RINGS OF INVARIANTS OF FINITE GROUPS
J. K. VERMA
1. Introduction
In these notes of lectures delivered at various places, we discuss invariant theory of finite groups.
We begin by proving the fundamental theorem on symmetric functions. We introduce the Schur
polynomials and derive the Jacobi-Trudi identity which expresses the Schur polynomials as determinants of matrices whose entries are complete homogeneous symmetric polynomials.
We shall prove the fundamental results of Hilbert and Noether for invariant subrings of finite
subgroups of the general linear groups in the non-modular case, i.e. when the field has characteristic
zero or coprime with the order of the group. We will also derive the Molien’s formula for the Hilbert
series of the ring of invariants. We will show, through examples, that the Molien’s formula helps
us to see when to stop computing the invariants. We will calculate, quite explicitly, the ring of
invariants of the dihedral and icosahedron groups. Finally we will present a rapid treatment of
Cohen-Macaulay graded rings. The Cohen-Macaulay property is one of the crucial properties of
the rings of invariants. We will show how this property helps us in presenting the ring of invariants
in an economical manner via the so called Hironaka decomposition.
2. Symmetric Polynomials
The ring of polynomials in n variables x1 , x2 , . . . , xn with coefficients in a field k will be denoted
by k[x1 , x2 , . . . , xn ] := k[x]. We shall assume throughout this section that the characteristic of k is
zero. The symmetric group on the set [n] = {1, 2, 3, . . . , n} will be denoted by Sn .
Definition 2.1. A polynomial f (x1 , x2 , . . . , xn ) ∈ k[x] is called symmetric if for all σ ∈ Sn ,
f (x1 , x2 , . . . , xn ) = f (xσ(1) , xσ(2) , . . . , xσ(n) ).
Symmetric polynomials form a subring of k[x]. This will be denoted by k[x]Sn . Let t be a new
variable. Consider the polynomial
g(t) = (t − x1 )(t − x2 ) . . . (t − xn ) = tn − e1 tn−1 + e2 tn−2 − . . . + (−1)n en .
The coefficients e1 , e2 , . . . , en are clearly symmetric. They are given by the equations
e1 = x1 + x2 + . . . + xn
P
e2 =
1≤i<j≤n xi xj , . . . ,
en = x1 x2 . . . xn .
1
2
J. K. VERMA
Definition 2.2. The polynomials e1 , e2 , . . . , en are called the elementary symmetric polynomials in
x1 , x2 , . . . , xn .
Theorem 2.3 (Fundamental theorem on symmetric polynomials). Every symmetric polynomial is
uniquely written as a polynomial in e1 , e2 , . . . , en .
Proof: The monomials in k[x] can be totally ordered by using the degree lexicographic order. Let
α = (α1 , α2 , . . . , αn ) and β = (β1 , β2 , . . . , βn ) ∈ IN n where IN = {0, 1, 2, . . .}. Put xα1 1 xα2 2 . . . xαnn =
xα and |α| = α1 + α2 + . . . + αn . Under the degree lexicographic order: xα < xβ iff either |α| < |β|
or |α| = |β| and there an r such that α1 = β1 , . . . , αr−1 = βr−1 and αr < βr .
We now describe an algorithm which enables us to write any symmetric polynomial as a polynomial in e1 , e2 , . . . , en . Write f as a unique linear combination of monomials. The largest monomial
in f with respect to the degree lexicographic order is denoted by in(f ). If xα occurs in f then σ(xα )
also occurs in f. Hence in(f ) = cxα1 1 xα2 2 . . . xαnn where c ∈ k and α1 ≥ α2 ≥ . . . ≥ αn .
α
n−1
Consider f = f − ceα1 1 −α2 eα2 2 −α3 . . . en−1
−αn αn
en .
For any (β1 , . . . , βn ) ∈ IN n ,
n
Y
β +β
+···+βn
in eβ1 1 eβ2 2 . . . eβnn =
xi i i+1
.
i=1
Hence in(f ) < in(f ). As there are only finitely many monomials less than a given monomial, this
algorithm terminates.
For uniqueness we need to show that e1 , e2 , . . . , en are algebraically independent. Suppose
g(y1 , y2 , . . . , yn ) ∈ k[y1 , y2 , . . . , yn ] such that g(e1 , e2 , . . . , en ) = 0. Let y1α1 y2α2 . . . ynαn be a term
in g. Then
n
Y
α +α
+...+αn
in(eα1 1 eα2 2 . . . eαnn ) =
xi i i+1
.
i=1
As the map φ :
IN n
→
IN n
where
φ(α1 , α2 , . . . , αn ) = (α1 + . . . + αn , α2 + α3 + . . . + αn , . . . , αn−1 + αn , αn )
is injective, the initial form in(eα1 1 eα2 2 . . . eαnn ), where y1α1 . . . ynαn is the largest monomial in g(y1 , y2 , . . . , yn ),
will not be cancelled by any other term in g(e1 , e2 , . . . , en ). This is a contradiction. Hence e1 , e2 , . . . , en
are algebraically independent.
Power sum symmetric polynomials
Definition 2.4. The polynomials pk = xk1 + xk2 + . . . + xkn for k = 1, 2, 3, . . . are called the power
sum symmetric polynomials.
Theorem 2.5 (Newton’s identities).
e1 pk−1 − e2 pk−2 + e3 pk−3 − · · · + (−1)k−2 p1 ek−1 + (−1)k−1 kek if k ≤ n
pk =
e1 pk−1 − e2 pk−2 + · · · + (−1)n−1 en pk−n
if k > n
RINGS OF INVARIANTS OF FINITE GROUPS
3
Proof: Put
E(t) = (1 − tx1 )(1 − tx2 ) . . . (1 − txn )
= 1 − e1 t + e2 t2 − . . . + (−1)n en tn
Then log E(t) =
n
P
log(1 − txi ). Hence
i=1
n
n
∞
i=1
i=1
k=0
∞
∞
n
X X
XX
X
E 0 (t) X
xi
k
−
=
=
xi
(txi )k =
xk+1
t
=
pk tk−1
i
E(t)
(1 − txi )
Hence E 0 (t) = −E(t)
∞
P
k=0 i=1
k=1
pk tk−1 . Equate the coefficients to get the identities.
k=1
Corollary 2.6. For k ≤ n,
pk = e1
1
0
···
2e2
e1
1
0 ···
e2
e1 1 · · ·
3e3
kek ek−1 ek−2
···
0
0
0
e1
and k!ek = p1
1
p2
p3
..
.
p1
p2
..
.
2
p1
3
pk pk−1 · · ·
···
.. . 0 .
k − 1 p1 Proof: By Newton’s identities
p1
p2
p3
p4
pk










1
p1
p2
p3
..
.
=
=
=
=
=
−2
−p1
−p2
e1
e1 p1 − 2e2
e1 p2 − e2 p1 + 3e3
e1 p3 − e2 p2 + e3 p1 − 4e4
e1 pk−1 − e2 pk−2 + . . . + (−1)k−1 kek .

