arXiv:0704.1700v1  [math.AC]  13 Apr 2007
RETRACT RATIONALITY AND NOETHER’S PROBLEM
Ming-chang Kang
Department of Mathematics
National Taiwan University
Taipei, Taiwan, Rep. of China
E-mail: kang@math.ntu.edu.tw
Abstract. Let K be any field and G be a finite group. Let G act on the rational
function fields K(xg : g ∈ G) by K-automorphisms defined by g · xh = xgh for any
g, h ∈ G. Denote by K(G) the fixed field K(xg : g ∈ G)
G. Noether’s problem asks
whether K(G) is rational (=purely transcendental) over K. We will prove that, if K
is any field, p an odd prime number, and G is a non-abelian group of exponent p with
|G| = p3 or p4 satisfying [K(ζp) : K] ≤ 2, then K(G) is rational over K. A notion of
retract rationality is introduced by Saltman in case K(G) is not rational. We will also
show that K(G) is retract rational if G belongs to a much larger class of p-groups. In
particular, generic G-polynomials of G-Galois extensions exist for these groups.
Mathematics Subject Classification 2000: Primary 12F12, 13A50, 11R32, 14E08.
Keywords and Phrases: Noether’s problem, rationality problem, retract rational, generic
polynomial, p-groups of exponent p, flabby class maps, projective modules.
http://arxiv.org/abs/0704.1700v1
§1. Introduction
Let K be any field and G be a finite group. Let G act on the rational function field
K(xg : g ∈ G) by K-automorphisms defined by g · xh = xgh for any g, h ∈ G. Denote
K(G) := K(xg : g ∈ G)
G the fixed subfield under the action of G. Noether’s problem
asks whether K(G) is rational (= purely transcendental) over K.
Noether’s problem for abelian groups was studied by Fischer, Furtwängler, K. Ma-
suda, Swan, Voskresenskii, S. Endo and T. Miyata, Lenstra, etc. It was known that
K(Zp) is rational if Zp is a cyclic group of order p with p = 3, 5, 7 or 11. The first
counter-example was found by Swan: Q(Z47) is not rational over Q [Sw1]. However
Saltman showed that Q(Zp) is retract rational over K for any prime number p, which
is enough to ensure the existence of a generic Galois G-extension and will fulfill the
original purpose of Emmy Noether [Sa1]. For the convenience of the reader, we recall
the definition of retract rationality.
Definition 1.1. ([Sa3]) Let K ⊂ L be a field extension. We say that L is re-
tract rational over K, if there is a K-algebra R contained in L such that (i) L is the
quotient field of R, and (ii) the identity map 1R : R → R factors through a local-
ized polynomial K-algebra, i.e. there is an element f ∈ K[x1, . . . , xn] the polynomial
ring over K and there are K-algebra homomorphisms ϕ : R → K[x1, . . . , xn][1/f ] and
ψ : K[x1, . . . , xn][1/f ] → R satisfying ψ ◦ ϕ = 1R.
It is not difficult to see that “rational” ⇒ “stably rational” ⇒ “retract rational”.
One of the motivation to study Noether’s problem arises from the inverse Galois
problem. If K is an infinite field, it is known that K(G) is retract rational over K if
and only if there exists a generic Galois G-extension over K [Sa1, Theorem 5.3; Sa3,
Theorem 3.12], which guarantees the existence of a Galois G-extension of K, provided
that K is a Hilbertian field. On the other hand, the existence of a generic Galois
G-extension over K is equivalent to the existence of a generic polynomial for G. For
the relationship among these notions, see [DM]. For a survey of Noether’s problem the
reader is referred to articles of Swan and Kersten [Sw2; Ke].
Although Noether’s problem for abelian groups was investigated extensively, our
knowledge for the non-abelian Noether’s problem was amazingly scarce (see, for exam-
ple, [Ka2]). We will list below some previous results of non-abelian Noether’s problem,
which are relevant to the theme of this article.
Theorem 1.2. (Saltman [Sa2]) For any prime number p and for any field K
with char K 6= p (in particular, K may be an algebraically closed field), there is a
meta-abelian p-group G of order p9 such that K(G) is not retract rational over K. In
particular, it is not rational.
Theorem 1.3. (Hajja [Ha]) Let G be a finite group containing an abelian normal
subgroup N such that G/N is a cyclic group of order < 23. Then C(G) is rational over
Theorem 1.4. (Chu and Kang [CK]) Let p be a prime number, G be a p-group
of order ≤ p4 with exponent pe. Let K be any field such that either char K = p or char
K 6= p and K contains a primitive pe-th root of unity. Then K(G) is rational over K.
Theorem 1.5. (Kang [Ka4]) Let G be a metacyclic p-group with exponent pe,
and let K be any field such that (i) char K = p, or (ii) char K 6= p and K contains a
primitive pe-th root of unity. Then K(G) is rational over K.
Note that, in Theorems 1.3—1.5, it is assumed that the ground field contains enough
roots of unity. We may wonder whether Q(G) is rational if G is a non-abelian p-group
of small order. The answer is rather optimistic when G is a group of order 8 or 16.
Theorem 1.6. ([CHK; Ka3]) Let K be any field and G be any non-abelian group
of order 8 or 16 other than the generalized quaternion group of order 16. Then K(G)
is always rational over K.
However Serre was able to show that Q(G) is not rational when G is the generalized
quaternion group [Se, p.441–442; GMS, Theorem 33.26 and Example 33.27, p.89–90].
On the other hand, if p is an odd prime number, Saltman proves the following
theorem.
Theorem 1.7. (Saltman [Sa4]) Let p be an odd prime number and G be a non-
abelian group of order p3. If K is a field containing a primitive p-th root of unity, then
K(G) is stably rational.
The above theorem may be generalized to the case of p-groups containing a maximal
cyclic subgroup, namely,
Theorem 1.8. (Hu and Kang [HuK]) Let p be a prime number and G be a non-
abelian p-group of order pn containing a cyclic subgroup of index p. If K is any field
containing a primitive pn−2-th root of unity, then K(G) is rational over K.
In this article we will prove the following theorem.
Theorem 1.9. Let p be an odd prime number, G be the non-abelian group of ex-
ponent p and of order p3 or p4. If K is a field with [K(ζp) : K] ≤ 2, then K(G) is
rational over K.
The rationality problem of K(G) seems rather intricate if the ground field K has no
enough root of unity. We don’t know the answer to the rationality of K(G) when the
assumption that [K(ζp) : K] ≤ 2 is waived in the above theorem. On the other hand,
as to the retract rationality of K(G), a lot of information may be obtained. Before
stating our results, we recall a theorem of Saltman first.
Theorem 1.10. (Saltman [Sa1, Theorem 3.5]) Let K be a field, G = A ⋊ G0
be a semi-direct product group where A is an abelian normal subgroup of G. Assume
that gcd{|A|, |G0|} = 1 and both K(A) and K(G0) are retract rational over K. Then
K(G) is retract rational over K.
Thus the main problem is to investigate the retract rationality for p-groups. We
will prove K(G) is retract rational for many p-groups G of exponent p.
Theorem 1.11. Let p be a prime number, K be any field, and G = A ⋊ G0 be a
semi-direct product group where A is a normal elementary p-group of G and G0 is a
cyclic group of order pm. If p = 2 and charK 6= 2, assume furthermore that K(ζ2m) is
a cyclic extension of K. Then K(G) is retract rational over K.
If p is an odd prime number, a p-group of exponent p containing an abelian normal
subgroup of index p certainly satisfies the assumption in Theorem 1.11. In particular,
a p-group of exponent p and of order p3 or p4 belongs to this class of p-groups (see
[CK]). There are six p-groups of exponent p and of order p5; only four of them contain
abelian normal subgroups of index p. Previously the retract rationality of K(G) for
non-abelian p-groups, i.e. the existence of generic polynomials for such groups G, is
known only when G is of order p3 and of exponent p.
Similarly if G = A⋊G0 is a semi-direct product of p-groups such that A is a normal
subgroup of order p, and G0 is a direct product of an elementary p-group with a cyclic
group of order pm, then G also satisfies the assumption in Theorem 1.11 provided that
the assumption that K(ζ2m) is a cyclic extension of K remains in force.
The above Theorem 1.11 is deduced from the following theorem.
Theorem 1.12. Let K be any field, and G = A⋊G0 be a semi-direct product group
where A is a normal abelian subgroup of exponent e and G0 is a cyclic group of order
m. Assume that
(i) either charK = 0 or charK > 0 with charK ∤ em, and
(ii) both K(ζe) and K(ζm) are cyclic extensions of K such that gcd{m, [K(ζe) :
K]} = 1.
Then K(G) is retract rational over K.
The idea of the proof of Theorem 1.12 is to add a primitive e-th root of unity to the
ground field and the question is reduced to a question of multiplicative group actions.
It is Voskresenskii who realizes that the multiplicative group action is related to the
birational classification of algebraic tori [Vo]. However, the multiplicative group action
arising in the present situation is not the function field of an algebraic torus; it is a
new type of multiplicative group actions. Thus we need a new criterion for retract
rationality. It is the following theorem.
Theorem 1.13. Let π1 and π2 be finite abelian groups, π = π1 × π2, and L be a
Galois extension of the field K with π1 = Gal(L/K). Regard L as a π-field through the
projection π → π1. Assume that
(i) gcd{|π1|, |π2|} = 1,
(ii) char K = 0 or charK > 0 with charK ∤ |π2|, and
(iii) K(ζm) is a cyclic extension of K where m is the exponent of π2.
If M is π-lattice such that ρπ(M) is an invertible π-lattice, than L(M)
π is retract
rational over K.
The reader will find that the above theorem is an adaptation of Saltman’s criterion
for retract rational algebraic tori [Sa3, Theorem 3.14] (see Theorem 2.5). We also
formulate another criterion for retract rationality of L(M)π when π is a semi-direct
product group (see Theorem 4.3). An amusing consequence of this criterion (when
compared with Theorem 1.3) is that, if G = A ⋊ H is a semi-direct product of an
abelian normal subgroup A and a cyclic subgroup H , then C(G) is always retract
rational (see Proposition 5.2).
We will organize this paper as follows. We recall some basic facts of multiplicative
group actions in Section 2. In particular, the flabby class map which was mentioned in
Theorem 1.13 will be defined. We will give additional tools for proving Theorem 1.9
and Theorems 1.11–1.13 in Section 3. In Section 4 Theorem 1.13 and its variants will
be proved. The proof of Theorem 1.11 and Theorem 1.12 will be given in Section 5.
Section 6 contains the proof of Theorem 1.9.
Acknowledgements. I am indebted to Prof. R. G. Swan for providing a simplified
proof in Step 7 of Case 1 of Theorem 1.9 (see Section 6). The proof in a previous version
of this paper was lengthy and complicated. I thank Swan’s generosity for allowing me
to include his proof in this article.
Notations and terminology. A field extension L over K is rational if L is purely
transcendental over K; L is stably rational over K if there exist y1, . . . , yN such that
y1, . . . , yN are algebraically independent over L and L(y1, . . . , yN) is rational over K.
More generally, two fields L1 and L2 are called stably isomorphic if L1(x1, . . . , xm) is
isomorphic to L2(y1, . . . , yn) where x1, . . . , xm and y1, . . . , yn are algebraically indepen-
dent over L1 and L2 respectively.
Recall the definition of K(G) at the beginning of this section: K(G) = K(xg : g ∈
G)G. If L is a field with a finite group G acting on it, we will call it a G-field. Two
G-fields L1 and L2 are G-isomorphic if there is an isomorphism ϕ : L1 → L2 satisfying
ϕ(σ · u) = σ · ϕ(u) for any σ ∈ G, any u ∈ L1.
We will denote by ζn a primitive n-th root of unity in some extension field of K
when char K = 0 or char K = p > 0 with p ∤ n. All the groups in this article are
finite groups. Zn will be the cyclic group of order n or the ring of integers modulo n
depending on the situation from the context. Z[π] is the group ring of a finite group
π over Z. Z(G) is the center of the group G. The exponent of a group G is the
least common multiple of the orders of elements in G. The representation space of the
regular representation of G over K is denoted by W =
g∈GK · x(g) where G acts on
W by g · x(h) = x(gh) for any g, h ∈ G.
§2. Multiplicative group actions
Let π be a finite group. A π-latticeM is a finitely generated Z[π]-module such that
M is a free abelian group when it is regarded as an abelian group.
For any field K and a π-lattice M , K[M ] will denote the Laurent polynomial ring
and K(M) is the quotient field of K[M ]. Explicitly, if M =
1≤i≤m Z · xi as a free
abelian group, then K[M ] = K[x±11 , . . . , x
m ] and K(M) = K(x1, . . . , xm). Since π
acts on M , it will act on K[M ] and K(M) by K-automorphisms, i.e. if σ ∈ π and
σ · xj =
1≤i≤m aijxi ∈ M , then we define the action of σ in K[M ] and K(M) by
σ · xj =
1≤i≤m x
The multiplicative action of π onK(M) is called a purely monomial action in [HK1].
If π is a group acting on the rational function field K(x1, . . . , xm) by K-automorphism
such that σ · xj = cj(σ) ·
1≤i≤m x
i where σ ∈ π, aij ∈ Z and cj(σ) ∈ K\{0}, such a
multiplicative group action is called a monomial action. Monomial actions arise when
studying Noether’s problem for non-split extension groups [Ha; Sa5].
We will introduce another kind of multiplicative actions. Let K ⊂ L be fields and
π be a finite group. Suppose that π acts on L by K-automorphisms (but it is not
assumed that π acts faithfully on L). Given a π-lattice M , the action of π on L can
be extended to an action of π on L(M) (= L(x1, . . . , xm) if M =
1≤i≤m Z · xi) by
K-automorphisms defined as follows: If σ ∈ π and σ · xj =
1≤i≤m aijxi ∈ M , then
the multiplication action in L(M) is defined by σ · xj =
1≤i≤m x
i for 1 ≤ j ≤ m.
When L is a Galois extension of K and π = Gal(L/K) (and therefore π acts
faithfully on L), the fixed subfield L(M)π is the function field of the algebraic torus
defined over K, split by L and with character group M (see [Vo]).
We recall some basic facts of the theory of flabby (flasque) π-lattices developed by
Endo and Miyata, Voskresenskii, Colliot-Thélène and Sansuc, etc. [Vo; CTS]. We refer
the reader to [Sw2; Sw3; Lo] for a quick review of the theory.
In the sequel, π denotes a finite group unless otherwise specified.
Definition 2.1. A π-lattice M is called a permutation lattice if M has a Z-basis
permuted by π. M is called an invertible (or permutation projective) lattice, if it is
a direct summand of some permutation lattice. A π-lattice M is called a flabby (or
flasque) lattice if H−1(π′,M) = 0 for any subgroup π′ of π. (Note that H−1(π′,M)
denotes the Tate cohomology group.) Similarly, M is called coflabby if H1(π′,M) = 0
for any subgroup π′ of π.
It is known that an invertible π-lattice is necessarily a flabby lattice [Sw2, Lemma
8.4; Lo, Lemma 2.5.1].
Theorem 2.2. (Endo and Miyata [Sw3, Theorem 3.4; Lo, 2.10.1]) Let π be
a finite group. Then any flabby π-lattice is invertible if and only if all Sylow subgroups
of π are cyclic.
Denote by Lπ the class of all π-lattices, and by Fπ the class of all flabby π-lattices.
Definition 2.3. We define an equivalence relation ∼ on Fπ: Two π-lattices E1
and E2 are similar, denoted by E1 ∼ E2, if E1 ⊕ P is isomorphic to E2 ⊕ Q for some
permutation lattices P and Q. The similarity class containing E will be denoted by
[E]. Define Fπ = Fπ/ ∼, the set of all similarity classes of flabby π-lattices.
Fπ becomes a commutative monoid if we define [E1]+[E2] = [E1⊕E2]. The monoid
Fπ is called the flabby class monoid of π.
Definition 2.4. We define a map ρ : Lπ → Fπ as follows. For any π-lattice M ,
there exists a flabby resolution, i.e. a short exact sequence of π-lattices 0 → M → P →
E → 0 where P is a permutation lattice and E is a flabby lattice [Sw2, Lemma 8.5].
We define ρπ(M) = [E] ∈ Fπ. The map ρπ : Lπ → Fπ is well-defined [Sw2, Lemma
8.7]; it is called the flabby class map. We will simply write ρ instead of ρπ, if the group
π is obvious from the context.
Theorem 2.5. (Saltman [Sa3, Theorem 3.14]) Let L be a Galois extension of
K with π = Gal(L/K) and M be a π-lattice. Then ρπ(M) is invertible if and only if
L(M)π is retract rational over K.
§3. Generalities
We recall several results which will be used later.
Theorem 3.1. ([HK2, Theorem 1]) Let L be a field and G be a finite group
acting on L(x1, . . . , xm), the rational function field of m variables over L. Suppose that
(i) for any σ ∈ G, σ(L) ⊂ L;
(ii) The restriction of the action of G to L is faithful;
(iii) for any σ ∈ G,

