# Z-Bases and $Z[1/2]$-bases for Washington's cyclotomic units of real cyclotomic fields and totally deployed fields

Rafik Souanef

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Rafik Souanef. Z-Bases and $Z[1/2]$-bases for Washington's cyclotomic units of real cyclotomic fields and totally deployed fields. 2025. hal-04630211v3

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# $\mathbb{Z}$-Bases and $\mathbb{Z}[1/2]$-bases for Washington's cyclotomic units of real cyclotomic fields and totally deployed fields

Rafik SOUANEF

Université de Franche-Comté, CNRS, LmB (UMR 6623)

16 route de Gray, 25000, Besançon, France

Email: rafik.souanef@ens-rennes.fr

Url: https://perso.eleves.ens-rennes.fr/people/rafik.souanef/

## Abstract

We present families of generators with minimal cardinality - we call such families bases - of the free abelian group $\mathbf{Was}(K)/\mathbf{Z}(K)$ for any real cyclotomic field $K = \mathbb{Q}(\zeta_n)^+$. If $K$ is a totally deployed abelian number field, we give a $\mathbb{Z}[1/2]$-basis of $\mathbf{Was}(K)/\mathbf{Z}(K) \otimes_{\mathbb{Z}} \mathbb{Z}[1/2]$. Here $\mathbf{Was}(K)$ refers to the group of Washington's cyclotomic units of $K$ and $\mathbf{Z}(K)$ refers to the group of roots of unity lying in $K$.

# 1 Introduction

Given an abelian number field $K$, one may construct the Galois module made of the cyclotomic units of $K$ and note that, in this article, we will consider problems that deal with its group aspect. One of the interest of cyclotomic units is their link with

2020 Mathematics Subject Classification: Primary 11R27, 11R18; Secondary 11R23.

Key words and phrases: Cyclotomic units, real cyclotomic fields, totally deployed fields, bases, basis, generators, circular units
the theory of $\mathbb{Z}_p$-extensions. For instance, the main conjecture of Iwasawa's theory can be roughly formulated in this way: a module that one can construct using cyclotomic units has the same characteristic ideal as the standard $p$-ramified module of Iwasawa's theory (see [1], proposition 4.5.7). Another reason why cyclotomic units are of interest is to approximate the whole group of units of any abelian number field. There are many different versions of cyclotomic units (see [9]) but in this article we will deal with two versions only that are the cyclotomic units of Washington and those of Sinnott. Washington's cyclotomic units are defined through Galois invariants and Sinnott's circular units can be defined by explicit generators that generate a subgroup of the group given by Washington: the drawback of having a smaller group - so that we may expect it to be a worse approximation of the group of units - is balanced by a better knowledge of the elements of this group. These two types of cyclotomic units can be constructed through two processes that have a common starting point that is to define in the same way the cyclotomic units of any cyclotomic field and then deduce a definition for the cyclotomic units of any abelian number field, using Kronecker-Weber theorem. A crucial article in the study of Sinnott's circular units is [14] in which it is proven that the group of Sinnott's circular units of any $K$ has finite index in the group of units of $K$ and that this index is linked to the class number of the maximal real subfield of $K$. On the other side, the group of Washington's cyclotomic units remains quite mysterious. There are two articles ([5] and [17]) in which these last units has been studied under some hypotheses on the considered number fields. These last two articles and the work that we present today aim to give us a better understanding of this group by giving explicit $\mathbb{Z}$-free systems of generators modulo roots of unity or $\mathbb{Z}[1/2]$-free systems of generators (we then say "basis") in the case of totally deployed fields. What is new in our present work is that we use different or more general hypotheses. For example, we do not consider real fields only. In particular, in our work, we consider totally deployed abelian number fields, that is we handle fields of the form

$$K = K_1 \cdots K_r$$

with $K_i \subset \mathbb{Q}(\zeta_{p_i}^{e_i})$ for some prime numbers $p_i$, some integers $e_i$ and we do not assume any hypothesis on whether the $K_i$'s are all real or all imaginary as in [17] or [10]. In other words, we are interested in abelian number fields that coincide with their
genus field.

More precisely, let $K$ be a number field. Suppose $K/Q$ is abelian. Then, recall that Kronecker-Weber theorem states there is an integer $n$ such that $K \subset Q(\exp(2i\pi/n))$. Define the conductor of $K$ to be the least integer $n$ that satisfies this last condition. From now on, suppose $K$ is an abelian number field with conductor $n$. Let $\zeta_n = \exp(2i\pi/n)$.

Let $\mathbb{Z}(K)$ denote the group of roots of unity of $K$. Let $\mathbf{Was}(K)$ denote the Galois module of Washington's cyclotomic units and let $\mathbf{Sin}(K)$ denote the Galois module of Sinnott's circular units (we will recall their definition later).

By abuse of language, when we talk about bases, we will rather talk about bases of $\mathbf{Was}(K)$ instead of $\mathbf{Was}(K)/\mathbb{Z}(K)$. When dealing with bases - and only in this case - if we write $\mathbf{Was}^{+}$, we do not mean to consider the invariant elements of $\mathbf{Was}(K)/\mathbb{Z}(K)$ under the complex conjugation but we mean $\mathbf{Was}(K^{+})/\{\pm 1\}$.

If $K = Q(\zeta_n)$ is a cyclotomic field, recall Gold and Kim have given bases of $\mathbf{Was}(Q(\zeta_n))$ (that is equal to $\mathbf{Sin}(Q(\zeta_n))$) and Kučera also did so (see [2], theorem 2 or [10], [6]). We base our work on Kučera's bases because it leads to an easier proof of theorem 22. Indeed, using Gold and Kim's bases is possible but it would impose to use induction arguments that would make the proof longer and more difficult to read.

Based on this work, we state theorems 11, 12 and 22 that all describe bases of $\mathbf{Was}(K)$ or $\mathbf{Was}(K) \otimes_{\mathbb{Z}} \mathbb{Z}[1/2]$ under different hypotheses on $K$. The first two theorems are easy consequences of proposition 9 that is itself a consequence of proposition 1. These two theorems give $\mathbb{Z}$-bases of $\mathbf{Was}(K)$ assuming there is an integer $n$ such that $K = Q(\zeta_n)^+$ is the real cyclotomic field of conductor $n$. In theorem 22, we suppose $K$ is totally deployed and we give a $\mathbb{Z}[1/2]$-basis of $\mathbf{Was}(K) \otimes_{\mathbb{Z}} \mathbb{Z}[1/2]$. The main idea in the proof of this theorem is to show we have a $\mathbb{Z}[1/2]$-basis by proving that the elements we consider generate a module that is a direct factor of $\mathbf{Was}(Q(\zeta_n)) \otimes_{\mathbb{Z}} \mathbb{Z}[1/2]$. This idea has also been used in [5] and [17] (with no tensor with $\mathbb{Z}[1/2]$). Divisibility relations arise from this last basis (see corollary 32).
## 2 Notation and preliminaries

Let $\mathbb{N}$ denote the set of all non negative integers and let $\mathbb{N}^* = \mathbb{N} \setminus \{0\}$ be the set of all positive integers.

### 2.1 On units

Let $A$ be a Galois module, that is $A$ is an abelian group with some $\text{Gal}(\mathbb{K}/\mathbb{Q})$ acting $\mathbb{Z}$-linearly on it (we consider extensions of $\mathbb{Q}$ only). Suppose $\mathbb{K}/\mathbb{Q}$ is an abelian extension, so that the complex conjugation is well defined as an element of $\text{Gal}(\mathbb{K}/\mathbb{Q})$. We let $A^+$ denote the Galois submodule of $A$ that consists of all the elements of $A$ on which the complex conjugation acts trivially. Later on, we will consider $A = \mathcal{O}_{\mathbb{K}}^{\times}$ the group of units of the ring of integers of $\mathbb{K}$.

If $x \in A$, then any $u \in \mathbb{Z}[\text{Gal}(\mathbb{K}/\mathbb{Q})]$ acts on $x$ and we denote by $ux$ or $u(x)$ or $x^u$ the image of $x$ under $u$.

Let $\zeta_n = \exp(2i\pi/n)$ for any $n \in \mathbb{N}^*$. From now on, let $n \ge 2$ satisfy $n \ne 2 \pmod 4$ (with no loss of generality because $\mathbb{Q}(\zeta_n) = \mathbb{Q}(\zeta_{2n})$ if $n$ is odd). If $p \in \mathbb{P}$ is a prime number, let $v_p(k)$ denote the $p$-valuation of any integer $k$. Let $n = \prod_{j=1}^r p_j^{e_j}$ and let $q_j = p_j^{e_j}$ for any $j \in [1, r]$.

We now recall that if $n$ is not a prime power, then $1 - \zeta_n$ is a unit of the ring of integers of $\mathbb{Q}(\zeta_n)$ (see [16] proposition 2.8). Now, if $n$ is a prime power, then $1 - \zeta_n$ is no longer a unit but $\frac{1-\zeta_n^\sigma}{1-\zeta_n}$ is a unit for all $\sigma \in \text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ (lemma 1.3 [16]).