3
p1
pk−1 pk−2 · · ·
−4
p1
(−1)k k
















e1
e2
e3
..
.
ek



 
 
=
 

p1
p2
..
.
pk



.

By Cramer’s rule and column exchanges we obtain the formula for k!ek . For the other formula we
use the system:
e1
2e2
3e3
..
.
= p1
= p1 e1 − p2
= p1 e2 − p2 e1 + p3
kek = p1 ek−1 − p2 ek−2 + . . . + pk (−1)k−1 .
4
J. K. VERMA
Hence







1
e1
e2
e3
ek−1
−1
−e1
1
−e2
e1
−ek−2 · · ·

−1
· · · (−1)k−1











p1
p2
..
.
pk



 
 
=
 

e2
2e2
3e3
..
.
kek




.


Solve for pk to get the formula.
Complete homogeneous polynomials
Definition 2.7. The complete homogeneous polynomials hr , r = 1, 2, 3, . . . are defined as hr =
P α1 α2
x1 x2 . . . xαnn . where the sum is over all α = (α1 , . . . , αn ) ∈ IN n such that |α| = r. By convention
we set h0 = 1, hr = 0 for r < 0.
Proposition 2.8. For r = 1, 2, . . . , n we have
h1
e1
1
0
· · · 0 h2 h1
e2
1
·
·
·
0
er = .
, hr = .
.
.
.
..
..
..
..
..
h h
e
h1
r
r−1 hr−2
r
1
e1
..
.
0
1
..
.
···
···
..
.
0
0
..
.
er−1 er−2 · · · e1
.
Proof: Consider the generating function
H(t) =
∞
X
r=0
n
n
Y
Y
2 2
hr t =
(1 + xi t + xi t + · · · ) =
(1 − xi t)−1 =
r
i=1
i=1
Thus H(t)E(t) = 1. Compare the coefficients of ti for i = 1, 2, . . . , r to get
−h1
−h2
−h3
−hr
=
=
=
=
−e1
−e1 h1 + e2
−e1 h2 + e2 h1 − e3
−e1 hr−1 + e2 hr−2 − . . . + (−1)r er .
Solve for er and hr by Cramer’s rule to get the desired formulas.
Proposition 2.9. For r = 1, 2, . . . , n.
p
−1
0
0 ···
1
p
p2 −2
···
2
p2
p1 −3 · · ·
p3
r!hr = .
..
pr−1 pr−2 · · ·
pr pr−1 · · ·
−(r − 1) pr
0
0
0
..
.
1
.
E(t)
RINGS OF INVARIANTS OF FINITE GROUPS
5
Proof: By logarithmic differentiation of
H(t) =
∞
X
hr tr = Πni=1 (1 − xi t)−1
r=D
we get H 0 (t) = H(t)
P∞
r=1 pr t
r−1 .
Hence
(h1 + 2h2 t + 3h3 t2 + . . .) = (1 + h1 t + h2 t2 + . . .)(p1 + p2 t + p3 t2 + . . .).
This leads to the system of equations
p1
p2
p3
p4
..
.
=
=
=
=
h1
−p1 h1 + 2h2
−p2 h1 − p1 h2 + 3h3
−p3 h1 − p2 h2 − p1 h3 + 4h4
pr = −pr−1 h1 − pr−2 h2 − . . . + rhr .
Hence







−1
p1
p2
..
.
0
−2
p1
0
0
−3
···
···
···
pr−1 pr−2 pr−3 · · ·
0
0
0
0
0
0
p1 −r












h1
h2
..
.
hr


 
 
=
 
−p1
−p2
..
.
−pr



.

