Table Of ContentAction of Dihedral Groups
Ming-chang Kang
Department of Mathematics
National Taiwan University
3
1 Taipei, Taiwan
0
2 E-mail: [email protected]
n
a
J
7
1
Abstract. Let K be any field and G be a finite group. Let G act on the rational
]
G
A function field K(x : g G) by K-automorphisms defined by g x = x for any
g h gh
∈ ·
.
h g, h G. Denote by K(G) the fixed field K(x : g G)G. Noether’s problem asks
t ∈ g ∈
a
m whether K(G) is rational (=purely transcendental) over K. We will give a brief
[
survey of Noether’s problem for abelian groups and dihedral groups, and will show
1
v
8 that Q(Dn) is rational over Q for n 10.
≤
9
9
3
.
1
0
3
1
:
v
i
X
Mathematics Subject Classification 2000: Primary 12F12, 13A50, 11R32, 14E08.
r
a
Keywords: Noether’s problem, rationality problem, dihedral groups.
This article was written for a local conference in 2005. It was circulated among a
few friends, but has not been published ever since. It is not difficult to adapt the
proof of this article so that the base field Q is replaced by a rather general field k.
1
2
1. Introduction
§
Let K be any field and G be a finite group. Let G act on the rational function
field K(x : g G) by K-automorphisms such that g x = x for any g, h G.
g h gh
∈ · ∈
Denote by K(G) the fixed field K(x : g G)G = f K(x : g G) : σ f =
g g
∈ { ∈ ∈ ·
f for any σ G . Noether’s problem asks whether K(G) is rational (=purely
∈ }
transcendental) over K [No].
Noether’s problem arose from the study of the inverse Galois problem. In
particular, if K(G) is rational and K is an infinite field, then a generic polyno-
mial for Galois G-extensions over K exists [DM; Sa1]. In case K is a Hilber-
tian field, i.e. Hilbert irreducibility theorem is valid for irreducible polynomials
f K[x ,x , ,x ] (for example, an algebraic number field or a field K which
1 2 n
∈ ···
is finitely generated over some field k so that trans deg K 1), the existence of
k
≥
a generic polynomial for Galois G-extensions over K will certainly guarantee an
infinite family of Galois field extensions of K with Galois groups isomorphic to G.
The first solution of Noether’s problem is provided by E. Fischer, a friend of
Emmy Noether introducing to her the then novel and abstract thinking of Dedekind
and Hilbert.
Theorem 1.1. (Fischer [Fi]) Let G be a finite abelian group with exponent e and
K be any field containing a primitive e-th root of unity. Then K(G) is rational over
K.
On the other hand only a handful of results for the rationality of Q(G) were
known before 1950’s. Samson Breuer was able to show that Q(Z ) and Q(Z ) are
3 6
rational over Q where Z is the cyclic group of order n [Br1]; he then showed
n
that Q(G) is rational for some transitive solvable subgroup G contained in S , the
p
symmetric group of degree p, if p = 5 or 7 [Br2]. Breuer’s results for transitive
3
solvable subgroups in S was extended by Furtw¨angler for p = 5, 7, 11 [Fr]; finally
p
Breuer himself extended these results for any prime number p 23 [Br3]. Several
≤
years later Gr¨obner proved that Q(G) is rational if G is the quaternion group of
order 8 [Gr].
Unfortunately almost all these results, except Fischer’s Theorem, were forgot-
ten after World War II. In 1955 H. Kuniyoshi and K. Masuda resumed this problem
; they called it a problem of Chevalley. Masuda rediscovered many previous re-
sults; in particular he proved that Q(Z ) is rational if p = 3, 5, 7, 11 [Ma]. To
p
the surprise of most people Swan constructed the first counter-example to Nother’s
problem in 1969 [Sw1]: Q(Z ) is not rational over Q if p = 47, 113, 233, . The
p
···
reader is referred to the survey articles [Sw1; Sw2; Sa2; Ke; Ka2] for subsequent
progress of Noether’s problem. We will remark that Noether’s problem for finite
abelian groups has been solved completely [Le].
