Let p ̸= 2 be a prime. In the following, we state some results of integral representations of dihedral groups of order 2p. Our reference is Lee [30].
Let S = Z or Z(2p) = {mn|n, m ∈ Z for k ∤ m, k = 2 and p}. Let θ be a primitive p-th roots of unity, K the field Q(θ) and K0 the field Q(θ + ¯θ), where ¯θ is the conjugate of θ. Let R and R0 be the integral closures of S in K and K0. Let G = ⟨a, b⟩, where a2 = bp = e, and ab = bp−1a.
Let Aut(K/K0) = ⟨α⟩ ∼= C2. Let Λ = R◦ C2 be a twisted group ring (cf. Definition 35). Note that R0 is fixed by α, and hence Λ is an R0-algebra via ϕ : R0 → Λ, where 17→ 1. However, Λ is not an R-algebra. Let ΛQ :=Q ⊗ZΛ, and then ΛQ is a K0-algebra.
Moreover, ΛQ is isomorphic to M2(K0) (cf. [10] Example 28.3).
The following is the motivation for studying projective Λ-modules of finite R-rank.
At first, SG is the twisted group ring S[b]◦ C2, where C2 =⟨a⟩ and C2 acts on S[b] by a(b) = aba−1 = bp−1 and a(1) = 1 (cf. Definition35). Let Ψp(X) = Xp−1+ Xp−2+ . . . + 1, the cyclotomic polynomial of degree p− 1. We have that
SG/Ψp(b)SG ∼= Λ by a7→ α and b 7→ θ.
Suppose M is a finitely generated S-free SG-module. Consider the submodule
M0 ={m ∈ M : Ψp(b)m = 0},
and we can regard M0 as an SG/Ψp(b)SG ∼= Λ-module. This is a finitely generated, R-torsion free module, hence an R-projective module. By Proposition 78, M0 is
Λ-projective. Since M /M0 is annihilated by (b− 1), M/M0 is an S[a]-module. Hence M is an extension of M /M0 by M0.
In [30], it is proved that the number of isomorphism classes of indecomposable ZG-modules is related to the class number of R0. We describe this relation between an S-free SG-indecomposable module and an ideal of R0 briefly.
Given an SG-module M , the module M is an extension of an S[a]-module M /M0 by M0, which is a Λ-module. Hence, we need to study Λ-modules M0, S[a]-modules M /M0 and extensions of M /M0 by M0.
Consider ϕ : R → Λ, 1 7→ 1, the natural map, and ϕ is a ring homomorphism. Hence a Λ-module can be regarded as an R-module.
Proposition 78. Every R-projective Λ-module is Λ-projective.
Proof. See [30] Proposition 1.1.
Definition 79. A ring R is said to be a left hereditary ring if every left ideal of R is a
left projective R-module.
Proposition 80. The ring Λ is a left hereditary ring.
Proof. See [30] Proposition 1.2.
Proposition 81. Suppose Γ is a left hereditary ring. Then any Γ-projective module is
Γ-isomorphic to an external direct sum of left ideals of Γ.
Proof. See [10] Proposition 4.3.
By Propositions 80, 81 we have that any Λ-projective module is Λ-isomorphic to an external direct sum of left ideals of Λ. Since Λ is an R-free module of rank 2, the submodules of Λ are of R-rank 1 or 2. Ideals of Λ which are of R-rank two are Λ-isomorphic to direct sums of two R-rank one ideals of Λ (see [30] Theorem 1.1). Hence any Λ-projective module is isomorphic to an external direct sum of left ideals of R-rank 1 of Λ.
Definition 82. An R-ideal I in K is said to be ambiguous if I = ¯I.
Suppose I is an ambiguous ideal in K. If x ∈ I, we have ¯x ∈ I. Therefore, we may regard I as a Λ-module of R-rank 1 by defining α· x = ¯x for x ∈ I.
Proposition 83. Two ambiguous ideals I1 and I2 of K are Λ-isomorphic if and only if I1 = xI2 for some x∈ K0×.
Proof. If I1 is Λ-isomorphic to I2, then it is R-isomorphic to I2. We have I1 = xI2 for some x ∈ K×. Since I1 and I2 are Λ-isomorphic, we have x(a· y) = α · (xy), ∀y ∈ I2. Therefore, we have x¯y = xy, so x∈ K0×. The converse is clear.
Proposition 84. The set of isomorphism classes of left ideals of Λ of R-rank 1 is in bijection with that of K0×-equivalence classes of fractional ambiguous ideals of K.
Proof. Suppose I is an ambiguous ideal in K. We know that I can be viewed as a Λ-module of R-rank 1. Since I is R-torsion free, I is R-projective, therefore, Λ-projective.
By Proposition 81, it is Λ-isomorphic to a left ideal of Λ.