We get the desired formula by Cramer’s rule.
Schur Polynomials
A partition is any finite sequence λ = (λ1 , λ2 , . . . , λn ) of nonnegative integers in decreasing order;
λ1 ≥ λ2 ≥ . . . ≥ λr . Two partitions which differ only in the terminal zeros are regarded same. The
nonzero λi are called parts of λ and their number is called the length of λ, denoted by l(λ). Finally
the sum of the parts is called the weight of λ and it is denoted by |λ|. If |λ| = n we say λ is a
partition of n. Partitions are often represented by a diagram. For example the partition (54311) is
represented by the diagram.
◦ ◦ ◦ ◦ ◦
◦ ◦ ◦ ◦
◦ ◦ ◦
◦
◦
6
J. K. VERMA
The conjugate λ0 of a partition λ is a partition
case it will be (53321).
◦ ◦
◦ ◦
◦ ◦
◦ ◦
◦
whose diagram is the transpose of λ. In the above
◦ ◦ ◦
◦
◦
1
2
n
Let α = (α1 , α2 , . . . , αn ) ∈ IN n and xα = xα1 1 xα2 2 . . . xαnn . Put σ(xα ) = xασ(1)
xασ(2)
. . . xασ(n)
. Consider
the polynomial
X
aα (x1 , x2 , . . . , xn ) = aα =
sign(σ)σ(xα )
σ∈Sn
If τ ∈ Sn then τ (aα ) = sign(τ )aα . Thus aα is skew-symmetric. Hence aα = 0 unless all α1 , α2 , . . . , αn
are distinct. Hence we may assume that α1 > α2 > . . . > αn ≥ 0. Write
α = (λ1 , λ2 , . . . , λn ) + (n − 1, n − 2, . . . 2, 1, 0)
where (λ1 , λ2 , . . . , λn ) = λ is a partition. Put δ = (n − 1, n − 2, . . . , 2, 1, 0). Then
Y
α
aα = det(xi j ) and aδ =
(xi − xj ).
1≤i≤j≤n
Since aδ |aλ+δ , Sλ = aλ+δ /aδ is a symmetric polynomial.
Definition 2.10. The symmetric polynomial Sλ (x1 , x2 , . . . , xn ) is called the Schur polynomial corresponding to the partition λ.
Remark 2.11. It is easy to see that deg Sλ = |λ|.
Example 2.12. Let λ = (1, 1, 1). Then δ = (2, 1, 0) and
3
x1 x21 x1 3
aλ+δ
= x2 x22 x2 ÷ (x1 − x2 )(x1 − x3 )(x2 − x3 ) = x1 x2 x3
aδ
x3 x2 x 3
3
3
Example 2.13. Let λ = (3, 0, 0). Then
aλ+δ
Hence Sλ = h3 .
5
x1 x1 1 5
= x2 x2 1 = aδ h3
x5 x 1 3
3
Proposition 2.14. The set {Sλ |λ = (λ1 , λ2 , . . . , λn ), |λ| = d} is a basis of the vector space of all
symmetric polynomials of degree d.
Proof: Let A be the vector space of all antisymmetric polynomials. Put aδ = D If f is
antisymmetric then f = gD, where g is symmetric. Define the vector space isomorphism
µ : k[x]Sn → A, µ(g) = gD.
RINGS OF INVARIANTS OF FINITE GROUPS
7
Under this isomorphism the Schur polynomials are mapped to all antisymmetrized monomials.
Since antisymmetrized monomials constitute a basis of A, the Schur polynomials form a basis of
k[x]Sn .
Invariants of the alternating group
Proposition 2.15. Every element of f ∈ k[x]An is uniquely written as f = g + hD where g and h
are symmetric polynomials.
Proof: For σ ∈ Sn write σ(f ) = f (xσ(1) , xσ(2) , . . . , xσ(n) ). Put τ = (12). Set
1
1
g = (f + τ (f )) and k = (f − τ (f )).
2
2
Then g is symmetric and k is antisymmetric. Thus D|k. Write k = Dh for some symmetric
polynomial h. Then f = g + hD. Suppose g1 and h1 are symmetric and f = g1 + h1 D. Then
(g − g1 ) = (h1 − h)D. Since 0 is the only polynomial which is both symmetric and antisymmetric,
we get g = g1 and h = h1 .
Jacobi-Trudi Identity
Theorem 2.16. Let λ = (λ1 , λ2 , . . . , λn ) be a partition with atmost n parts. Let λ0 be the conjugate
partition corresponding to λ. Suppose that l(λ0 ) = number of nonzero parts of λ0 ≤ m. Then
Sλ = det(hλi −i+j )1≤i,j≤n = det(eλ0i −i+j )1≤i,j≤m .
(k)
(k)
(k)
Proof: We work in k[x1 , x2 , . . . , xn ]. Let e1 , e2 , . . . , en−1 denote the elementary symmetric
functions of the variables {x1 , x2 , . . . , x∧
k , . . . , xn }. These are obtained from e1 , e2 , . . . , en by putting
xk = 0. Consider the generating function
(k)
(k)
(k)
E (k) (t) = 1 + e1 t + e2 t2 + . . . + en−1 tn−1
= Πi6=k (1 + xi t)
Recall that H(t) =
∞
P
r=0
hr tr = Πni=1 (1 − xi t)−1 .
Hence
(1)
H(t)E (k) (−t) = (1 − xk t)−1 .
Let α = (α1 , α2 , . . . , αn ) ∈ IN n . Put



Aα = (xαj i ) = 

xα1 1
xα1 2
..
.
xα2 1
xα2 2
xα1 n
xα2 n

. . . xαn1
. . . xαn2 

.