In the remaining of this article we will focus on the rationality of Q(D ) where
n
D =< σ, τ : σn = τ2 = 1, τστ−1 = σ−1 > is the dihedral group of order 2n. As
n
mentioned before, the rationality of Q(D ) was proved for prime numbers p 23 by
p
≤
S. Breuer and Furtw¨angler. It seems strange that the answer to the rationality of
Q(D ), 3 n 10, is difficult to locate in the literature. The rationality of Q(D )
n 8
≤ ≤
can be found in [CHK, Theorem 3.1]; an easier case, the rationality of Q(D ), is
4
given in [CHK, Proposition 2.6]. In fact, K(G) is rational for any field K and any
non-abelian group G of order 8 or 16, except for the case when G is the generalized
quaternion group of order 16 [CHK; Ka2]. The task of this article is to study
the rationality problem of Q(D ), Q(D ) and Q(D ). What we will prove is the
6 9 10
following.
Theorem 1.2. Q(D ) is rational over Q for 3 n 10.
n
≤ ≤
4
Needless to say, the method in proving the above theorem may be adapted for
a more general context, which will be embodied in a separate article. One of the
purposes of this article is to illustrate some techniques of solving Noether’s problem
through concrete cases.
The article is organized as follows: Some basic tools will be recalled in Section
2. The rationality of Q(D ) (resp. Q(D ) and Q(D )) will be proved in Section
9 6 10
3 (resp. Section 4). The proof of Theorem 1.2 will be finished once we obtain
Theorem 3.1, Theorem 4.1 and Theorem 4.2.
2. Preliminaries
§
We recall some basic results which will be used in Section 3 and Section 4.
Theorem 2.1. ([CHK, Theorem 2.1]) Let L be a field and G be a finite group
acting on L(x , ,x ), the rational function field of m variables over L. Suppose
1 n
···
that
(i) for any σ G, σ(L) L;
∈ ⊂
(ii) the restriction of the action of G on L is faithful;
(iii) for any σ G,
∈
σ(x ) x
1 1
. .
.. = A(σ) .. +B(σ)
σ(xm) xm
where A(σ) GL (L) and B(σ) is an m 1 matrix over L. Then there exists
m
∈ ×
z , ,z L(x , ,x ) with L(x , ,x ) = L(z , ,z ) such that σ(z ) =
1 m 1 m 1 m 1 m i
··· ∈ ··· ··· ···
z for any σ G, any 1 i m.
i
∈ ≤ ≤
Theorem 2.2. ([CHK,Theorem 2.4]) Let G be any group whose order may be finite
or infinite. Suppose that G acts on L(x), the rational function field of one variable
5
over a field L. Assume that, for any σ G, σ(L) L, and σ(x) = a x+b for
σ σ
∈ ⊂ ·
some a ,b L with a = 0. Then L(x)G is rational over LG.
σ σ σ
∈ 6
Theorem 2.3. ([CHK, Theorem 2.3]) Let K be any field, a, b K 0 and
∈ \ { }
σ : K(x,y) K(x,y) be a K-automorphism defined by σ(x) = a/x, σ(y) = b/y.
−→
Then K(x,y)<σ> = K(u,v) where
a b
x y
− x − y
u = , v = .
ab ab
xy xy
− xy − xy
Theorem 2.4. (Hajja [Ha]) Let K be any field and G be a finite group. Suppose
that G acts on the rational function field K(x ,x ) by K-automorphisms such that,
1 2
for any σ G, for 1 i 2, σ(x ) = a (σ) xmi(σ)xni(σ) where a (σ) K 0
∈ ≤ ≤ i i · 1 2 i ∈ \ { }
and m (σ), n (σ) Z. Then K(x ,x )G is rational over K.
i i 1 2
∈
3. Q(D )
9
§
Let D =< σ, τ : σn = τ2 = 1,τστ−1 = σ−1 > be the dihedral group of order
n
2n. Let V = Q x(g) be the regular representation space of D over Q, i.e.
Lg∈Dn · n
g x(h) = x(gh) for any g, h D .
n
· ∈
Define x = x(σi)+x(σiτ) for 0 i n 1. Then
i
≤ ≤ −
σ : x x x x ,
0 1 n−1 0
7→ 7→ ··· 7→ 7→
τ : x x
i −i
7→
where the index of x is taken modulo n.
i
Clearly Q x is a faithful D -subspace of V. By Theorem 2.1, if
L0≤i≤n−1 · i n
Q(x , ,x )Dn is rational over Q, then Q(D ) = Q(x : g D )Dn is also
0 n−1 n g n
··· ∈
rational over Q.