σ(x1)
σ(xm)

= A(σ)


+B(σ)
where A(σ) ∈ GLm(L) and B(σ) is an m× 1 matrix over L.
Then L(x1, . . . , xm) = L(z1, . . . , zm) where σ(zi) = zi for all σ ∈ G, and for any
1 ≤ i ≤ m. In fact, z1, . . . , zm can be defined by


= A ·


for some A ∈ GLm(L) and for some B which is an m× 1 matrix over L. Moreover, if
B(σ) = 0 for all σ ∈ G, we may choose B = 0 in defining z1, . . . , zm.
Theorem 3.2. (Kuniyoshi [CHK, Theorem 2.5]) LetK be a field with charK =
p > 0 and G be a p-group. Then K(G) is always rational over K.
Proposition 3.3. Let π be a finite group and L be a π-field. Suppose that 0 →
M1 → M2 → N → 0 is a short exact sequence of π-lattices satisfying (i) π acts
faithfully on L(M1), and (ii) N is an invertible π-lattice. Then the π-fields L(M2) and
L(M1 ⊕N) are π-isomorphic.
Proof. We follow the proof of [Le, Proposition 1.5]. Denote L(M1)
× = L(M1)\{0}.
Consider the exact sequence of π-modules:
0 → L(M1)
× → L(M1)
× ·M2 → N → 0.
By Hilbert Theorem 90, we find that H1(π′, L(M1)
×) = 0 for any subgroup π′ ⊂ π.
Applying [Le, Proposition 1.2] we find that the above exact sequence splits. The
resulting π-morphism N → L(M1)
× · M2 provides the required π-isomorphism form
L(M2) to L(M1 ⊕N). �
Lemma 3.4. Let the assumptions be the same as in Proposition 3.3. Assume fur-
thermore that N is a permutation π-lattice. Then L(M2)
π is rational over L(M1)
Proof. By Proposition 3.3, L(M2) = L(M1)(N). Since π acts faithfully on L(M1),
we may apply Theorem 3.1 and find u1, . . . , un ∈ L(M2) such that L(M2) = L(M1)(u1,
. . . , un) with σ(ui) = ui for any σ ∈ π, any 1 ≤ i ≤ n where n = rank(N). Hence
L(M2)
π = L(M1)
π(u1, . . . , un). �
Lemma 3.5. Let π be a finite abelian group of exponent e and K be a field such
that char K = 0 or charK > 0 with charK ∤ e. If P is a permutation π-lattice, then
K(P )π = K(ζe)(M)
π0 where π0 = Gal(K(ζe)/K) and M is some π0-lattice.
Proof. We follow the standard approach to solving Noether’s problem for abelian
groups [Sw1; Sw2; Le].
Note that K(P )π = {K(ζe)(P )
〈π0〉}〈π〉 = K(ζe)(P )
〈π,π0〉 where the action of π is
extended to K(ζe)(P ) by defining g(ζe) = ζe for any g ∈ π, and the action of π0 is
extended to K(ζe)(P ) by requiring that π0 acts trivially on P .
Since π is abelian of exponent e, we may diagonalize its action on P , i.e. we may
find x1, . . . , xn ∈ K(ζe)(P ) such that n = rank(P ), g(xi)/xi ∈ 〈ζe〉 for any g ∈ π, and
K(ζe)(P ) = K(ζe)(x1, . . . , xn).
Thus K(ζe)(P )
〈π〉 = K(ζe)(y1, . . . , yn) where y1, . . . , yn are monomials in x1, . . . , xn.
LetM be the multiplicative subgroup generated by y1, . . . , yn in K(ζe)(y1, . . . , yn)\{0}.
Then M is a π0-lattice and K(ζe)(y1, . . . , yn)
π0 = K(ζe)(M)
π0. �
Proposition 3.6. Let π and K be the same as in Lemma 3.5. If K(ζe) is a cyclic
extension of K, then K(π) is retract rational over K.
Proof. We may regard the regular representation of π is given by a permutation
π-lattice. Thus K(π) = K(ζe)(M)
π0 where π0 = Gal(K(ζe)/K). Since π0 is assumed
cyclic, thus we may apply Theorem 2.2 and Theorem 2.5. �
§4. Proof of Theorem 1.13
Lemma 4.1. Let π be a finite group, M be a π-lattice. Suppose that π0 is a normal
subgroup of π and π0 acts trivially on M . Thus we may regard M as a lattice over
π/π0.
(1) M is a permutation π-lattice ⇔ So is it as a π/π0-lattice.
(2) M is an invertible π-lattice ⇔ So is it as a π/π0-lattice.
(3) M is a flabby π-lattice ⇔ So is it as a π/π0-lattice.
(4) If 0 → M → P → E → 0 is a flabby resolution of M as a π/π0-lattice, this
short exact sequence is also a flabby resolution of M as a π-lattice.
(5) ρπ(M) is an invertible π-lattice ⇔ ρπ/π0(M) is an invertible π/π0-lattice.
Proof. The properties (1)–(4) can be found in [CTS, Lemma 2, p.179–180]. As
to (5), the direction “⇐” is obvious by applying (4). For the other direction, assume
ρπ(M) is an invertible π-lattice. Let 0 → M → P → E → 0 be a flabby resolution
of M as a π-lattice. Then 0 → Mπ0 → P π0 → Eπ0 → 0 is a flabby resolution of
M =Mπ0 in the category of π/π0-lattices by [CTS, Lemma (xi), p.180]. It remains to
show that Eπ0 is invertible. Since [E] = ρπ(M) is invertible, we can find a π-lattice
N such that E ⊕N = N ′ is a permutation π-lattice. Note that N ′
π0 is a permutation
π/π0-lattice by [CTS, Lemma 2(i), p.180]. We find that E
π0 is invertible because
Eπ0 ⊕Nπ0 = (E ⊕N)π0 = N ′
π0 . �
Lemma 4.2. Let the assumptions be the same as in Theorem 1.13. If P is a
permutation π-lattice, then L(P )π is retract rational over K.
Proof. Since π1 and π2 are abelian groups with gcd{|π1|, |π2|} = 1, every subgroup
π′ of π can be written as π′ = ρ × λ where ρ is a subgroup of π1 and λ is a subgroup
of π2.
As a permutation π-lattice, we may write P =
Z[π/π(i)] where π(i) is a subgroup
of π. Write π(i) = ρi × λi where ρi ⊂ π1, λi ⊂ π2. Hence Z[π/π
(i)] = Z[π/(ρi × λi)] =
Z[(π1/ρi)× (π2/λi)].
It is not difficult to see that Z[π/π(i)] =
Z · u
kl where 1 ≤ k ≤ t = |π1/ρi|,
1 ≤ l ≤ r = |π2/λi|. Moreover, if g ∈ π1 and g
′ ∈ π2, then g · u
kl = u
g(k),l
, g′ ·
kl = u
k,g′(l)
and the homomorphisms π1 → St and π2 → Sr are induced from the
permutation representations associated to π/π(i) = (π1/ρi)× (π2/λi) where St and Sr
are the symmetric groups of degree t and r respectively.
Since π1 is faithful on L, we may apply Theorem 3.1. Explicitly, for any 1 ≤ l ≤ r,
we may find A(i) ∈ GLt(L) and define v
kl by


= A(i)