. . . xαnn
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J. K. VERMA
Equate the coefficient of tαi on either side of (1):
xαk i
=
n
X
(k)
hαi −(n−j) (−1)n−j en−j .
j=1
Define
(k)
Hα = (hαi −(n−j) )1≤i,j≤n and M = ((−1)n−i en−i )1≤i,k≤n .
Then Aα = Hα M. Put α = δ = (n − 1, n − 2, . . . , 2, 1, 0). As det Hδ = 1, det M = det Aδ = aδ.
Hence aα = aδ det(Hα ). Put α = λ + δ to get
Sλ = aλ+δ /aδ = det(Hλ+δ ) = det(hλi −i+j ).
3. Theorems of Hilbert and Noether
Let k[x] denote the polynomial ring in the indeterminates x1 , x2 , . . . , xn . Let G < GL(n, k) be a
finite subgroup. The ring of invariants of G is the ring
k[x]G = {f ∈ k[x] | M (f ) = f for all M ∈ G}.
Any M ∈ GL(n, k) acts on the indeterminates linearly by the rule (x1 , x2 , . . . , xn )t 7−→ M (x1 , x2 , . . . , xn )t .
This action extends to all f ∈ k[x] so that f 7−→ M (f ) is an automorphism of k[x]. The fundamental problems of invariant theory are the following:
(1) Find a set of generators of C[x]G . These are called the fundamental invariants.
(2) Describe the relations among the fundamental invariants. These are called the syzygies.
(3) Write an arbitrary invariant as a polynomial in the fundamental invariants.
(4) Relate geometric properties with invariants and vice-versa.
We have seen that k[x]Sn is generated by n elementary symmetric polynomials. They are also
algebraically independent. Hence k[x]Sn has transcendence degree n over k. We now show that
this property is shared by k[x]G for all finite subgroups G of GL(n, k).
Proposition 3.1. The ring k[x]G has transcendence degree n over k.
Proof: Let trdeg denote the transcendence degree. Then trdegk k[x] = n. Thus it is enough to
show that x1 , x2 , . . . , xn are algebraic over k[x]G . Define for i = 1, 2, . . . , n;
Y
Pi (t) =
(t − g(xi )).
g∈G
k[x]G .
The coefficients of Pi (t) are in
As Pi (xi ) = 0, xi is integral over k[x]G for i = 1, 2, . . . , n.
Hence k[x] and k[x]G have same transcendence degree over k.
Let R be any ring, G a finite group of automorphisms of R such that |G| is invertible in R. Put
S = RG , the ring of invariants of G acting on R. Consider the map ρ : R −→ S defined as:
1 X
ρ(r) =
g(r).
|G|
g∈G
Then (i) ρ is S-linear (ii) ρ|S = idS . Such a map ρ : R −→ S is called the Reynolds operator of
the pair S ⊂ R.
RINGS OF INVARIANTS OF FINITE GROUPS
9
Proposition 3.2. Let S be a subring of a ring R and ρ : R → S be a Reynolds operator. Then,
(i) IR ∩ S = I for all ideals I of S.
(ii) If R is Noetherian then so is S.
Proof: (i) Let
n
P
ai ri = a ∈ S where a1 , . . . , an ∈ I and r1 , r2 , . . . , rn ∈ R. Then a = ρ(a) =
i=1
P
i ρ(ri )ai
∈ I.
(ii) Let I1 ⊂ I2 ⊂ . . . be an ascending chain of ideals in S. Then In R = In+1 R = . . . for some n
as R is Noetherian. Hence
In = In R ∩ S = In+1 R ∩ S = In+1 = . . . .
Thus S is Noetherian.
Theorem 3.3 (Hilbert’s finiteness theorem). Let G be a subgroup of GL(n, k) acting linearly
on k[x] = R. Put S = k[x]G . Suppose that there is a Reynolds operator ρ : R → S. Then S is a
finitely generated k-algebra.
Proof: Let M be the maximal ideal of S generated by homogeneous elements of positive degree.
Since R is Noetherian M R has finitely many generators. Let these be homogeneous elements
f1 , f2 , . . . , fs ∈ M.
We claim that R = k[x]G = k[f1 , f2 , . . . , fs ]. Let f ∈ R be homogeneous of degree d. Apply
induction on d. If d = 0, f ∈ k. Suppose that d > 0. Then f ∈ M. Hence ∃ g1 , . . . , gs ∈ R such
that f = g1 f1 + · · · + gs fs . Apply ρ to get f = ρ(g1 )f1 + · · · + ρ(gs )fs . We may assume that gi are
homogeneous. Then deg ρ(gi ) = deg gi = deg f − deg fi < deg f. Since ρ(gi ) are of smaller degree
than deg f, by induction ρ(g1 ), . . . , ρ(gs ) ∈ k[f1 , f2 , . . . fs ]. Hence f ∈ k[f1 , f2 , . . . , fs ].
Corollary 3.4. Let G be a finite subgroup of GL(n, k) acting linearly on k[x]. Suppose that
(|G|, char k) = 1. Then k[x]G is finitely generated k-algebra.
Theorem 3.5 (Noether’s bound). Let G ⊂ GL(n, k)
be a finite subgroup of order g such that
(g, char k) = 1. Then k[x]G is generated by atmost n+g
invariants of degree atmost g.
n
1 P
M (xα ). The
g M ∈G
action of M on k[x] gives rise to an automorphism of k[x]. Note that M (xi ) is the ith entry
P
of the column vector M (x1 , x2 , . . . , xn )t . Let f (x) =
fα xα , fα ∈ k be an invariant and put
Proof: For an integral vector (α1 , α2 , . . . , αn ) ∈ IN n we have ρ(xα ) =
M (xα ) = M (xα1 1 ) . . . M (xαnn ). Then
α
1X
fα M (x1 )α1 . . . M (xn )αn .
g
α,M
P
Thus each invariant is a k-linear combination of the invariants M M (xα ) := Jα . Let u1 , u2 , . . . , un
be another set of variables. In the polynomial
X
Sd (u) =
(u1 M (x1 ) + · · · + un M (xn ))d , d = |α|.
f = ρ(f (x1 , x2 , . . . , xn )) =
M
10
J. K. VERMA
Jα appears upto a constant factor as the coefficient of uα1 1 uα2 2 . . . uαnn . The polynomial Sd (u) is the
dth power sum in the g polynomials u1 M (x1 ) + · · · + un M (xn ). By Newton’s identities we know
that Sd (u) are polynomials in the first g power sums S1 (u), S2 (u), . . . , Sg (u). Hence all invariants
Jα where |α| > g are in the subring k[Jα | |α| ≤ g]. Hence
k[x]G = k[Jα | |α| ≤ g],
which shows that k[x]G can be generated by n+g
invariants ρ(xα ) where |α| ≤ g.
g
The next example indicates that Noether’s bound is the best possible under the given assumptions.
Theorem 3.6. Let p be a prime number and n ≥ 2. Consider the cyclic group
G = { diag(wk , wk , . . . , wk ) : k = 0, 1, . . . , p − 1}
where w = e2πi/p . Then C[x]G = C[{xα | |α| = p}].
Proof: Let ρ : C[x] → C[x]G be the Reynolds operator. Then
p−1
p−1
p−1
1 X k|α| α xα X k|α|
1 X k α1
k
αn
(w x1 ) · · · (w xn ) =
w x =
w .
ρ(x ) =
p
p
p
α
k=0
k=0
If (p, |α|) = 1 then w|α| is a primitive pth complex root of 1. Hence
k=0
p−1
P
wk|α| = 0. If p divides |α|,
k=0
then ρ(xα ) = xα . Hence C[x]G is generated as a C-algebra by all monomials of total degree p.
4. Molien’s Theorem
Theorem 4.1. Let G be a finite subgroup of GL(n, C) acting linearly on R = C[x]. Let C[x]G
i
denote the vector space generated by all homogeneous invariants of degrees i. Put H(C[x]G , λ) =
∞
P
i
dim C[x]G
i λ . Then
i=0
H(C[x]G , λ) =
1 X
1
.
|G|
det(I − λM )
M ∈G
Proof: Put g := |G|. The Reynolds operator ρ : R → RG is a C-linear map. Hence ρ induces a
linear map ρ|Ri = ρi : Ri → (RG )i where ( )i denotes elements of degree i. Clearly ρ2i = ρi . Hence
0 and 1 are the only eigenvalues of ρi . Thus rank(ρi ) = tr(ρi ) = dim(RG )i . Therefore
!
∞
∞
∞
∞
X
X
1 X X
1 X X
G
G
i
i
i
i
H(R , λ) =
dim(R )i λ =
tr(ρi )λ =
trM |Ri λ =
trM |Ri λ
g
g
i=0
M ∈G i=0
i=0
We now prove that
∞
X
i=0
trM |Ri λi =
1
.
det(I − λM )
M ∈G
i=0
RINGS OF INVARIANTS OF FINITE GROUPS
11
Since C is algebraically closed and each M has finite order, M can be diagonalised. Let
x1 , x2 , . . . , xn be a basis of eigenvectors in C[x]1 whose eigenvalues are λ1 , λ2 , . . . , λn . A basis of
C[x]i consists of all monomials of degree i in C[x]. Hence eigenvalues of M |Ri are λα1 1 . . . λαnn where
α = (α1 , α2 , . . . , αn ) ∈ IN n and |α| = i. Thus
X
trM |Ri =
λα1 1 λα2 1 . . . λαnn .
α1 +···+αn =i
Hence
n
Y
λ−1
1
det M −1
1
j
=
=
=
.
trM |Ri λ =
−1
−1
(1 − λj λ)
det(M − λI)
det(I − λM )
(λj − λ)
j=1
i=0
j=1
∞
X
n
Y
i
1
1 P
.
|G| M ∈G det(I − λM )
Therefore H(RG , λ) =
Example 4.2. Consider the quaternion group Q8 acting on C[x, y]. The Molien series is calculated
as follows
det(I − λM )
M
1 0
0 1
i 0
0 −i
0 1
−1 0
0 i
i 0
det(I − λM )
M
−1 0
0 −1
(1 −
λ)2
1+
λ2
−i 0
0 i
1 + λ2
1+
λ2
0 −1
1 0
1 + λ2
1 + λ2
0 −i
−i 0
1 + λ2
(1 + λ)2
Hence,
Q8
H(C[x, y]
1
1
1
6
1 + λ6
, λ) =
+
+
=
= 1 + 2λ4 + λ6 + · · · .
8 (1 − λ)2 (1 + λ)2 1 + λ2
(1 − λ4 )2
Hence there are 2 invariants of degree 4 and one of degree 6. The orbit of x and y is {±x, ±y, ±ix, ±iy}.
Hence
1
1 4
ρ(x4 ) =
x + x4 + y 4 + y 4 + x4 + x4 + y 4 + y 4 = (x4 + y 4 ),
8
2
ρ(x2 y 2 ) = x2 y 2 .
12
J. K. VERMA
Let J denote the Jacobian. If f1 , f2 , . . . , fn ∈ C[x1 , x2 , . . . , xn ] and M ∈ GL(n, C) then by chain
rule:
J(M (f1 ), M (f2 ), . . . M (fn )) = det M −1 J(f1 , . . . , fn ).
In the present situation, J(x4 + y 4 , x2 y 2 ) = 8(x5 y − xy 5 ). Hence α = x4 + y 4 , β = x2 y 2 , γ =
x5 y − xy 5 ∈ C[x, y]Q8 . Hence C[α, β, γ] ∈ C[x, y]Q8 . It can be verified that γ 2 = α2 β − 4β 3 . Hence
C[α, β, γ] '
C[u, v, w]
(w2
− u2 v + 4v 3 )
.
where deg u = deg v = 4 and deg w = 6. Therefore the Hilbert series is
H(C[α, β, γ], λ) =
1 − λ12
1 + λ6
=
.
(1 − λ4 )2 (1 − λ6 )
(1 − λ4 )2
Therefore C[α, β, γ] = C[x, y]Q8 .
Example 4.3. The dihedral group G = D12 has a representation in GL(3, C), namely
G = {σ i δ j |i = 0, 1, j = 0, 1, . . . , 5}
where
√



