}}\\}$
$=(\alpha_{i+1,n}^{p}-\sum_{j=1}^{i}\xi_{j}\theta_{i+1,j-1})w_{1}^{p^{n+1}}-[(\sigma-1)(z)]^{p-1}(\alpha_{i+1,n}-\sum_{j=1}^{i}\eta_{j}\theta_{i+1,j-1})w_{1}^{p^{n}}$.

Recall that by (9) $(\tau_{i+1}-1)(y_{i})=\theta_{i+1,i}w_{1}^{p^{i}}$. If
$\theta_{i+1,i}=0$ then
$S(W)^{<\tau_{1},...,\tau_{i}>}=F[w_{1},y_{i}]\subseteq
S(W)^{<\tau_{1},...,\tau_{i+1}>}$ in violation of (6).

Set
$\xi_{i+1}:=\frac{\alpha_{i+1,n}^{p}-\sum_{j=1}^{i}\xi_{j}\theta_{i+1,j-1}}{\theta_{i+1,i}}$,
$\eta_{i+1}:=\frac{\alpha_{i+1,n}-\sum_{j=1}^{i}\eta_{j}\theta_{i+1,j-1}}{\theta_{i+1,i}}$
and:
$b_{i+1}:=\xi_{i+1}w_{1}^{p^{n+1}-p^{i}}y_{i}-\eta_{i+1}[(\sigma-1)(z)]^{p-1}y_{i}w_{1}^{p^{n}-p^{i}}.$
So it follows by (12) that
$(\tau_{i+1}-1)(b_{i+1})=(\tau_{i+1}-1)(a-b_{1}-\cdots-b_{i})$, and therefore
$(\tau_{i+1}-1)(a-b_{1}-\cdots-b_{i}-b_{i+1})=0$. So the inductive step is
verified. Note that $b_{i+1}=0$ is also possible.

Consequently $a-b_{1}-\cdots-b_{i}\in
S(V)^{<\sigma,\epsilon_{1},...,\epsilon_{n},\tau_{1},...,\tau_{i}>}$ for each
$i=1,...,m$. Also $\deg(a-b_{1}-\cdots-b_{i})=p^{n+1}$, $y_{i}\in
S(V)^{<\sigma,\epsilon_{1},...,\epsilon_{n},\tau_{1},...,\tau_{i}>}$ and
$\deg(y_{i})=p^{i}$. Now $\sigma_{12}\neq 0$, so (5) implies that
$|<\sigma,\epsilon_{1},...,\epsilon_{n}>|=p^{n+1}$.

We shall next observe that
$<\tau_{1},...,\tau_{m}>\cap<\sigma,\epsilon_{1},...,\epsilon_{n}>=1$. Indeed
if
$\tau_{1}^{e_{1}}\cdots\tau_{m}^{e_{m}}\in<\sigma,\epsilon_{1},...,\epsilon_{n}>$,
with $0\leq e_{i}\leq p-1$, then this element acts trivially on $S(W)$. Hence
$\tau_{1|W}^{e_{1}}\cdots\tau_{m|W}^{e_{m}}=0$, implying by (6) that
$e_{1}=\cdots=e_{m}=0$. Consequently
$<\sigma,\epsilon_{1},...,\epsilon_{n},\tau_{1},...,\tau_{i}>=p^{n+1}p^{i}$.
Since $\\{w_{1}$, $y_{i}$, $a-b_{1}-\cdots-b_{i}\\}$ are $F$-algebraically
independent, it follows by Proposition 2.14 that
$F[w_{1},y_{i},a-b_{1}-\cdots-
b_{i}]=S(V)^{<\sigma,\epsilon_{1},...,\epsilon_{n},\tau_{1},...,\tau_{i}>}$,
for $i=1,...,m$. Therefore $S(V)^{T}=F[w_{1},y,X]$, where $X:=a-b_{1}-\cdots-
b_{m}$. $\Box$

The next result extends Proposition 3.1. The basic idea is as in Proposition
2.9.