Let $\mathbb{K}$ be an abelian number field of conductor $n$. We say $\mathbb{K}$ is totally deployed when $\text{Gal}(\mathbb{K}/\mathbb{Q})$ is the direct product of its inertia subgroups (see the introduction of [3]). As we supposed $\mathbb{K}$ to be abelian, this previous condition is equivalent to

$$ \mathbb{K} = \mathbb{K}_1 \cdots \mathbb{K}_r $$

with $\mathbb{K}_j \subset \mathbb{Q}(\zeta_{q_j})$. Note the fact that one can state more results on cyclotomic units when $\mathbb{K}$ is totally deployed appears in [14], [17] and [5].

Let $\mathbf{E}(\mathbb{K})$ be the group of units (of the ring of integers $\mathcal{O}_{\mathbb{K}}$) of $\mathbb{K}$. Let $\mathcal{C}_n$ be the Galois module generated by the roots of unity of $\mathbb{Q}(\zeta_n)$ and by the $1 - \zeta_d$'s for $d \mid n, d > 1$.
Let $\mathbf{Was}(\mathbb{K}) = \mathbf{E}(\mathbb{K}) \cap \mathcal{C}_n$ and let $\mathbf{Sin}(\mathbb{K})$ be the intersection of $\mathbf{E}(\mathbb{K})$ with the Galois module generated by the roots of unity lying in $\mathbb{K}$ and the

$$N_{\mathbb{Q}(\zeta_d)/\mathbb{K}\cap\mathbb{Q}(\zeta_d)}(1 - \zeta_d)'s$$

where $d > 1$.

One can show (see [8]) that the group $\mathbf{Sin}(\mathbb{K})$ is generated by:

* the roots of unity of $\mathbb{K}$, which form a group that we will denote by $\mathbf{Z}(\mathbb{K})$

* the $N_{\mathbb{Q}(\zeta_d)/\mathbb{K}\cap\mathbb{Q}(\zeta_d)}(1 - \zeta_d^\sigma)$'s with $d \mid n$ such that $d$ is not a prime power, $d \neq 1$ and $d \wedge (n/d) = 1$ and $\sigma \in \mathrm{Gal}(\mathbb{Q}(\zeta_d)/\mathbb{Q})$

* the $N_{\mathbb{Q}(\zeta_d)/\mathbb{K}\cap\mathbb{Q}(\zeta_d)}(1 - \zeta_d)^{1-\sigma}$'s with $d$ being a prime power dividing $n$ such that $d \wedge (n/d) = 1$ and $\sigma \in \mathrm{Gal}(\mathbb{Q}(\zeta_d)/\mathbb{Q})$.

It is known that both $\mathbf{Was}(\mathbb{K})$ and $\mathbf{Sin}(\mathbb{K})$ have finite index in $\mathbf{E}(\mathbb{K})$, that is they both have maximal rank as $\mathbb{Z}$-submodules of $\mathbf{E}(\mathbb{K})$ and that their index is linked to the class number of the maximal real subfield $\mathbb{K}^+$ of $\mathbb{K}$ (see [14], see [16] theorem 8.2).

When the situation makes it clear, we will omit writing $\mathbb{K}$. For example, we will write $\mathbf{Was}$ instead of writing $\mathbf{Was}(\mathbb{K})$ or $\mathbf{Was}^+$ instead of $\mathbf{Was}(\mathbb{K}^+)$. Also, we will note $\mathbf{Was}_2 = \mathbf{Was} \otimes_{\mathbb{Z}} \mathbb{Z}[1/2]$.

We now recall the following relations (see [15] lemma 2.1):

$$1 - \zeta_n^a = -\zeta_n^a(1 - \zeta_n^{-a}) \qquad (1)$$

$$N_{\mathbb{Q}(\zeta_n)/\mathbb{Q}(\zeta_d)}(1 - \zeta_n) = \left( \prod_{\substack{p|n \\ p \nmid d}} (1 - \mathrm{Frob}(p))^{-1} \right) (1 - \zeta_d) \qquad (2)$$

where $d \mid n$ is such that $d > 1$, the integers $p$ are prime and $\mathrm{Frob}(p)$ denotes the Frobenius of $\mathbb{Q}(\zeta_d)$ that is defined by $\zeta_d \mapsto \zeta_d^p$. We will refer to this second relation as "norm relation". We will call this relation "norm relation along $\sigma_i$" (we will define $\sigma_i$ later) to mean we consider this norm relation with $d = n/q_i$.

We recall a property of Hasse's unit index.
**Proposition 1.** We have

$$[\mathbf{E} : \mathbf{ZE}^+] \in \{1; 2\}.$$

Moreover, if $\mathbb{K} = \mathbb{Q}(\zeta_n)$, this index is 1 if and only if $n$ is a prime power.

*Proof.* See [16] theorem 4.12 and corollary 4.13. □

We also recall Dirichlet's units theorem: the abelian group $\mathbf{E}(\mathbb{K})$ is finitely generated, its torsion part is $\mathbf{Z}(\mathbb{K})$ and it has rank $r_1 + r_2 - 1$ (with usual notation).

We now introduce some of the notation we will use to work with bases of **Was** (most of this notation comes from [2], [17] and [10]). Keep in mind that, when talking about bases, we write **Was** instead of **Was**/$\mathbb{Z}$.

Recall $n = \prod_{j=1}^{r} p_j^{e_j}$ and $q_j = p_j^{e_j}$. For any $j \in [1, r]$, let $J_j$ denote the complex conjugation considered as an element of $\text{Gal}(\mathbb{Q}(\zeta_{q_j})/\mathbb{Q})$. If $p_j$ is odd, let $\sigma_j$ be a generator of $\text{Gal}(\mathbb{Q}(\zeta_{q_j})/\mathbb{Q})$. If $p_j = 2$, let $\sigma_j$ be such that $\text{Gal}(\mathbb{Q}(\zeta_{q_j})/\mathbb{Q})$ is generated by $\sigma_j$ and $J_j$ (so that $\text{Gal}(\mathbb{Q}(\zeta_{q_j})/\mathbb{Q})$ is the direct product of $\langle J_j \rangle$ and $\langle \sigma_j \rangle$).

From now on, for any $j \in [1, r]$, see the elements of $\text{Gal}(\mathbb{Q}(\zeta_{q_j})/\mathbb{Q})$ as elements of $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ by letting them act trivially on $\mathbb{Q}(\zeta_{n/q_j})$.

Now, the complex conjugation $J_j$ of $\text{Gal}(\mathbb{Q}(\zeta_{q_j})/\mathbb{Q})$ is considered as an element of $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$. Let $J = J_1 \cdots J_r$ be the complex conjugation considered as an element of $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$.

Define (see [16] lemma 8.1) (up to a sign because of the following square root)

$$\xi_{q_j} = \sqrt{\zeta_{q_j}^{1-\sigma_j} \frac{1-\zeta_{q_j}^{\sigma_j}}{1-\zeta_{q_j}}} \in \mathbf{Was}^{+}(\mathbb{Q}(\zeta_{q_j})).$$

Next, we construct some sets and set the notation to recall the basis of $\mathbf{Was}(\mathbb{Q}(\zeta_n))$ given in [10] and [7].

**Definition 2.** For any $i \in [1, r]$, the set $\mathcal{R}_i$ is defined in [10] (lemma 1.1) in the following way. If $p_i \neq 2$, let $z \in \text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{Q})$ be such that $z$ generates the 2-Sylow of $\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{Q})$ and let $H$ be the non 2-part of $\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{Q})$ (that is $H$ is the
product of all $l$-Sylow for all prime integers $l > 2$). Let $a$ be minimal with respect to $z^{2^a} = J_i$. Let

$$ \mathcal{R}_i = \{z^k h : 0 \le k < 2^a, h \in H\}. $$

If $p_i = 2$, let $\mathcal{R}_i = \langle \sigma_i \rangle$.

**Remark 3.** For any $i \in [1, r]$, the set $\mathcal{R}_i$ is a set of representatives of $\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{Q})$ modulo $\langle J_i \rangle$ and we have $1 \in \mathcal{R}_i$

**Definition 4.** Let $\Omega = \{i\} \subset [1, r]$ for some $i$. Let $Y_\Omega$ denote $\mathcal{R}_i \setminus \{1\}$.

**Definition 5.** Let $\Omega = \{i_1, \dots, i_s\} \subset [1, r]$ for some $s \ge 2$ and $i_1 < \dots < i_s$. Let $Y_\Omega$ be the set of all $u_1 \cdots u_k$ with $k \in [1, s]$, satisfying $u_k \in \mathcal{R}_{i_k} \setminus \{1\}$ and

$$ \forall j \in [1, k-1], \quad u_j \in \text{Gal}(\mathbb{Q}(\zeta_{q_{i_j}})/\mathbb{Q}) \setminus \{J_{i_j}\}. $$

If $|\Omega|$ is even, then add 1 to $Y_\Omega$.

**Definition 6.** For any non empty set $\Omega \subset [1, r]$, let $n_\Omega = \prod_{j \in \Omega} q_j$, let $\zeta_\Omega = \zeta_{n_\Omega}$ and let

$$ c_\Omega = \begin{cases} \xi_{n_\Omega} & \text{if } |\Omega| = 1 \\ 1 - \zeta_\Omega & \text{if } |\Omega| > 1. \end{cases} $$

Let

$$ C_{\Omega} = \{c_{\Omega}^{u} : u \in Y_{\Omega}\}. $$

**Theorem 7.** The family $C := \bigcup_\Omega C_\Omega$ where $\Omega$ runs over the set of all non empty subset of $[1, r]$ is a basis of $\text{Was}(\mathbb{Q}(\zeta_n))$.