1
0
0
√1/2 − 3/2 0
σ =  0 −1
0  and δ =  3/2
1/2
0 .
0
0 −1
0
0
1

Thus the Molien series of C[x]G is
1
2
1
2
7
MG (t) =
+
+
+
.
12 (1 − t)3 (1 − t)(t2 − t − 1) (1 − t)(t2 + t + 1) (1 − t)(1 + t)2
Hence
1 + t7
= 1 + 2t2 + 3t4 + 5t6 + t7 + 7t8 + · · ·
(1 − t2 )2 (1 − t6 )
Thus there are two invariants of degree 2. Since

√
  

   1
3
x
x
x
x
−
y
2
2
 √

δ  y  =  3x + 1y  ,
σ  y  =  −y  ,
2
2
z
−z
z
z
MG (t) =
the polynomials f = x2 + y 2 and g = z 2 are two algebraically independent invariants. As f 2 , f g, g 2
are linearly independent, we have the desired number of degree 4 invariants. There is an invariant
of degree 6 which does not lie in C[f, g]. Using Reynolds operator applied to degree 6 monomials
in x, y, z, we find that h = x6 − 6x4 y 2 + 9x2 y 4 is an invariant. Since J(f, g, h) is
2x 0 6x5 − 24x3 y 2 + 18xy 4 2y 0
−12x4 y + 36x2 y 3 = −2z(−24x5 y + 72x3 y 3 − 12x5 y + 48x3 y 3 − 36xy 5 )
0 2z
0
is nonzero, f, g, h are algebraically independent. Let S = C[f, g, h]. Then
1
H(S, t) =
.
(1 − t2 )2 (1 − t6 )
Thus
MG (t) − H(S, t) = t7 + · · · .
RINGS OF INVARIANTS OF FINITE GROUPS
13
Hence we need an invariant of degree 7. The polynomial p = 3x5 yz − 10x3 y 3 z + 3xy 5 z is a degree
7 invariant. Let F, G, H, P be indeterminates. Consider the ideal
I = (f − F, g − G, h − H, p − P ) ⊂ C[x, y, z, F, G, H, P ].
Let G be a Gröbner basis of I. Then
G ∩ C[F, G, H, P ] = {P 2 − F 3 GH + GH 2 }.
Hence there is a degree 14 syzygy among the invariants constructed above. Therefore
T := C[f, g, h, p] ' C[F, G, H, P ]/(P 2 − F 3 GH + GH 2 ),
H(T, t) =
1 − t14
1 + t7
=
.
(1 − t2 )2 (1 − t6 )(1 − t7 )
(1 − t2 )2 (1 − t6 )
Therefore H(T, t) = H(C[x]G , t). This shows that the ring of invariants is C[f, g, h, p].
5. Invariants of the dihedral group D2k
The dihedral group D2k is the group of symmetries of a regular k-gon centerd at the origin. As
a subgroup of GL(2, C), it is generated by
cos 2π/k − sin 2π/k
1
0
ρ=
,r =
.
sin 2π/k cos 2π/k
0 −1
Thus D2k = ri ρj i = 0, 1; j = 1, 2, . . . , k − 1 . The matrix ρ is diagonalizable with eigenvalues
λ = e2πi/k and λ−1 = λ. Hence det(1 − ρi t) = (1 − λi t)(1 − λ−i t). The reflections rρi are all
diagonalizable with eigenvalues 1 and −1. Hence det(1 − rρi t) = (1 − t)(1 + t) = (1 − t2 ). By
Molien’s theorem
(
)
k−1
X
1
k
1
H C[x, y]D2k , t =
+
.
2k 1 − t2
(1 − λi t)(1 − λ−i t)
i=0
We now calculate the sum
k−1
X
i=0
1
(1 −
λi t)(1
− λ−i t)
by invariant theory. Consider the cyclic groups Ck generated by ρ. The matrix of ρ can be diagonalized and its diagonal form is diag(λ, λ−1 ). We have
λx
λ 0
x
=
.
0 λ−1
y
λ−1 y
Hence monomials xa y b are mapped to monomials (λx)a (λ−1 y)b = λa−b xa y b . Hence ρ(xa y b ) =
P i(a−b) a b
1 k−1
λ
x y . Therefore the monomial xa y b is an invariant if and only if (λi x)a (λ−1 y)b =
k i=0
λi(a−b) xa y b = xa y b if and only if k|a − b. Write a = ka1 + r1 , b = kb1 + s1 where 0 ≤ r, s ≤ k − 1.
14
J. K. VERMA
As a − b = k(a1 − b1 ) + (r1 − s1 ), r1 = s1 . Therefore invariant monomials are xka y kb (xy)c where
0 ≤ c ≤ k − 1 and a, b ∈ IN. Thus
Ck
C[x, y]
=
k−1
M
C[xk , y k ](xy)i .
i=0
Hence
H(C[x, y]Ck , t) =
k−1
X
i=0
t2i
.
(1 − tk )2
Therefore
k−1
1X
1
1 + t2 + t4 + · · · + t2k−2
=
.
k
(1 − λi t)(1 − λ−i t)
(1 − tk )2
i=0
We substitute the above expression into the Molien series of D2k to get
1 k(1 + t2 + · · · + t2k−2 )
k
D2k
H(C[x, y] , t) =
+
2k
1 − t2
(1 − tk )2
1
=
.
(1 − tk )(1 − t2 )
Hence the Molien series is the Hilbert series of a polynomial algebra with generators of degree 2
and degree k. The matrices in D2k are orthogonal. Hence they preserve f = x2 + y 2 . Now we look
2π
2π
for a degree k invariant. The vertices (cos p, sin p) are permuted by the action of the matrices
k
k
2π
in D2k . Put θ =
. Then for p = 0, 1, . . . , k − 1,
k
ρ(x cos pθ + y sin pθ) = (x cos θ − y sin θ) cos pθ + (x sin θ + y cos θ) sin pθ
= x cos(p − 1)θ + y sin(p − 1)θ
r(x cos pθ + y sin pθ) = x cos(−pθ) + y sin(−pθ).
Hence h =
k−1
Y
x cos pθ + y sin pθ is an invariant of degree k. The invariants f and h are algebraically
p=0
independent by the fact that
2x 2y
fx fy
det
= hx hy
h x hy
= 2(xhy − yhx ) 6= 0.
6. Invariants of the icosahedron group
The group I of symmetries of an icosahedron in R3 centerd at origin is isomorphic to the alternating group of order 60. The icosahedron has 20 faces each of which is an equilateral triangle.
The group I ⊂ SO(3) ⊂ GL(3, C). Each matrix in I represents a rotation about an axis. There
are 12 vertices and hence 6 lines joining pairs of opposite vertices. There are 15 axes joining mid
points of opposite edges and 10 axes joining the centers of opposite faces. There are 4 rotations of
angle 2πk/5, k = 1, 2, 3, 4 about the axes joining opposite vertices. There are 15 rotations through
π about the axes joining mid points of the edges. There are 2 rotations of angles 2π/3 and 4π/3
about the axes joining pairs of faces. These add upto 1 + 10.2 + 6.4 + 15.1 = 60 rotations. There
RINGS OF INVARIANTS OF FINITE GROUPS
15
are 4 conjugacy classes of rotations. Hence there are 4 types of characteristic polynomials of the
matrices in I. Put w = exp(2πi/5) and ζ = exp(2πi/3).
diagonal form
det(I − M t)

