Lemma 9.6. Let G be a finite group, let : (Z/2Z)[G] → (Z/2Z)[G] be the ring anti-automorphism that maps each element of G to its inverse, let
p1 : (Z/2Z)[G] → Z/2Z be the nonzero additive group homomorphism that
maps each element ofGdifferent from the identity to 0, and let αbe the map
α: (Z/2Z)[G]×(Z/2Z)[G]→Z/2Z,
(f, g)7→p1(f g).
Then the triple ((Z/2Z)[G],Z/2Z, α)is a skew abelian group.
Proof. Sinceαis an antisymmetric perfect pairing, the statement of Lemma 9.6 follows.
Remark 9.7. Let the notation be as in Lemma 9.6. Then the underlying set of Gis a basis of the (Z/2Z)-vector space (Z/2Z)[G] and for allσ, τ ∈Gone
has
α(σ, τ) =
(
0 ifσ6=τ,
1 ifσ=τ.
Lemma 9.8. Let the notation be as in Lemma 9.6. Let Φ be the group of
(Z/2Z)[G]-module isomorphisms (Z/2Z)[G] −∼→ (Z/2Z)[G] that are isomor-
phisms of skew abelian groups ((Z/2Z)[G],Z/2Z, α)−∼→((Z/2Z)[G],Z/2Z, α).
Then there is a group isomorphism
Φ→ {u∈(Z/2Z)[G]∗:uu= 1}
given by
ϕ7→ϕ(1).
Proof. LetRbe the ring (Z/2Z)[G]. The opposite ringRopofRis isomorphic
to the endomorphism ring of the left R-moduleR by the mapping each ele- mentr∈Ropto the endomorphism given by right multiplication byr. Hence,
there is a group isomorphism {R-module isomorphismsR→R} →R∗ given by ϕ 7→ ϕ(1)−1. Let ϕ be an R-module isomorphism R → R. By definition
9.2. Group rings
of isomorphism of skew abelian groups, we have ϕ∈ Φ if and only if for all
f, g ∈R we have α(f, g) =α(ϕ(f), ϕ(g)), that is p1(f g) and p1(f ϕ(1)ϕ(1)g)
are equal. Hence ϕ ∈ Φ is equivalent to ϕ(1)ϕ(1) = 1. The statement of Lemma 9.8 follows.
Remark 9.9. LetKbe a field and letGbe a group. The subgroup
{u∈K[G]∗:uu= 1}
of the group of units of K[G] is often called the unitary subgroup. See [69] by Szak´acs for a description of the structure of the unitary subgroup of the group of units ofK[G] whenGis a finite abelian group.
Theorem 9.10. Let G be a finite abelian group of odd order, let M be a free (Z/2Z)[G]-module of rank 1, let δ : M ×M → Z/2Z be an antisym- metric perfect pairing that for each σ ∈ G and for all x, y ∈ M satisfies
δ(σ(x), σ(y)) =δ(x, y), and let((Z/2Z)[G],Z/2Z, α)be the skew abelian group
defined as in Lemma 9.6. Then there exists a(Z/2Z)[G]-module isomorphism
M → (Z/2Z)[G] that is an isomorphism of skew abelian groups from
(M,Z/2Z, δ)to((Z/2Z)[G],Z/2Z, α).
Proof. Let R be the ring (Z/2Z)[G] and let e ∈ M be a generator of M as
anR-module. Since the (Z/2Z)-dual ofM is a freeR-module of rank 1, there
exists u∈R that for eachz∈Rsatisfiesδ(ze, e) =p1(zu). Letu∈R be such
an element. For eachσ∈Gand for eachw∈Rwe have
δ(we, σe) =δ(σwe, e) =p1(σwu) =p1(wuσ).
By linearity of δin the second argument, for allw, z∈Rwe get
δ(we, ze) =p1(wuz).
Since δ is a perfect pairing, we have u ∈ R∗. By symmetry of δ we get
u=u. Since by Maschke’s Theorem the group ringRis semisimple, the Artin– Wedderburn theorem implies that it is isomorphic to a product of finite fields of characteristic 2. Hence u has odd order. Let d be the order of u and let
v=ud+12 . Since we havevv=u, for allw, z∈R we get
δ(we, ze) =α(wv, zv).
Hence, theR-module isomorphismM →Rthat mapsetovis an isomorphism of skew abelian groups from (M,Z/2Z, δ) to ((Z/2Z)[G],Z/2Z, α).
Lemma 9.11. LetGbe a finite abelian group of odd order, letebe the exponent of G, and let α : (Z/2Z)[G]×(Z/2Z)[G] → Z/2Z be the map defined in
Lemma 9.6. Suppose that the residue class of−1in(Z/eZ)∗is contained in the subgroup generated by the residue class of2. Then(Z/2Z)[G]has no nontrivial self-annihilating sub-(Z/2Z)[G]-modules with respect toα.
Chapter 9. Cyclic number fields
Proof. Let k be an integer satisfying the congruence 2k ≡ −1 mode. Since by Maschke’s Theorem the group ring (Z/2Z)[G] is semisimple, the Artin– Wedderburn theorem implies that it is isomorphic to a product of finite fields of characteristic 2. The ring isomorphism : (Z/2Z)[G] → (Z/2Z)[G] that maps each element ofGto its inverse acts as an automorphism on each finite field in the product, because for each g ∈ G we have g2k = g−1. Since α
is a perfect pairing, on each component isomorphic to a finite field the map
p1 defined in Lemma 9.6 is not the zero map. The statement of Lemma 9.11 follows.