*Proof.* See [10], corollary 4.3. □

In the following, we may write $C(\mathbb{Q}(\zeta_d))$ to talk about the basis that is given by theorem 7 for $\mathbb{Q}(\zeta_d)$, $d$ being any positive integer that satisfies $d \ne 2 \mod 4$. We now make some remarks on this theorem.

* We will keep in mind that there is a one-to-one correspondence between all the elements of $C$ and all the tuples $(\Omega, u_1 \cdots u_k)$ with $u_1 \cdots u_k \in Y_\Omega$. We say $\Omega$ or $n_\Omega$ is the level of the element of $C$ that corresponds to $(\Omega, u_1 \cdots u_k)$.
* This theorem comes with an algorithm to compute the expression of any element of **Was**(Q(ζ_n)) in the basis C that we will now present as it will be useful to prove our theorem 22. This following algorithm is quite similar to the one that is presented in [2] and [10]. More precisely, if we consider c_n^{u_1...u_r} with u_1...u_r /∈ ⟨J_1, ..., J_r⟩, we can apply the following algorithm. Note that we will suppose n is not a prime power - that is we will not explain how to decompose terms with some prime power level - since we will not need it.

If one - and exactly one - of the $u_i$'s is $J_i$, we can get rid of it by using the norm relation along $\sigma_i$ (Eq. (2)). Indeed, this norm relation allows us to decompose $c_n^{u_1 \cdots u_r}$ with some $c_n^{v_1 \cdots v_r}$'s such that $v_j \neq J_j$ for all $j$ and terms with lower level that can be treated by induction (on $r$):

$$c_n^{u_1 \cdots u_r} = c_{n/q_r}^{(1-\text{Frob}(p_r)^{-1})u_1 \cdots u_r} \prod_{v_i \in \text{Gal}(\mathbb{Q}(\zeta_i)/\mathbb{Q}) \setminus \{J_i\}} c_n^{-u_1 \cdots u_r v_i}.$$

If many of the $u_i$'s satisfy $u_i = J_i$ then we may use norm relations in a row along each of those $\sigma_i$ to get to handle terms of the form $c_n^{u_1 \cdots u_r}$ with $u_i \neq J_i$ for all $i$ (and terms with lower level). Let $b = c_n^{u_1 \cdots u_r}$, suppose $u_1 \cdots u_r \neq 1$, $u_i \neq J_i$ for all $i$ and let us now explain how the algorithm works for such $b$. This algorithm works on induction on $r$ so that we will act like if we do not have to consider terms with lower level - that is we will not mention those terms - but some of those do appear (every time we use a norm relation).

- Suppose $u_r \neq 1$. We distinguish two cases. If $u_r \in \mathcal{R}_r$, we have nothing to do (that is $b \in C$ so that $b$ is already decomposed in $C$). Now, suppose $u_r \in J_r \mathcal{R}_r$. As we supposed $u_r \neq J_r$, if we apply Eq. (1), then we get to handle a term of the form $v_1 \cdots v_{r-1} J_r u_r$ (with $v_i = J_i u_i$ for all $i$) so that we just have to get rid of the $v_i$'s that satisfy $v_i = J_i$ (that happens when $u_i = 1$) to get back to the previous case $u_r \in \mathcal{R}_r \setminus \{1\}$. In both cases, we see $b$ decomposes with terms with lower level and terms associated to some $v_1 \cdots v_r \in Y_{\ll 1, r]}$. Moreover, suppose $u_r \in \mathcal{R}_r$ or $u_1 \neq 1, \dots, u_{r-1} \neq 1$. Then, if the term associated to some $v_1 \cdots v_r \in Y_{\ll 1, r]}$ appears in the decomposition of $b$, then we have

$$v_1 \cdots v_r = \begin{cases} u_1 \cdots u_r & \text{if } u_r \in \mathcal{R}_r \setminus \{1\} \\ J_1 u_1 \cdots J_r u_r & \text{if } u_r \in J_r \mathcal{R}_r \setminus \{J_r\} \end{cases}$$
and it appears with exponent 1.

- Suppose $u_r = 1$ and $u_{r-1} \neq 1$. Again, distinguish two cases. If $u_{r-1} \in \mathcal{R}_{r-1}$, then we are done. Else, if $u_{r-1} \in J_{r-1}\mathcal{R}_{r-1} \setminus \{J_{r-1}\}$, use the norm relation along $\sigma_r$ (Eq. (2)) to get

$$b = c_n^{u_1 \cdots u_{r-1}} = c_{n/q_r}^{(1-\text{Frob}(p_r)^{-1})u_1 \cdots u_{r-1}} \prod_{v_r \in \text{Gal}(\mathbb{Q}(\zeta_{q_r})/\mathbb{Q}) \setminus \{1\}} c_n^{-u_1 \cdots u_{r-1}v_r}$$

to get back to the cases $u_1 \cdots u_{r-1}v_r$ with $v_r \in \mathcal{R}_r \setminus \{1\}$ or $v_r \in J_r\mathcal{R}_r \setminus \{J_r\}$ - those cases were treated before - and one more case with $v_r = J_r$. This last case can be treated using Eq. (1) before getting rid of the $J_i$'s that potentially just appeared after using Eq. (1) (depending on whether some $u_i$'s are 1's). We see $b$ decomposes with lower level terms and terms associated to some $v_1 \cdots v_k$ with $k \ge r-1$. Moreover, suppose $u_{r-1} \in \mathcal{R}_{r-1}$ or $u_1, \dots, u_{r-2} \neq 1$. Then, if the term associated to some $v_1 \cdots v_{r-1} \in Y_{\ll 1, r}$ appears in the decomposition of $b$, we have

$$v_1 \cdots v_{r-1} = \begin{cases} u_1 \cdots u_{r-1} & \text{if } u_{r-1} \in \mathcal{R}_{r-1} \setminus \{1\} \\ J_1 u_1 \cdots J_{r-1} u_{r-1} & \text{if } u_{r-1} \in J_{r-1} \mathcal{R}_{r-1} \setminus \{J_{r-1}\} \end{cases}$$

and it appears with exponent $-1$ if $u_{r-1} \in J_{r-1}\mathcal{R}_{r-1} \setminus \{J_{r-1}\}$.

- More generally, if $u_r = 1, \dots, u_{i+1} = 1$ and $u_i \neq 1$, we can proceed by induction on $i$ as follows (note there is such $i$ as we supposed $u_1 \cdots u_r \neq 1$). If $u_i \in \mathcal{R}_i$, we have nothing to do. If $u_i \in J_i\mathcal{R}_i \setminus \{J_i\}$, using the norm relation along $\sigma_{i+1}$ (Eq. (2)) brings us back to the previous cases and one more case with $u_1 \cdots u_i J_{i+1}$. For this last case, use the norm relation along $\sigma_{i+2}$. This leads us to consider terms of the form $u_1 \cdots u_i J_{i+1} v_{i+2}$ with $v_{i+2} \in \text{Gal}(\mathbb{Q}(\zeta_{q_{i+2}})/\mathbb{Q}) \setminus \{1\}$. If $v_{i+2} \in \mathcal{R}_{i+2} \setminus \{1\}$ or $v_{i+2} \in J_{i+2}\mathcal{R}_{i+2} \setminus \{J_{i+2}\}$, we just have to get rid of the $J_j$'s (that is we have to get rid of $J_{i+1}$ here) to get back the previous cases. If $v_{i+2} = J_{i+2}$, we use the norm relation along $\sigma_{i+3}$ and repeat the same trichotomy over and over until we have to consider the case $u_1 \cdots u_i J_{i+1} \cdots J_r$ for which we simply use Eq. (1) before getting rid of the $J_i$'s that may just have appeared after using Eq. (1). We see $b$ decomposes with lower level terms and terms associated
to some $v_1 \cdots v_k$ with $k \ge i$. Moreover, suppose $u_i \in \mathcal{R}_i$ or $u_1, \dots, u_{i-1} \ne 1$. Then, if the term associated to some $v_1 \cdots v_i \in Y_{[1,r]}$ appears in the decomposition of $b$, we have

$$v_1 \cdots v_i = \begin{cases} u_1 \cdots u_i & \text{if } u_i \in \mathcal{R}_i \setminus \{1\} \\ J_1 u_1 \cdots J_i u_i & \text{if } u_i \in J_i \mathcal{R}_i \setminus \{J_i\} \end{cases}$$

and it appears with exponent $(-1)^{r-i}$ if $u_i \in J_i \mathcal{R}_i \setminus \{J_i\}$.

* Observe that we have $C(Q(\zeta_d)) \subset C(Q(\zeta_{d'}))$ for any $d | d'$ such that $d'/d \wedge d = 1$ so that any element of $\mathbf{Was}(Q(\zeta_d))$ decomposes in $C(Q(\zeta_{d'}))$ with terms that have lower or equal level to $d$.

* Later, we will need the following notation. Let

$$\mathbb{L}_n = Q(\zeta_{q_1})^{+} \cdots Q(\zeta_{q_r})^{+}.$$

If $r \ge 2$, there is a root of unity $\eta \in Q(\zeta_n)$ (see [17] 2-ii) such that

$$\eta_n := \eta N_{Q(\zeta_n)/Q(\zeta_{q_1}) + \cdots + Q(\zeta_{q_{r-1}}) + Q(\zeta_{q_r})} (1 - \zeta_n) \in Q(\zeta_{q_1})^{+} \cdots Q(\zeta_{q_r})^{+}$$

and $\eta_n^2 = N_{Q(\zeta_n)/Q(\zeta_{q_1}) + \cdots + Q(\zeta_{q_{r-1}}) + Q(\zeta_{q_r})} (1 - \zeta_n)$.