1 0 0
 0 1 0 
0 0 1
(1 − t)3
1
1
(1 − t)(1 − t2 )
15
2
1 − t3
20
3
24
5



0 1 0
 1 0 0 
0 0 1


ζ
0 0
 0 ζ −1 0 
0 0 1
No. of matrices order

wi
0 0
 0 w−i 0  (1 − t)[1 − t(wi + w−i ) + t2 ]
0
0 1
i = 1, 2, 3, 4

Hence the Molien series is given by
H(C[x, y, z]I , t) =
X
4
1
1
15
20
6
+
+
+
.
60 (1 − t)3 (1 − t)(1 − t2 ) 1 − t3
(1 − t)(1 − t(wi + w−i ) + t2 )
i=1
Consider the sum
4
X
i=1
1
i
(1 − t(w + w−i ) + t2 )
=
4
X
i=0
=
1
(1 −
wi t)(1
−
w−i t)
−
1
(1 − t)2
5(1 + t2 + t4 + t6 + t7 )
1
−
5
2
(1 − t )
(1 − t)2
Hence
H(C[x, y, z]I , t) =
1 + t15
.
(1 − t2 )(1 − t6 )(1 − t10 )
This indicates that the ring of invariants has algebraically independent elements of degree 2,6 and
10 and an invariant of degree 15. The polynomial f = x2 + y 2 + z 2 is an invariant as I consists of
orthogonal matrices. The polar axes are permuted among themselves under the action of I. They
form 3 orbits A, B, C consisting of 6, 10, and 15 axes consisting of rotations of order 5, 3 and 2.
Take the equations of 6 planes perpendicular to the axes in A and multiply them to get a degree 6
invariant. Similarly we obtain invariants of degree 10 and 15.
16
J. K. VERMA
7. Cohen-Macaulay graded rings
In this section we present a rapid treatment of Cohen-Mecaulay graded rings. Among various
classes of rings studied in commutative Algebra, these rings are most useful in applications.
Let k be an infinite field and R = ⊕∞
n=0 Rn be a Noetherian graded R0 = k-algebra. Elements
of Rn are called homogeneous of degree n. Let R+ denote the ideal generated by homogeneous
elements of positive degree and put H(R+ ) = ∪∞
n=1 Rn . The Krull dimension of R is defined to be
the maximum number of algebraically independent elements of R over k. A system of homogeneous
elements θ1 , θ2 , . . . , θd in R where d = dim R is called a homogeneous system of parameters of R
if the ideal rad(θ1 , θ2 , . . . , θd ) = R+ = the unique maximal homogeneous ideal of R. It is easy to
see that θ1 , θ2 , . . . , θd is a homogeneous system of parameters (hsop) iff R/(θ1 , . . . , θd )R is a finite
dimensional k-vector space iff R is a finite module over k[θ1 , θ2 , . . . , θd ].
Definition 7.1. A sequence a1 , a2 , . . . , an in a commutative ring R is called an R-regular sequence
if (a1 , a2 , . . . , an )R 6= R and ai is a nonzerodivisor on R/(a1 , a2 , . . . , ai−1 )R for i = 1, 2, . . . , n.
Lemma 7.2. Let b1 , b2 , . . . , bn be a regular sequence in R and b1 r1 + b2 r2 + . . . + bn rn = 0 for some
r1 , r2 , . . . , rn ∈ R. Then ri ∈ (b1 , b2 , . . . , bn ) for i = 1, 2, . . . , n.
Proof: Induct on n. The n = 1 case is clear. Suppose that b1 r1 + . . . + bn rn = 0. Then
rn ∈ (b1 , b2 , . . . , bn−1 ) : bn = (b1 , . . . , bn−1 ). Hence rn = b1 s1 + b2 s2 + . . . + bn−1 sn−1 for some
s1 , . . . , sn−1 ∈ R. Therefore b1 r1 + b2 r2 + . . . + bn (b1 s1 + . . . + bn−1 sn−1 ) = 0 Hence b1 (r1 + bn s1 ) +
b2 (r2 + bn s2 ) + · · · + bn−1 (rn−1 + bn sn−1 ) = 0 By induction hypothesis ri + bn si ∈ (b1 , b2 , . . . , bn−1 )
for i = 1, 2, . . . , n − 1. Thus (r1 , r2 , . . . , rn ) ⊆ (b1 , b2 , . . . , bn ).
Proposition 7.3. Let r1 , r2 , . . . , rn ∈ R. Then r1 , r2 , . . . , rg is a regular sequence if and only if
n
r1n1 , r2n2 , . . . , rg g is a regular sequence for any positive integers n1 , n2 , . . . , ng .
Proof: Apply induction on g. Suppose r is a nonzerodivisor in R and n is an integer. If brn = 0
then (brn−1 )r = 0. Hence brn−1 = 0. Hence induction on n shows that rn is regular for all n ≥ 1.
Conversely let rn be regular for some n ≥ 1. If br = 0 then brn = 0. Hence b = 0. Thus r is regular.
It is enough to prove the proposition for n1 = n, n2 = n3 = . . . = ng = 1. Suppose that s ∈
(r1n , r2 , . . . , ri−1 ) : ri for some i ≥ 2. Then sri = t1 r1n + t2 r2 + . . . + ti−1 ri−1 for some t1 , t2 , . . . , ti−1 ∈
R. Since r1n−1 , r2 , . . . , ri−1 is a regular sequence, s ∈ (r1n−1 , r2 , . . . , ri−1 ). Write s = r1n−1 u1 + . . . +
ri−1 ui−1 for some u1 , u2 , . . . , ui−1 ∈ R. Hence
ri (r1n−1 u1 + r2 u2 + . . . + ri−1 ui−1 ) = t1 r1n + t2 r2 + . . . + ti−1 ri−1 .
Therefore
r1n−1 (u1 ri − t1 r1 ) + r2 (u2 ri − t2 ) + . . . + ri−1 (ri ui−1 − ti−1 ) = 0.
Thus u1 ri − t1 r1 ∈ (r1n−1 , r2 , . . . , ri−1 ). Hence u1 ∈ (r1 , . . . , ri−1 ) which proves that r1n , r2 , . . . , ri is
regular sequence for all i and all n.
Lemma 7.4. Let R be a one dimensional graded k-algebra. Let a1 , a2 , . . . , an are homogeneous
elements of equal positive degree in R such that R+ = rad (a1 , . . . , an ). Then there is a k-linear
combination λ1 a1 + λ2 a2 + . . . + λn an = a such that rad (aR) = R+ .
RINGS OF INVARIANTS OF FINITE GROUPS
17
Proof: Put S = k[a1 , a2 , . . . , an ]. Then R is a finite S-module. Thus dim S = dim R = 1. As S
is a standard graded k-algebra, there exist λ1 , λ2 , . . . , λn ∈ k such that a = λ1 a1 + . . . + λn an is a
homogeneous parameter for S. Thus rad(aR) = R+ .
Lemma 7.5. Let b = (b1 , b2 , . . . , bd ) and a = (a1 , a2 , . . . , ad ) be hsops for a graded Noetherian
k-algebra R. Suppose a1 , a2 , . . . , ad have equal degree. Then there is a k-linear combination a =
λ1 a1 + λ2 + . . . + λd ad such that b1 , b2 , . . . , bd−1 , a is an hsop for R.