Lemma 9.12. Let pbe an odd prime, letGa cyclic group of orderp, letσbe a generator of G, and let : (Z/2Z)[G]→(Z/2Z)[G] be the ring isomorphism
that maps σ toσ−1. Suppose that the residue class of2 in the group (Z/pZ)∗
has even order and let dbe its order. Then one has
|{u∈(Z/2Z)[G]∗:uu= 1}|= (2d/2+ 1) p−1
d .
Proof. The group ring (Z/2Z)[G] is isomorphic to the ring (Z/2Z)[X]/(Xp−1) by mappingσto the residue class ofX. LetF2dbe a finite field of 2delements.
The factorization ofXp−1 in (
Z/2Z)[X] and the Chinese remainder theorem
give a ring isomorphism
(Z/2Z)[G]−→∼ Z/2Z×(F2d) p−1
d .
Since we have 2d/2 ≡ −1 modp and X = X−1, we get X ≡ X2d/2 mod Xp−1. Hence, the ring isomorphism : (
Z/2Z)[G]→(Z/2Z)[G] maps every
element in the ring to its 2d/2-th power and acts as an automorphism on each
component isomorphic to the field F2d. Since the equation y2 d/2+1
= 1 has 2d/2+ 1 solutions in F2d, the statement of Lemma 9.12 follows.
Corollary 9.13. Let the notation and hypotheses be as in Lemma 9.12. Then the following are equivalent.
(i) One has {u∈(Z/2Z)[G]∗:uu= 1}=G.
(ii) One has |{u∈(Z/2Z)[G]∗:uu= 1}|=p.
(iii) The prime pequals either 3 or5.
Proof. The inclusion G⊆ {u∈(Z/2Z)[G]∗ :uu= 1} implies the equivalence
between (i) and (ii). Since the pairs (d, p) of integers satisfying the equalities
2d/2+ 1 =p p−1 d = 1
are (2,3) and (4,5), by Lemma 9.12 we get the equivalence between (ii) and (iii).
9.2. Group rings
Corollary 9.14. Let pbe either 3 or 5, let G be a cyclic group of order p, let ϕ: (Z/2Z)[G] →(Z/2Z)[G] be a(Z/2Z)[G]-module isomorphism, and let
α: (Z/2Z)[G]×(Z/2Z)[G]→Z/2Zbe the map defined in Lemma 9.6. Then the following are equivalent.
(i) The mapϕis an isomorphism of skew abelian groups
((Z/2Z)[G],Z/2Z, α)−→∼ ((Z/2Z)[G],Z/2Z, α).
(ii) The mapϕpermutes the basis{g:g∈G}of the vector space(Z/2Z)[G]
overZ/2Z.
Proof. This follows from Lemma 9.8 and Corollary 9.13.
Lemma 9.15. Let n be a positive integer, let F2n be a finite field of 2n ele-
ments, let Tr :F2n→Z/2Zbe the trace map, and letδ be the map
δ:F2n×F2n→Z/2Z,
(a, b)7→Tr(ab).
Then the triple (F2n,Z/2Z, δ)is a skew abelian group.
Proof. The mapδis a perfect pairing, because the field extensionF2n/(Z/2Z)
is separable. Since by definition it is symmetric and the fieldZ/2Zhas charac-
teristic 2, it is antisymmetric. Hence, the triple (F2n,Z/2Z, δ) is a skew abelian
group.
Definition 9.16 (Self-dual basis). Letnbe a positive integer, let L/K be a Galois field extension of degree n, and let Tr :L →K be the trace map. A basis {ei : i ∈ Z/nZ} of the K-vector space L is a self-dual basis if for all
i, j∈Z/nZone has
Tr(eiej) =
(
0 ifi6=j,
1 ifi=j.
Theorem 9.17. Let the notation be as in Lemma 9.15, let Gbe the Galois group of the field extension F2n/(Z/2Z), and let ((Z/2Z)[G],Z/2Z, α) be the
skew abelian group defined as in Lemma 9.6. Suppose thatnis odd. Then there exists a (Z/2Z)[G]-module isomorphism F2n →(Z/2Z)[G] that is an isomor-
phism of skew abelian groups from (F2n,Z/2Z, δ)to((Z/2Z)[G],Z/2Z, α).
Proof. The fieldF2nis a free (Z/2Z)[G]-module of rank 1. Since for eachσ∈G
and for alla, b∈F2n we have
δ(σ(a), σ(b)) =δ(a, b),
Chapter 9. Cyclic number fields
Corollary 9.18. Every finite extension ofZ/2Zof odd degree has a self-dual normal basis.
Proof. This follows from Theorem 9.17 and Remark 9.7.
One may wonder which finite field extensions have a self-dual normal basis. A complete answer was given by Lempel and Weinberger [40]. Theorem 9.19 extends their result to finite abelian extensions of arbitrary fields.
Theorem 9.19 (Bayer-Fluckiger and Lenstra [4]). Letnbe a positive integer and letL/K be an abelian field extension of degree n.
(a) IfK does not have characteristic2, thenLhas a self-dual normal basis overK if and only ifnis odd.
(b) IfK has characteristic2, then Lhas a self-dual normal basis overK if and only if the exponent of the Galois group of L/K is not divisible by4. Proof. See Theorem 6.1 in [4] by Bayer-Fluckiger and Lenstra.