For any $\mathbb{L} \subset \mathbb{L}_n$ with conductor $n$, let

$$e_{\mathbb{L}} = N_{\mathbb{L}_n/\mathbb{L}}(\eta_n) \in \mathbb{L}.$$

We also define similar objects $\eta_d$ and $e_{\mathbb{L}}$ by swapping $n$ with any of its non trivial positive divisors $d$.

## 2.2 On the convolution product

Through this section, we recall - in the needed context only - several facts that are stated in a more general context in [12] and [4] and that deal with Möbius functions.

Let $E$ be a finite set. Define $\mathcal{F}(E)$ to be the set of functions

$$f: \mathcal{P}(E) \to \mathbb{C}.$$
This set has a law of addition and a convolution product defined in the following way

$$ \forall f, g \in \mathcal{F}(E), \forall \Omega \subset E, \quad f * g(\Omega) = \sum_{X \subset \Omega} f(X)g(\Omega \setminus X). $$

One can show $(\mathcal{F}(E), +, *)$ is a ring whose identity element is the function that maps $\emptyset$ to 1 and any subset $\Omega \neq \emptyset$ to 0.

Denote by $\mathbf{1}$ the element of $\mathcal{F}(E)$ that maps any $\Omega \subset E$ to 1. One can show $\mathbf{1}$ is a unit and we let $\mu$ denote its inverse. We have (see [4] equation 3.3)

$$ \forall \Omega \subset E, \quad \mu(\Omega) = (-1)^{|\Omega|}. $$

In particular, we have the following theorem.

**Theorem 8.** Let $f, g \in \mathcal{F}(E)$. We have

$$ \forall \Omega \subset E, \quad \sum_{X \subset \Omega} f(X) = g(\Omega) \iff \forall \Omega \subset E, \quad f(\Omega) = \sum_{X \subset \Omega} (-1)^{|\Omega|-|X|} g(X). $$

*Proof.* See [12] proposition 2. $\square$

Later, we will use this convolution product with $E = [1, r]$.

# 3 From imaginary fields to real fields

In this section, we aim to give $\mathbb{Z}$-bases of $\mathbf{Was}^+(\mathbb{Q}(\zeta_n))$ (recall we talk about $\mathbf{Was}^+(\mathbb{Q}(\zeta_n))$ instead of talking about the quotient $\mathbf{Was}^+(\mathbb{Q}(\zeta_n))/\mathbb{Z}^+(\mathbb{Q}(\zeta_n)))$.

More precisely, for any abelian number field $K$, we give a way to construct a basis of $\mathbf{Was}^+(K)$ given a basis of $\mathbf{Was}(K)$ (proposition 9) and we then apply this method when $K$ is a cyclotomic field (theorems 11 and 12).

## 3.1 Abelian fields

**Proposition 9.** Let $K$ be an abelian number field. Let $(x_1, \dots, x_r)$ be a $\mathbb{Z}$-basis of $\mathbf{Was}(K)$. With no loss of generality, suppose there is $r' \in [0; r]$ such that $x_1, \dots, x_{r'}$ have order 2 in the quotient group $E/ZE^+$ and $x_{r'+1}, \dots, x_r$ have order 1. Then, the family $(|x_1||x_1|, \dots, |x_1||x_{r'}|, |x_{r'+1}|, \dots, |x_r|)$ is a basis of $\mathbf{Was}(K^+)$.
Proof. First, if $x \in \mathbb{Z}\mathbf{E}^{+}$, observe we have $|x| \in \mathbf{E}^{+}$ and, if we also suppose $x \in \mathbf{Was}$, then $|x| \in \mathbf{Was}^{+}$. Indeed, write $x = zu \in \mathbb{Z}\mathbf{E}^{+}$. Then, we have $|x| = \pm u$ and this proves $|x| \in \mathbf{E}^{+}$. Now, suppose we also have $x \in \mathbf{Was}$. Then, we have $u = z^{-1}x \in \mathbf{Was} \cap \mathbf{E}^{+}$ and this proves $|x| \in \mathbf{Was}^{+}$. Hence, the family $(|x_1||x_1|, \dots, |x_1||x_{r'}|, |x_{r'+1}|, \dots, |x_r|)$ is made of elements of $\mathbf{Was}^{+}$. Now, let us show these elements generate $\mathbf{Was}^{+}$ (modulo $\pm 1$).

Let $x \in \mathbf{Was}^{+}$ and write

$$x = \zeta \prod_{i=1}^{r} x_i^{a_i}$$

for some $a_i \in \mathbb{Z}$, $\zeta \in \mathbb{Z}$. In particular, we have $x \in \mathbf{E}^{+}$ so there is an even number of elements of the form $x_i$ with $i \le r'$ (thanks to proposition 1), that is we have:

$$\sum_{i=1}^{r'} a_i \in 2\mathbb{Z}.$$

Let

$$A = a_1 - \sum_{i=2}^{r'} a_i \in 2\mathbb{Z}.$$

Therefore, we have (thanks to proposition 1 again)

$$\begin{aligned} x &= \zeta x_1^A \left( \prod_{i=2}^{r'} (x_1 x_i)^{a_i} \right) \left( \prod_{i>r'} x_i^{a_i} \right) \\ &= \zeta' |x_1|^A \left( \prod_{i=2}^{r'} |x_1 x_i|^{a_i} \right) \left( \prod_{i>r'} |x_i|^{a_i} \right) \end{aligned}$$

for some root of unity $\zeta'$. As we have $x \in \mathbf{E}^{+}$ and $|x_i| \in \mathbb{R}$, we have $\zeta' = \pm 1$, which proves the considered family is a generating family. Hence, it is a basis because of Dirichlet's theorem. $\square$

## 3.2 Cyclotomic field with odd conductor

**Remark 10.** To understand the following theorems better, observe the set $C_{\lfloor 1, r \rfloor}$ is non empty whenever $r \ge 2$: if $r$ is even then there is $c_{\lfloor 1, r \rfloor}$ in this set and if $r$ is odd then there is $k$ such that $q_k \ne 4, 3$ so that there is $u_k \in \mathcal{R}_k \setminus \{1\}$ (as 4 and 3 are the only integers $a$ that satisfy $\varphi(a) = 2$).
**Theorem 11.** Suppose $n$ is odd. If $n$ is not a prime power, let $x_1 \in C_{\llbracket 1, r \rrbracket}$. A basis of $\mathbf{Was}^+(\mathbb{Q}(\zeta_n))$ is given by:

i) the Galois conjugates of $\xi_{p^{vp(n)}}$ with $p$ running over the set of all prime divisors of $n$

ii) the $|x_1| |x|$'s with $\Omega \subset \llbracket 1, r \rrbracket$, $|\Omega| \ge 2$ and $x \in C_{\Omega}$.

*Proof.* We will apply proposition 9 to the basis of $\mathbf{Was}(\mathbb{Q}(\zeta_n))$ given in theorem 7.

We have $\xi_{p^{vp(n)}} \in \mathbf{Z}(\mathbb{Q}(\zeta_{p^{vp(n)}}))\mathbf{E}^+(\mathbb{Q}(\zeta_{p^{vp(n)}}))$ from proposition 1 and then the same statement can be made for its conjugates. For the other generators, observe we have, for any divisor $d | n$, $d \ne 1$ and for any $a \in \mathbb{Z}$

$$1 - \zeta_d^a = \zeta_{2d}^a (\zeta_{2d}^{-a} - \zeta_{2d}^a).$$

As $d$ is odd, we have $\zeta_{2d} \in \mathbb{Q}(\zeta_d)$, so that the previous decomposition takes place in $\mathbb{Q}(\zeta_d)$. Moreover, we have $(\zeta_{2d}^{-a} - \zeta_{2d}^a) = \pm i |1 - \zeta_d^a|$. We then have the following decomposition

$$1 - \zeta_d^a = \pm i \zeta_{2d}^a |1 - \zeta_d^a|. \qquad (3)$$

which shows that $1 - \zeta_d^a$ has order 2 in $\mathbf{E}(\mathbb{Q}(\zeta_n))/\mathbf{Z}(\mathbb{Q}(\zeta_n))\mathbf{E}^+(\mathbb{Q}(\zeta_n))$ whenever $a$ is prime to $d$, otherwise we would have $i \in \mathbb{Q}(\zeta_n)$ and that is not the case. $\square$

### 3.3 Cyclotomic field with even conductor

**Theorem 12.** Suppose $n$ is even and write $n = 2^{e_1} p_2^{e_2} \cdots p_r^{e_r} \in \mathbb{N}$ (recall $n \ne 2$ mod 4 so that we have $e_1 \ge 2$). If $n$ is not a prime power, let $x_1 \in C_{\llbracket 1, r \rrbracket}$. A basis of $\mathbf{Was}^+(\mathbb{Q}(\zeta_n))$ is given by:

i) the Galois conjugates of $\xi_{p^{vp(n)}}$ with $p$ being any prime divisor of $n$

ii) the $|x|$'s where $\Omega \subset \llbracket 1, r \rrbracket$ satisfies $|\Omega| \ge 2$ and $x \in C_{\Omega}$ has some odd level $d$

iii) the $|x_1| |x|$'s where $\Omega \subset \llbracket 1, r \rrbracket$ satisfies $|\Omega| \ge 2$ and $x \in C_{\Omega}$ has some even level $d$, that is $v_2(d) = e_1$

*Proof.* We just apply proposition 9 to the basis of $\mathbf{Was}(\mathbb{Q}(\zeta_n))$ given in theorem 7.
For the first group of generators, see our previous proof.