Proof: Apply induction on d. The case d = 1 is clear. Let m be the maximal homogeneous ideal
of R. Put S = R/(b1 , b2 , . . . , bd−1 )R. Then mS = rad (a1 , a2 , . . . , ad )S where “ − ” denotes images
in S. As dim S = 1, there is a k-linear combination λ1 a1 + . . . + λd ad = a such that a is an hsop
for S. Hence (a1 , a2 , . . . , ad−1 , a) is an hsop for R.
Theorem 7.6. Let R be a graded Noetherian k-algebra of dimension d. Let a = (a1 , a2 , . . . , ad )
and b = (b1 , b2 , . . . , bd ) be two hsops of R. If a is a regular sequence then so is b.
ar1
Proof: Induction on d. Let d = 1. Then rad (a1 ) = rad (b1 ). Hence there is an r ≥ 1 such that
= b1 c for some c ∈ R. As a1 is regular, ar1 is regular. Hence b1 is regular.
We may assume that a1 , a2 , . . . , ad have equal degree. Then there is a k-linear combination a =
λ1 a1 + · · · + λd ad such that a1 , a2 , . . . , ad−1 , a and b1 , b2 , . . . , bd−1 , a are hsops. Let S = R/aR. Then
(a1 , a2 , . . . , ad−1 ) and (b1 , b2 , . . . , bd−1 ) are hsops in S. As a1 , a2 , . . . , ad−1 is regular in S, b1 , . . . , bd−1
is regular in S. Hence a is regular in R. As a is regular on T = R/(b1 , b2 , . . . , bd−1 )R, bd is regular
on T. Hence b1 , b2 , . . . , bd is a regular sequence.
Definition 7.7. A Noetherian graded k-algebra R is called Cohen-Macaulay, abbreviated as CM,
if there is an hsop in R which is a regular sequence.
Theorem 7.8. R is CM if and only if for any hsop a1 , a2 , . . . , ad in R, R is a finitely generated
free S = k[a1 , a2 , . . . , ad ]− module.
Proof: Let R be a free S - module. As a1 , a2 , . . . , ad are algebraically independent, they
constitute an S-sequence. For, freeness of R over S yields (I : a)R = IR : aR for any ideal I of S
and a ∈ S.
Conversely let R be CM. If d = dim R = 0 then R is a vector space and S = k. Hence R is
free S-module. Let d > 0. We show that the natural map f : S/a1 S → R/a1 R is injective. As
R/a1 R is integral over f (S/a1 S), dim f (S/a1 S) = dim R/a1 R = d − 1. But S/a1 S is a polynomial
ring of dimension d − 1, hence f is injective. Therefore S/a1 S is a graded Noether normalization
of R/a1 R. Hence R/a1 R a free, finite S/a1 S-module by induction hypothesis. Pick a homogeneous
0
0
0
basis {e1 , e2 , . . . , en } of R/a1 R as an S/a1 S-module.
We prove that {e1 , . . . , en } is a free S-basis of R. First we show that R = Se1 + . . . + Sen
0
0
0
by inducting on degree. Let b ∈ R and degree b = d. Then b = s1 e1 + . . . + sn en for certain
0
homogeneous elements s1 , . . . , sn ∈ S . Hence b − s1 e1 − . . . − sn en = a1 c for some c ∈ R. Clearly
n1
P
deg c < deg b. By induction c ∈
Sei .
i=1
18
J. K. VERMA
n1
P
n1
P
P
0 0
si ei = 0. Hence si = a1 ti for some ti ∈ S. Hence a1 ti ei = 0.
i=1
i=1 P
But a1 is a nonzero divisor in R. Hence ti ei = 0. Induction on max{deg si |i = 1, 2, . . . , n} finishes
the argument.
Suppose that
si ei = 0. Then
8. Cohen-Macaulayness of rings of invariants
Theorem 8.1. Let G ⊂ GL(n, k) be a finite group acting linearly on k[x]. Suppose that (char(k), |G|) =
1. Then the ring of invariants k[x]G is CM.
Proof: Put R = k[x] and S = k[x]G and let ρ : R → S be the Reynolds operator. Let I be an
ideal of S. Then we claim that IR ∩ S = I. Suppose a1 , a2 , . . . , an ∈ I and r1 , r2 , . . . , rn ∈ R such
0
that S = a1 r1 + . . . + an rn ∈ S . Then ρ(s) = s = a1 ρ(r1 ) + . . . + an ρ(rn ) ∈ I.
Let k[a1 , a2 , . . . , an ] be a graded Noether normalization of S. Then (a1 , a2 , . . . , an )R is primary
for R+ . As R is CM, a = a1 , a2 , . . . , an is an R-regular sequence. We claim that a is also an
S-regular sequence.
Suppose that 1 ≤ i ≤ n and sai ∈ (a1 , a2 , . . . , ai−1 )S. Then sai ∈ (a1 , a2 , . . . , ai−1 )R. Hence
s ∈ (a1 , a2 , . . . , ai−1 )R ∩ S = (a1 , a2 , . . . , ai−1 ). Thus a is an S-regular sequence. Therefore S is
Cohen-Macaulay.
As k[x]G is CM, it is finite free module over any graded Noether normalization S = k[a1 , a2 , . . . , an ]
of k[x]G . Therefore there are invariants b1 , b2 , . . . , br such that
G
k[x] =
r
M
Sbi .
i=1
The polynomials a1 , a2 , . . . , an are called primary invariants and b1 , b2 , . . . , br are called secondary
invariants. The above decomposition of k[x]G is called a Hironaka decomposition.
Proposition 8.2. Let d1 , d2 , . . . , dn be degrees of a collection of primary invariants of a finite
matrix group G ⊆ GL(n, C). Then
(i) the number of secondary invariants is r = d1 d2 . . . dn /|G| and
(ii) the degrees (together with multiplicities) of the secondary invariants are the exponents of the
generating function
H(C[x]G , t) Πni=1 (1 − tdi ) = te1 + te2 + . . . + ter .
Proof: By Molien’s theorem
H(C[x]G , t) =
1 X
1
te1 + te2 + . . . + ter
=
.
|G|
det(I − M t)
Πni=1 (1 − tdi )
M ∈G
Hence
(1 − t)n
te1 + te2 + . . . + ter
1 X
= n
.
|G|
det(I − M t)
Πi=1 (1 + t + t2 + . . . + tdi −1 )
M ∈G
RINGS OF INVARIANTS OF FINITE GROUPS
19
Put t = 1 on both sides. On the right hand side we get r/d1 d2 . . . dn . On the left hand side the
only nonzero term after putting t = 1 is the one for M = I. Hence the result.
(ii) Clear.
Example 8.3. Consider the matrix group