Conversely, suppose L is a left ideal of Λ of R-rank one, and then LQ is a left ideal of ΛQ of K-rank one. Recall that we have an isomorphism ΛQ ∼= M2(K0). Hence LQ is a simple ΛQ-module. We can embedd K ,→ ΛQ as ΛQ-modules by sending r to r· (1 + α). Since any two simple modules over ΛQ are isomorphic, we have a ΛQ-isomorphism LQ ≃ K.
Therefore, we have L ,→ K as a Λ-embedding; hence L is Λ-isomorphic to an ambiguous ideal of K. By Proposition 83, two ambigiuous ideals of R are Λ-isomorphic if and only if they are K0×-equivalent. The proof is complete.
Proposition 85. An ideal I in R is ambiguous if and only if I can be written as the form (1− θ)εW R, where W is an ideal of R0 and ε = 0 or 1.
Proof. Suppose I is an ambiguous ideal of R. We can decompose I into a product of prime ideals, say, I = Q1Q2· · · Qr. Since ¯I = I, Qi breaks into 2 cases. For some Qi,
there exists Qj in the decomposition of I such that Qj = Qi, and the other Qi = Qi. We may write I as (Q1Q¯1)· · · (QmQ¯m)(Q2m+1· · · Qr), where the first 2m factors are of the first case and the remnants of the factors are of the second case. Note that since Q(θ) is a field extension of K0 of degree 2, given an ideal P of R0, we have that
Also since K is the p-th cyclotomic extension, we have that P is ramified only if P is the prime ideal lying over (p). The prime ideal lying over ramified P is (1− θ). For 2m + 1≤ i ≤ r, Qi is either ramified or inert. So we have that
Proof. By Proposition 83, two ambiguous ideals (1− θ)εW R and (1− θ)ε′W′R are
Theorem 87. Let h be the class number of R0. There are 2h non-isomorphic, indecom-posable, projective, Λ-modules of R-rank 1.
Proof. Suppose M is a projective and indecomposable Λ-module of R-rank 1. Then M is isomorphic to a left ideal of Λ of R-rank 1 by Proposition 84. Then this follows from Propositions 83, 85 and 86.
We may choose a set of representatives {Ui} for ideal classes of R0, and express the ambiguous ideals of R by (θ− ¯θ)εUiR up to Λ-isomorphism.
Proposition 88. If M is a projective Λ-module, then M is isomorphic to a direct sum of UiR and (¯θ− θ)UiR.
Proof. This follows from Propositions 81and 85.
Proposition 89. Suppose M is a projective Λ-module of R-rank N , and
M ∼=
Proof. See [30] Theorem 1.3.
Let X and Y be two SG-modules. Let F ∈ HomS(SG, HomS(Y, X)) be an element, then F (a) and F (b) are elements in HomS(Y, X). Write Fa and Fb for F (a) and F (b), respectively. We will use F to construct an SG-module MF which is an extension of Y by X. Let MF = X ⊕ Y as S-modules. We define the action of G on MF as follows:
a· (x, y) = (a · x + Fa(y), a· y), b · (x, y) = (b · x + Fb(y), b· y).
We can find out what conditions of F should be satisfied to order to make M an SG-module. For example, since
(aa)· (x, y) = a · (a(x, y)) = a(a · x + Fa(y), a· y)
= ((aa)· x + a · Fa(y) + Fa(a· y), (aa) · y) = 1 · (x, y) = (x, y),
we have
a· Fa(y) + Fa(a· y) = 0. (4.4)
Similarly, we have
p−1
∑
i=0
bp−1−iFb(bi· y) = 0 (4.5)
and
aFb(y) + Fa(b· y) = bp−1· Fa(y)− bp−1Fb(bp−1a· y). (4.6)
These are all restrictions on F for making M an SG-module. We denote MF by (X, Y ; F ).
Conversely, suppose we have an SG-module M and M is an extension of Y by X. Since
Y is S-projective, we can choose a decomposition of S-modules M = X ⊕ Y . This decomposition induces an element F ∈ HomS(SG, HomS(Y, X)) which satisfies above conditions.
Denote by S the trivial representation of G, and denote by S′ := S the SG-module with G-action given by a· s = −s and b · s = s, for s ∈ S. Let L be a free S-module with basis {e1, e2}, together with G-action given by a · e1 = e2 and a· e2 = e1 and b· ei = ei for i = 1 or 2. Let {Ui} be a representative system of the ideal class group of R0. Let Ai denote the SG-module UiR with G-action given by a · x = ¯x and b · x = θx, for x ∈ UiR. Let A′i denote the SG-module with set elements in UiR but with G-action given by a· x = −¯x and b · x = θx for x ∈ UiR. According to the discussion at the beginning of this section, together with Theorem 89 and Example 77, every SG-module M is an extension of
Sr⊕ S′s⊕ Lt by An1 ⊕ · · · ⊕ Anα⊕ A′m1 ⊕ · · · ⊕ A′mβ.
Since the functor Hom commutes with the functor oplus, we may consider only F ∈ HomS(SG, HomS(Y, X)) with Y = S, S′ or L, and X = Ai or A′i first.