such that g · v
k,l = v
k,l for any g ∈ π1, and L(Z[π/π
(i)]) = L(u
kl : 1 ≤ k ≤ t, 1 ≤ l ≤ r.
If g′ ∈ π2, from the relation g
′ · u
kl = u
k,g′(l)
and Formula (1), we find that g′ · v
k,g′(l)
Since L(P )π1 = L(u
π1 = L(v
π1 = K(v
kl ) where i, k, l runs over index sets
which are understood, it follows that L(P )π = K(v
π2. Note that π2 acts on {v
by permutations.
By Lemma 3.5 K(v
π2 = K(ζm)(M)
π0 where m is the exponent of π2, π0 =
Gal(K(ζm)/K) and M is some π0-lattice.
By our assumption, π0 is a cyclic group. Hence ρπ0(M) is invertible by Theorem
2.2. Apply Theorem 2.5. We find that K(ζm)(M)
π0 is retract rational over K. �
Proof of Theorem 1.13 ———————————–
Step 1. Suppose that M is a π-lattice such that ρπ(M) is invertible.
Define π0 = {g ∈ π2 : g acts trivially on L(M)}. Then π/π0 acts faithfully on
L(M). Moreover, ρπ/π0(M) is invertible by Lemma 4.1.
In other words, without loss of generality we may assume that π is faithfully on
L(M) . Thus we will keep in force this assumption in the sequel.
Step 2. Since ρπ(M) is invertible, by [Sa3, Theorem 2.3, p.176], we may find π-
lattice M ′, P , Q such that P and Q are permutation lattices, 0 →M →M ′ → Q→ 0
is exact, and the inclusion map M →M ′ factors through P , i.e. the following diagram
commutes
0 ✲ M ✲ M ′ ✲ Q ✲ 0.
The remaining proof proceeds quite similar to that of [Sa3, Theorem 3.14, p.189].
Step 3. We get a commutative diagram of K-algebra morphisms from the diagram
in (2), i.e.
L[M ]π
L[P ]π
L[M ]π ✲ L[M ′]π ✲ L[Q]π
Step 4. The quotient field of L[P ]π is L(P )π, which is retract rational over K by
Lemma 4.2. Thus the identity map 1 : L[P ]π → L[P ]π factors rationally by [Sa3,
Lemma 3.5], i.e. there is a localized polynomial ring K[x1, . . . , xn][1/f ] and K-algebra
maps ϕ : L[P ]π → K[x1, . . . , xn][1/f ], ψ : K[x1, . . . , xn][1/f ] → L[P ]
π such that
ψ ◦ ϕ = 1.
It follows that the composite map g : L[M ]π → L[P ]π → L[M ′]π also factors ratio-
nally, i.e. there areK-algebra ϕ′ : L[M ]π → K[x1, . . . , xn][1/f ], ψ
′ : K[x1, . . . , xn][1/f ] →
L[M ′]π such that g = ψ′ ◦ ϕ′.
Step 5. By Lemma 3.4 L(M ′)π is rational over L(M)π. (This is the only one step
we use the assumption that π is faithful on L(M).)
Now we may apply [Sa3, Proposition 3.6(b), p.183] where, in the notation of [Sa3],
we take S = T = L[M ]π, ϕ is the identity map on L(M)π. We conclude that
1 : L[M ]π → L[M ]π factors rationally, i.e. L(M)π is retract rational over K. �
Here is a variant of Theorem 1.13.
Theorem 4.3. Let π be a finite group, 0 → π1 → π → π2 → 1 is a group extension,
and L be a Galois extension of the field K with π2 = Gal(L/K). Let π act on L through
the projection π → π2. Assume that
(i) π1 is an abelian group of exponent e with ζe ∈ L,
(ii) the extension 0 → π1 → π → π2 → 1 splits, and
(iii) every Sylow subgroup of π2 is cyclic.
If M is a π-lattice such that ρπ(M) is an invertible lattice, then L(M)
π is retract
rational over K.
Proof. The proof is very similar to the proof of Theorem 1.13.
We claim that L(P )π is retract rational for any permutation π-lattice P .
For the proof, we will use [Sa5, Theorem 2.1, p.546]. We will show that (c) of
[Sa5, Theorem 2.1] is valid, which will guarantee that L(P )π is retract rational. By
assumption (iii), part (d) of [Sa5, Theorem 2.1] is valid by Theorem 2.2. It remains to
check that the embedding problem of L/K and the extension 0 → π1 → π → π2 → 1
is solvable. But this is the well-known split embedding problem [ILF, Theorem 1.9,
p.12].
Now define π0 = {g ∈ π : g acts trivially on L(M)}. Note that π0 ⊂ π1. The
remaining proof is the same as in the proof of Theorem 1.13 and is omitted. �
Corollary 4.4. Let π be an abelian group of exponent e, K be a field with ζe ∈ K.
Suppose that M is a π-lattice and π acts on K(M) by K-automorphisms. If ρπ(M) is
an invertible module, than K(M)π is retract rational over K.
§5. Proof of Theorems 1.11 and 1.12
Proof of Theorem 1.12 ———————————-
Step 1. Write G = A ⋊ G0 where G0 =< σ > is a cyclic group of order m, and
A = A1 ×A2 × · · · ×Ar with each Ai =< ρi > being a cyclic group of order ei, e = e1
and er | er−1, · · · , e2 | e1.
Define Bi =< ρi, · · · , ρ̂i, · · · , ρr > to be the subgroup of A by deleting the i-th direct
factor.
Let Gal(K(ζe)/K) =< τ > be a cyclic group of order n. Write ζ = ζe and τ ·ζ = ζ
for some integer a.
For 1 ≤ i ≤ r, choose ζi ∈< ζ > such that ζi is a primitive ei-th root of unity. Note
that τ · ζi = ζ
Let ni = [K(ζi) : K] for 1 ≤ i ≤ r. Note that n = n1.
Step 2. Let W =
g∈GK · x(g) be the representation space of the regular repre-
sentation of G. Then
K(G) = K(W )G = {K(ζ)(W )<τ>}G = K(ζ)(W )<G,τ>
where the action of G and τ are extended to K(ζ)(W ) by requiring that G and τ act
trivially on K(ζ) and W respectively.
Step 3. For 1 ≤ i ≤ r, define
l∈Zei , g∈Bi
ζ li · x(ρ
i · g).
Note that ρi · ui = ζ
i ui and, if j 6= i, then ρi · ui = ui.
For 1 ≤ i ≤ r, write χi to be the character χi : A −→ K(ζ)
× such that g · ui =
χi(g) · ui. Since χ1, · · · , χr are distinct characters on A, it follows that u1, · · · , ur are
linearly independent vectors in K(ζ)⊗K W .
Moreover, the subgroup A acts faithfully on
1≤i≤rK(ζ)ui.
Step 4. For 1 ≤ i ≤ r, j ∈ Zni , define
vij =
l∈Zei , g∈Bi
i · x(ρ
i · g).
Thus we have elements v1,1, · · · , v1,n1 , v2,1, · · · , v2,n2, · · · , vr,nr ; in total we have n1 +
n2 + · · ·+ nr such elements.
As in Step 3, these elements vij are linearly independent. Note that τ · vij = vi,j+1.
Step 5. For 1 ≤ i ≤ r, j ∈ Zni , k ∈ Zm. define
xi,j,k =
l∈Zei , g∈Bi
i · x(σ
k · ρli · g).
We have in total m(n1 + n2 + · · ·+ nr) such elements xi,j,k.
Note that xi,j,0 = vij , and the vector xi,j,k is just a “translation” of the vector xi,j,0
in the space
g∈GK(ζ)x(g) (with basis x(g), g ∈ G). Thus these vectors xi,j,k are
linearly independent. Note that σ · vi,j,k = vi,j,k+1. Moreover, the group < G, τ > acts
faithfully on
K(ξ) · xi,j,k.
Apply Theorem 3.1. We find K(G) = K(ζ)(xijk : 1 ≤ i ≤ r, j ∈ Zni, k ∈
Zm)(w1, · · · , ws) where s = |G| −m(n1 + n2 + · · ·+ nr), and λ(wi) = wi for any i, any
λ ∈< G, τ >.
Step 6. We will consider the fixed fieldK(ζ)(xijk : 1 ≤ i ≤ r, j ∈ Zni , k ∈ Zm)
<G,τ>.
Let π =< σ, τ >, Λ = Z[π]. Let N =< xijk : 1 ≤ i ≤ r, j ∈ Zni , k ∈ Zm > be the
multiplicative subgroup generated by these xijk in K(ζ)(xijk : 1 ≤ i ≤ r, j ∈ Zni , k ∈
Zm) \ {0}. Note that N is a Λ-module. In fact, N is a permutation π-lattice.
Define
Φ : N −→ Zn1 × Zn2 × · · · × Znr
by Φ(x) = (l̄1, l̄2, · · · , l̄r) ∈ Zn1 × Zn2 × · · · × Znr , if x =
ijk (with bijk ∈ Z) and
ρ1(x) = ζ
1 x, ρ2(x) = ζ
2 x, · · ·, ρr(x) = ζ
Define M = Ker(Φ). We find that K(ζ)(xijk : 1 ≤ i ≤ r, j ∈ Zni , k ∈ Zm)
K(ζ)(M). Thus it remains to find K(ζ)(M)π. Note that M is a π-lattice.
Step 7. Since gcd{m, n} = 1, it follows that π is a cyclic group. Hence ρπ(M) is
invertible by Theorem 2.2.
Apply Theorem 1.13. We find that K(ζ)(M)π is retract rational over K. Since
K(G) is rational over K(ζ)(M)π, it follows that K(G) is also retract rational over K
by [Sa3, Proposition 3.6(a), p.183]. �
Proof of Theorem 1.11 ——————————-
If charK = p, apply Theorem 3.2. Thus K(G) is rational. In particular it is retract
rational.
¿From now on we will assume that charK 6= p. It is not difficult to verify that all
the assumptions of Theorem 1.12 are valid in the present situation. Hence the result.
Corollary 5.1. Let K be a field, p be any prime number, G = A⋊ G0 be a semi-
direct product of p-groups where A is a normal abelian subgroup of exponent pe and G0
is a cyclic group of order pm. If charK 6= p, assume that (i) both K(ζpe) and K(ζpm)
are cyclic extensions of K and (ii) p ∤ [K(ζpe) : K]. Then K(G) is retract rational over
Proposition 5.2. Let K be a field and G = A ⋊ G0 be a semi-direct product.
Assume that
(i) A is an abelian normal subgroup of exponent e in G and G0 is a cyclic group of
order m.
(ii) either charK = 0 or charK > 0 with charK ∤ em, and
(iii) ζe ∈ K and K(ζm) is a cyclic extension of K.
If G −→ GL(V ) is a finite-dimensional linear representation of G over K, then
K(V )G is retract rational over K.
Proof. Decompose V into V =
Vχ where χ : A −→ K \ {0} runs over linear
characters of A and Vχ = {v ∈ V : g · v = χ(g) · v for all g ∈ A}
It is easy to see that σ(Vχ) = Vχ′ for some χ
′. Suppose Vχ1, Vχ2, · · · , Vχs is an orbit
of σ, i.e. σ(Vχj ) = Vχj+1 and σ(Vχs) = Vχ1. Choose a basis v1, · · · , vt of Vχ1. It follows
that {σj(vi) : 1 ≤ i ≤ t, 0 ≤ j ≤ s− 1} is a basis of ⊕1≤j≤sVχj .
In this way, we may find a basis w1, · · · , wn in V such that (i) for any g ∈ A, any
1 ≤ i ≤ r, g · wi = αwi for some α ∈ K; and (ii) σ · wi = wj for some j.
It follows that K(V )A = K(w1, · · · , wn)
A = K(u1, · · · , un) where u1, · · · , un are
monomials in w1, · · · , wn. Thus K(u1, · · · , un) = K(M) for some G0-lattice M . It
follows that K(V )G = {K(V )A}G0 = K(M)<σ>.
By Theorem 2.2, ρG0(M) is invertible. Apply Theorem 1.12. �
Remark. In the above theorem it is essential that G is a is a split extension group.
For non-split extension groups, monomial actions (instead of merely purely monomial
actions) may arise; see the proof of Theorem 1.3 [Ha] and also [Sa5].
§6. Proof of Theorem 1.9
We will prove Theorem 1.9 in this section.
If charK = p, apply Theorem 3.2. We find that K(G) is rational. In particular, it
is retract rational.
Thus we will assume that charK 6= p from now on till the end of this section.
If ζp ∈ K, we apply Theorem 1.4 to find that K(G) is rational. Hence we will
consider only the situation [K(ζp) : K] = 2 in the sequel.
Since p be an odd prime number, there is only one non-abelian p-group of order
p3 with exponent p, and there are precisely two non-isomorphic non-abelian p-groups
of order p4 with exponent p (see [CK, Section 2 and Section 3]). We will solve the
rationality problem of these three groups separately.
Let ζ = ζp, Gal(K(ζ)/K) = 〈τ〉 with τ · ζ = ζ
Case 1. G = 〈σ1, σ2, σ3 : σ
i = 1, σ1 ∈ Z(G), σ2σ3 = σ3σ1σ2〉.
Step 1. Let W =
g∈GK · x(g) be the representation space of the regular repre-
sentation of G. Note that
K(G) = K(x(g) : g ∈ G)G = {K(ζ)(x(g) : g ∈ G)〈τ〉}G = K(ζ)(x(g) : g ∈ G)〈G,τ〉.
Step 2. For i ∈ Zp, define x0,i, x1,i ∈
g∈GK(ζ) · x(g) by
x0,i =
j,k∈Zp
ζ−j−kx(σi3σ
x1,i =
j,k∈Zp
ζj+k · x(σi3σ
We find that
σ1 : x0,i 7→ ζx0,i, x1,i 7→ ζ
−1x1,i,
σ2 : x0,i 7→ ζ
i+1x0,i, x1,i 7→ ζ
−i−1x1,i,
σ3 : x0,i 7→ x0,i+1, x1,i 7→ x1,i+1,
τ : x0,i ↔ x1,i.
The restriction of the action of 〈σ1, σ2〉 toK(ζ)·x0,i andK(ζ)·x1,i is given by distinct
characters of 〈σ1, σ2〉 to K(ζ) \ {0}. Hence x0,0, x0,1, x0,2, . . . , x0,p−1, x1,0, . . . , x1,p−1 are
linearly independent vectors. Moreover, the action of 〈G, τ〉 on K(ζ)(x0,i, x1,i : i ∈ Zp)
is faithful.
By Theorem 3.1 K(ζ)(x(g) : g ∈ G) = K(ζ)(x0,i, x1,i : i ∈ Zp)(X1, . . . , Xl) where
l = p3 − 2p and ρ(Xj) = Xj for any 1 ≤ j ≤ l, any ρ ∈ 〈G, τ〉.
Step 3. Define x0 = x
0,0, y0 = x0,0x1,0 and xi = x0,i · x
0,i−1, yi = y0,i · y
0,i−1 for
1 ≤ i ≤ p − 1. It follows that K(ζ)(x0,i, x1,i : i ∈ Zp)
〈σ1〉 = K(ζ)(xi, yi : i ∈ Zp).
Moreover, the actions of σ2, σ3, τ are given as
σ2 : x0 7→ x0, y0 7→ y0, xi 7→ ζxi, yi 7→ ζ
−1yi for 1 ≤ i ≤ p− 1,
σ3 : x0 7→ x0x
1, x1 7→ x2 7→ · · · 7→ xp−1 7→ (x1x2 · · ·xp−1)
y0 7→ y0y1x1, y1 7→ y2 7→ · · · 7→ yp−1 7→ (y1y2 · · · yp−1)
τ : x0 7→ y
0 , y0 7→ y0, xi ↔ yi for 1 ≤ i ≤ p− 1.
Step 4. Define X = x0y
−(p−1)/2
0 , Y = x
(p+1)/2
Then K(ζ)(xi, yi : i ∈ Zp) = K(ζ)(X, Y, xi, yi : 1 ≤ i ≤ p − 1) and σ2(X) = X ,
σ2(Y ) = Y , σ3(X) = αX , σ3(Y ) = βY , τ : X ↔ Y where α, β ∈ K(ζ)(xi, yi : 1 ≤ i ≤
p− 1)\{0}.
Apply Theorem 3.1. We may find X̃ , Ỹ so that K(ζ)(xi, yi : i ∈ Zp) = K(ζ)(xi, yi :
1 ≤ i ≤ p − 1)(X̃, Ỹ ) with ρ(X̃) = X̃ , ρ(Ỹ ) = Ỹ for any ρ ∈ 〈σ2, σ3, τ〉. Thus it
remains to consider K(ζ)(xi, yi : 1 ≤ i ≤ p− 1)
〈σ2,σ3,τ〉.
Step 5. Define u0 = x
1, v0 = x1y1, ui = xi+1x
i , vi = yi+1y
i for 1 ≤ i ≤ p− 2.
It follows that K(ζ)(xi, yi : 1 ≤ i ≤ p − 1)
〈σ2〉 = K(ζ)(ui, vi : 0 ≤ i ≤ p − 2). The
actions of σ3 and τ are given by
σ3 : u0 7→ u0u
1, v0 7→ v0v1u1, u1 7→ u2 7→ · · · 7→ up−2 7→ (u0u
2 · · ·u
v1 7→ v2 7→ · · · 7→ vp−2 7→ u0(v
1 · · · v
τ : u0 7→ u
0, v0 7→ v0, ui ↔ vi for 1 ≤ i ≤ p− 2.
Step 6. Define up−1 = (u0u
2 · · ·u
−1, wi = v0v1 · · · viu1u2 · · ·ui for 1 ≤ i ≤
p − 2, and wp−1 = (v
1 · · · vp−2u
2 · · ·up−2)
−1. We find that K(ui, vi : 0 ≤
i ≤ p− 2) = K(ui, wi : 1 ≤ i ≤ p− 1) and
σ3 : u1 7→ u2 7→ · · · 7→ up−1 7→ (u1u2 · · ·up−1)
w1 7→ w2 7→ · · · 7→ wp−1 7→ (w1w2 · · ·wp−1)
τ : ui 7→ wi(uiwi−1)
−1, wi 7→ wi for 1 ≤ i ≤ p− 1.
where we write w0 = v0 for convenience. (Granting that w0 is defined as above, we
have a relation w0w1 · · ·wp−1 = 1. But we don’t have the relation u0u1 · · ·up−1 = 1
because we define u0, u1, . . . , up−2 first and up−1 is defined by another way.)
We will study whether K(ζ)(ui, wi : 1 ≤ i ≤ p− 1)
〈σ3,τ〉 is rational or not.
Step 7. The multiplicative action in Step 6 can be formulated in terms of π-lattices
where π = 〈τ, σ3〉 as follows.
LetM = (
1≤i≤p−1 Z·ui)⊕(
1≤i≤p−1Z·wi) and define w0 = −w1−w2−· · ·−wp−1.
Define a Z[π]-module structure on M by