Any element $x$ of the second group of generators can be written as $1 - \zeta_d^a$ for some $d | n$ with $d$ being odd and $a \wedge d = 1$. Moreover, Eq. (3) shows we have $1 - \zeta_d^a \in \mathbb{Z}(\mathbb{Q}(\zeta_n))\mathbf{E}^+(\mathbb{Q}(\zeta_n))$ as expected.

Any element $x$ of the third group of generators can be written as $1 - \zeta_d^a$ for some $d | n$ satisfying $v_2(d) = e_1$ and $a \wedge d = 1$. The same equation as before shows that we have $1 - \zeta_d^a \notin \mathbb{Z}(\mathbb{Q}(\zeta_n))\mathbf{E}^+(\mathbb{Q}(\zeta_n))$, otherwise we would have some primitive $2^{1+e_1}$-th root of unity lying in $\mathbb{Q}(\zeta_n)$, which is not the case. $\square$

# 4 Totally deployed fields

Recall we let $\mathbf{Was}_2(\mathbb{K}) = \mathbf{Was}(\mathbb{K}) \otimes_{\mathbb{Z}} \mathbb{Z}[1/2]$. Through this section, we aim to give a $\mathbb{Z}[1/2]$-basis of $\mathbf{Was}_2(\mathbb{K})$ through theorem 22 assuming $\mathbb{K}$ is a totally deployed abelian number field (recall $\mathbf{Was}$ means $\mathbf{Was}/\mathbb{Z}$ when we talk about bases). In particular, we will have a family that is a $\mathbb{Z}_p$-basis of $\mathbf{Was}(\mathbb{K}) \otimes_{\mathbb{Z}_p}$ for any prime integer $p > 2$. For now, we suppose $\mathbb{K}$ is a totally deployed abelian number field, with conductor $n$ and we write

$$ \mathbb{K} = \mathbb{K}_1 \cdots \mathbb{K}_r $$

with $\mathbb{K}_i \subset \mathbb{Q}(\zeta_{q_i})$ for all $i \in [1, r]$. To simplify the proof of theorem 22, if there is $i$ such that $p_i = 2$ and $\mathbb{K}_i$ is imaginary, suppose $i = r$.

To construct our basis, we will consider a family of elements of $\mathbb{K}$ that has $r_1 + r_2 - 1$ elements and that generates a direct factor of $\mathbf{Was}_2(\mathbb{Q}(\zeta_n))$. It is not hard to see that this property makes this family generate $\mathbf{Was}_2(\mathbb{K})$ so that this family is a basis. More precisely, we will construct a basis of $\mathbf{Was}_2(\mathbb{K})$ that can be completed with some terms from the basis $C$ from theorem 7 to form a basis of $\mathbf{Was}_2(\mathbb{Q}(\zeta_n))$. This idea has already been used in [17], [5]. Actually, in order to prove proposition 2 from [17], the author proves the following fact.

**Lemma 13.** Let $\mathbb{L}$ be an abelian number field with conductor $n$. Let $H$ be a group such that $\mathbb{Z}(\mathbb{L}) \subset H \subset \mathbf{Was}(\mathbb{L})$. Assume $H$ is a direct factor of $\mathbf{Was}(\mathbb{Q}(\zeta_n))/\mathbb{Z}(\mathbb{Q}(\zeta_n))$ and suppose $H$ has the same $\mathbb{Z}$-rank as $\mathbf{Was}(\mathbb{L})$. Then we have $H = \mathbf{Was}(\mathbb{L})$.
Proof. See the proof of proposition 2 from [17]. □

It is clear that a similar statement can be made with $\mathbb{Z}[1/2]$-modules instead of abelian groups.

Recall $\mathcal{R}_i$ is the set of representatives of $\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{Q})$ modulo $J_i$ given by lemma 1.1 in [10]. We now introduce the notation we will use to state our next theorem 22. To make it easier to understand, we divided it into many definitions. The reader may not understand the following definition items as independent definitions but instead think of this separation as a help to read the following more easily.

**Definition 14.** With no loss of generality, let $t$ be such that $\mathbb{K}_1, \dots, \mathbb{K}_{t-1}$ are real and $\mathbb{K}_t, \dots, \mathbb{K}_r$ are imaginary. If $2 \mid n$ and $\mathbb{K} \cap \mathbb{Q}(\zeta_{2^{v_2(n)}})$ is imaginary, we will suppose $p_r = 2$.

**Definition 15.** For any $i \in [1, t-1]$, let $(\mathcal{R}_{i,1}(\mathbb{K}), \mathcal{T}_i(\mathbb{K}))$ be such that $\mathcal{R}_{i,1}(\mathbb{K})$ is a set of representatives of $\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{Q})$ modulo $\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{K}_i)$ with $1 \in \mathcal{R}_{i,1}(\mathbb{K})$ and $\mathcal{T}_i(\mathbb{K})$ is a set of representatives of $\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{K}_i)$ modulo $J_i$ such that $\mathcal{T}_i \cdot \mathcal{R}_{i,1}(\mathbb{K}) \subset \mathcal{R}_i$.

For instance, we can construct $\mathcal{R}_{i,1}(\mathbb{K})$, $\mathcal{T}_i(\mathbb{K})$ as follows. First, if $p_i = 2$ then the construction is clear as $\langle J_i \rangle$ is a direct factor of $\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{Q})$. Next, suppose $p_i$ is odd. Let $z$ denote a generator of the 2-Sylow of $\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{Q})$ and let $m \in \mathbb{N}$ be minimal with respect to $z^{2m} \in \text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{K}_i)$. Let $a \in \mathbb{N}$ be minimal with respect to $z^{2a} = J_i$. Let

$$ \begin{aligned} \mathcal{T}_i(\mathbb{K}) &= \{z^{k2^m}h : k \in [0, 2^{a-m}[, h \in \text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{K}_i) \text{ has odd order}\} \\ \mathcal{R}_{i,1}(\mathbb{K}) &= \{z^k h : 0 \le k < 2^m \text{ and } h \in H\} \end{aligned} $$

where $H$ denotes any set of representatives of the non 2-part of $\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{Q})/\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{K}_i)$ that lies in the non 2-part of $\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{Q})$.

Now, to shorten definition 19, swap 1 with $J_i$ in $\mathcal{R}_{i,1}(\mathbb{K})$.

**Definition 16.** For any $i \in [t, r]$, let $\mathcal{R}_{i,1}(\mathbb{K})$ be a set of representatives of $\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{Q})$ modulo $\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{K}_i)$ with $1, J_i \in \mathcal{R}_{i,1}(\mathbb{K})$. If $p_i \ne 2$, observe $\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{K}_i)$ acts on $\mathcal{R}_i$ by multiplication. Then, let $\mathcal{R}_{i,2}(\mathbb{K})$ be a set of representatives of $\mathcal{R}_i$ modulo $\text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{K}_i)$ with $1 \in \mathcal{R}_{i,2}(\mathbb{K})$ so that $\mathcal{R}_{i,2}(\mathbb{K})$ is a set of representatives of
Gal(Q(ζ_q_i)/Q) modulo ⟨J_i, Gal(Q(ζ_q_i)/K_i)⟩. If p_i = 2, observe Gal(Q(ζ_q_i)/K_i) still acts on R_i and define R_{i,2}(K) as before (the action is given by a transport of structure through the canonical bijection R_i ≃ Gal(Q(ζ_q_i)/Q)/⟨J_i⟩ as Gal(Q(ζ_q_i)/K_i) acts on this last quotient by multiplication). The set R_{i,2}(K) is still a set of representatives of Gal(Q(ζ_q_i)/Q) modulo ⟨J_i, Gal(Q(ζ_q_i)/K_i)⟩ but we can no longer assume Gal(Q(ζ_q_i)/K_i) ⋅ R_i ⊂ R_i.