 
 
 

0 1 0
−1 0 0
0 −1 0
 1 0 0

G =  0 1 0  ,  −1 0 0  ,  0 −1 0  ,  1 0
0 


0 0 1
0 0 −1
0
0 1
0 0 −1
G is a cyclic group of order 4. The ring of invariants C[x1 , x2 , x2 ]G = R has the Hilbert series
1
1
2
1
H(R, t) =
+
+
4 (1 − t)3 (1 + t)(1 + t2 ) (1 + t)2 (1 − t)
=
1 − t + t2 + t3
(1 + t)2 (1 + t2 )(1 − t)3
= 1 + 2t2 + 2t3 + 5t4 + 4t5 + · · ·
The polynomials θ1 = x21 + x22 , θ2 = x23 , θ3 = x41 + x42 are invariants. They are algebraically
independent by Jacobian criterion. Hence these are primary invariants. The number of secondary
invariants is deg θ1 deg θ2 deg θ3 /4 = 4. To find their degrees we find
H(C[x]G , t)(1 − t2 )2 (1 − t4 ) =
1 − t + t2 + t3
· (1 − t2 )2 (1 − t4 )
(1 + t)2 (1 + t2 )(1 − t)3
= 1 + 2t3 + t4 .
Hence e1 = 1, e2 = e3 = 3 and e4 = 4. Apply Reynolds operator to get the secondary invariants:
η1 = 1, η2 = x1 x2 x3 , η3 = x21 x3 − x22 x3 , η4 = x31 x2 − x1 x32 .
It is easy to verify that a Hironaka decomposition of C[x]G is given by
C[x]G =
4
M
C[θ1 , θ2 , θ3 ]ηi .
i=1
References
[1] W. Bruns and J. Herzog, Cohen-Macaulay rings, Cambridhe studies in advanced mathematics 39, Revised
edition, Cambridge University Press, 1998.
[2] H. Derksen and G. Kemper, Computational Invariant Theory, Springer-Verlag, Berlin, 2002.
[3] L. Smith, Polynomial invariants of finite groups, A. K. Peters, Wellesley, Mass. 1995.
[4] R. P. Stanley, Invariant theory of finite groups and their applications to combinatorics, Bulletin of Amer. Math.
Soc, New Series, vol 1 (1979),475-511.
[5] B. Sturmfels, Algorithms in invariant theory, Springer-Verlag, New York, 1993.
Department of Mathematics. Indian Institute of Technology Bombay Powai, Mumbai 400 076 INDIA
E-mail address: [email protected]