We will prove that Q(x , ,x )Dn is rational over Q for n = 9,6,10.
0 n−1
···
6
Theorem 3.1. Q(x , ,x )D9 is rational over Q.
0 8
···
Proof. Step 1. Let ζ be a primitive 9-th root of unity and π = Gal(Q(ζ)/Q). Let
ρ π such that ρ(ζ) = ζ2. Then ρ is a generator of π.
∈
ExtendtheactionsofD andπ toQ(ζ)(x , ,x )by requiringσ(ζ) = τ(ζ) =
9 0 8
···
ζ, ρ(x ) = x for 0 i 8.
i i
≤ ≤
Define y = ζ−ijx . Then
i 0≤j≤8 j
P
σ : y ζiy ,
i i
7→
τ : y y ,
i −i
7→
ρ : y y
i 2i
7→
We find that Q(x , ,x )D9 = Q(ζ)(x , ,x )<ρ> D9 = Q(ζ)(x , ,
0 8 0 8 0
··· { ··· } ···
x )<D9,ρ> = Q(ζ)(y , ,y )<σ,τ,ρ>. Moreover τ ρ3(y ) = y for any 0 i 8.
8 0 8 i i
··· · ≤ ≤
ByTheorem2.1,ifQ(ζ)(y : i Z×)<σ,τ,ρ> isrationaloverQ, soisQ(ζ)(y , ,
i ∈ 9 0 ···
y )<σ,τ,ρ>. Hence it suffices to prove that Q(ζ)(y : i Z×)<σ,τ,ρ> is rational over
8 i ∈ 9
Q.
Step 2. Let < y ,y ,y ,y ,y ,y > be the multiplicative subgroup of Q(ζ)(y ,
1 2 4 5 7 8 0
,y ) 0 generated by y where i Z×. As an abelian group, the group
··· 8 \ { } i ∈ 9
< y ,y ,y , y ,y ,y > is isomorphic to the free abelian group Z6.
1 2 4 5 7 8
There is a natural π-modulestructure on< y ,y ,y ,y ,y ,y >. Asa module
1 2 4 5 7 8
over Λ := Z[π], < y ,y ,y ,y ,y ,y > is isomorphic to Λ. In fact,we may identify
1 2 4 5 7 8
y ,y with 1,ρ Λ respectively and write Λ =< y ,y ,y ,y ,y ,y >.
1 2 1 2 4 5 7 8
∈
Define a map
Φ : Λ =< y ,y ,y ,y ,y ,y > Z
1 2 4 5 7 8 9
−→
y j
7−→
if σ(y) = ζjy where y = ya1ya2ya4ya5ya7ya8 with a Z.
1 2 4 5 7 8 i ∈
7
Define a π-module structure on Z by ρ j = 2j. Thus Φ becomes a π-
9
·
equivariant map, i.e. Φ(λ y) = λ Φ(y) for any λ π.
· · ∈
Let M be the kernel of Φ. Then Q(ζ)(y ,y ,y ,y ,y ,y )<σ> = Q(ζ)(M).
1 2 4 5 7 8
Note that M is an ideal of Λ with [Λ : M] = 9.
By [Le, (3.3) Proposition, p. 311], M is a projective Λ-module.
Step 3. We will prove that M is a free module over Λ. In fact, we will show
that any rank-one Λ-projective module is free.
Write Λ = Z[T]/ < T6 1 >, Λ = Z[T]/ < T3+1 >, Λ = Z[T]/ < T3 1 >.
1 2
− −
(Note that Λ Λ .) It is not difficult to prove that all the Picard groups of
1 2
≃
Λ, Λ , Λ are zero by using Mayer-Vietoris sequences of K-groups [Mi, Theorem
1 2
3.3, p. 28] for the Cartesian squares
Λ Λ Λ Z[ζ ]
1 2 3
−−−−→ −−−−→
Λy Z [T]/<yT3 1 > yZ Zy
2 2 3
−−−−→ − −−−−→
where ζ is a primitive 3rd root of unity.
