About the Kernel of the Augmentation of finitely generated

Z-modules. by Marc Conrad

Abstract: Let M be a free finitely generated Z-module with basis B and ΔM the kernel of the homomorphism M → Z which maps B to 1. A basis of ΔM can be easily constructed from the basis B of M. Let further R be a submodule of M such that N = M/R is free. Subject of investigation is the module ΔN = (ΔM + R)/R. We compute the index [N : ΔN] and construct bases of ΔN with the help of a basis of N. Finally the results are applied to a special class of modules which is connected with the group of cyclotomic units.

1 Introduction

Well known in the context of group rings is the augmentation of an group ring element which is the homomorphism obtained by mapping the group elements to 1. The augnmentation defines the augmentation ideal of the group ring which denotes the kernel of the augmentation [3]. Similarly, in a free Z-module M each basis B defines an homomorphism aug : M → Z, ∑_{b∈B} α_{b}b ↦ ∑_{b∈B} α_{b}. We denote the kernel of aug by ΔM. We consider further the module N = M/R where R is a submodule of M such that N is free, and let ΔN = (ΔM + R)/R. In the following we assume that the module M is finitely generated. It is easy to see that the index [M : ΔM] is infinite. In Theorem 2.1 we identify the index [N : ΔN] as the greatest common divisor of the augmentation of the elements of R. It can be seen straightforwardly that for a fixed b_{0} ∈ B the set

B_{0} = {b − b_{0}; b ∈ B, b ≠ b_{0}} (1) is a basis of ΔM. A similar result is obtained for ΔN in Theorem 3.2. In Section 4 we will apply this result to a class of modules which is connected to the group of cyclotomic units. This group plays an important role in the theory of cyclotomic fields [4].

2 The index of ΔN

We use in the following the notation of the introduction.

Theorem 2.1 We have [N : ΔN] = gcd aug(R) where the greatest common divisor of {0} is defined as ∞.

Proof: Let b_{0} ∈ B. Because b ≡ b_{0} mod ΔM for all b ∈ B we see that N/ΔN is cyclic and generated by b_{0}. The index is the smallest positive number such that mb_{0} ∈ R + ΔM. Note that aug(mb_{0}) = m for m ∈ Z. In the case R ⊆ ΔM we have aug R = {0}. From mb_{0} ∉ ΔM for all m ≠ 0 we see [N : ΔN] = ∞ as it was claimed in the Theorem. For R ⊈ ΔM there exists an element r ∈ R with minimal positive augmentation ρ. Noting that ρb_{0} ≡ r mod ΔM it follows [N : ΔN] ≤ ρ. If we have on the other hand k ∈ Z and r′ ∈ R such that kb_{0} ≡ r′ mod ΔM it follows ρ ≤ k = aug(r′) because of the minimality of ρ, and we obtain [N : ΔN] = ρ. It remains to show that ρ = gcd aug(R). Suppose there exists r′ ∈ R such that ρ is not a divisor of ρ′ = aug(r′). Then we find by computing δ = gcd(ρ, ρ′) numbers α, β ∈ Z with δ = αρ + βρ′. But αr + βr′ ∈ R is an element with positive augmentation δ < ρ which is a contradiction to the minimality of ρ. □ We show in the next lemma how the index [N : ΔN] can be explicitly computed.

Lemma 2.2 If E ⊆ R generates R then gcd aug(R) = gcd aug(E).

Proof: For [N : ΔN] = ∞ there is nothing to show. In the case that ρ = [N : ΔN] < ∞ the claim follows from the existence of r ∈ R and α_{e} ∈ Z such that ρ = aug(r) = ∑_{e∈E} α_{e} aug(e). With this we obtain

(2) gcd aug(R) = ρ = gcd(aug(E) ∪ {ρ}) = gcd aug(E).