**Definition 17.** For any non-empty $\Omega = \{i_1, \dots, i_s\} \subset [1, r]$, let

$$K_{\Omega} = K_{i_1} \cdots K_{i_s}$$

$$\Omega_{\mathbb{R}} = \Omega \cap [1, t-1]$$

$$\Omega_{\mathbb{C}} = \Omega \cap [t, r].$$

**Definition 18.** For any $\Omega = \{j\} \subset [1, r]$, let $Y_{\Omega}(K)$ be $R_{j,2}(K) \setminus \{1\}$ if $K_j$ is imaginary and let $Y_{\Omega}(K)$ be $R_{j,1}(K) \setminus \{J_j\}$ if $K_j$ is real. Let

$$C_{\Omega}(K) = \left\{ \mathrm{N}_{\mathbb{Q}(\zeta_{q_j})^{+} / \mathbb{K}_j^{+}}(\xi_{q_j})^u : u \in Y_{\Omega}(K) \right\}.$$

**Definition 19.** For any $\Omega = \{i_1, \dots, i_s\} \subset [1, r]$ with $s \ge 2$, such that $i_1 < \dots < i_s$ and $K_{i_s}$ is imaginary (that is $K_{\Omega}$ decomposes with at least 1 imaginary field), let $t_{\Omega}$ be the integer such that $K_{i_1}, \dots, K_{i_{t_{\Omega}-1}}$ are real and $K_{i_{t_{\Omega}}}, \dots, K_{i_s}$ are imaginary. Let $Y_{\Omega}(K)$ be the set of all $u_1 \cdots u_k$ with $k \in [t_{\Omega}, s]$, satisfying $u_k \in R_{i_k, 2}(K) \setminus \{1\}$ and

$$\forall j \in [1, k-1], \quad u_j \in R_{i_j, 1}(K) \setminus \{J_{i_j}\}.$$

If $|\Omega_{\mathbb{C}}|$ is even, then add to $Y_{\Omega}(K)$ all the products $u_1 \cdots u_{t_{\Omega}-1}$ with $u_j \in R_{i_j, 1}(K) \setminus \{J_{i_j}\}$ for all $j \in [1, t_{\Omega}-1]$ (if $t_{\Omega} = 1$, understand we add 1 to $Y_{\Omega}$). Let

$$C_{\Omega}(K) = \left\{ N_{\mathbb{Q}(\zeta_{n_{\Omega}})/K_{\Omega}} (1 - \zeta_{n_{\Omega}})^u : u \in Y_{\Omega}(K) \right\}.$$

We will call these last elements of $Y_{\Omega}(K)$ - or their corresponding elements in $C_{\Omega}(K)$ - "problematic terms".

**Definition 20.** If $K_{i_s}$ is real (that is $K_{\Omega}$ decomposes with real fields only), let $Y_{\Omega}(K)$ be the set of all $u_1 \cdots u_s$ such that $u_j \in R_{i_j, 1}(K) \setminus \{J_{i_j}\}$ for all $j \in [1, s]$. Let

$$C_{\Omega}(K) = \{e_{K_{\Omega}}^u : u \in Y_{\Omega}(K)\}$$

(recall $e_{K_{\Omega}}$ is defined at the end of section 2.1).
**Definition 21.** For all non empty set $\Omega \subset [1, r]$, let

$$c_{\Omega}(\mathbb{K}) = \begin{cases} N_{\mathbb{Q}(\zeta_{\Omega})^{+}/\mathbb{K}_{\Omega}^{+}}(\xi_{n_{\Omega}}) & \text{if } |\Omega| = 1 \\ N_{\mathbb{Q}(\zeta_{\Omega})/\mathbb{K}_{\Omega}}(1 - \zeta_{\Omega}) & \text{if } |\Omega_{\mathbb{C}}| \ge 1 \\ e_{\mathbb{K}_{\Omega}} & \text{in any other case.} \end{cases}$$

**Theorem 22.** *Recall $\text{Was}_2(\mathbb{K}) = \text{Was}(\mathbb{K})/\mathbb{Z}(\mathbb{K}) \otimes \mathbb{Z}[1/2]$ in this context. The family $C(\mathbb{K}) = \cup_{\Omega} C_{\Omega}(\mathbb{K})$ where $\Omega$ runs over the set of all non-empty subsets of $[1, r]$ is a $\mathbb{Z}[1/2]$-basis of $\text{Was}_2(\mathbb{K})$. Moreover, $\text{Was}_2(\mathbb{K})$ is a direct factor of $\text{Was}_2(\mathbb{Q}(\zeta_n))$. More precisely, the family $C(\mathbb{K})$ can be completed to a basis of $\text{Was}_2(\mathbb{Q}(\zeta_n))$ with some terms from the basis $C$ of theorem 7.*

Before proving this theorem, we may state and prove the following lemma.

**Lemma 23.** *For any $i \in [1, r]$, let $d_i$ denote the degree of $K_i/\mathbb{Q}$. For all non-empty subset $\Omega \subset [1, r]$, let*

$$\begin{align*} f_{\mathbb{C}}(\Omega) &= \frac{1}{2} \left( \prod_{i \in \Omega} d_i - 1 \right) + \frac{(-1)^{|\Omega|}}{2} \\ f_{\mathbb{R}}(\Omega) &= \left( \prod_{i \in \Omega} d_i - 1 \right) \\ g_{\mathbb{C}}(\Omega) &= \frac{1}{2} \prod_{i \in \Omega} d_i \\ g_{\mathbb{R}}(\Omega) &= \prod_{i \in \Omega} d_i. \end{align*}$$

*Say each of these functions maps $\emptyset$ to 1. We have*

$$\sum_{\substack{\Omega \subset [1, r] \\ \Omega \neq \emptyset}} f_{\mathbb{C}}(\Omega) = g_{\mathbb{C}}([1, r]) - 1 \quad (4)$$

$$\sum_{\substack{\Omega \subset [1, r] \\ \Omega \neq \emptyset}} f_{\mathbb{R}}(\Omega) = g_{\mathbb{R}}([1, r]) - 1. \quad (5)$$

*Proof.* Let us prove the lemma first.
**Case 1** We may start with Eq. (5).

We have to prove

$$1 * f_{\mathbb{R}}(\llbracket 1, r \rrbracket) = g_{\mathbb{R}}(\llbracket 1, r \rrbracket)$$

but instead, we will show we have,

$$\forall \Omega \subset \llbracket 1, r \rrbracket, \quad f_{\mathbb{R}}(\Omega) = \mu * g_{\mathbb{R}}(\Omega)$$

and the expected result will then be proven (see our section on the convolution product). We have

$$\begin{aligned}
\mu * g_{\mathbb{R}}(\Omega) &= \sum_{X \subset \Omega} (-1)^{|\Omega|-|X|} g_{\mathbb{R}}(X) \\
&= \sum_{X \subset \Omega} (-1)^{|\Omega|-|X|} \prod_{i \in X} d_i \\
&= (-1)^{|\Omega|} + \sum_{k=1}^{|\Omega|} (-1)^{|\Omega|-k} \sum_{\substack{i_1, \dots, i_k \in \Omega \\ i_1 < \dots < i_k}} d_{i_1} \dots d_{i_k}.
\end{aligned}$$

Using Vieta's formulas, we can see that this last expression matches the evaluation of the polynomial $(-1)^{|\Omega|} \prod_{i \in \Omega} X - d_i$ at $X = 1$, hence

$$\mu * g_{\mathbb{R}}(\Omega) = (-1)^{|\Omega|} \prod_{i \in \Omega} (1 - d_i) = f_{\mathbb{R}}(\Omega).$$

**Case 2** In a similar way, we now consider Eq. (4). We have

$$\begin{aligned}
\mu * g_{\mathbb{C}}(\Omega) &= \sum_{X \subset \Omega} (-1)^{|\Omega|-|X|} g_{\mathbb{C}}(X) \\
&= (-1)^{\Omega} + \sum_{\substack{X \subset \Omega \\ X \neq \emptyset}} (-1)^{|\Omega|-|X|} \frac{1}{2} \prod_{i \in X} d_i \\
&= (-1)^{\Omega} + \frac{1}{2} \sum_{k=1}^{|\Omega|} (-1)^{|\Omega|-k} \sum_{\substack{i_1, \dots, i_k \in \Omega \\ i_1 < \dots < i_k}} d_{i_1} \dots d_{i_k}.
\end{aligned}$$
Using Vieta's formulas, we can see that this last term on the right side matches the evaluation of the polynomial

$$ \frac{(-1)^{|\Omega|}}{2} \left( \left( \prod_{i \in \Omega} X - d_i \right) - X^{|\Omega|} \right) $$

at $X = 1$, hence

$$ \mu * g_{\mathbb{C}}(\Omega) = (-1)^{|\Omega|} + \frac{(-1)^{|\Omega|}}{2} \left( \left( \prod_{i \in \Omega} 1 - d_i \right) - 1 \right) = f_{\mathbb{R}}(\Omega). $$

□

We may now prove the previously stated theorem.

*Proof.* First, we may prove $C(\mathbb{K})$ has cardinality $r_1 + r_2 - 1$, that is

$$ |C(\mathbb{K})| = \begin{cases} \frac{1}{2} \left( \prod_{i \in [1, r]} d_i \right) - 1 & \text{if } [1, r]_{\mathbb{C}} \neq \emptyset \\ \left( \prod_{i \in [1, r]} d_i \right) - 1 & \text{if not.} \end{cases} $$

This can also be stated in the following way. For any non-empty subset $\Omega \subset [1, r]$, we denote by $f(\Omega)$ the number of elements of $C_{\Omega}(\mathbb{K})$ and we let

$$ g(\Omega) = \begin{cases} \frac{1}{2} \prod_{i \in \Omega} d_i & \text{if } \Omega_{\mathbb{C}} \neq \emptyset \\ \prod_{i \in \Omega} d_i & \text{if not.} \end{cases} $$

Also, say these functions both map $\Omega = \emptyset$ to 1. Then, we have to show

$$ 1 * f([1, r]) = g([1, r]). $$

Again, we rather show

$$ \forall \Omega \subset [1, r], \quad f(\Omega) = \mu * g(\Omega). $$
If $|\Omega| \le 1$, there is nothing to prove so we may suppose $|\Omega| > 1$. We separate three cases.