Proposition 90. After fixing (Y, X) such that (Y, X)̸= (S, Ai) or (S′, A′i), there exists a unique indecomposable extension of Y by X up to SG-isomorphism. In other words, the extension of Y by X is either trivial (decomposable) or the unique indecomposable SG-module up to isomorphism.
Proof. See [30] Proposition 2.2.
Proposition 91. For (Y, X) = (S, Ai) or (S′, A′i), there does not exist any indecom-posable extension.
Proof. See [30] Lemma 2.1.
By Proposition90, we may drop F from the notation (X, Y ; F ) and denote the unique indecomposable extension by (X, Y ). The following is a more explicit expression of these indecomposable extensions.
Let Fa, Fb ∈ HomS(S, A′i), and define
Fb(s) = sn0, and Fa(s) =
−F¯b(1) + ¯θFb(1)
(¯θ− 1) = s(− ¯n0+ ¯θn0)
θ¯− 1 (4.7)
for n0 ∈ A′i and n0 ̸∈ (θ − 1)A′i. For Fa, Fb ∈ HomS(L, Ai), we define
Fa(e1) = n0 and Fb(e2) = n0
where n0 is an element in A′i and n0 ̸∈ (θ − 1)A′i and for any P , P|2R, n0 ̸∈ P . Using the conditions (4.4) and (4.6), we have that
Fa(e2) = Fa(e1) and ¯n0+ Fa(e2) = ¯θn0− ¯θFb(e1). (4.8)
By Proposition46and the fact that M0 isZ-free, we can reduce finding the indecom-posable ZG-modules to finding the indecomposable Z(2p)G-modules. Now we denote by R and R0 the integral closures of Z(2p) in K and K0, respectively. Since Z(2p) has only finitely many maximal ideals (in fact two), R and R0 have only finitely many maximal ideals as well. Thus, R0 and R are Dedekind domains having only finitely many maximal ideals, and hence they are PID. Lee proves that any indecomposable Z(2p)G-modules is one of the following 5 forms:
(A′,Z(2p)), (A,Z′(2p)), (A, L), (A′, L) and (A′+ A, L).
The former 4 types are proved by a matrix calculation, and the last one follows from a result of Swan [39]. Once all indecomposableZ(2p)G-modules are known, all
indecompos-able ZG-modules are known.
For S = Z or Z(2p), there are 5 types of indecomposable SG-extensions. Also there are Ai, A′i and S, S′, L as SG-modules. Eventually, all integral representations of G are of these 7h + 3 types, where h is the class number of the integral closure of S in K0.
Theorem 92. For S =Z or Z(2p) and R0 be the integral closure of S in K0. Let h be the class number of R0. These 7h + 3 isomorphism classes of indecomposable SG-modules are all isomorphism classes of indecomposable SG-modules.
Proof. See [30] Theorem 2.1.
Since the Krull-Schmidt Theorem holds for Z(p)G- and Z(2)G-modules (see [30] The-orem 3.1.), we can consider the unique decompositions of Z(2p)G-modules into indecom-posable Z(p)G- and Z(2)G-modules after scalar extensions. For two Z(2p)G-modules M and N , by Proposition 47M ∼= N if and only if M(p) ∼= N(p) and M(2) ∼= N(2).
If a Z(2p)G-module M decomposes as
M ∼=Zs(2p)1 ⊕ Z′(2p)
s2 ⊕ Ll⊕ Ar1 ⊕ A′r2 ⊕ (A, Z′(2p))u1 ⊕ (A′,Z(2p))u2
⊕ (A, L)v1 ⊕ (A′, L)v2 ⊕ (A + A′, L)t,
then
s1+ u2, s2+ u1, r1, r2, u1+ v1+ t, u2+ v2+ t, s1+ l + v1, s2+ l + v2
are invariants of M and determine M up to Z(2p)G-isomorphic. This follows from Propo-sition 47. Below is the chart of decompositions of indecomposable Z(2p)G-modules.
Z(2p)G-module Z(2)G-module Z(p)G-module Table 4.1: Decompositions of Z(2p)Dp modules with coefficients Z(p) and Z(2)
Theorem 93. Let M be a ZG-module, and write
M ∼=Zs1⊕ Z′s2 ⊕ Ll⊕ UiδRr1⊕ (¯θ − θ)UiϵRr2 ⊕ (UiζR,Z′)u1 ⊕ (¯θ − θ)UiηR,Z)u2
⊕ (UiλR, L)v1 ⊕ ((¯θ − θ)UiµR, L)v2 ⊕ (R + (¯θ − θ)UiνR, L)t,
then we have s1+ u2, s2+ u1, r1, r2, u1 + v1+ t, u2 + v2+ t, s1+ l + v1, s2 + l + v2 are invariants of M and they determine M up to Z(2p)G-isomorphism. The class of
(∏
3)/2. Fortunately, R0 =Z is PID. Hence, there are 7 + 3 = 10 isomorphism classes of indecomposable ZS3-modules. We will write down these indecomposableZS3-modules explicitly.