□

Remark 2.3 Similarly to ΔM we can identify ΔN as a kernel of an homomorphism. With k = [N : ΔN] for finite and k = 0 for infinite index we have an homomorphism

(3) \overline{aug} : N → Z/kZ, a + R ↦ aug(a) + kZ

and ΔN = ker_{N} \overline{aug}.

3 Construction of a Basis of ΔN

In the following, let C ⊆ M induce a basis of N, i. e. let {c + R; c ∈ C} be a basis of N. We assume that there exist γ ∈ Z such that aug(c) = γ for all c ∈ C. Note that this is no restriction to the module N. In Algorithm 3.4 we will show how such a basis can be constructed from an arbitrary basis of N. Let ρ = [N : ΔN]. In the case ρ = ∞ it is easy to see that similarly to (1) for a fixed c_{0} ∈ C the set C_{0} = {c − c_{0}; c ∈ C, c ≠ c_{0}} is a basis of ΔN. We assume in the following ρ < ∞ and show in the next Lemma and the subsequent Theorem how to construct bases of ΔN in this case.

Lemma 3.1 Let c_{1} ∈ C. Then

C_{1} = {c − c_{1}; c ∈ C, c ≠ c_{1}} ∪ {ρc_{1}} (4)

induces a basis of ΔN.

Proof: We show that the elements b − b_{0} with b, b_{0} ∈ B are modulo R generated by C_{1}. Because C is a basis of N we have α_{c}, β ∈ Z such that

b − b_{0} = r + βc_{1} + ∑_{c∈C, c≠c_{1}} α_{c}(c − c_{1}). (5)

The application of aug to (5) and reducing modulo ρ gives βγ ≡ 0 mod ρ. We show in the rest of the proof that gcd(γ, ρ) = 1. Then we have ρ|β and the claim of the Lemma follows. We can write any b ∈ B as b = c + r with c ∈ ⟨C⟩ and r ∈ R. This gives 1 = aug(c) + aug(r) = νγ + μρ with ν, μ ∈ Z which leads to gcd(γ, ρ) = 1. □ Compared with the basis B_{0} of ΔM in (1) the basis C_{1} from (4) has the extra element ρc_{1} added to the expected elements c−c_{1}. In the next theorem we give a basis which looks more similar to B_{0}.

Theorem 3.2 Let c_{0} ≡ c′ mod R such that c′ ∈ ⟨C⟩ and aug(c′) = (1−ρ)γ. Then C_{0} = {c − c_{0}; c ∈ C} (6) induces a basis of ΔN.

Proof: Let c_{1} as in Lemma 3.1 and C′ = {c − c_{1}; c ∈ C, c ≠ c_{1}} such that C_{1} = C′ ∪ {ρc_{1}} induces a basis of ΔN. Because of c_{1} − c′ ≡ ρc_{1} mod ⟨C′⟩ we can replace ρc_{1} by c_{1} − c′ in C_{1}. By replacing the other elements of C_{1} using the relation c−c′ = c−c_{1}+(c_{1}−c′) for c ∈ C we obtain {c−c′; c ∈ C} as a basis of ΔN. With c_{0} ≡ c′ mod R we get the claim. □

Remark 3.3 If we choose c_{0} = c′ + γr with r ∈ R such that aug(r) = ρ we obtain aug(c_{0}) = γ and therefore C_{0} ⊆ ΔM. So, with C_{0}, we obtain directly a basis of ΔM/(ΔM ∩ R) (which is of course isomorphic to ΔN).

In Lemma 3.1 and Theorem 3.2 we assume that there is a basis C ⊆ M of N with aug(c) = γ for all c ∈ C. We give here an Euclidean-like algorithm which shows how to construct such a basis from an arbitrary basis.

Algorithm 3.4 Let C ⊆ M induce a basis of N. The algorithm leads to aug(c) = γ for all c ∈ C by successively replacing elements of C. If aug(c) = 0 for all c ∈ C it remains nothing to do. Otherwise we chose first c′ with aug(c′) ≠ 0 and replace each c ∈ C \ {c′} by c + λc′ with λ ∈ Z such that aug(c) > 0. If aug(c′) < 0 we have also to replace c′ by −c′. After that we perform the following steps.