######  Proposition 3.2.

Let $dim_{F}V=3$, $p=charF>0$ and $T\subset GL(V)$ the finite transvection
group,
$T=<\sigma_{r},...,\sigma_{1},\tau_{1},...,\tau_{m},\epsilon_{1},...,\epsilon_{n}>$,
where with respect to a fixed basis $\\{v,w_{2},w_{1}\\}$, we have:

  1. (1)

$\tau_{i}:=\begin{bmatrix}1&0&\tau_{13}^{(i)}\\\ 0&1&\tau_{23}^{(i)}\\\
0&0&1\end{bmatrix}$, $\tau_{23}^{(i)}\neq 0$, $i=1,...,m$,

  2. (2)

$\sigma_{i}:=\begin{bmatrix}1&\sigma_{12}^{(i)}&\sigma_{13}^{(i)}\\\ 0&1&0\\\
0&0&1\end{bmatrix}$, $\sigma_{12}^{(i)}\neq 0$, $i=1,...,r$,

  3. (3)

$\epsilon_{i}:=\begin{bmatrix}1&0&\epsilon_{13}^{(i)}\\\ 0&1&0\\\
0&0&1\end{bmatrix}$, $\epsilon_{13}^{(i)}\neq 0$, $i=1,...,r$.

Suppose further:

  1. (4)

$(H,H)\subseteq<\epsilon_{1},...,\epsilon_{n}>$, where
$H:=<\sigma_{r},...,\sigma_{1},\tau_{1},...,\tau_{m}>$,

  2. (5)

$\\{\epsilon_{1},...,\epsilon_{n}\\}$ is a minimal generating set of
$<\epsilon_{1},...,\epsilon_{n}>$.

Then $S(V)^{T}$ is a polynomial ring.

Proof: We may also assume:

  1. (6)

$\\{\tau_{1|W},...,\tau_{m|W}\\}$ is a minimal generating set of
$<\tau_{1|W},....,\tau_{m|W}>$. Equivalently $\\{\tau_{23}^{(i)}|i=1,...,m\\}$
is $\mathbb{F}_{p}$-linearly independent.

Indeed say $\\{\tau_{1|W},...,\tau_{r|W}\\}$, $r<m$ is a minimal generating
set for this subgroup. So for $i>r$ we have
$\tau_{i|W}\in<\tau_{1|W},...,\tau_{r|W}>$. Therefore
$\tau_{i|W}=\tau_{1|W}^{e_{1}}\cdots\tau_{r|W}^{e_{r}}$, $0\leq e_{j}\leq
p-1$, $1\leq j\leq r$ implying that
$\delta_{i}=\tau_{i}\cdot\tau_{1}^{-e_{1}}\cdots\tau_{r}^{-e_{r}}\in\begin{bmatrix}1&0&F\\\
0&1&0\\\ 0&0&1\end{bmatrix}$. Consequently we can replace $\tau_{i}$ by adding
$\delta_{i}$ to $\\{\epsilon_{1},...,\epsilon_{n}\\}$ in the generating set.

Similarly we may assume:

  1. (7)

$\\{(\sigma_{1})_{|V/Fw_{1}},...,(\sigma_{r})_{|V/Fw_{1}}\\}$ is a minimal
generating set of $<(\sigma_{1})_{|V/Fw_{1}},...,(\sigma_{r})_{|V/Fw_{1}}>$.
Equivalently $\\{\sigma_{12}^{(i)}|i=1,...,r\\}$ are $\mathbb{F}_{p}$-linearly
independent.

As a result of (4), (5), (6) and (7) we have
$|<\sigma_{j},...,\sigma_{1},\tau_{1},...,\tau_{m},\epsilon_{1},...,\epsilon_{n}>|=p^{j}\cdot
p^{m}\cdot p^{n}$, for $j=1,...,r$.

By Proposition 3.1 we have
$S(V)^{<\sigma_{1},\tau_{1},...,\tau_{m},\epsilon_{1},...,\epsilon_{n}>}=F[w_{1},y,X_{1}]$,
where $y\in S(W)^{T}$, $X_{1}=a-b_{1}-\cdots-b_{m}$ of Proposition 3.1,
$\deg(X_{1})=p^{n+1}$.

Define by induction $X_{j+1}:=X_{j}^{p}-(\sigma_{j+1}-1)(X_{j})^{p-1}X_{j}$,
$j=1,...,r-1$. Assume by induction that
$F[w_{1},y,X_{i}]=S(V)^{<\sigma_{i},...,\sigma_{1},\tau_{1},...,\tau_{m},\epsilon_{1},...,\epsilon_{n}>}$.