3
Step 4. By Step 3, we may find z < y ,y ,y ,y ,y ,y > such that Q(M) =
0 1 2 4 5 7 8
∈
Q(z ,z , ,z )where z = ρi(z ) for 1 i 5. Clearly ρ : z z z z
0 1 5 i 0 0 1 2 3
··· ≤ ≤ 7→ 7→ 7→ 7→
z z z and τρ3 is the identity map on Q(z , ,z ).
4 5 0 0 5
7→ 7→ ···
Now Q(ζ)(y ,y ,y ,y ,y ,y )<τρ3> = Q(ζ)(z , ,z )<τρ3> = Q(η)(z , ,
1 2 4 5 7 8 0 5 0
··· ···
z ) where η = ζ +ζ−1. It remains to show that Q(η)(z , ,z )<ρ> is rational.
5 0 5
···
Define u = z z , v = z +z for 0 i 2. We find that
i i i+3 i i i+3
− ≤ ≤
ρ : u u u u , v v v v .
0 1 2 0 0 1 2 0
7→ 7→ 7→ − 7→ 7→ 7→
By Theorem 2.1, it suffices to show that Q(η)(u ,u ,u )<ρ> is rational over
0 1 2
Q.
8
Step 5. Note that ρ3(u ) = u . Define v = u2, v = u /u , v = u /u .
i − i 0 0 1 1 0 2 2 1
Then Q(η)(u ,u ,u )<ρ3> = Q(η)(v ,v ,v ). Moreover,
0 1 2 0 1 2
ρ : v v v2, v v 1/(v v ).
0 7→ 0 1 1 7→ 2 7→ − 1 2
ByTheorem 2.1, ifQ(η)(v ,v )<ρ> isrationalover Q, then Q(η)(v ,v ,v )<ρ>
1 2 0 1 2
is also rational over Q.
Step 6. Define w = 1/(1 v + v v ), w = v /(1 v + v v ). Then
1 1 1 2 2 1 1 1 2
− − −
Q(η)(v ,v ) = Q(η)(w ,w ) and ρ : w w w w +1. By Theorem 2.1,
1 2 1 2 1 2 1 2
7→ 7→ − −
Q(η)(w ,w ) = Q(η)(X,Y) for some X, Y such that ρ(X) = X, ρ(Y) = Y. Hence
1 2
Q(η)(w ,w )<ρ> = Q(η)(X,Y)<ρ> = Q(η)<ρ>(X,Y) = Q(X,Y) (cid:3)
1 2
Remark. We may prove that Q(D ) is rational when n = 3, 5, 6, 7 by the same
n
method as above. For the case when n = 6, see Theorem 4.1 for another proof.
4. Q(D ) and Q(D )
6 10
§
We will prove that Q(x , ,x )Dn is rational for n = 6, 10 where σ : x
0 n−1 0
··· 7→
x x x , τ : x x .
1 n−1 0 i −i
7→ ··· 7→ 7→ 7→
Let n = 2m where m is an odd integer. (In case n =6, 10, m is actually an
odd integer.) Define y = x x , y′ = x +x where 0 i m 1. We get
i i − m+i i i m+i ≤ ≤ −
σ : y y y y , y′ y′ y′ y′,
0 7→ 1 7→ ··· 7→ m−1 7→ − 0 0 7→ 1 7→ ··· 7→ m−1 7→ 0
τ : y y , y′ y′, y y , y′ y′
0 7→ 0 0 7→ 0 i 7→ − −i i 7→ −i
where the index of y is taken modulo m.
i
By Theorem 2.1, it suffices to prove the rationality of Q(y , ,y )Dn.
0 m−1
···
Theorem 4.1. Q(y ,y ,y )D6 is rational over Q.
0 1 2
Proof. Recall that σ : y y y y , τ : y y , y y , y y .
0 1 2 0 0 0 1 2 2 1
7→ 7→ 7→ − 7→ 7→ − 7→ −
9
Define z = y /y , z = y /y . We find
1 1 0 2 2 1
σ : y y z , z z 1/(z z ),
0 0 1 1 2 1 2
7→ 7→ 7→ −
τ : y y , z z z , z 1/z .