σ3 : u1 7→ u2 7→ · · · 7→ up−1 7→ −u1 − u2 − · · · − up−1,
w1 7→ w2 7→ · · · 7→ wp−1 7→ −w1 − w2 − · · · − wp−1,
τ : ui 7→ −ui + wi − wi−1, wi 7→ wi for 1 ≤ i ≤ p− 1.
We claim that M ≃ Z[π1]
Z[π2]/Φp(σ3) with π1 = 〈τ〉, π2 = 〈σ3〉.
Throughout this step, we will write σ = σ3 and π =< σ, τ >. Define ρ = στ . Then
ρ is a generator of π with order 2p where p is an odd prime number.
Let Φp(T ) ∈ Z[T ] be the p-th cyclotomic polynomial. Note that Φp(σ)(ui) =
Φp(σ)(wi) = 0. It follows that Φp(σ) ·M = 0.
Since Φp(σ
0≤i≤p−1 σ
i = Φp(σ), we know that Φp(σ
2) ·M = 0. From ρ2 = σ2,
we find that Φp(ρ
2) ·M = 0. It is well-known that Φp(T
2) = Φp(T )Φ2p(T ) (see [Ka1,
Theorem 1.1] for example). We conclude that Φp(ρ)Φ2p(ρ) ·M = 0. In other words,
we may regard M as a module over Λ = Z[π]/ < Φp(ρ)Φ2p(ρ) >.
Clearly Λ is isomorphic to Z[π1]
Z[π2]/Φp(σ) where π1 = 〈τ〉, π2 = 〈σ〉.
It remains to show that M is isomorphic to Λ as a Λ-module.
It is not difficult to verify that Φ2p(ρ)(u1) = f(ρ)(w1) where f(T ) = Φ2p(T )−T
Define v = u1 − w1. We find that Φ2p(ρ)(v) = −w0.
Since M is a Λ-module generated by u1 and w0, it follows that, as a Λ-module,
M =< u1, w0 >=< u1 − σw0, w0 > =< u1 − w1, w0 >=< v,w0 > =< v,w0 +
Φ2p(ρ)(v) >=< v,w0 − w0 >=< v >, i.e. M is a cyclic Λ-module generated by
v. Thus we get an epimorphism Λ → M , which is an isomorphism by counting the
Z-ranks of both sides. Hence the result.
Step 8. By Step 7, M ≃ Z[π1]
Z[π2]/Φp(σ3) where π1 = 〈τ〉, π2 = 〈σ3〉.
Thus we may choose a Z-basis Y1, Y2, . . . , Yp−1, Z1, . . . , Zp−1 for M such that
σ3 : Y1 7→ Y2 7→ · · · 7→ Yp−1 7→ −Y1 − · · · − Yp−1,
Z1 7→ Z2 7→ · · · 7→ Zp−1 7→ −Z1 − · · · − Zp−1,
τ : Yi ↔ Zi for 1 ≤ i ≤ p− 1.
Hence K(ζ)(M) = K(ζ)(Yi, Zi : 1 ≤ i ≤ p− 1). We emphasize that σ3 acts on Yi,
Zi by multiplicative actions on the field K(ζ)(M) and σ3 · ζ = ζ , τ · ζ = ζ
Step 9. In the field K(ζ)(M), define
s0 = 1 + Y1 + Y1Y2 + Y1Y2Y3 + · · ·+ Y1Y2 · · ·Yp−1,
t0 = 1 + Z1 + Z1Z2 + · · ·+ Z1Z2 · · ·Zp−1,
s1 = (1/s0)− (1/p), s2 = (Y1/s0)− (1/p), . . . , sp−1 = (Y1Y2 · · ·Yp−2/s0)− (1/p),
t1 = (1/t0)− (1/p), t1 = (Z1/t0)− (1/p), . . . , tp−1 = (Z1Z2 · · ·Zp−2/t0)− (1/p).
It is easy to verify that K(ζ)(M) = K(ζ)(si, ti : 1 ≤ i ≤ p− 1) and
σ3 : s1 7→ s2 7→ · · · 7→ sp−1 7→ −s1 − s2 − · · · − sp−1,
t1 7→ t2 7→ · · · 7→ tp−1 7→ −t1 − t2 − · · · − tp−1,
τ : si ↔ ti.
Step 10. Define ri = si + ti for 1 ≤ i ≤ p − 1. Then K(ζ)(si, ti : 1 ≤ i ≤ p− 1) =
K(ζ)(si, ri : 1 ≤ i ≤ p− 1). Note that
σ3 : r1 7→ r2 7→ · · · 7→ rp−1 7→ −r1 − r2 − · · · − rp−1,
τ : ri 7→ ri, si 7→ −si + ri.
Apply Theorem 3.1. We find that K(ζ)(si, ri : 1 ≤ i ≤ p− 1) = K(ζ)(ri : 1 ≤ i ≤
p−1)(A1, . . . , Ap−1) for some A1, . . . , Ap−1 with σ3(Ai) = τ(Ai) = Ai for 1 ≤ i ≤ p−1.
Thus K(ζ)(ri, si : 1 ≤ i ≤ p − 1)
〈τ〉 = K(ζ)(ri : 1 ≤ i ≤ p − 1)
〈τ〉(A1, . . . , Ap−1) =
K(r1, . . . , rp−1, A1, . . . , Ap−1).
It remains to find K(r1, . . . , rp−1)
〈σ3〉.
Step 11. Write π2 = 〈σ3〉. The π2-fieldsK(r1, . . . , rp−1, A1) andK(B0, B1, . . . , Bp−1)
are π2-isomorphic where σ3 : B0 7→ B1 7→ · · · 7→ Bp−1 7→ B0. For, we may define
B = B0 +B1 + · · ·+Bp−1 and Ci = Bi − (B/p) for 0 ≤ i ≤ p− 1.
In other words K(r1, . . . , rp−1, A1, . . . , Ap−1)
〈σ3〉 = K(B0, B1, . . . , Bp−1)
〈σ3〉(A2, . . .,
Ap−1) = K(Zp)(A2, . . . , Ap−1).
By Lemma 3.5, K(Zp) = K(ζ)(N)
π1 where ζ = ζp, π1 = Gal(K(ζ)/K) = 〈τ〉 with
τ · ζ = ζ−1, and N is some π1-lattice.
By Reiner’s Theorem [Re], the π1-lattice N is a direct sum of lattices of three types:
(1) τ : z 7→ −z, (2) τ : z 7→ z, (3) τ : z1 ↔ z2. Thus we find K(ζ)(N)
〈τ〉 = K(ζ)(z1,
. . . , za, z
2, . . . , z
1 , w
1 , . . . , z
c , w
c ) where τ : z1 7→ 1/z1, . . ., za 7→ 1/za, z
1 7→ z
. . ., z′b 7→ z
1 ↔ w
1 , . . ., z
c ↔ z
By Theorem 3.1 we may “neglect” the roles of z′1, . . . , z
1 , w
1 , . . . , z
c , w
c . We may
linearize the action on z1, . . . , za by defining wi = 1/(1 + zi) when 1 ≤ i ≤ a. Then
τ : wi 7→ −wi + 1. Thus we may “neglect” the roles of wi by Theorem 3.1 again. In
conclusion, K(ζ)(z1, . . . , za, z
1, . . . , z
1 , w
1 , . . . , z
c , w
〈τ〉 is rational over K. �
Case 2. G = 〈σ1, σ2, σ3, σ4 : σ
i = 1, σ1, σ2 ∈ Z(G), σ
4 σ3σ4 = σ1σ3〉.
The proof is very similar to Case 1.
Step 1. For i ∈ Zp, define x0,i, x1,i, y0,i, y1,i ∈
g∈GK(ζ) · x(g) by
x0,i =
j,k,l∈Zp
ζ−j−k−lx(σi4σ
x1,i =
j,k,l∈Zp
ζj+k+lx(σi4σ
y0,i =
j,k,l∈Zp
ζ−j−k+lx(σi4σ
y1,i =
j,k,l∈Zp
ζj+k−lx(σi4σ
The action of 〈G, τ〉 is given by
σ1 : x0,i 7→ ζx0,i, x1,i 7→ ζ
−1x1,i, y0,i 7→ ζy0,i, y1,i 7→ ζ
−1y1,i,
σ2 : x0,i 7→ ζx0,i, x1,i 7→ ζ
−1x1,i, y0,i 7→ ζ
−1y0,i, y1,i 7→ ζy1,i,
σ3 : x0,i 7→ ζ
i+1x0,i, x1,i 7→ ζ
−i−1x1,i, y0,i 7→ ζ
i+1y0,i, y1,i 7→ ζ
−i−1y1,i,
σ4 : x0,i 7→ x0,i+1, x1,i 7→ x1,i+1, y0,i 7→ y0,i+1, y1,i 7→ y1,i+1,
τ : x0,i ↔ x1,i, y0,i ↔ y1,i.
Note that x0,i, x1,i, y0,i, y1,i where i ∈ Zp are linearly independent vectors in
g∈GK(ζ) · x(g). Apply Theorem 3.1. It suffices to consider the rationality prob-
lem of K(ζ)(x0,i, x1,i, y0,i, y1,i : i ∈ Zp)
〈G,τ〉.
Step 2. Define x0 = x
0,0, y0 = x0,0x1,0, X0 = y0,0(x0,0)
−1, Y0 = y1,0(x1,0)
−1; for
1 ≤ i ≤ p − 1, define xi = x0,i(x0,i−1)
−1, yi = x1,i(x1,i−1)
−1, Xi = y0,i(y0,i−1)
−1, Yi =
y1,i(y1,i−1)
−1. Then K(ζ)(x0,i, x1,i, y0,i, y1,i : i ∈ Zp)
〈σ1〉 = K(ζ)(xi, yi, Xi, Yi : i ∈ Zp)
and the actions are given by
σ2 : x0 7→ x0, y0 7→ y0, X0 7→ ζ
−2X0, Y0 7→ ζ
All the other generators are fixed by σ2;
σ3 : x0 7→ x0, y0 7→ y0, X0 7→ X0, Y0 7→ Y0,
xi 7→ ζxi, yi 7→ ζ
−1yi, Xi 7→ ζXi, Yi 7→ ζ
−1Yi,
σ4 : x0 7→ x0x
1, y0 7→ y0y1x1, X0 7→ X0X1x
1 , Y0 7→ Y0Y1y
x1 7→ x2 7→ x3 7→ · · · 7→ xp−1 7→ (x1x2 · · ·xp−1)
Similarly for y1, y2, . . . , yp−1, X1, . . . , Xp−1, Y1, . . . , Yp−1;
τ : x0 7→ y
0 , y0 7→ y0, X0 7→ Y0 7→ X0,
xi ↔ yi, Xi ↔ Yi for 1 ≤ i ≤ p− 1.
Step 3. Define x̃ = x0y
−(p−1)/2
0 , ỹ = x
(p+1)/2
Since x̃, ỹ are fixed by both σ2 and σ3, while σ4 : x̃ 7→ αx̃, ỹ 7→ βỹ, τ : x̃ ↔ ỹ
where α, β ∈ K(x1, y1, X1, Y1)\{0}, we may apply Theorem 3.1. Thus the roles of x̃, ỹ
may be “neglected”. It suffices to consider whether K(ζ)(X0, Y0, xi, yi, Xi, Yi : 1 ≤ i ≤
p− 1)〈σ2,σ3,σ4,τ〉 is rational over K.
Step 4. Define X̃ = X
0 , Ỹ = X0Y0. Then K(ζ)(X0, Y0, xi, yi, Xi, Yi : 1 ≤ i ≤
p− 1)〈σ2〉 = K(ζ)(X̃, Ỹ , xi, yi, Xi, Yi : 1 ≤ i ≤ p− 1). Moreover, the actions on X̃ and
Ỹ are given by
σ3 : X̃ 7→ X̃, Ỹ 7→ Ỹ ,
σ4 : X̃ 7→ X̃X
1 , Ỹ 7→ Ỹ X1Y1x
τ : X̃ 7→ X̃−1Ỹ p, Ỹ 7→ Ỹ .
Define X ′ = X̃Ỹ −(p−1)/2, Y ′ = X̃−1Ỹ (p+1)/2. As in Step 3, we may apply The-
orem 3.1 and “neglect” the roles of X ′ and Y ′. It remains to make sure whether
K(ζ)(xi, yi, Xi, Yi : 1 ≤ i ≤ p− 1)
〈σ3,σ4,τ〉 is rational.
Define u0 = x
1, v0 = x1y1; for 1 ≤ i ≤ p−2, define ui = xi+1(x
i ), vi = yi+1 · (y
and for 1 ≤ i ≤ p−1, define Ui = Xix
i , Vi = Yiy
i . It follows that K(ζ)(xi, yi, Xi, Yi :
1 ≤ i ≤ p− 1)〈σ3〉 = K(ζ)(u0, v0, Up−1, Vp−1, ui, vi, Ui, Vi : 1 ≤ i ≤ p− 2). Note that the
actions of σ4 and τ are given by
σ4 : u0 7→ u0u
1, v0 7→ v0v1u1, u1 7→ u2 7→ · · · 7→ up−2 7→ (u0u
2 · · ·u
v1 7→ v2 7→ · · · 7→ vp−2 7→ u0(v
1 · · · v
U1 7→ U2 7→ · · · 7→ Up−1 7→ (U1U2 · · ·Up−1)
V1 7→ V2 7→ · · · 7→ Vp−1 7→ (V1V2 · · ·Vp−1)
τ : u0 7→ u
0, v0 7→ v0, ui ↔ vi for 1 ≤ i ≤ p− 2,
Ui ↔ Vi for 1 ≤ i ≤ p− 1.
Note that the actions of σ4 and τ on U1, U2, . . . , Up−1, V1, . . . , Vp−1 may be linearized
by the same method as in Step 9 of Case 1. Hence we may apply Theorem 3.1 and
“neglect” the roles of Ui, Vi for 1 ≤ i ≤ p− 1.
In conclusion, all we need is to prove that K(ζ)(ui, vi : 0 ≤ i ≤ p−2)
〈σ4,τ〉 is rational
over K.
Compare the present situation with that of Step 5 of Case 1. We have the same
generators and the same actions (and even the same notation). Thus we finish the
proof. �
Case 3. G = 〈σ1, σ2, σ3, σ4 : σ
i = 1, σ1 ∈ Z(G), σ2σ3 = σ3σ2, σ
4 σ2σ4 =
σ1σ2, σ
4 σ3σ4 = σ2σ3〉.
Step 1. For i ∈ Zp, define x0,i, x1,i, y0,i, y1,i ∈
g∈GK(ζ) · x(g) by
x0,i =
j,k,l∈Zp
ζ−j−k−lx(σi4σ
x1,i =
j,k,l∈Zp
ζj+k+lx(σi4σ
y0,i =
j,k,l∈Zp
ζ−j−k+lx(σi4σ
y1,i =
j,k,l∈Zp
ζj+k−lx(σi4σ
The action of 〈G, τ〉 is given by
σ1 : x0,i 7→ ζx0,i, x1,i 7→ ζ
−1x1,i, y0,i 7→ ζ
−1y0,i, y1,i 7→ ζy1,i,
σ2 : x0,i 7→ ζ
i+1x0,i, x1,i 7→ ζ
−i−1x1,i, y0,i 7→ ζ
−i+1y0,i, y1,i 7→ ζ
i−1y1,i,
σ3 : x0,i 7→ ζ
i+1x0,i, x1,i 7→ ζ
−i−1x1,i, y0,i 7→ ζ
i+1y0,i, y1,i 7→ ζ
−i−1y1,i,
σ4 : x0,i 7→ x0,i+1, x1,i 7→ x1,i+1, y0,i 7→ y0,i+1, y1,i 7→ y1,i+1,
τ : x0,i ↔ x1,i, y0,i ↔ y1,i.
Checking the restriction of 〈σ1, σ2, σ3〉 as in Step 1 of Case 2, we find that these
vectors x0,i, x1,i, y0,i, y1,i are linearly independent except possibly for the case x0,p−2
and y1,0, and the case x1,p−2 and y0,0. But it is easy to see that these vectors are linearly
independent, because their ”supports” are disjoint. Apply Theorem 3.1. It suffices to
consider K(ζ)(x0,i, x1,i, y0,i, y1,i : i ∈ Zp)
〈G,τ〉.
Step 2. Define x0 = x
0,0, y0 = x0,0x1,0, X0 = x0,0 · y0,0, Y0 = x1,0y1,0; and for
1 ≤ i ≤ p − 1, define xi = x0,i(x0,i−1)
−1, yi = x1,i(x1,i−1)
−1, Xi = y0,i(y0,i−1)
−1, Yi =
y1,i(y1,i−1)
−1. Then K(ζ)(x0,i, x1,i, y0,i, y1,i : i ∈ Zp)
〈σ1〉 = K(ζ)(xi, yi, Xi, Yi : i ∈ Zp)
and the actions are given by
σ2 : x0 7→ x0, y0 7→ y0, X0 7→ ζ
2X0, Y0 7→ ζ
−2Y0,
xi 7→ ζxi, yi 7→ ζ
−1yi, Xi 7→ ζ
−1Xi, Yi 7→ ζYi,
σ3 : x0 7→ x0, y0 7→ y0, X0 7→ ζ
2X0, Y0 7→ ζ
−2Y0,
xi 7→ ζxi, yi 7→ ζ
−1yi, Xi 7→ ζXi, Yi 7→ ζ
−1Yi,
σ4 : x0 7→ x0x
1, y0 7→ y0y1X1, X0 7→ X0X1x1, Y0 7→ Y0Y1y1,
x1 7→ x2 7→ x3 7→ · · · 7→ xp−1 7→ (x1x2 · · ·xp−1)
Similarly for y1, y2, . . . , yp−1, X1, . . . , Xp−1, Y1, . . . , Yp−1.
τ : x0 7→ y
0 , y0 7→ y0, X0 7→ Y0 7→ X0,
xi ↔ yi, Xi ↔ Yi for 1 ≤ i ≤ p− 1.
Step 3. Define x̃ = x0y
−(p−1)/2
0 , ỹ = x
(p+1)/2
0 . Apply Theorem 3.1 again. It
suffices to consider K(ζ)(X0, Y0, xi, yi, Xi, Yi : 1 ≤ i ≤ p− 1)
〈σ2,σ3,σ4,τ〉.
We may apply Theorem 3.1 again to “neglect” X0 and Y0. Thus it suffices to
consider K(ζ)(xi, yi, Xi, Yi : 1 ≤ i ≤ p− 1)
〈σ2,σ3,σ4,τ〉.
Step 4. Define u0 = x
1, v0 = x1y1, U0 = x1X1, V0 = y1Y1; and for 1 ≤ i ≤ p−2, de-
fine ui = x
i xi+1, vi = y
i yi+1, Ui = X
i Xi+1, Vi = Y
i Yi+1. ThenK(ζ)(xi, yi, Xi, Yi :
1 ≤ i ≤ p− 1)〈σ2〉 = K(ζ)(ui, vi, Ui, Vi : 0 ≤ i ≤ p− 2). Note that the actions of σ3, σ4,
τ are given by
σ3 : u0 7→ u0, v0 7→ v0, U0 7→ ζ
2U0, V0 7→ ζ
−2U0,
All the other generators are fixed by σ3.
σ4 : u0 7→ u0u
1, v0 7→ v0v1u1, U0 7→ U0U1u1, V0 7→ V0V1v1,
u1 7→ u2 7→ · · · 7→ up−2 7→ (u0u
2 · · ·u
v1 7→ v2 7→ · · · 7→ vp−2 7→ u0(v
1 · · · v
U1 7→ U2 7→ · · · 7→ Up−2 7→ u0(U
1 · · ·U
V1 7→ V2 7→ · · · 7→ Vp−2 7→ u
1 · · ·V
τ : u0 7→ v
0 , v0 7→ v0, U0 7→ V0 7→ U0,
ui ↔ vi, Ui ↔ Vi for 1 ≤ i ≤ p− 2.
Step 5. Define R0 = U
0 , S0 = U0V0; and for 1 ≤ i ≤ p− 2, define Ri = Ui, Si = Vi.
Then K(ζ)(ui, vi, Ui, Vi : 0 ≤ i ≤ p − 2)
〈σ3〉 = K(ζ)(ui, vi, Ri, Si : 0 ≤ i ≤ p − 2). We
will write the actions of σ4 and τ on Ri, Si as follows.
σ4 : R0 7→ R0R
1, S0 7→ S0S1R1u1v1,
R1 7→ R2 7→ · · · 7→ Rp−2 7→ u0(R0R
2 · · ·R
S1 7→ S2 7→ · · · 7→ Sp−2 7→ u
0R0(S
1 · · ·S
τ : R0 7→ S
0 , S0 7→ S0, Ri ↔ Si for 1 ≤ i ≤ p− 2.
Step 6. Imitate the change of variables in Step 6 of Case 1. We define up−1 =
2 · · ·u
−1, Rp−1 = u0(R0R
2 · · ·R
−1; and for 1 ≤ i ≤ p − 2,
define wi = v0v1 · · · viu1u2 · · ·ui, Ti = S0S1 · · ·SiR1R2 · · ·Ri; define wp−1 = (v
1 · · · vp−2u
2 · · ·up−2)
−1, Tp−1 = u1v1v
1 · · ·Sp−2R
2 · · ·Rp−2)
We find that K(ui, vi, Ri, Si : 0 ≤ i ≤ p− 2) = K(ui, wi, Ri, Ti : 1 ≤ i ≤ p− 1) and
σ4 : u1 7→ u2 7→ · · · 7→ up−1 7→ (u1u2 · · ·up−1)
w1 7→ w2 7→ · · · 7→ wp−1 7→ (w1w2 · · ·wp−1)
R1 7→ R2 7→ · · · 7→ Rp−1 7→ (R1R2 · · ·Rp−1)
T1 7→ T2 7→ · · · 7→ Tp−1 7→ (T1T2 · · ·Tp−1)
τ : wi 7→ wi, Ti 7→ Ti, ui 7→ wi(uiwi−1)
−1, Ri 7→ Ti(RiTi−1)
where we write w0 = v0, T0 = S0 for convenience.
Step 7. The multiplicative action in Step 6 can be formulated as follows.
Let π = 〈τ, σ4〉 and define a π-lattice as in Step 7 of Case 1 (but σ3 should be
replaced by σ4 in the present situation). Then K(ζ)(ui, wi, Ri, Ti : 1 ≤ i ≤ p − 1)
K(ζ)(M ⊕M)π where M is the same lattice in Step 7 of Case 1.
The structure ofM has been determined in Step 7 of Case 1. Thus M⊕M ≃ Λ⊕Λ
where Λ ≃ Z[π1] ⊗Z Z[π2]/Φp(σ4) where π1 = 〈τ〉 and π2 = 〈σ4〉. It follows that we
can find elements Y1, . . . , Yp−1, Z1, . . . , Zp−1, W1, . . . ,Wp−1, Q1, . . . , Qp−1 in the field
K(ζ)(ui, wi, Ri, Ti : 1 ≤ i ≤ p − 1) so that K(ζ)(ui, wi, Ri, Ti : 1 ≤ i ≤ p − 1) =
K(ζ)(Y1, . . . , Yp−1, Z1, . . . , Zp−1,W1, . . . ,Wp−1, Q1, . . . , Qp−1) and the actions of σ4 and
τ are given by
σ4 : Y1 7→ · · · 7→ Yp−1 7→ (Y1Y2 · · ·Yp−1)
Similarly for Z1, . . . , Zp−1, W1, . . . ,Wp−1, Q1, . . . , Qp−1;
τ : Yi ↔ Zi, Wi ↔ Qi.
Step 8. We can linearize the actions of σ4 and τ by the same method in Step 9 of
Case 1. Apply Theorem 3.1. We may “neglect” the roles of Wi, Qi. Thus the present
situation is the same as in Step 8 of Case 1. �
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  Let K be any field and G be a finite group. We will prove that, if K is any
field, p an odd prime number, and G is a non-abelian group of exponent p with
|G|=p^3 or p^4 satisfying [K(\zeta_p):K] <= 2, then K(G) is rational over K. We
will also show that K(G) is retract rational if G belongs to a much larger
class of p-groups. In particular, generic G-polynomials of G-Galois extensions
exist for these groups.