1. If all elements of C have the same augmentation the algorithm is finished.

2. Pick c, c′ ∈ C such that aug(c) < aug(c′) and replace c′ by c′ − c.

3. Go to Step 1.

The algorithm terminates because ∑_{c∈C} aug(c) ∈ N decreases in every run of Step 2.

4 A special class of modules

For a finite set A we denote by ΣA the sum ∑_{a∈A} a in the free module ⟨A⟩ generated by A. For i = 1, …, r let A_{i} be a finite set with an involution σ operating nontrivially on each element. So we have sets H_{i} such that B_{i} = H_{i} ∪ σH_{i} and H_{i} ∩ σH_{i} = ∅ for i = 1, …, r. We define further the module Z = ⟨B_{1}⟩/⟨ΣB_{1}⟩ ⊗ ⋯ ⊗ ⟨B_{r}⟩/⟨ΣB_{r}⟩. (7)

The involution on the B_{i} defines an involution on Z and we may interpret Z also as a Z[σ]-module. Subject of investigation is the module N = Z/ ker_{Z}(σ + 1).

Remark 4.1 The module N is directly connected with the group of cyclotomic units C^{(n)}. Let ϵ_{n} be a primitive nth root of unity. Then C^{(n)} is defined as the multiplicative subgroup of D^{(n)} which are units of Z[ϵ_{n}]. The group D^{(n)} is generated by the elements 1 − ϵ_{n}^{a} with 1 ≤ a < n modulo torsion. With Ĉ^{(n)} = C^{(n)}/L^{(n)} where L^{(n)} = ∏_{d|n, d≠n} C^{(d)} we have for n = p_{1} ⋯ p_{r} odd, square free and not a prime an isomorphism N ≅ Ĉ^{(n)} when we choose B_{i} = {1, …, p_{i} − 1}. For general n we have similar isomorphisms (see [1]).

Let M = ⟨B_{1} × ⋯ × B_{r}⟩ and let S be the module generated by the sums

s_{i}(a_{1}, …, a_{r}) = ∑_{b∈B_{i}}(a_{1}, …, a_{i−1}, b, a_{i+1}, …, a_{r}), i = 1, …, r (8)

where a_{j} ∈ B_{j} for j = 1, …, r. By [2] we have then for r even

N ≅ M/(S + (1 − σ)M) (9)

and for r odd N ≅ M/(S + (1 − σ)M + ⟨e⟩) (10) with e = Σ(H_{1} × ⋯ × H_{r}).

Theorem 4.2 For i = 1, …, r let β_{i} = |B_{i}|, the number of elements of B_{i}, and β = gcd(β_{1}, …, β_{r}). Then we have for ρ = [N : ΔN] that

ρ = \{ \begin{matrix}β/2 & if r = 1 or \\ & r odd and β_{i} ≡ 2 mod 4 for i = 1, …, r, \\ β & else. \end{matrix}

Proof: The claim follows for r even and r = 1 from Lemma 2.2 and the isomorphisms (9) and (10). For r odd we additionally use aug(e) = ∏_{i=1}^{r}(β_{i}/2).

□

A basis of N can be constructed with weak σ-bases according to [1]. We get the following result.