We shall now verify the equality
$F[w_{1},y,X_{i+1}]=S(V)^{<\sigma_{i+1},...,\sigma_{1},\tau_{1},...,\tau_{m},\epsilon_{1},...,\epsilon_{n}>}$.
It follows from (4) that
$(T,T)\subseteq<\epsilon_{1},...,\epsilon_{n}>\subseteq Z(T)$ and therefore
$<\sigma_{i},...,\sigma_{1},\tau_{1},...,\tau_{m},\epsilon_{1},...,\epsilon_{n}>$
is normal in $T$. Consequently
$S(V)^{<\sigma_{i},...,\sigma_{1},\tau_{1},...,\tau_{m},\epsilon_{1},...,\epsilon_{n}>}$
is $\sigma_{i+1}$-stable, so $\sigma_{i+1}(X_{i})\in F[w_{1},y,X_{i}]$. Since
$\deg(\sigma_{i+1}(X_{i}))=\deg(X_{i})=p^{(n+1)+i-1}>p^{m}=\deg(y)$, it
follows that $\sigma_{i+1}(X_{i})=\alpha X_{i}+t$, $t\in F[w_{1},y]$,
$\alpha\in F$. Since $\sigma_{i+1}^{p}=1$ and $F[w_{1},y]\subseteq
F[w_{1},w_{2}]$ is fixed by $\sigma_{i+1}$, it follows by iterations that
$\alpha^{p}=1$, so $\alpha=1$. Therefore
$(\sigma_{i+1}-1)((\sigma_{i+1}-1)(X_{i}))=(\sigma_{i+1}-1)(t)=0$. Hence:

$\begin{split}(\sigma_{i+1}-1)(X_{i+1})&=(\sigma_{i+1}-1)[(X_{i}^{p}-(\sigma_{i+1}-1)(X_{i})^{p-1}X_{i}]=(\sigma_{i+1}-1)[X_{i}^{p}-t^{p-1}X_{i}]\\\
&=[(\sigma_{i+1}-1)(X_{i})]^{p}-t^{p-1}(\sigma_{i+1}-1)(X_{i})=t^{p}-t^{p}=0.\end{split}$

Therefore $\\{X_{i+1},y,w_{1}\\}\subset
S(V)^{<\sigma_{i}+1,...,\sigma_{1},\tau_{1},...,\tau_{m},\epsilon_{1},...,\epsilon_{n}>}$
are algebraically independent and
$(\deg(X_{i+1}))(\deg(y))(\deg(w_{1}))=p^{(n+1)+i}\cdot p^{m}\cdot
1=|<\sigma_{i+1},...,\sigma_{1},\tau_{1},...,\tau_{m},\epsilon_{1},...,\epsilon_{n}>|$.
So again by Proposition 2.14
$F[w_{1},y,X_{i+1}]=S(V)^{<\sigma_{i+1},...,\sigma_{1},\tau_{1},...,\tau_{m},\epsilon_{1},...,\epsilon_{n}>}$.
Hence $S(V)^{T}=F[w_{1},y,X]$, where $X:=X_{n}$.$\Box$

We finally arrive at the main result of this section.

######  Theorem 3.3.

Suppose $T\subset GL(V)$ is a finite transvection group, $dim_{F}V=3$ and
$Fw_{1}\subset W=Fw_{2}+Fw_{1}\subset V$ is a flag of $T$-modules. Then
$S(V)^{T}$ is a polynomial ring. Moreover $S(V)^{T}=F[w_{1},y,X]$, with $y\in
S(W)$, if $T\not\subset\begin{bmatrix}1&0&F\\\ 0&1&F\\\ 0&0&1\end{bmatrix}$,
$\begin{bmatrix}1&F&F\\\ 0&1&0\\\ 0&0&1\end{bmatrix}$.