Suppose we have $\Omega_{\mathbb{C}} = \emptyset$. Then, lemma 23 gives

$$ 
\begin{aligned}
\mu * g(\Omega) &= \sum_{X \subset \Omega} (-1)^{|\Omega|-|X|} g(X) \\
&= \sum_{X \subset \Omega} (-1)^{|\Omega|-|X|} \prod_{i \in X} d_i \\
&= \prod_{i \in \Omega} (d_i - 1)
\end{aligned}
$$

and it remains to observe

$$ f(\Omega) = \prod_{i \in \Omega} (d_i - 1) $$

since we supposed $\Omega_{\mathbb{C}} = \emptyset$.

Now suppose $\Omega_{\mathbb{R}} = \emptyset$. Again, lemma 23 gives

$$ \mu * g(\Omega) = f_{\mathbb{C}}(\Omega). $$

For any integer $k$, let $C_{\Omega}^k(\mathbb{K})$ denote the terms from $C_{\Omega}$ obtained with elements of the form $u_1 \cdots u_k$.

If $|\Omega|$ is odd, we have

$$ f(\Omega) = \sum_{k=1}^{s} |C_{\Omega}^{k}(\mathbb{K})| = \sum_{k=1}^{s} (\frac{1}{2} d_{i_k} - 1)(d_{i_{k-1}} - 1) \cdots (d_{i_1} - 1) $$

and by induction on $N \in [1, s]$, one can show

$$ \sum_{k=1}^{N} (\frac{1}{2} d_{i_k} - 1)(d_{i_{k-1}} - 1) \cdots (d_{i_1} - 1) = \frac{1}{2}(d_{i_N} - 1) \cdots (d_{i_1} - 1) - \frac{1}{2}. $$

Taking $N = s$, we get

$$ \mu * g(\Omega) = f_{\mathbb{C}}(\Omega) = f(\Omega). $$

In the same way, if $|\Omega|$ is even, we have

$$ f(\Omega) = 1 + \sum_{k=1}^{r} |C_{\Omega}^{k}(\mathbb{K})| = 1 + \sum_{k=1}^{r} (\frac{1}{2} d_{i_k} - 1)(d_{i_{k-1}} - 1) \cdots (d_{i_1} - 1) $$
and we get the same conclusion.

Now suppose we have $\Omega_{\mathbb{C}} \neq \emptyset$ and $\Omega_{\mathbb{R}} \neq \emptyset$. We have

$$
\begin{align*}
\mu * g(\Omega) &= \sum_{X \subset \Omega} (-1)^{|\Omega|-|X|} g(X) \\
&= \sum_{\substack{X_1 \subset \Omega_{\mathbb{R}} \\ X_2 \subset \Omega_{\mathbb{C}} \\ X_2 \neq \emptyset}} (-1)^{|\Omega|-|X_1|-|X_2|} \times \frac{1}{2} \prod_{i \in X_1 \cup X_2} d_i + \sum_{X_1 \subset \Omega_{\mathbb{R}}} (-1)^{|\Omega|-|X_1|} \prod_{i \in X_1} d_i \\
&= \sum_{\substack{X_1 \subset \Omega_{\mathbb{R}} \\ X_2 \subset \Omega_{\mathbb{C}}}} (-1)^{|\Omega|-|X_1|-|X_2|} g_{\mathbb{C}}(X_1 \cup X_2) \\
&\quad - \sum_{X_1 \subset \Omega_{\mathbb{R}}} (-1)^{|\Omega_{\mathbb{C}}|+|\Omega_{\mathbb{R}}|-|X_1|} g_{\mathbb{C}}(X_1) \\
&\quad + \sum_{X_1 \subset \Omega_{\mathbb{R}}} (-1)^{|\Omega_{\mathbb{C}}|+|\Omega_{\mathbb{R}}|-|X_1|} g_{\mathbb{R}}(X_1) \\
&= f_{\mathbb{C}}(\Omega) - (-1)^{|\Omega_{\mathbb{C}}|} f_{\mathbb{C}}(\Omega_{\mathbb{R}}) + (-1)^{|\Omega_{\mathbb{C}}|} f_{\mathbb{R}}(\Omega_{\mathbb{R}}) \\
&= \frac{1}{2} \prod_{i \in \Omega} (d_i - 1) + \frac{(-1)^{|\Omega_{\mathbb{C}}|}}{2} \prod_{i \in \Omega_{\mathbb{R}}} (d_i - 1).
\end{align*}
$$

Separate cases depending on whether $|\Omega_{\mathbb{C}}|$ is even or not and one can show (using a similar induction argument as before) that we have

$$f(\Omega) = \frac{1}{2} \prod_{i \in \Omega} (d_i - 1) + \frac{(-1)^{|\Omega_{\mathbb{C}}|}}{2} \prod_{i \in \Omega_{\mathbb{R}}} (d_i - 1).$$

This conclude the proof of the fact $C(\mathbb{K})$ has the expected cardinality.

We may now show the elements of $C(\mathbb{K})$ generate a direct factor of $\mathrm{Was}_2(\mathbb{Q}(\zeta_n))$.

To this aim, we will decompose every element of $C(\mathbb{K})$ in the basis $C$ of theorem 7 and we will associate a term $\phi(c) \in C$ to each $c \in C(\mathbb{K})$ such that $\phi(c)$ appears with exponent 1 or 2 in the decomposition of $c$ in the basis $C$. Then, we will observe that the decomposition of the elements of $C(\mathbb{K})$ are almost pairwise disjoint so that, if we order the terms of $C(\mathbb{K}) \cup (C \setminus \{\phi(c) : c \in C(\mathbb{K})\})$ properly, the matrix of
this last family in the basis $C$ is invertible in $\mathbb{Z}[1/2]$ because it is triangular with diagonal coefficients lying in $\{1, 2\}$. To ease the reading, we will handle elements of $C_{\llbracket 1, r \rrbracket}(\mathbb{K})$ only but it is easily seen that the same kind of arguments works for any other element of $C(\mathbb{K})$.

Let $u \in Y_{\llbracket 1, r \rrbracket}(\mathbb{K})$ and let $c = c_{\llbracket 1, r \rrbracket}(\mathbb{K})^u$. We will show that we can let $\phi(c) = 1 - \zeta_n^u$. In each of the following cases, we will then compute the exponent of $1 - \zeta_n^u$ in the decomposition of $c$ and we will investigate the decomposition of $c$.

Suppose $r = 1$. This case has already been considered in the proof of proposition 2 from [17] and in the proof of theorem 2.1 from [5]. More precisely, $\phi(c) = c_{\llbracket 1, r \rrbracket}^u$ is such that $\phi(c)$ appears with exponent 1 in the decomposition of $c$ in the basis $C$ and the decomposition of all the elements of $C_{\llbracket 1, r \rrbracket}(\mathbb{K})$ are pairwise disjoint.

From now on, suppose $r \ge 2$.

Suppose $\llbracket 1, r \rrbracket_{\mathbb{C}} = \emptyset$. This case has already been considered in [17] (see proposition 2 and remark 4) and we now write it down for the convenience of the reader. In this case, we may show that, if $u = u_1 \cdots u_r$ with $u_1 \in \mathcal{R}_{1,1}(\mathbb{K}) \setminus \{J_1\}, \dots, u_r \in \mathcal{R}_{r,1}(\mathbb{K}) \setminus \{J_r\}$, we can associate $1 - \zeta_n^u$ to $c$.

Modulo roots of unity of $\mathbb{Q}(\zeta_n)$, we have

$$c = N_{\mathbb{L}_n/\mathbb{K}}(\eta_n^u) = \prod_{v_1 \in \mathcal{T}_1(\mathbb{K})} \cdots \prod_{v_r \in \mathcal{T}_r(\mathbb{K})} \prod_{\varepsilon_1, \dots, \varepsilon_{r-1} \in \{0;1\}} 1 - \zeta_n^{J_1^{\varepsilon_1} u_1 v_1 \cdots J_{r-1}^{\varepsilon_{r-1}} u_{r-1} v_{r-1} u_r v_r}$$

and this is the decomposition of $c$ in the basis of theorem 7. Indeed, we have $J_i^{\varepsilon_i} u_i v_i \in \text{Gal}(\mathbb{Q}(\zeta_i)/\mathbb{Q}) \setminus \{J_i\}$ for all $i < r$ and $u_r v_r \in \mathcal{R}_r \setminus \{1\}$.

As expected, we see $\phi(c) = 1 - \zeta_n^u$ works and appears with exponent 1. Also, observe all the decompositions of all the elements of $C_{\llbracket 1, r \rrbracket}(\mathbb{K})$ are pairwise disjoint in this case.

Suppose $\llbracket 1, r \rrbracket_{\mathbb{R}} = \emptyset$. Suppose $2 \nmid n$ (we will explain what to do if $2 \mid n$ later). We go through two cases depending on $u$ (and the parity of $r$).
Suppose $u \neq 1$. Then we have

$$N_{\mathbb{Q}(\zeta_n)/\mathbb{K}}(1 - \zeta_n^u) = \prod_{s_1 \in \text{Gal}(\mathbb{Q}(\zeta_{q_1})/\mathbb{K}_1)} \cdots \prod_{s_r \in \text{Gal}(\mathbb{Q}(\zeta_{q_r})/\mathbb{K}_r)} (1 - \zeta_n^{us_1 \cdots s_r}) \quad (6)$$

and observe we have $us_1 \cdots s_r \in Y_{\llbracket 1, r \rrbracket}$. Then, we see $\phi(c) = 1 - \zeta_n^u$ works, appears with exponent 1 and we may note that the other terms $1 - \zeta_n^v$ that appear in this decomposition all satisfy $v = u$ modulo $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{K})$ so that the decompositions of such elements of $C(\mathbb{K})$ are pairwise disjoint.