Introduction
Let K be any field and G be a finite group. Let G act on the rational function field
K(xg : g ∈ G) by K-automorphisms defined by g · xh = xgh for any g, h ∈ G. Denote
K(G) := K(xg : g ∈ G)
G the fixed subfield under the action of G. Noether’s problem
asks whether K(G) is rational (= purely transcendental) over K.
Noether’s problem for abelian groups was studied by Fischer, Furtwängler, K. Ma-
suda, Swan, Voskresenskii, S. Endo and T. Miyata, Lenstra, etc. It was known that
K(Zp) is rational if Zp is a cyclic group of order p with p = 3, 5, 7 or 11. The first
counter-example was found by Swan: Q(Z47) is not rational over Q [Sw1]. However
Saltman showed that Q(Zp) is retract rational over K for any prime number p, which
is enough to ensure the existence of a generic Galois G-extension and will fulfill the
original purpose of Emmy Noether [Sa1]. For the convenience of the reader, we recall
the definition of retract rationality.
Definition 1.1. ([Sa3]) Let K ⊂ L be a field extension. We say that L is re-
tract rational over K, if there is a K-algebra R contained in L such that (i) L is the
quotient field of R, and (ii) the identity map 1R : R → R factors through a local-
ized polynomial K-algebra, i.e. there is an element f ∈ K[x1, . . . , xn] the polynomial
ring over K and there are K-algebra homomorphisms ϕ : R → K[x1, . . . , xn][1/f ] and
ψ : K[x1, . . . , xn][1/f ] → R satisfying ψ ◦ ϕ = 1R.
It is not difficult to see that “rational” ⇒ “stably rational” ⇒ “retract rational”.
One of the motivation to study Noether’s problem arises from the inverse Galois
problem. If K is an infinite field, it is known that K(G) is retract rational over K if
and only if there exists a generic Galois G-extension over K [Sa1, Theorem 5.3; Sa3,
Theorem 3.12], which guarantees the existence of a Galois G-extension of K, provided
that K is a Hilbertian field. On the other hand, the existence of a generic Galois
G-extension over K is equivalent to the existence of a generic polynomial for G. For
the relationship among these notions, see [DM]. For a survey of Noether’s problem the
reader is referred to articles of Swan and Kersten [Sw2; Ke].
Although Noether’s problem for abelian groups was investigated extensively, our
knowledge for the non-abelian Noether’s problem was amazingly scarce (see, for exam-
ple, [Ka2]). We will list below some previous results of non-abelian Noether’s problem,
which are relevant to the theme of this article.
Theorem 1.2. (Saltman [Sa2]) For any prime number p and for any field K
with char K 6= p (in particular, K may be an algebraically closed field), there is a
meta-abelian p-group G of order p9 such that K(G) is not retract rational over K. In
particular, it is not rational.
Theorem 1.3. (Hajja [Ha]) Let G be a finite group containing an abelian normal
subgroup N such that G/N is a cyclic group of order < 23. Then C(G) is rational over
Theorem 1.4. (Chu and Kang [CK]) Let p be a prime number, G be a p-group
of order ≤ p4 with exponent pe. Let K be any field such that either char K = p or char
K 6= p and K contains a primitive pe-th root of unity. Then K(G) is rational over K.
Theorem 1.5. (Kang [Ka4]) Let G be a metacyclic p-group with exponent pe,
and let K be any field such that (i) char K = p, or (ii) char K 6= p and K contains a
primitive pe-th root of unity. Then K(G) is rational over K.
Note that, in Theorems 1.3—1.5, it is assumed that the ground field contains enough
roots of unity. We may wonder whether Q(G) is rational if G is a non-abelian p-group
of small order. The answer is rather optimistic when G is a group of order 8 or 16.
Theorem 1.6. ([CHK; Ka3]) Let K be any field and G be any non-abelian group
of order 8 or 16 other than the generalized quaternion group of order 16. Then K(G)
is always rational over K.
However Serre was able to show that Q(G) is not rational when G is the generalized
quaternion group [Se, p.441–442; GMS, Theorem 33.26 and Example 33.27, p.89–90].
On the other hand, if p is an odd prime number, Saltman proves the following
theorem.
Theorem 1.7. (Saltman [Sa4]) Let p be an odd prime number and G be a non-
abelian group of order p3. If K is a field containing a primitive p-th root of unity, then
K(G) is stably rational.
The above theorem may be generalized to the case of p-groups containing a maximal
cyclic subgroup, namely,
Theorem 1.8. (Hu and Kang [HuK]) Let p be a prime number and G be a non-
abelian p-group of order pn containing a cyclic subgroup of index p. If K is any field
containing a primitive pn−2-th root of unity, then K(G) is rational over K.
In this article we will prove the following theorem.
Theorem 1.9. Let p be an odd prime number, G be the non-abelian group of ex-
ponent p and of order p3 or p4. If K is a field with [K(ζp) : K] ≤ 2, then K(G) is
rational over K.
The rationality problem of K(G) seems rather intricate if the ground field K has no
enough root of unity. We don’t know the answer to the rationality of K(G) when the
assumption that [K(ζp) : K] ≤ 2 is waived in the above theorem. On the other hand,
as to the retract rationality of K(G), a lot of information may be obtained. Before
stating our results, we recall a theorem of Saltman first.
Theorem 1.10. (Saltman [Sa1, Theorem 3.5]) Let K be a field, G = A ⋊ G0
be a semi-direct product group where A is an abelian normal subgroup of G. Assume
that gcd{|A|, |G0|} = 1 and both K(A) and K(G0) are retract rational over K. Then
K(G) is retract rational over K.
Thus the main problem is to investigate the retract rationality for p-groups. We
will prove K(G) is retract rational for many p-groups G of exponent p.
Theorem 1.11. Let p be a prime number, K be any field, and G = A ⋊ G0 be a
semi-direct product group where A is a normal elementary p-group of G and G0 is a
cyclic group of order pm. If p = 2 and charK 6= 2, assume furthermore that K(ζ2m) is
a cyclic extension of K. Then K(G) is retract rational over K.
If p is an odd prime number, a p-group of exponent p containing an abelian normal
subgroup of index p certainly satisfies the assumption in Theorem 1.11. In particular,
a p-group of exponent p and of order p3 or p4 belongs to this class of p-groups (see
[CK]). There are six p-groups of exponent p and of order p5; only four of them contain
abelian normal subgroups of index p. Previously the retract rationality of K(G) for
non-abelian p-groups, i.e. the existence of generic polynomials for such groups G, is
known only when G is of order p3 and of exponent p.
Similarly if G = A⋊G0 is a semi-direct product of p-groups such that A is a normal
subgroup of order p, and G0 is a direct product of an elementary p-group with a cyclic
group of order pm, then G also satisfies the assumption in Theorem 1.11 provided that
the assumption that K(ζ2m) is a cyclic extension of K remains in force.
The above Theorem 1.11 is deduced from the following theorem.
Theorem 1.12. Let K be any field, and G = A⋊G0 be a semi-direct product group
where A is a normal abelian subgroup of exponent e and G0 is a cyclic group of order
m. Assume that
(i) either charK = 0 or charK > 0 with charK ∤ em, and
(ii) both K(ζe) and K(ζm) are cyclic extensions of K such that gcd{m, [K(ζe) :
K]} = 1.
Then K(G) is retract rational over K.
The idea of the proof of Theorem 1.12 is to add a primitive e-th root of unity to the
ground field and the question is reduced to a question of multiplicative group actions.
It is Voskresenskii who realizes that the multiplicative group action is related to the
birational classification of algebraic tori [Vo]. However, the multiplicative group action
arising in the present situation is not the function field of an algebraic torus; it is a
new type of multiplicative group actions. Thus we need a new criterion for retract
rationality. It is the following theorem.
Theorem 1.13. Let π1 and π2 be finite abelian groups, π = π1 × π2, and L be a
Galois extension of the field K with π1 = Gal(L/K). Regard L as a π-field through the
projection π → π1. Assume that
(i) gcd{|π1|, |π2|} = 1,
(ii) char K = 0 or charK > 0 with charK ∤ |π2|, and
(iii) K(ζm) is a cyclic extension of K where m is the exponent of π2.
If M is π-lattice such that ρπ(M) is an invertible π-lattice, than L(M)
π is retract
rational over K.
The reader will find that the above theorem is an adaptation of Saltman’s criterion
for retract rational algebraic tori [Sa3, Theorem 3.14] (see Theorem 2.5). We also
formulate another criterion for retract rationality of L(M)π when π is a semi-direct
product group (see Theorem 4.3). An amusing consequence of this criterion (when
compared with Theorem 1.3) is that, if G = A ⋊ H is a semi-direct product of an
abelian normal subgroup A and a cyclic subgroup H , then C(G) is always retract
rational (see Proposition 5.2).
We will organize this paper as follows. We recall some basic facts of multiplicative
group actions in Section 2. In particular, the flabby class map which was mentioned in
Theorem 1.13 will be defined. We will give additional tools for proving Theorem 1.9
and Theorems 1.11–1.13 in Section 3. In Section 4 Theorem 1.13 and its variants will
be proved. The proof of Theorem 1.11 and Theorem 1.12 will be given in Section 5.
Section 6 contains the proof of Theorem 1.9.
Acknowledgements. I am indebted to Prof. R. G. Swan for providing a simplified
proof in Step 7 of Case 1 of Theorem 1.9 (see Section 6). The proof in a previous version
of this paper was lengthy and complicated. I thank Swan’s generosity for allowing me
to include his proof in this article.
Notations and terminology. A field extension L over K is rational if L is purely
transcendental over K; L is stably rational over K if there exist y1, . . . , yN such that
y1, . . . , yN are algebraically independent over L and L(y1, . . . , yN) is rational over K.
More generally, two fields L1 and L2 are called stably isomorphic if L1(x1, . . . , xm) is
isomorphic to L2(y1, . . . , yn) where x1, . . . , xm and y1, . . . , yn are algebraically indepen-
dent over L1 and L2 respectively.
Recall the definition of K(G) at the beginning of this section: K(G) = K(xg : g ∈
G)G. If L is a field with a finite group G acting on it, we will call it a G-field. Two
G-fields L1 and L2 are G-isomorphic if there is an isomorphism ϕ : L1 → L2 satisfying
ϕ(σ · u) = σ · ϕ(u) for any σ ∈ G, any u ∈ L1.
We will denote by ζn a primitive n-th root of unity in some extension field of K
when char K = 0 or char K = p > 0 with p ∤ n. All the groups in this article are
finite groups. Zn will be the cyclic group of order n or the ring of integers modulo n
depending on the situation from the context. Z[π] is the group ring of a finite group
π over Z. Z(G) is the center of the group G. The exponent of a group G is the
least common multiple of the orders of elements in G. The representation space of the
regular representation of G over K is denoted by W =
g∈GK · x(g) where G acts on
W by g · x(h) = x(gh) for any g, h ∈ G.
§2. Multiplicative group actions
Let π be a finite group. A π-latticeM is a finitely generated Z[π]-module such that
M is a free abelian group when it is regarded as an abelian group.
For any field K and a π-lattice M , K[M ] will denote the Laurent polynomial ring
and K(M) is the quotient field of K[M ]. Explicitly, if M =
1≤i≤m Z · xi as a free
abelian group, then K[M ] = K[x±11 , . . . , x
m ] and K(M) = K(x1, . . . , xm). Since π
acts on M , it will act on K[M ] and K(M) by K-automorphisms, i.e. if σ ∈ π and
σ · xj =
1≤i≤m aijxi ∈ M , then we define the action of σ in K[M ] and K(M) by
σ · xj =
1≤i≤m x
The multiplicative action of π onK(M) is called a purely monomial action in [HK1].
If π is a group acting on the rational function field K(x1, . . . , xm) by K-automorphism
such that σ · xj = cj(σ) ·
1≤i≤m x
i where σ ∈ π, aij ∈ Z and cj(σ) ∈ K\{0}, such a
multiplicative group action is called a monomial action. Monomial actions arise when
studying Noether’s problem for non-split extension groups [Ha; Sa5].
We will introduce another kind of multiplicative actions. Let K ⊂ L be fields and
π be a finite group. Suppose that π acts on L by K-automorphisms (but it is not
assumed that π acts faithfully on L). Given a π-lattice M , the action of π on L can
be extended to an action of π on L(M) (= L(x1, . . . , xm) if M =
1≤i≤m Z · xi) by
K-automorphisms defined as follows: If σ ∈ π and σ · xj =
1≤i≤m aijxi ∈ M , then
the multiplication action in L(M) is defined by σ · xj =
1≤i≤m x
i for 1 ≤ j ≤ m.
When L is a Galois extension of K and π = Gal(L/K) (and therefore π acts
faithfully on L), the fixed subfield L(M)π is the function field of the algebraic torus
defined over K, split by L and with character group M (see [Vo]).
We recall some basic facts of the theory of flabby (flasque) π-lattices developed by
Endo and Miyata, Voskresenskii, Colliot-Thélène and Sansuc, etc. [Vo; CTS]. We refer
the reader to [Sw2; Sw3; Lo] for a quick review of the theory.
In the sequel, π denotes a finite group unless otherwise specified.
Definition 2.1. A π-lattice M is called a permutation lattice if M has a Z-basis
permuted by π. M is called an invertible (or permutation projective) lattice, if it is
a direct summand of some permutation lattice. A π-lattice M is called a flabby (or
flasque) lattice if H−1(π′,M) = 0 for any subgroup π′ of π. (Note that H−1(π′,M)
denotes the Tate cohomology group.) Similarly, M is called coflabby if H1(π′,M) = 0
for any subgroup π′ of π.
It is known that an invertible π-lattice is necessarily a flabby lattice [Sw2, Lemma
8.4; Lo, Lemma 2.5.1].
Theorem 2.2. (Endo and Miyata [Sw3, Theorem 3.4; Lo, 2.10.1]) Let π be
a finite group. Then any flabby π-lattice is invertible if and only if all Sylow subgroups
of π are cyclic.
Denote by Lπ the class of all π-lattices, and by Fπ the class of all flabby π-lattices.
Definition 2.3. We define an equivalence relation ∼ on Fπ: Two π-lattices E1
and E2 are similar, denoted by E1 ∼ E2, if E1 ⊕ P is isomorphic to E2 ⊕ Q for some
permutation lattices P and Q. The similarity class containing E will be denoted by
[E]. Define Fπ = Fπ/ ∼, the set of all similarity classes of flabby π-lattices.
Fπ becomes a commutative monoid if we define [E1]+[E2] = [E1⊕E2]. The monoid
Fπ is called the flabby class monoid of π.
Definition 2.4. We define a map ρ : Lπ → Fπ as follows. For any π-lattice M ,
there exists a flabby resolution, i.e. a short exact sequence of π-lattices 0 → M → P →
E → 0 where P is a permutation lattice and E is a flabby lattice [Sw2, Lemma 8.5].
We define ρπ(M) = [E] ∈ Fπ. The map ρπ : Lπ → Fπ is well-defined [Sw2, Lemma
8.7]; it is called the flabby class map. We will simply write ρ instead of ρπ, if the group
π is obvious from the context.
Theorem 2.5. (Saltman [Sa3, Theorem 3.14]) Let L be a Galois extension of
K with π = Gal(L/K) and M be a π-lattice. Then ρπ(M) is invertible if and only if
L(M)π is retract rational over K.
§3. Generalities
We recall several results which will be used later.
Theorem 3.1. ([HK2, Theorem 1]) Let L be a field and G be a finite group
acting on L(x1, . . . , xm), the rational function field of m variables over L. Suppose that
(i) for any σ ∈ G, σ(L) ⊂ L;
(ii) The restriction of the action of G to L is faithful;
(iii) for any σ ∈ G,