Proof: Let $\theta\in T$ be a transvection. Then $\theta(w_{1})=w_{1}$. Also
$\theta_{|W}$ is a transvection on $W$ (or $1_{W}$) implying that
$\theta_{|W}\in\begin{bmatrix}1&F\\\ 0&1\end{bmatrix}$ (with respect to the
basis $\\{w_{2},w_{1}\\}$). Since $det\theta=1$ it follows that
$\theta\in\begin{bmatrix}1&F&F\\\ 0&1&F\\\ 0&0&1\end{bmatrix}$, with respect
to the basis $\\{v,w_{2},w_{1}\\}$ where $v\in V-W$. Since
$rank(\theta-1_{V})=1$, it follows that either
$\theta\in\begin{bmatrix}1&F&F\\\ 0&1&0\\\ 0&0&1\end{bmatrix}$ or
$\theta\in\begin{bmatrix}1&0&F\\\ 0&1&F\\\ 0&0&1\end{bmatrix}$. Therefore each
transvection generator of $T$ has the form of either (1), (2) or (3) of
Proposition 3.2.

If $\sigma_{r}=\cdots=\sigma_{1}=1$, then by Proposition 2.9 $S(V)^{T}$ is a
polynomial ring. Similarly if $\tau_{1}=\cdots=\tau_{m}=1$, then by
Proposition 2.7 $S(V)^{T}$ is again a polynomial ring. So we may assume that
$T$ is generated by transvections as described in Proposition 3.2.

Keeping the notation of Proposition 3.2, then since $(H,H)$ is a finite
subgroup of $(T,T)\subseteq\begin{bmatrix}1&0&F\\\ 0&1&0\\\
0&0&1\end{bmatrix}$ it follows that the generators of $(H,H)$ can be added to
$<\epsilon_{1},...,\epsilon_{n}>$, if they are not already there, so condition
(4) of Proposition 3.2 can be arranged. Finally condition (5) of Proposition
3.2 can also be arranged by dropping any redundant generator. Thus all the
requirements of Proposition 3.2 are in place, and $S(V)^{T}$ is therefore a
polynomial ring.$\Box$

The next result is a direct consequence.

######  Theorem 3.4.

Let $T\subset GL(V)$ be a finite $p$-group which is generated by
transvections, where $p=\operatorname{char}F>0$ and $\dim_{F}V=3$. Then
$S(V)^{T}$ is a polynomial ring.

Proof: By a standard result for $p$-groups, there is a flag of $T$-modules
$Fw_{1}\subset W=Fw_{2}+Fw_{1}\subset V$. Therefore Theorem 3.3 is
applicable.$\Box$

##  4\. Appendix

In order to better explain the proof of Theorem 2.15 we present below a
detailed sketch of the special case $\dim_{F}V=4$. It avoids the somewhat
complicated induction appearing in the proof, resulting in a much shorter
proof where most crucial ingredients are still presented.

The proof of Theorem 2.15 when $\dim_{F}V=4$, $F=F^{p}$

We assume that $S(V)^{G}$ is Cohen-Macaulay and satisfies Serre’s $R_{2}$
condition. $G$ is generated by transvections by Lemma 2.5. By a standard
result there is a non-zero element $v_{1}\in V$ with $g(v_{1})=v_{1}$, for
each $g\in G$.

Let $H_{1}:=\\{g\in G|(g-1)(V)\subseteq Fv_{1}\\}$, $V_{2}:=V/Fv_{1}$. Then
since $H_{1}=\ker(G\rightarrow G_{|V_{2}})$, $H_{1}$ is normal in $G$ and
$G/H_{1}:=G_{2}$ acts faithfully and linearly on $V_{2}$. It is a consequence
of Proposition 2.9 that $S(V)^{H_{1}}=F[v_{1},x_{2},x_{3},x_{4}]$ is a
polynomial ring, where $\\{x_{2},x_{3},x_{4}\\}$ are homogenous elements,
$\deg x_{i}=p^{f_{i}}$, and $x_{i}-u_{i}^{p^{f_{i}}}\in v_{1}S(V)$, $u_{i}\in
V$, $i=2,3,4$. Consequently by Proposition 2.14 $|H_{1}|=p^{f_{2}}\cdot
p^{f_{2}}\cdot p^{f_{4}}$.