0 0 1 1 2 2 2
7→ 7→ − 7→
By Theorem 2.2, it suffices to prove that Q(z ,z ) is rational over Q. But
1 2
Q(z ,z )<σ,τ> is rational by Theorem 2.4. (cid:3)
1 2
Theorem 4.2. Q(y , ,y )D10 is rational over Q.
0 4
···
Proof. Recall that σ : y y y y y y , τ : y y , y
0 1 2 3 4 0 0 0 1
7→ 7→ 7→ 7→ 7→ − 7→ 7→
y , y y , y y , y y .
4 2 3 3 2 4 1
− 7→ − 7→ − 7→ −
Step 1. Let ζ be a primitive 5-th root of unity and π = Gal(Q(ζ)/Q). Let
ρ π such that ρ(ζ) = ζ2. Then ρ is a generator of π.
∈
Extend the actions of D and π to Q(ζ)(y , ,y ) by requiring σ(ζ) =
10 0 4
···
τ(ζ) = ζ, ρ(y ) = y for 0 i 4. Note that Q(y , ,y )<σ,τ> = Q(ζ)(y ,
i i 0 4 0
≤ ≤ ···
,y )<σ,τ,ρ>.
4
···
For 0 i 4, define z = (σ +ζj)(y ). We find that
≤ ≤ i Qj6=i 0
σ : z ζiz ,
i i
7→ −
τ : z ζ−2iz ,
i −i
7→
ρ : ζ ζ2, z z
i 2i
7→ 7→
where the index of z is taken modulo 5.
i
By Theorem 2.1, it suffices to prove that Q(ζ)(z ,z ,z ,z )<σ,τ,ρ> is rational
1 2 3 4
over Q.
Step 2. For 2 i 4, define u = z /z . Then Q(ζ)(z ,z ,z ,z ) =
i i i−1 1 2 3 4
≤ ≤
Q(ζ)(z ,u ,u ,u ) and
1 2 3 4
σ : z ζz , u ζu for 2 i 4,
1 1 i i
7→ − 7→ ≤ ≤
τ : z ζ3z u u u , u ζ3/u , u ζ3/u , u ζ3/u ,
1 1 2 3 4 2 4 3 3 4 2
7→ 7→ 7→ 7→
ρ : z z u , u u u , u 1/(u u u ), u u u .
1 1 2 2 3 4 3 2 3 4 4 2 3
7→ 7→ 7→ 7→
10
By Theorem 2.2, it suffices to prove that Q(ζ)(u ,u ,u )<σ,τ,ρ> is rational
2 3 4
over Q.
Step 3. Define v = u5, v = u /u , v = u /u . Then Q(ζ)(u ,u ,u )<σ> =
1 2 2 4 2 3 3 2 2 3 4
Q(ζ)(v ,v ,v ) and
1 2 3
τ : v 1/(v v5), v v , v v /v ,
1 7→ 1 2 2 7→ 2 3 7→ 2 3
ρ : v v2v5v5, v 1/v , v 1/(v v2v2).
1 7→ 1 2 3 2 7→ 2 3 7→ 1 2 3
Clearly Q(ζ)(v ,v ,v )<τρ2> = Q(η)(v ,v ,v ) where η = ζ +ζ−1. Note that
1 2 3 1 2 3
Q(η) = Q(√5). It remains to find Q(√5)(v ,v ,v )<ρ>.
1 2 3
Step 4. Define t = 1/v , x = v v v2, y = v . Then Q(√5)(v ,v ,v ) =
2 1 2 3 3 1 2 3
Q(√5)(t,x,y) and
ρ : √5 √5, t 1/t, x y t/x 1/(ty) x,
7→ − 7→ 7→ 7→ 7→ 7→
ρ2 : t t, x t/x, y 1/(ty).
7→ 7→ 7→
Define
a b
x y
− x − y
u = , v =
ab ab
xy xy
− xy − xy
where a = t, b = 1/t. Apply Theorem 2.3. We get Q(√5)(t,x,y)<ρ2> = Q(√5)
(t,u,v).
Step 5. The action of ρ on u, v is given by
b a
y (x )
− y − − x
ρ : u , v .
7→ ay bx 7→ ay bx
x − y x − y
Define w = u/(tv). Then Q(√5)(t,u,v) = Q(√5)(t,v,w) and
ρ : √5 √5, w 1/w, v λ/v
7→ − 7→ − 7→