Lemma 4.3 For each i = 1, …, r we fix h_{i} ∈ H_{i}. Let H_{i}^{♭} = H_{i} \ {h_{i}} and A_{i}^{♭} = A_{i} \ {h_{i}}. Then we obtain C = F^{0} ∪ F^{+} as a basis of N where

F^{0} = ⋃_{i=1}^{r}{h_{1}} × ⋯ × {h_{i−1}} × H_{i}^{♭} × B_{i+1}^{♭} × ⋯ × B_{r}^{♭} (11)

and

(12) F^{+} = \{ \begin{matrix}∅ & for r odd, \\ \{h_{1}\} × ⋯ × \{h_{r}\} & for r even. \end{matrix}

We see here that the basis C can be chosen as a subset of B = B_{1} ×⋯×B_{r}. So, all elements of C have augmentation 1, and we may apply Theorem 3.2 with γ = 1 and c_{0} = (1 − ρ)c where c is any element of C. This leads to a basis of ΔN as in (6). However, we might ask a stronger question: Can we find a basis C ⊆ B of N such that we can choose c_{0} ∈ B? Up to now there is no general answer to this. We will discuss in the rest of this section some special cases where the answer is affirmative. In the following, we call a basis C_{0} of ΔN which has the form C_{0} = {c − c_{0}; c ∈ C} with C ⊆ B and c_{0} ∈ B a handsome basis of ΔN.

Theorem 4.4 If there exists a j ∈ {1, …, r} such that β_{j} = ρ then ΔN has a handsome basis.

Proof: In the case F_{0} ≠ ∅ we rearrange the sets B_{i} such that j = r. Let (a_{1}, …, a_{r−1}, a_{r}) be any element of F^{0}. Then (a_{1}, …, a_{r−1}, b) ∈ F^{0} for b ≠ h_{r}. So we may choose in Theorem 3.2 c′ = − ∑_{b∈B^{♭}}(a_{1}, …, a_{r−1}, b), and the claim follows with c_{0} = (a_{1}, …, a_{r−1}, h_{r}). In the case F_{0} = ∅ we take c_{0} = (h_{1}, …, h_{r−1}, σh_{r}). □ The converse of Theorem 3.2 is not true. Even if a basis of the form C = F^{0} ∪F^{+} as in Lemma 4.3 cannot be used for the construction of a handsome basis we may have more success when starting with a different basis. We will give an example in the next Lemma.

Lemma 4.5 Let B_{1} = {a, b, σa, σb} and B_{2} = {a, b, c, σa, σb, σc} two sets of four respectively six elements with σ acting nontrivially on B_{1} and B_{2}. The module N is as in (9) given as the free module M generated over B = B_{1} × B_{2} modulo (1 − σ)M and the relations described in (8). With

C = {(a, b), (b, c), (σb, a), (σb, b), (σb, c), (a, σa), (a, σb), (σa, c)}

and c_{0} = (a, a) we obtain {c − c_{0}; c ∈ C} as a basis of ΔN.

Proof: We show first that C is a basis of N. By Lemma 4.3 we obtain rank N = 8 = |C| and it is sufficient to show that C generates N. Because of c ≡ σc mod (1 − σ)M we see that σC is generated by C. The elements of B \ (C ∪ σC) can then be generated by C ∪ σC directly by relations of S. Theorem 2.1 gives [N : ΔN] = 2. We will now apply Theorem 3.2 in order to construct a basis of ΔN. Let c′ = c_{0} + r with

r = ∑_{x∈B_{1}}(x, c) − ∑_{y∈B_{2}}(a, y) − (1 − σ)(σa, c). (13)

Because c′ satisfies the conditions in Theorem 3.2 the claim follows. □ Without going into details we note that a construction as in Lemma 4.5 can be generalized to more complicated cases. However, the problem to give a general algorithm for the construction of a handsome basis remains open.

References

[1] M. Conrad, Construction of Bases for the Group of Cyclotomic Units, J. Number Theory, to appear.

[2] M. Conrad, On Explicit Relations between Cyclotomic Numbers, submitted.

[3] G. Karpilovski, Commutative group algebras, Dekker, New York, 1983.

[4] L. C. Washington, Introduction to Cyclotomic Fields, Springer, New York, 1982.

Marc Conrad Universität des Saarlandes Postf. 151150 D-66041 Saarbrücken Germany marc@math.uni-sb.de http://simath.math.uni-sb.de/∼marc