Since $S(V_{2})\supset
F[\bar{u}_{2}^{p^{f_{2}}},\bar{u}_{3}^{p^{f_{3}}},\bar{u}_{4}^{p^{f_{4}}}]=S(V)^{H_{1}}/v_{1}S(V)^{H_{1}}$
and the latter is a polynomial ring in $3$ variables, it follows that
$\\{\bar{u}_{2},\bar{u}_{3},\bar{u}_{4}\\}$ is a basis of $V_{2}$. This also
shows that $G_{2}$ acts faithfully on $S(V)^{H_{1}}/v_{1}S(V)^{H_{1}}$. Indeed
if $g\in G$ with $(g-1)(S(V)^{H_{1}})\subseteq v_{1}S(V)^{H_{1}}$, then
$(g-1)(u_{i})^{p^{f_{i}}}\in v_{1}S(V)$ and since the latter is prime ideal
$(g-1)(u_{i})\in v_{1}S(V)\cap V=Fv_{1}$, implying, since
$\\{v_{1},u_{2},u_{3},u_{4}\\}$ is a basis of $V$, that $g\in H_{1}$.
Consequently by Proposition 2.13, using the Cohen-Macaulay and $R_{2}$
properties, we have that
$S(V)^{G}/(v_{1})=[S(V)^{H_{1}}/v_{1}S(V)^{H_{1}}]^{G_{2}}$.

We may assume that $f_{1}\leq f_{2}\leq f_{3}$. In order to exploit the
$3$-dimensional result of $S(V_{2})^{G_{2}}$ (Theorem A),
$\\{\bar{u}_{2},\bar{u}_{3},\bar{u}_{4}\\}$ should have been a triangulating
basis for the action of $G_{2}$ on $V_{2}$. We therefore change, by
Proposition 2.11, the generators $\\{x_{2},x_{3},x_{4}\\}$ into
$\\{y_{2},y_{3},y_{4}\\}$ having the properties:

  1. (1)

$S(V)^{H_{1}}=F[v_{1},y_{2},y_{3},y_{4}]$, $\deg y_{i}=p^{f_{i}}$,
$y_{i}-w_{i}^{p^{f_{i}}}\in v_{1}S(V)$, $i=2,3,4$, and

  2. (2)

$(g-1)(y_{2})\in Fv_{1}^{p^{f_{2}}}$, $(g-1)(y_{i})\in
F[v_{1},y_{2},...,y_{i-1}]_{+}$, $i=3,4$, $\forall g\in G$.

Hence $(g-1)(\bar{w}_{2})^{p^{f_{2}}}=0$,
$(g-1)(\bar{w}_{i})^{p^{f_{i}}}\in(\bar{w}_{2},...,\bar{w}_{i-1})S(V_{2})$,
$i=3,4$, $\forall g\in G_{2}$. Since the latter is a prime ideal in
$S(V_{2})$, we have by degree consideration:

$g(\bar{w}_{2})=\bar{w}_{2},\text{
}(g-1)(\bar{w}_{i})\in(\bar{w}_{2},...,\bar{w}_{i-1})S(V_{2})\cap
V_{2}=F\bar{w}_{2}+\cdots+F\bar{w}_{i-1},\text{ }i=3,4,\text{ }\forall g\in
G_{2}.$

$G_{2}$ is generated by transvections by Lemma 2.6. We can use now the
$3$-dimensional result (Theorem A) according to the following separation into
cases, where all matrices are given with respect to the $G_{2}$-triangulating
basis $\\{\bar{w}_{2},\bar{w}_{3},\bar{w}_{4}\\}$ of $V_{2}$.

$\textbf{1: }G_{2}\subset\begin{bmatrix}1&0&F\\\ 0&1&F\\\
0&0&1\end{bmatrix},\textbf{ 2: }G_{2}\subset\begin{bmatrix}1&F&F\\\ 0&1&0\\\
0&0&1\end{bmatrix},\textbf{ 3: }G_{2}\not\subset\begin{bmatrix}1&0&F\\\
0&1&F\\\ 0&0&1\end{bmatrix},\begin{bmatrix}1&F&F\\\ 0&1&0\\\
0&0&1\end{bmatrix}.$

Case 1: This case is translated into $(g-1)(\bar{w}_{4})$,
$(g-1)(\bar{w}_{3})\in F\bar{w}_{2}$, $g(\bar{w}_{2})=\bar{w}_{2}$ $\forall
g\in G_{2}$. Therefore by Corollary 2.10:

$S(V)^{G}/(v_{1})=F[\bar{y}_{2},\bar{y}_{3},\bar{y}_{4}]^{G_{2}}=F[\bar{w}_{2}^{p^{f_{2}}},\bar{w}_{3}^{p^{f_{3}}},\bar{w}_{4}^{p^{f_{4}}}]^{G_{2}}=F[\bar{w}_{2}^{p^{f_{2}}},y,z].$

Taking $\\{x^{\prime},y^{\prime},z^{\prime}\\}\subset S(V)^{G}$, (homogenous)
preimages of $\\{\bar{w}_{2}^{p^{f_{2}}},y,z\\}$, we get that
$S(V)^{G}=F[v_{1},x^{\prime},y^{\prime},z^{\prime}]$ as needed.