σ(x1)
σ(xm)

= A(σ)


+B(σ)
where A(σ) ∈ GLm(L) and B(σ) is an m× 1 matrix over L.
Then L(x1, . . . , xm) = L(z1, . . . , zm) where σ(zi) = zi for all σ ∈ G, and for any
1 ≤ i ≤ m. In fact, z1, . . . , zm can be defined by


= A ·


for some A ∈ GLm(L) and for some B which is an m× 1 matrix over L. Moreover, if
B(σ) = 0 for all σ ∈ G, we may choose B = 0 in defining z1, . . . , zm.
Theorem 3.2. (Kuniyoshi [CHK, Theorem 2.5]) LetK be a field with charK =
p > 0 and G be a p-group. Then K(G) is always rational over K.
Proposition 3.3. Let π be a finite group and L be a π-field. Suppose that 0 →
M1 → M2 → N → 0 is a short exact sequence of π-lattices satisfying (i) π acts
faithfully on L(M1), and (ii) N is an invertible π-lattice. Then the π-fields L(M2) and
L(M1 ⊕N) are π-isomorphic.
Proof. We follow the proof of [Le, Proposition 1.5]. Denote L(M1)
× = L(M1)\{0}.
Consider the exact sequence of π-modules:
0 → L(M1)
× → L(M1)
× ·M2 → N → 0.
By Hilbert Theorem 90, we find that H1(π′, L(M1)
×) = 0 for any subgroup π′ ⊂ π.
Applying [Le, Proposition 1.2] we find that the above exact sequence splits. The
resulting π-morphism N → L(M1)
× · M2 provides the required π-isomorphism form
L(M2) to L(M1 ⊕N). �
Lemma 3.4. Let the assumptions be the same as in Proposition 3.3. Assume fur-
thermore that N is a permutation π-lattice. Then L(M2)
π is rational over L(M1)
Proof. By Proposition 3.3, L(M2) = L(M1)(N). Since π acts faithfully on L(M1),
we may apply Theorem 3.1 and find u1, . . . , un ∈ L(M2) such that L(M2) = L(M1)(u1,
. . . , un) with σ(ui) = ui for any σ ∈ π, any 1 ≤ i ≤ n where n = rank(N). Hence
L(M2)
π = L(M1)
π(u1, . . . , un). �
Lemma 3.5. Let π be a finite abelian group of exponent e and K be a field such
that char K = 0 or charK > 0 with charK ∤ e. If P is a permutation π-lattice, then
K(P )π = K(ζe)(M)
π0 where π0 = Gal(K(ζe)/K) and M is some π0-lattice.
Proof. We follow the standard approach to solving Noether’s problem for abelian
groups [Sw1; Sw2; Le].
Note that K(P )π = {K(ζe)(P )
〈π0〉}〈π〉 = K(ζe)(P )
〈π,π0〉 where the action of π is
extended to K(ζe)(P ) by defining g(ζe) = ζe for any g ∈ π, and the action of π0 is
extended to K(ζe)(P ) by requiring that π0 acts trivially on P .
Since π is abelian of exponent e, we may diagonalize its action on P , i.e. we may
find x1, . . . , xn ∈ K(ζe)(P ) such that n = rank(P ), g(xi)/xi ∈ 〈ζe〉 for any g ∈ π, and
K(ζe)(P ) = K(ζe)(x1, . . . , xn).
Thus K(ζe)(P )
〈π〉 = K(ζe)(y1, . . . , yn) where y1, . . . , yn are monomials in x1, . . . , xn.
LetM be the multiplicative subgroup generated by y1, . . . , yn in K(ζe)(y1, . . . , yn)\{0}.
Then M is a π0-lattice and K(ζe)(y1, . . . , yn)
π0 = K(ζe)(M)
π0. �
Proposition 3.6. Let π and K be the same as in Lemma 3.5. If K(ζe) is a cyclic
extension of K, then K(π) is retract rational over K.
Proof. We may regard the regular representation of π is given by a permutation
π-lattice. Thus K(π) = K(ζe)(M)
π0 where π0 = Gal(K(ζe)/K). Since π0 is assumed
cyclic, thus we may apply Theorem 2.2 and Theorem 2.5. �
§4. Proof of Theorem 1.13
Lemma 4.1. Let π be a finite group, M be a π-lattice. Suppose that π0 is a normal
subgroup of π and π0 acts trivially on M . Thus we may regard M as a lattice over
π/π0.
(1) M is a permutation π-lattice ⇔ So is it as a π/π0-lattice.
(2) M is an invertible π-lattice ⇔ So is it as a π/π0-lattice.
(3) M is a flabby π-lattice ⇔ So is it as a π/π0-lattice.
(4) If 0 → M → P → E → 0 is a flabby resolution of M as a π/π0-lattice, this
short exact sequence is also a flabby resolution of M as a π-lattice.
(5) ρπ(M) is an invertible π-lattice ⇔ ρπ/π0(M) is an invertible π/π0-lattice.
Proof. The properties (1)–(4) can be found in [CTS, Lemma 2, p.179–180]. As
to (5), the direction “⇐” is obvious by applying (4). For the other direction, assume
ρπ(M) is an invertible π-lattice. Let 0 → M → P → E → 0 be a flabby resolution
of M as a π-lattice. Then 0 → Mπ0 → P π0 → Eπ0 → 0 is a flabby resolution of
M =Mπ0 in the category of π/π0-lattices by [CTS, Lemma (xi), p.180]. It remains to
show that Eπ0 is invertible. Since [E] = ρπ(M) is invertible, we can find a π-lattice
N such that E ⊕N = N ′ is a permutation π-lattice. Note that N ′
π0 is a permutation
π/π0-lattice by [CTS, Lemma 2(i), p.180]. We find that E
π0 is invertible because
Eπ0 ⊕Nπ0 = (E ⊕N)π0 = N ′
π0 . �
Lemma 4.2. Let the assumptions be the same as in Theorem 1.13. If P is a
permutation π-lattice, then L(P )π is retract rational over K.
Proof. Since π1 and π2 are abelian groups with gcd{|π1|, |π2|} = 1, every subgroup
π′ of π can be written as π′ = ρ × λ where ρ is a subgroup of π1 and λ is a subgroup
of π2.
As a permutation π-lattice, we may write P =
Z[π/π(i)] where π(i) is a subgroup
of π. Write π(i) = ρi × λi where ρi ⊂ π1, λi ⊂ π2. Hence Z[π/π
(i)] = Z[π/(ρi × λi)] =
Z[(π1/ρi)× (π2/λi)].
It is not difficult to see that Z[π/π(i)] =
Z · u
kl where 1 ≤ k ≤ t = |π1/ρi|,
1 ≤ l ≤ r = |π2/λi|. Moreover, if g ∈ π1 and g
′ ∈ π2, then g · u
kl = u
g(k),l
, g′ ·
kl = u
k,g′(l)
and the homomorphisms π1 → St and π2 → Sr are induced from the
permutation representations associated to π/π(i) = (π1/ρi)× (π2/λi) where St and Sr
are the symmetric groups of degree t and r respectively.
Since π1 is faithful on L, we may apply Theorem 3.1. Explicitly, for any 1 ≤ l ≤ r,
we may find A(i) ∈ GLt(L) and define v
kl by


= A(i)