Case 2: This case is translated into $g(\bar{w}_{2})=\bar{w}_{2}$,
$g(\bar{w}_{3})=\bar{w}_{3}$, $(g-1)(\bar{w}_{4})\in
F\bar{w}_{2}+F\bar{w}_{3}$, $\forall g\in G_{2}$. Therefore by Proposition 2.7
$F[\bar{w}_{2},\bar{w}_{3},\bar{w}_{4}]^{G_{2}}=F[\bar{w}_{2},\bar{w}_{3},a]$,
where $\deg a=|G_{2}|$. Therefore, using $f_{2}\leq f_{3}\leq f_{4}$,
$\\{\bar{w}_{2}^{p^{f_{2}}},\bar{w}_{3}^{p^{f_{3}}},a^{p^{f_{4}}}\\}$ is a
s.o.p. in
$S(V)^{G}/(v_{1})=F[\bar{w}_{2}^{p^{f_{2}}},\bar{w}_{3}^{p^{f_{3}}},\bar{w}_{4}^{p^{f_{4}}}]^{G_{2}}$.
Taking $\\{x^{\prime},y^{\prime},z^{\prime}\\}$ in $S(V)^{G}$, which are
homogenous preimages of
$\\{\bar{w}_{2}^{p^{f_{2}}},\bar{w}_{3}^{p^{f_{3}}},a^{p^{f_{4}}}\\}$, we get
that $\\{v_{1},x^{\prime},y^{\prime},z^{\prime}\\}$ is a s.o.p. in $S(V)^{G}$
and:

$(\deg v_{1})(\deg x^{\prime})(\deg y^{\prime})(\deg z^{\prime})=1\cdot
p^{f_{2}}\cdot p^{f_{3}}\cdot p^{f_{4}}(\deg
a)=|H_{1}||G_{2}|=|H_{1}||G/H_{1}|=|G|.$

Hence by Proposition 2.14
$S(V)^{G}=F[v_{1},x^{\prime},y^{\prime},z^{\prime}]$.

Case 3: By Theorem 3.3
$S(V_{2})^{G_{2}}=F[\bar{w}_{2},\bar{w}_{3},\bar{w}_{4}]^{G_{2}}=F[\bar{w}_{2},a,b]$,
where $a\in F[\bar{w}_{2},\bar{w}_{3}]^{G_{2}}$. Hence $(\deg a)(\deg
b)=|G_{2}|$. The argument now follows as in the previous case, using
$f_{2}\leq f_{3}\leq f_{4}$. We have that
$\\{\bar{w}_{2}^{p^{f_{2}}},a^{p^{f_{3}}},b^{p^{f_{4}}}\\}$ is a s.o.p. in
$F[\bar{w}_{2}^{p^{f_{2}}},\bar{w}_{3}^{p^{f_{3}}},\bar{w}_{4}^{p^{f_{4}}}]^{G_{2}}=S(V)^{G}/(v_{1})$.
Taking $\\{x^{\prime},y^{\prime},z^{\prime}\\}$, homogenous preimages in
$S(V)^{G}$ of $\\{\bar{w}_{2}^{p^{f_{1}}},a^{p^{f_{3}}},b^{p^{f_{4}}}\\}$, we
get that $\\{v_{1},x^{\prime},y^{\prime},z^{\prime}\\}$ is a s.o.p. in
$S(V)^{G}$ with:

$(\deg v_{1})(\deg x^{\prime})(\deg y^{\prime})(\deg z^{\prime})=1\cdot
p^{f_{2}}(\deg a)(p^{f_{3}})(\deg b)p^{f_{4}}=|H_{1}||G_{2}|=|G|.$

Hence $S(V)^{G}=F[v_{1},x^{\prime},y^{\prime},z^{\prime}]$.

##  5\. Acknowledgements

I thank Gregor Kemper for discussions regarding [10] and Lemma 2.4.

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