such that g · v
k,l = v
k,l for any g ∈ π1, and L(Z[π/π
(i)]) = L(u
kl : 1 ≤ k ≤ t, 1 ≤ l ≤ r.
If g′ ∈ π2, from the relation g
′ · u
kl = u
k,g′(l)
and Formula (1), we find that g′ · v
k,g′(l)
Since L(P )π1 = L(u
π1 = L(v
π1 = K(v
kl ) where i, k, l runs over index sets
which are understood, it follows that L(P )π = K(v
π2. Note that π2 acts on {v
by permutations.
By Lemma 3.5 K(v
π2 = K(ζm)(M)
π0 where m is the exponent of π2, π0 =
Gal(K(ζm)/K) and M is some π0-lattice.
By our assumption, π0 is a cyclic group. Hence ρπ0(M) is invertible by Theorem
2.2. Apply Theorem 2.5. We find that K(ζm)(M)
π0 is retract rational over K. �
Proof of Theorem 1.13 ———————————–
Step 1. Suppose that M is a π-lattice such that ρπ(M) is invertible.
Define π0 = {g ∈ π2 : g acts trivially on L(M)}. Then π/π0 acts faithfully on
L(M). Moreover, ρπ/π0(M) is invertible by Lemma 4.1.
In other words, without loss of generality we may assume that π is faithfully on
L(M) . Thus we will keep in force this assumption in the sequel.
Step 2. Since ρπ(M) is invertible, by [Sa3, Theorem 2.3, p.176], we may find π-
lattice M ′, P , Q such that P and Q are permutation lattices, 0 →M →M ′ → Q→ 0
is exact, and the inclusion map M →M ′ factors through P , i.e. the following diagram
commutes
0 ✲ M ✲ M ′ ✲ Q ✲ 0.
The remaining proof proceeds quite similar to that of [Sa3, Theorem 3.14, p.189].
Step 3. We get a commutative diagram of K-algebra morphisms from the diagram
in (2), i.e.
L[M ]π
L[P ]π
L[M ]π ✲ L[M ′]π ✲ L[Q]π
Step 4. The quotient field of L[P ]π is L(P )π, which is retract rational over K by
Lemma 4.2. Thus the identity map 1 : L[P ]π → L[P ]π factors rationally by [Sa3,
Lemma 3.5], i.e. there is a localized polynomial ring K[x1, . . . , xn][1/f ] and K-algebra
maps ϕ : L[P ]π → K[x1, . . . , xn][1/f ], ψ : K[x1, . . . , xn][1/f ] → L[P ]
π such that
ψ ◦ ϕ = 1.
It follows that the composite map g : L[M ]π → L[P ]π → L[M ′]π also factors ratio-
nally, i.e. there areK-algebra ϕ′ : L[M ]π → K[x1, . . . , xn][1/f ], ψ
′ : K[x1, . . . , xn][1/f ] →
L[M ′]π such that g = ψ′ ◦ ϕ′.
Step 5. By Lemma 3.4 L(M ′)π is rational over L(M)π. (This is the only one step
we use the assumption that π is faithful on L(M).)
Now we may apply [Sa3, Proposition 3.6(b), p.183] where, in the notation of [Sa3],
we take S = T = L[M ]π, ϕ is the identity map on L(M)π. We conclude that
1 : L[M ]π → L[M ]π factors rationally, i.e. L(M)π is retract rational over K. �
Here is a variant of Theorem 1.13.
Theorem 4.3. Let π be a finite group, 0 → π1 → π → π2 → 1 is a group extension,
and L be a Galois extension of the field K with π2 = Gal(L/K). Let π act on L through
the projection π → π2. Assume that
(i) π1 is an abelian group of exponent e with ζe ∈ L,
(ii) the extension 0 → π1 → π → π2 → 1 splits, and
(iii) every Sylow subgroup of π2 is cyclic.
If M is a π-lattice such that ρπ(M) is an invertible lattice, then L(M)
π is retract
rational over K.
Proof. The proof is very similar to the proof of Theorem 1.13.
We claim that L(P )π is retract rational for any permutation π-lattice P .
For the proof, we will use [Sa5, Theorem 2.1, p.546]. We will show that (c) of
[Sa5, Theorem 2.1] is valid, which will guarantee that L(P )π is retract rational. By
assumption (iii), part (d) of [Sa5, Theorem 2.1] is valid by Theorem 2.2. It remains to
check that the embedding problem of L/K and the extension 0 → π1 → π → π2 → 1
is solvable. But this is the well-known split embedding problem [ILF, Theorem 1.9,
p.12].
Now define π0 = {g ∈ π : g acts trivially on L(M)}. Note that π0 ⊂ π1. The
remaining proof is the same as in the proof of Theorem 1.13 and is omitted. �
Corollary 4.4. Let π be an abelian group of exponent e, K be a field with ζe ∈ K.
Suppose that M is a π-lattice and π acts on K(M) by K-automorphisms. If ρπ(M) is
an invertible module, than K(M)π is retract rational over K.
§5. Proof of Theorems 1.11 and 1.12
Proof of Theorem 1.12 ———————————-
Step 1. Write G = A ⋊ G0 where G0 =< σ > is a cyclic group of order m, and
A = A1 ×A2 × · · · ×Ar with each Ai =< ρi > being a cyclic group of order ei, e = e1
and er | er−1, · · · , e2 | e1.
Define Bi =< ρi, · · · , ρ̂i, · · · , ρr > to be the subgroup of A by deleting the i-th direct
factor.
Let Gal(K(ζe)/K) =< τ > be a cyclic group of order n. Write ζ = ζe and τ ·ζ = ζ
for some integer a.
For 1 ≤ i ≤ r, choose ζi ∈< ζ > such that ζi is a primitive ei-th root of unity. Note
that τ · ζi = ζ
Let ni = [K(ζi) : K] for 1 ≤ i ≤ r. Note that n = n1.
Step 2. Let W =
g∈GK · x(g) be the representation space of the regular repre-
sentation of G. Then
K(G) = K(W )G = {K(ζ)(W )<τ>}G = K(ζ)(W )<G,τ>
where the action of G and τ are extended to K(ζ)(W ) by requiring that G and τ act
trivially on K(ζ) and W respectively.
Step 3. For 1 ≤ i ≤ r, define
l∈Zei , g∈Bi
ζ li · x(ρ
i · g).
Note that ρi · ui = ζ
i ui and, if j 6= i, then ρi · ui = ui.
For 1 ≤ i ≤ r, write χi to be the character χi : A −→ K(ζ)
× such that g · ui =
χi(g) · ui. Since χ1, · · · , χr are distinct characters on A, it follows that u1, · · · , ur are
linearly independent vectors in K(ζ)⊗K W .
Moreover, the subgroup A acts faithfully on
1≤i≤rK(ζ)ui.
Step 4. For 1 ≤ i ≤ r, j ∈ Zni , define
vij =
l∈Zei , g∈Bi
i · x(ρ
i · g).
Thus we have elements v1,1, · · · , v1,n1 , v2,1, · · · , v2,n2, · · · , vr,nr ; in total we have n1 +
n2 + · · ·+ nr such elements.
As in Step 3, these elements vij are linearly independent. Note that τ · vij = vi,j+1.
Step 5. For 1 ≤ i ≤ r, j ∈ Zni , k ∈ Zm. define
xi,j,k =
l∈Zei , g∈Bi
i · x(σ
k · ρli · g).
We have in total m(n1 + n2 + · · ·+ nr) such elements xi,j,k.
Note that xi,j,0 = vij , and the vector xi,j,k is just a “translation” of the vector xi,j,0
in the space
g∈GK(ζ)x(g) (with basis x(g), g ∈ G). Thus these vectors xi,j,k are
linearly independent. Note that σ · vi,j,k = vi,j,k+1. Moreover, the group < G, τ > acts
faithfully on
K(ξ) · xi,j,k.
Apply Theorem 3.1. We find K(G) = K(ζ)(xijk : 1 ≤ i ≤ r, j ∈ Zni, k ∈
Zm)(w1, · · · , ws) where s = |G| −m(n1 + n2 + · · ·+ nr), and λ(wi) = wi for any i, any
λ ∈< G, τ >.
Step 6. We will consider the fixed fieldK(ζ)(xijk : 1 ≤ i ≤ r, j ∈ Zni , k ∈ Zm)
<G,τ>.
Let π =< σ, τ >, Λ = Z[π]. Let N =< xijk : 1 ≤ i ≤ r, j ∈ Zni , k ∈ Zm > be the
multiplicative subgroup generated by these xijk in K(ζ)(xijk : 1 ≤ i ≤ r, j ∈ Zni , k ∈
Zm) \ {0}. Note that N is a Λ-module. In fact, N is a permutation π-lattice.
Define
Φ : N −→ Zn1 × Zn2 × · · · × Znr
by Φ(x) = (l̄1, l̄2, · · · , l̄r) ∈ Zn1 × Zn2 × · · · × Znr , if x =
ijk (with bijk ∈ Z) and
ρ1(x) = ζ
1 x, ρ2(x) = ζ
2 x, · · ·, ρr(x) = ζ
Define M = Ker(Φ). We find that K(ζ)(xijk : 1 ≤ i ≤ r, j ∈ Zni , k ∈ Zm)
K(ζ)(M). Thus it remains to find K(ζ)(M)π. Note that M is a π-lattice.
Step 7. Since gcd{m, n} = 1, it follows that π is a cyclic group. Hence ρπ(M) is
invertible by Theorem 2.2.
Apply Theorem 1.13. We find that K(ζ)(M)π is retract rational over K. Since
K(G) is rational over K(ζ)(M)π, it follows that K(G) is also retract rational over K
by [Sa3, Proposition 3.6(a), p.183]. �
Proof of Theorem 1.11 ——————————-
If charK = p, apply Theorem 3.2. Thus K(G) is rational. In particular it is retract
rational.
¿From now on we will assume that charK 6= p. It is not difficult to verify that all
the assumptions of Theorem 1.12 are valid in the present situation. Hence the result.
Corollary 5.1. Let K be a field, p be any prime number, G = A⋊ G0 be a semi-
direct product of p-groups where A is a normal abelian subgroup of exponent pe and G0
is a cyclic group of order pm. If charK 6= p, assume that (i) both K(ζpe) and K(ζpm)
are cyclic extensions of K and (ii) p ∤ [K(ζpe) : K]. Then K(G) is retract rational over
Proposition 5.2. Let K be a field and G = A ⋊ G0 be a semi-direct product.
Assume that
(i) A is an abelian normal subgroup of exponent e in G and G0 is a cyclic group of
order m.
(ii) either charK = 0 or charK > 0 with charK ∤ em, and
(iii) ζe ∈ K and K(ζm) is a cyclic extension of K.
If G −→ GL(V ) is a finite-dimensional linear representation of G over K, then
K(V )G is retract rational over K.
Proof. Decompose V into V =
Vχ where χ : A −→ K \ {0} runs over linear
characters of A and Vχ = {v ∈ V : g · v = χ(g) · v for all g ∈ A}
It is easy to see that σ(Vχ) = Vχ′ for some χ
′. Suppose Vχ1, Vχ2, · · · , Vχs is an orbit
of σ, i.e. σ(Vχj ) = Vχj+1 and σ(Vχs) = Vχ1. Choose a basis v1, · · · , vt of Vχ1. It follows
that {σj(vi) : 1 ≤ i ≤ t, 0 ≤ j ≤ s− 1} is a basis of ⊕1≤j≤sVχj .
In this way, we may find a basis w1, · · · , wn in V such that (i) for any g ∈ A, any
1 ≤ i ≤ r, g · wi = αwi for some α ∈ K; and (ii) σ · wi = wj for some j.
It follows that K(V )A = K(w1, · · · , wn)
A = K(u1, · · · , un) where u1, · · · , un are
monomials in w1, · · · , wn. Thus K(u1, · · · , un) = K(M) for some G0-lattice M . It
follows that K(V )G = {K(V )A}G0 = K(M)<σ>.
By Theorem 2.2, ρG0(M) is invertible. Apply Theorem 1.12. �
Remark. In the above theorem it is essential that G is a is a split extension group.
For non-split extension groups, monomial actions (instead of merely purely monomial
actions) may arise; see the proof of Theorem 1.3 [Ha] and also [Sa5].
§6. Proof of Theorem 1.9
We will prove Theorem 1.9 in this section.
If charK = p, apply Theorem 3.2. We find that K(G) is rational. In particular, it
is retract rational.
Thus we will assume that charK 6= p from now on till the end of this section.
If ζp ∈ K, we apply Theorem 1.4 to find that K(G) is rational. Hence we will
consider only the situation [K(ζp) : K] = 2 in the sequel.
Since p be an odd prime number, there is only one non-abelian p-group of order
p3 with exponent p, and there are precisely two non-isomorphic non-abelian p-groups
of order p4 with exponent p (see [CK, Section 2 and Section 3]). We will solve the
rationality problem of these three groups separately.
Let ζ = ζp, Gal(K(ζ)/K) = 〈τ〉 with τ · ζ = ζ
Case 1. G = 〈σ1, σ2, σ3 : σ
i = 1, σ1 ∈ Z(G), σ2σ3 = σ3σ1σ2〉.
Step 1. Let W =
g∈GK · x(g) be the representation space of the regular repre-
sentation of G. Note that
K(G) = K(x(g) : g ∈ G)G = {K(ζ)(x(g) : g ∈ G)〈τ〉}G = K(ζ)(x(g) : g ∈ G)〈G,τ〉.
Step 2. For i ∈ Zp, define x0,i, x1,i ∈
g∈GK(ζ) · x(g) by
x0,i =
j,k∈Zp
ζ−j−kx(σi3σ
x1,i =
j,k∈Zp
ζj+k · x(σi3σ
We find that
σ1 : x0,i 7→ ζx0,i, x1,i 7→ ζ
−1x1,i,
σ2 : x0,i 7→ ζ
i+1x0,i, x1,i 7→ ζ
−i−1x1,i,
σ3 : x0,i 7→ x0,i+1, x1,i 7→ x1,i+1,
τ : x0,i ↔ x1,i.
The restriction of the action of 〈σ1, σ2〉 toK(ζ)·x0,i andK(ζ)·x1,i is given by distinct
characters of 〈σ1, σ2〉 to K(ζ) \ {0}. Hence x0,0, x0,1, x0,2, . . . , x0,p−1, x1,0, . . . , x1,p−1 are
linearly independent vectors. Moreover, the action of 〈G, τ〉 on K(ζ)(x0,i, x1,i : i ∈ Zp)
is faithful.
By Theorem 3.1 K(ζ)(x(g) : g ∈ G) = K(ζ)(x0,i, x1,i : i ∈ Zp)(X1, . . . , Xl) where
l = p3 − 2p and ρ(Xj) = Xj for any 1 ≤ j ≤ l, any ρ ∈ 〈G, τ〉.
Step 3. Define x0 = x
0,0, y0 = x0,0x1,0 and xi = x0,i · x
0,i−1, yi = y0,i · y
0,i−1 for
1 ≤ i ≤ p − 1. It follows that K(ζ)(x0,i, x1,i : i ∈ Zp)
〈σ1〉 = K(ζ)(xi, yi : i ∈ Zp).
Moreover, the actions of σ2, σ3, τ are given as
σ2 : x0 7→ x0, y0 7→ y0, xi 7→ ζxi, yi 7→ ζ
−1yi for 1 ≤ i ≤ p− 1,
σ3 : x0 7→ x0x
1, x1 7→ x2 7→ · · · 7→ xp−1 7→ (x1x2 · · ·xp−1)
y0 7→ y0y1x1, y1 7→ y2 7→ · · · 7→ yp−1 7→ (y1y2 · · · yp−1)
τ : x0 7→ y
0 , y0 7→ y0, xi ↔ yi for 1 ≤ i ≤ p− 1.
Step 4. Define X = x0y
−(p−1)/2
0 , Y = x
(p+1)/2
Then K(ζ)(xi, yi : i ∈ Zp) = K(ζ)(X, Y, xi, yi : 1 ≤ i ≤ p − 1) and σ2(X) = X ,
σ2(Y ) = Y , σ3(X) = αX , σ3(Y ) = βY , τ : X ↔ Y where α, β ∈ K(ζ)(xi, yi : 1 ≤ i ≤
p− 1)\{0}.
Apply Theorem 3.1. We may find X̃ , Ỹ so that K(ζ)(xi, yi : i ∈ Zp) = K(ζ)(xi, yi :
1 ≤ i ≤ p − 1)(X̃, Ỹ ) with ρ(X̃) = X̃ , ρ(Ỹ ) = Ỹ for any ρ ∈ 〈σ2, σ3, τ〉. Thus it
remains to consider K(ζ)(xi, yi : 1 ≤ i ≤ p− 1)
〈σ2,σ3,τ〉.
Step 5. Define u0 = x
1, v0 = x1y1, ui = xi+1x
i , vi = yi+1y
i for 1 ≤ i ≤ p− 2.
It follows that K(ζ)(xi, yi : 1 ≤ i ≤ p − 1)
〈σ2〉 = K(ζ)(ui, vi : 0 ≤ i ≤ p − 2). The
actions of σ3 and τ are given by
σ3 : u0 7→ u0u
1, v0 7→ v0v1u1, u1 7→ u2 7→ · · · 7→ up−2 7→ (u0u
2 · · ·u
v1 7→ v2 7→ · · · 7→ vp−2 7→ u0(v
1 · · · v
τ : u0 7→ u
0, v0 7→ v0, ui ↔ vi for 1 ≤ i ≤ p− 2.
Step 6. Define up−1 = (u0u
2 · · ·u
−1, wi = v0v1 · · · viu1u2 · · ·ui for 1 ≤ i ≤
p − 2, and wp−1 = (v
1 · · · vp−2u
2 · · ·up−2)
−1. We find that K(ui, vi : 0 ≤
i ≤ p− 2) = K(ui, wi : 1 ≤ i ≤ p− 1) and
σ3 : u1 7→ u2 7→ · · · 7→ up−1 7→ (u1u2 · · ·up−1)
w1 7→ w2 7→ · · · 7→ wp−1 7→ (w1w2 · · ·wp−1)
τ : ui 7→ wi(uiwi−1)
−1, wi 7→ wi for 1 ≤ i ≤ p− 1.
where we write w0 = v0 for convenience. (Granting that w0 is defined as above, we
have a relation w0w1 · · ·wp−1 = 1. But we don’t have the relation u0u1 · · ·up−1 = 1
because we define u0, u1, . . . , up−2 first and up−1 is defined by another way.)
We will study whether K(ζ)(ui, wi : 1 ≤ i ≤ p− 1)
〈σ3,τ〉 is rational or not.
Step 7. The multiplicative action in Step 6 can be formulated in terms of π-lattices
where π = 〈τ, σ3〉 as follows.
LetM = (
1≤i≤p−1 Z·ui)⊕(
1≤i≤p−1Z·wi) and define w0 = −w1−w2−· · ·−wp−1.
Define a Z[π]-module structure on M by
σ3 : u1 7→ u2 7→ · · · 7→ up−1 7→ −u1 − u2 − · · · − up−1,
w1 7→ w2 7→ · · · 7→ wp−1 7→ −w1 − w2 − · · · − wp−1,
τ : ui 7→ −ui + wi − wi−1, wi 7→ wi for 1 ≤ i ≤ p− 1.
We claim that M ≃ Z[π1]
Z[π2]/Φp(σ3) with π1 = 〈τ〉, π2 = 〈σ3〉.
Throughout this step, we will write σ = σ3 and π =< σ, τ >. Define ρ = στ . Then
ρ is a generator of π with order 2p where p is an odd prime number.
Let Φp(T ) ∈ Z[T ] be the p-th cyclotomic polynomial. Note that Φp(σ)(ui) =
Φp(σ)(wi) = 0. It follows that Φp(σ) ·M = 0.
Since Φp(σ
0≤i≤p−1 σ
i = Φp(σ), we know that Φp(σ
2) ·M = 0. From ρ2 = σ2,
we find that Φp(ρ
2) ·M = 0. It is well-known that Φp(T
2) = Φp(T )Φ2p(T ) (see [Ka1,
Theorem 1.1] for example). We conclude that Φp(ρ)Φ2p(ρ) ·M = 0. In other words,
we may regard M as a module over Λ = Z[π]/ < Φp(ρ)Φ2p(ρ) >.
Clearly Λ is isomorphic to Z[π1]
Z[π2]/Φp(σ) where π1 = 〈τ〉, π2 = 〈σ〉.
It remains to show that M is isomorphic to Λ as a Λ-module.
It is not difficult to verify that Φ2p(ρ)(u1) = f(ρ)(w1) where f(T ) = Φ2p(T )−T
Define v = u1 − w1. We find that Φ2p(ρ)(v) = −w0.
Since M is a Λ-module generated by u1 and w0, it follows that, as a Λ-module,
M =< u1, w0 >=< u1 − σw0, w0 > =< u1 − w1, w0 >=< v,w0 > =< v,w0 +
Φ2p(ρ)(v) >=< v,w0 − w0 >=< v >, i.e. M is a cyclic Λ-module generated by
v. Thus we get an epimorphism Λ → M , which is an isomorphism by counting the
Z-ranks of both sides. Hence the result.
Step 8. By Step 7, M ≃ Z[π1]
Z[π2]/Φp(σ3) where π1 = 〈τ〉, π2 = 〈σ3〉.
Thus we may choose a Z-basis Y1, Y2, . . . , Yp−1, Z1, . . . , Zp−1 for M such that
σ3 : Y1 7→ Y2 7→ · · · 7→ Yp−1 7→ −Y1 − · · · − Yp−1,
Z1 7→ Z2 7→ · · · 7→ Zp−1 7→ −Z1 − · · · − Zp−1,
τ : Yi ↔ Zi for 1 ≤ i ≤ p− 1.
Hence K(ζ)(M) = K(ζ)(Yi, Zi : 1 ≤ i ≤ p− 1). We emphasize that σ3 acts on Yi,
Zi by multiplicative actions on the field K(ζ)(M) and σ3 · ζ = ζ , τ · ζ = ζ
Step 9. In the field K(ζ)(M), define
s0 = 1 + Y1 + Y1Y2 + Y1Y2Y3 + · · ·+ Y1Y2 · · ·Yp−1,
t0 = 1 + Z1 + Z1Z2 + · · ·+ Z1Z2 · · ·Zp−1,
s1 = (1/s0)− (1/p), s2 = (Y1/s0)− (1/p), . . . , sp−1 = (Y1Y2 · · ·Yp−2/s0)− (1/p),
t1 = (1/t0)− (1/p), t1 = (Z1/t0)− (1/p), . . . , tp−1 = (Z1Z2 · · ·Zp−2/t0)− (1/p).
It is easy to verify that K(ζ)(M) = K(ζ)(si, ti : 1 ≤ i ≤ p− 1) and
σ3 : s1 7→ s2 7→ · · · 7→ sp−1 7→ −s1 − s2 − · · · − sp−1,
t1 7→ t2 7→ · · · 7→ tp−1 7→ −t1 − t2 − · · · − tp−1,
τ : si ↔ ti.
Step 10. Define ri = si + ti for 1 ≤ i ≤ p − 1. Then K(ζ)(si, ti : 1 ≤ i ≤ p− 1) =
K(ζ)(si, ri : 1 ≤ i ≤ p− 1). Note that
σ3 : r1 7→ r2 7→ · · · 7→ rp−1 7→ −r1 − r2 − · · · − rp−1,
τ : ri 7→ ri, si 7→ −si + ri.
Apply Theorem 3.1. We find that K(ζ)(si, ri : 1 ≤ i ≤ p− 1) = K(ζ)(ri : 1 ≤ i ≤
p−1)(A1, . . . , Ap−1) for some A1, . . . , Ap−1 with σ3(Ai) = τ(Ai) = Ai for 1 ≤ i ≤ p−1.
Thus K(ζ)(ri, si : 1 ≤ i ≤ p − 1)
〈τ〉 = K(ζ)(ri : 1 ≤ i ≤ p − 1)
〈τ〉(A1, . . . , Ap−1) =
K(r1, . . . , rp−1, A1, . . . , Ap−1).
It remains to find K(r1, . . . , rp−1)
〈σ3〉.
Step 11. Write π2 = 〈σ3〉. The π2-fieldsK(r1, . . . , rp−1, A1) andK(B0, B1, . . . , Bp−1)
are π2-isomorphic where σ3 : B0 7→ B1 7→ · · · 7→ Bp−1 7→ B0. For, we may define
B = B0 +B1 + · · ·+Bp−1 and Ci = Bi − (B/p) for 0 ≤ i ≤ p− 1.
In other words K(r1, . . . , rp−1, A1, . . . , Ap−1)
〈σ3〉 = K(B0, B1, . . . , Bp−1)
〈σ3〉(A2, . . .,
Ap−1) = K(Zp)(A2, . . . , Ap−1).
By Lemma 3.5, K(Zp) = K(ζ)(N)
π1 where ζ = ζp, π1 = Gal(K(ζ)/K) = 〈τ〉 with
τ · ζ = ζ−1, and N is some π1-lattice.
By Reiner’s Theorem [Re], the π1-lattice N is a direct sum of lattices of three types:
(1) τ : z 7→ −z, (2) τ : z 7→ z, (3) τ : z1 ↔ z2. Thus we find K(ζ)(N)
〈τ〉 = K(ζ)(z1,
. . . , za, z
2, . . . , z
1 , w
1 , . . . , z
c , w
c ) where τ : z1 7→ 1/z1, . . ., za 7→ 1/za, z
1 7→ z
. . ., z′b 7→ z
1 ↔ w
1 , . . ., z
c ↔ z
By Theorem 3.1 we may “neglect” the roles of z′1, . . . , z
1 , w
1 , . . . , z
c , w
c . We may
linearize the action on z1, . . . , za by defining wi = 1/(1 + zi) when 1 ≤ i ≤ a. Then
τ : wi 7→ −wi + 1. Thus we may “neglect” the roles of wi by Theorem 3.1 again. In
conclusion, K(ζ)(z1, . . . , za, z
1, . . . , z
1 , w
1 , . . . , z
c , w
〈τ〉 is rational over K. �
Case 2. G = 〈σ1, σ2, σ3, σ4 : σ
i = 1, σ1, σ2 ∈ Z(G), σ
4 σ3σ4 = σ1σ3〉.
The proof is very similar to Case 1.
Step 1. For i ∈ Zp, define x0,i, x1,i, y0,i, y1,i ∈
g∈GK(ζ) · x(g) by
x0,i =
j,k,l∈Zp
ζ−j−k−lx(σi4σ
x1,i =
j,k,l∈Zp
ζj+k+lx(σi4σ
y0,i =
j,k,l∈Zp
ζ−j−k+lx(σi4σ
y1,i =
j,k,l∈Zp
ζj+k−lx(σi4σ
The action of 〈G, τ〉 is given by
σ1 : x0,i 7→ ζx0,i, x1,i 7→ ζ
−1x1,i, y0,i 7→ ζy0,i, y1,i 7→ ζ
−1y1,i,
σ2 : x0,i 7→ ζx0,i, x1,i 7→ ζ
−1x1,i, y0,i 7→ ζ
−1y0,i, y1,i 7→ ζy1,i,
σ3 : x0,i 7→ ζ
i+1x0,i, x1,i 7→ ζ
−i−1x1,i, y0,i 7→ ζ
i+1y0,i, y1,i 7→ ζ
−i−1y1,i,
σ4 : x0,i 7→ x0,i+1, x1,i 7→ x1,i+1, y0,i 7→ y0,i+1, y1,i 7→ y1,i+1,
τ : x0,i ↔ x1,i, y0,i ↔ y1,i.
Note that x0,i, x1,i, y0,i, y1,i where i ∈ Zp are linearly independent vectors in
g∈GK(ζ) · x(g). Apply Theorem 3.1. It suffices to consider the rationality prob-
lem of K(ζ)(x0,i, x1,i, y0,i, y1,i : i ∈ Zp)
〈G,τ〉.
Step 2. Define x0 = x
0,0, y0 = x0,0x1,0, X0 = y0,0(x0,0)
−1, Y0 = y1,0(x1,0)
−1; for
1 ≤ i ≤ p − 1, define xi = x0,i(x0,i−1)
−1, yi = x1,i(x1,i−1)
−1, Xi = y0,i(y0,i−1)
−1, Yi =
y1,i(y1,i−1)
−1. Then K(ζ)(x0,i, x1,i, y0,i, y1,i : i ∈ Zp)
〈σ1〉 = K(ζ)(xi, yi, Xi, Yi : i ∈ Zp)
and the actions are given by
σ2 : x0 7→ x0, y0 7→ y0, X0 7→ ζ
−2X0, Y0 7→ ζ
All the other generators are fixed by σ2;
σ3 : x0 7→ x0, y0 7→ y0, X0 7→ X0, Y0 7→ Y0,
xi 7→ ζxi, yi 7→ ζ
−1yi, Xi 7→ ζXi, Yi 7→ ζ
−1Yi,
σ4 : x0 7→ x0x
1, y0 7→ y0y1x1, X0 7→ X0X1x
1 , Y0 7→ Y0Y1y
x1 7→ x2 7→ x3 7→ · · · 7→ xp−1 7→ (x1x2 · · ·xp−1)
Similarly for y1, y2, . . . , yp−1, X1, . . . , Xp−1, Y1, . . . , Yp−1;
τ : x0 7→ y
0 , y0 7→ y0, X0 7→ Y0 7→ X0,
xi ↔ yi, Xi ↔ Yi for 1 ≤ i ≤ p− 1.
Step 3. Define x̃ = x0y
−(p−1)/2
0 , ỹ = x
(p+1)/2
Since x̃, ỹ are fixed by both σ2 and σ3, while σ4 : x̃ 7→ αx̃, ỹ 7→ βỹ, τ : x̃ ↔ ỹ
where α, β ∈ K(x1, y1, X1, Y1)\{0}, we may apply Theorem 3.1. Thus the roles of x̃, ỹ
may be “neglected”. It suffices to consider whether K(ζ)(X0, Y0, xi, yi, Xi, Yi : 1 ≤ i ≤
p− 1)〈σ2,σ3,σ4,τ〉 is rational over K.
Step 4. Define X̃ = X
0 , Ỹ = X0Y0. Then K(ζ)(X0, Y0, xi, yi, Xi, Yi : 1 ≤ i ≤
p− 1)〈σ2〉 = K(ζ)(X̃, Ỹ , xi, yi, Xi, Yi : 1 ≤ i ≤ p− 1). Moreover, the actions on X̃ and
Ỹ are given by
σ3 : X̃ 7→ X̃, Ỹ 7→ Ỹ ,
σ4 : X̃ 7→ X̃X
1 , Ỹ 7→ Ỹ X1Y1x
τ : X̃ 7→ X̃−1Ỹ p, Ỹ 7→ Ỹ .
Define X ′ = X̃Ỹ −(p−1)/2, Y ′ = X̃−1Ỹ (p+1)/2. As in Step 3, we may apply The-
orem 3.1 and “neglect” the roles of X ′ and Y ′. It remains to make sure whether
K(ζ)(xi, yi, Xi, Yi : 1 ≤ i ≤ p− 1)
〈σ3,σ4,τ〉 is rational.
Define u0 = x
1, v0 = x1y1; for 1 ≤ i ≤ p−2, define ui = xi+1(x
i ), vi = yi+1 · (y
and for 1 ≤ i ≤ p−1, define Ui = Xix
i , Vi = Yiy
i . It follows that K(ζ)(xi, yi, Xi, Yi :
1 ≤ i ≤ p− 1)〈σ3〉 = K(ζ)(u0, v0, Up−1, Vp−1, ui, vi, Ui, Vi : 1 ≤ i ≤ p− 2). Note that the
actions of σ4 and τ are given by
σ4 : u0 7→ u0u
1, v0 7→ v0v1u1, u1 7→ u2 7→ · · · 7→ up−2 7→ (u0u
2 · · ·u
v1 7→ v2 7→ · · · 7→ vp−2 7→ u0(v
1 · · · v
U1 7→ U2 7→ · · · 7→ Up−1 7→ (U1U2 · · ·Up−1)
V1 7→ V2 7→ · · · 7→ Vp−1 7→ (V1V2 · · ·Vp−1)
τ : u0 7→ u
0, v0 7→ v0, ui ↔ vi for 1 ≤ i ≤ p− 2,
Ui ↔ Vi for 1 ≤ i ≤ p− 1.
Note that the actions of σ4 and τ on U1, U2, . . . , Up−1, V1, . . . , Vp−1 may be linearized
by the same method as in Step 9 of Case 1. Hence we may apply Theorem 3.1 and
“neglect” the roles of Ui, Vi for 1 ≤ i ≤ p− 1.
In conclusion, all we need is to prove that K(ζ)(ui, vi : 0 ≤ i ≤ p−2)
〈σ4,τ〉 is rational
over K.
Compare the present situation with that of Step 5 of Case 1. We have the same
generators and the same actions (and even the same notation). Thus we finish the
proof. �
Case 3. G = 〈σ1, σ2, σ3, σ4 : σ
i = 1, σ1 ∈ Z(G), σ2σ3 = σ3σ2, σ
4 σ2σ4 =
σ1σ2, σ
4 σ3σ4 = σ2σ3〉.
Step 1. For i ∈ Zp, define x0,i, x1,i, y0,i, y1,i ∈
g∈GK(ζ) · x(g) by
x0,i =
j,k,l∈Zp
ζ−j−k−lx(σi4σ
x1,i =
j,k,l∈Zp
ζj+k+lx(σi4σ
y0,i =
j,k,l∈Zp
ζ−j−k+lx(σi4σ
y1,i =
j,k,l∈Zp
ζj+k−lx(σi4σ
The action of 〈G, τ〉 is given by
σ1 : x0,i 7→ ζx0,i, x1,i 7→ ζ
−1x1,i, y0,i 7→ ζ
−1y0,i, y1,i 7→ ζy1,i,
σ2 : x0,i 7→ ζ
i+1x0,i, x1,i 7→ ζ
−i−1x1,i, y0,i 7→ ζ
−i+1y0,i, y1,i 7→ ζ
i−1y1,i,
σ3 : x0,i 7→ ζ
i+1x0,i, x1,i 7→ ζ
−i−1x1,i, y0,i 7→ ζ
i+1y0,i, y1,i 7→ ζ
−i−1y1,i,
σ4 : x0,i 7→ x0,i+1, x1,i 7→ x1,i+1, y0,i 7→ y0,i+1, y1,i 7→ y1,i+1,
τ : x0,i ↔ x1,i, y0,i ↔ y1,i.
Checking the restriction of 〈σ1, σ2, σ3〉 as in Step 1 of Case 2, we find that these
vectors x0,i, x1,i, y0,i, y1,i are linearly independent except possibly for the case x0,p−2
and y1,0, and the case x1,p−2 and y0,0. But it is easy to see that these vectors are linearly
independent, because their ”supports” are disjoint. Apply Theorem 3.1. It suffices to
consider K(ζ)(x0,i, x1,i, y0,i, y1,i : i ∈ Zp)
〈G,τ〉.
Step 2. Define x0 = x
0,0, y0 = x0,0x1,0, X0 = x0,0 · y0,0, Y0 = x1,0y1,0; and for
1 ≤ i ≤ p − 1, define xi = x0,i(x0,i−1)
−1, yi = x1,i(x1,i−1)
−1, Xi = y0,i(y0,i−1)
−1, Yi =
y1,i(y1,i−1)
−1. Then K(ζ)(x0,i, x1,i, y0,i, y1,i : i ∈ Zp)
〈σ1〉 = K(ζ)(xi, yi, Xi, Yi : i ∈ Zp)
and the actions are given by
σ2 : x0 7→ x0, y0 7→ y0, X0 7→ ζ
2X0, Y0 7→ ζ
−2Y0,
xi 7→ ζxi, yi 7→ ζ
−1yi, Xi 7→ ζ
−1Xi, Yi 7→ ζYi,
σ3 : x0 7→ x0, y0 7→ y0, X0 7→ ζ
2X0, Y0 7→ ζ
−2Y0,
xi 7→ ζxi, yi 7→ ζ
−1yi, Xi 7→ ζXi, Yi 7→ ζ
−1Yi,
σ4 : x0 7→ x0x
1, y0 7→ y0y1X1, X0 7→ X0X1x1, Y0 7→ Y0Y1y1,
x1 7→ x2 7→ x3 7→ · · · 7→ xp−1 7→ (x1x2 · · ·xp−1)
Similarly for y1, y2, . . . , yp−1, X1, . . . , Xp−1, Y1, . . . , Yp−1.
τ : x0 7→ y
0 , y0 7→ y0, X0 7→ Y0 7→ X0,
xi ↔ yi, Xi ↔ Yi for 1 ≤ i ≤ p− 1.
Step 3. Define x̃ = x0y
−(p−1)/2
0 , ỹ = x
(p+1)/2
0 . Apply Theorem 3.1 again. It
suffices to consider K(ζ)(X0, Y0, xi, yi, Xi, Yi : 1 ≤ i ≤ p− 1)
〈σ2,σ3,σ4,τ〉.
We may apply Theorem 3.1 again to “neglect” X0 and Y0. Thus it suffices to
consider K(ζ)(xi, yi, Xi, Yi : 1 ≤ i ≤ p− 1)
〈σ2,σ3,σ4,τ〉.
Step 4. Define u0 = x
1, v0 = x1y1, U0 = x1X1, V0 = y1Y1; and for 1 ≤ i ≤ p−2, de-
fine ui = x
i xi+1, vi = y
i yi+1, Ui = X
i Xi+1, Vi = Y
i Yi+1. ThenK(ζ)(xi, yi, Xi, Yi :
1 ≤ i ≤ p− 1)〈σ2〉 = K(ζ)(ui, vi, Ui, Vi : 0 ≤ i ≤ p− 2). Note that the actions of σ3, σ4,
τ are given by
σ3 : u0 7→ u0, v0 7→ v0, U0 7→ ζ
2U0, V0 7→ ζ
−2U0,
All the other generators are fixed by σ3.
σ4 : u0 7→ u0u
1, v0 7→ v0v1u1, U0 7→ U0U1u1, V0 7→ V0V1v1,
u1 7→ u2 7→ · · · 7→ up−2 7→ (u0u
2 · · ·u
v1 7→ v2 7→ · · · 7→ vp−2 7→ u0(v
1 · · · v
U1 7→ U2 7→ · · · 7→ Up−2 7→ u0(U
1 · · ·U
V1 7→ V2 7→ · · · 7→ Vp−2 7→ u
1 · · ·V
τ : u0 7→ v
0 , v0 7→ v0, U0 7→ V0 7→ U0,
ui ↔ vi, Ui ↔ Vi for 1 ≤ i ≤ p− 2.
Step 5. Define R0 = U
0 , S0 = U0V0; and for 1 ≤ i ≤ p− 2, define Ri = Ui, Si = Vi.
Then K(ζ)(ui, vi, Ui, Vi : 0 ≤ i ≤ p − 2)
〈σ3〉 = K(ζ)(ui, vi, Ri, Si : 0 ≤ i ≤ p − 2). We
will write the actions of σ4 and τ on Ri, Si as follows.
σ4 : R0 7→ R0R
1, S0 7→ S0S1R1u1v1,
R1 7→ R2 7→ · · · 7→ Rp−2 7→ u0(R0R
2 · · ·R
S1 7→ S2 7→ · · · 7→ Sp−2 7→ u
0R0(S
1 · · ·S
τ : R0 7→ S
0 , S0 7→ S0, Ri ↔ Si for 1 ≤ i ≤ p− 2.
Step 6. Imitate the change of variables in Step 6 of Case 1. We define up−1 =
2 · · ·u
−1, Rp−1 = u0(R0R
2 · · ·R
−1; and for 1 ≤ i ≤ p − 2,
define wi = v0v1 · · · viu1u2 · · ·ui, Ti = S0S1 · · ·SiR1R2 · · ·Ri; define wp−1 = (v
1 · · · vp−2u
2 · · ·up−2)
−1, Tp−1 = u1v1v
1 · · ·Sp−2R
2 · · ·Rp−2)
We find that K(ui, vi, Ri, Si : 0 ≤ i ≤ p− 2) = K(ui, wi, Ri, Ti : 1 ≤ i ≤ p− 1) and
σ4 : u1 7→ u2 7→ · · · 7→ up−1 7→ (u1u2 · · ·up−1)
w1 7→ w2 7→ · · · 7→ wp−1 7→ (w1w2 · · ·wp−1)
R1 7→ R2 7→ · · · 7→ Rp−1 7→ (R1R2 · · ·Rp−1)
T1 7→ T2 7→ · · · 7→ Tp−1 7→ (T1T2 · · ·Tp−1)
τ : wi 7→ wi, Ti 7→ Ti, ui 7→ wi(uiwi−1)
−1, Ri 7→ Ti(RiTi−1)
where we write w0 = v0, T0 = S0 for convenience.
Step 7. The multiplicative action in Step 6 can be formulated as follows.
Let π = 〈τ, σ4〉 and define a π-lattice as in Step 7 of Case 1 (but σ3 should be
replaced by σ4 in the present situation). Then K(ζ)(ui, wi, Ri, Ti : 1 ≤ i ≤ p − 1)
K(ζ)(M ⊕M)π where M is the same lattice in Step 7 of Case 1.
The structure ofM has been determined in Step 7 of Case 1. Thus M⊕M ≃ Λ⊕Λ
where Λ ≃ Z[π1] ⊗Z Z[π2]/Φp(σ4) where π1 = 〈τ〉 and π2 = 〈σ4〉. It follows that we
can find elements Y1, . . . , Yp−1, Z1, . . . , Zp−1, W1, . . . ,Wp−1, Q1, . . . , Qp−1 in the field
K(ζ)(ui, wi, Ri, Ti : 1 ≤ i ≤ p − 1) so that K(ζ)(ui, wi, Ri, Ti : 1 ≤ i ≤ p − 1) =
K(ζ)(Y1, . . . , Yp−1, Z1, . . . , Zp−1,W1, . . . ,Wp−1, Q1, . . . , Qp−1) and the actions of σ4 and
τ are given by
σ4 : Y1 7→ · · · 7→ Yp−1 7→ (Y1Y2 · · ·Yp−1)
Similarly for Z1, . . . , Zp−1, W1, . . . ,Wp−1, Q1, . . . , Qp−1;
τ : Yi ↔ Zi, Wi ↔ Qi.
Step 8. We can linearize the actions of σ4 and τ by the same method in Step 9 of
Case 1. Apply Theorem 3.1. We may “neglect” the roles of Wi, Qi. Thus the present
situation is the same as in Step 8 of Case 1. �
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