Suppose $u = 1$ (this case has to be considered when $r$ is even only). We have the same equation as before and the same observation can be made (because $1 - \zeta_n \in C_{\llbracket 1, r \rrbracket}$ as we supposed $r$ even).

If $2 | n$, we have to do more manipulations to get the decomposition of $N_{\mathbb{Q}(\zeta_n)/\mathbb{K}}(1 - \zeta_n^u)$. First, recall we suppose $p_r = 2$ in this case. We may write $u = u_1 \cdots u_k \in Y_{\llbracket 1, r \rrbracket}(\mathbb{K})$ and then let $u_i = 1$ for all $i > k$ so that $u = u_1 \cdots u_r$. If $u_r s_r \in \mathcal{R}_r$ then we still have $1 - \zeta_n^{us_1 \cdots s_r} \in C$ so that there is nothing to do - in particular, this happens when $s_r = 1$. When $u_r s_r \notin \mathcal{R}_r$, we can show that $1 - \zeta_n^{us_1 \cdots s_r}$ decomposes with terms of $C$ having lower level and terms of the form $1 - \zeta_n^{s_1' \cdots s_{r-1}'}$ with $s_i' \in \text{Gal}(\mathbb{Q}(\zeta_{q_i})/\mathbb{Q}) \setminus \{J_i\}$ and $s_r' = J_r u_r s_r \in \mathcal{R}_r$ as it was explained in the remarks that follow theorem 7. Indeed, if $u_i s_i \neq \{1\}$ for all $i$, then it suffices to use Eq. (1). More generally, we may use Eq. (1) before applying norm relations (Eq. (2)) in a row along all the $\sigma_i$'s that are such that $u_i s_i = 1$. Then, we conclude again that we can let $\phi(c) = 1 - \zeta_n^u$, it appears with exponent 1 but, this time, the decompositions of those $N_{\mathbb{Q}(\zeta_n)/\mathbb{K}}(1 - \zeta_n^u)$ are pairwise disjoint if we consider only the part of those decompositions that lie in $C_{\llbracket 1, r \rrbracket}$. We will still have a triangular matrix at the end since $\phi(c)$ is not involved in the decomposition of the other terms of $C(\mathbb{K})$ as we just observed that if $v = v_1 \cdots v_k$ is involved in the decomposition of $N_{\mathbb{Q}(\zeta_n)/\mathbb{K}}(1 - \zeta_n^u)$ then we have one of the following two cases

i) $v = u$ modulo $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{K})$

ii) $v_r \neq u_r$ and $v_r = u_r$ modulo $\text{Gal}(\mathbb{Q}(\zeta_{q_r})/\mathbb{K}_r^+)$

- again writing $v = v_1 \cdots v_r$ as we did for $u$.
Suppose $[[1, r]]_C \neq \emptyset$ and $[[1, r]]_R \neq \emptyset$. Suppose $u$ is not of the form $u_1 \cdots u_{t-1}$ with $u_i \in \mathcal{R}_{i,1}(K) \setminus \{J_i\}$ (that is $u$ is not a problematic term). We still have the same equation 6 and the same statements can be made. More precisely, we can let $\phi(c) = 1 - \zeta_n^u$ and it appears with exponent 1 in the decomposition of $c$. If $2 \nmid n$ or $K \cap \mathbb{Q}(\zeta_{2^{v_2(n)}})$ is real, then, the decompositions of such $c(K)^u$'s are pairwise disjoint and the terms $1 - \zeta_n^v$ that appear all satisfy $v = u$ modulo $\text{Gal}(\mathbb{Q}(\zeta_n)/K)$. If $2 \mid n$ and $K \cap \mathbb{Q}(\zeta_{2^{v_2(n)}})$ is imaginary, then the parts of the decomposition of those elements that lie in $C_{[[1,r]]}$ are pairwise disjoint and if $1 - \zeta_n^v$ appears then we have one of the following cases:

i) $v = u$ modulo $\text{Gal}(\mathbb{Q}(\zeta_n)/K)$

ii) $v_r \neq u_r$ and $v_r = u_r$ modulo $\text{Gal}(\mathbb{Q}(\zeta_{q_r})/K_r^+)$.

Let us say this last paragraph forms our key fact number one.

Now, suppose $u$ is of the form $u_1 \cdots u_{t-1}$ with $u_i \in \mathcal{R}_{i,1}(K) \setminus \{J_i\}$ for all $i \in [[1, t-1]]$ (this case has to be considered when $|[[1, r]]_C|$ is even only). We may show $\phi(c) = 1 - \zeta_n^u$ still works. Again, we have equation 6. If one of the $s_i$'s is non trivial for some $i \in [[t, r]]$, then we have $1 - \zeta_n^{us_1 \cdots s_r} \in C$ if $2 \nmid n$ - so that the decomposition of $1 - \zeta_n^{us_1 \cdots s_r}$ does not involve $1 - \zeta_n^u$. If $2 \mid n$ and $K \cap \mathbb{Q}(\zeta_{2^{v_2(n)}})$ is imaginary, we may not have $1 - \zeta_n^{us_1 \cdots s_r} \in C$ given $s_i \neq 1$ for some $i \in [[t, r]]$ but, as explained in the remarks that follow theorem 7, we can see $1 - \zeta_n^u$ is not involved in the decomposition of $1 - \zeta_n^{us_1 \cdots s_r}$. Then, we now just have to consider the decomposition of the following product

$$ \prod_{s_1 \in \text{Gal}(\mathbb{Q}(\zeta_{q_1})/K_1)} \cdots \prod_{s_{t-1} \in \text{Gal}(\mathbb{Q}(\zeta_{q_{t-1}})/K_{t-1})} 1 - \zeta_n^{us_1 \cdots s_{t-1}} $$

For now, let $s_1 \in \text{Gal}(\mathbb{Q}(\zeta_{q_1})/K_1), \dots, s_{t-1} \in \text{Gal}(\mathbb{Q}(\zeta_{q_{t-1}})/K_{t-1})$. If $u_{t-1}s_{t-1} \in \mathcal{R}_{t-1}$ (that is if $s_{t-1} \in \mathcal{T}_{t-1}$), then $1 - \zeta_n^{us_1 \cdots s_{t-1}} \in C$. Under this condition, observe $1 - \zeta_n^u$ appears if and only if $s_1 = 1, \dots, s_{t-1} = 1$ and it appears with exponent 1 if this last condition is satisfied. Else (that is if $s_{t-1} \in J_{t-1} \mathcal{T}_{t-1}$), we have to use multiple norm relations in a row as explained in the remarks that follow theorem 7: first, the norm relation along $\sigma_t$ (Eq. (2)) gives

$$ 1 - \zeta_n^{us_1 \cdots s_{t-1}} = \prod_{s_t \in \text{Gal}(\mathbb{Q}(\zeta_{q_t})/\mathbb{Q}) \setminus \{1\}} (1 - \zeta_n^{us_1 \cdots s_{t-1}s_t})^{-1} $$
modulo terms having lower level (note the power $-1$).

If $s_t \in \mathcal{R}_t$, then $1 - \zeta_n^{us_1 \cdots s_{t-1}s_t} \in C$ and there is nothing to do. Else, we have to use the norm relation along $\sigma_{t+1}$ (note this will transform the power $-1$ to $+1$)

$$ (1 - \zeta_n^{us_1 \cdots s_{t-1}s_t})^{-1} = \prod_{s_{t+1} \in \text{Gal}(\mathbb{Q}(\zeta_{q_t})/\mathbb{Q}) \setminus \{1\}} (1 - \zeta_n^{us_1 \cdots s_{t+1}})^{+1} $$

modulo terms having lower level.

If $s_{t+1} \in \mathcal{R}_{t+1}$, then we only have to get rid of $s_t$ if $s_t = J_t$ (see the remarks that follow theorem 7) and we are done with the term associated to $us_1 \cdots s_{t+1}$. Else, we must repeat this process over and over, up to the moment we call the norm relation along $\sigma_r$. At this time, we are led to consider terms of the following form (with exponent $+1$ because we supposed $|\llbracket t, r \rrbracket| = |\llbracket 1, r \rrbracket|_{\mathbb{C}}$ is even)

$$ \prod_{s_r \in \text{Gal}(\mathbb{Q}(\zeta_{q_r})/\mathbb{Q}) \setminus \{1\}} 1 - \zeta_n^{us_1 \cdots s_r} $$

for some $s_t \in \text{Gal}(\mathbb{Q}(\zeta_{q_t})/\mathbb{Q}) \setminus \mathcal{R}_t, \dots, s_{r-1} \in \text{Gal}(\mathbb{Q}(\zeta_{q_{r-1}})/\mathbb{Q}) \setminus \mathcal{R}_{r-1}$. If we have $s_r \in \mathcal{R}_r$, then we have to get rid of the $s_i$'s satisfying $s_i = J_i$ using norm relations along those $\sigma_i$'s. Else, use Eq. (1) to transform $1 - \zeta_n^{us_1 \cdots s_r}$ into the term of $C$ that corresponds to $uJ_1s_1 \cdots J_r s_r$. Observe the term $1 - \zeta_n^u$ does not appear, unless we have $s_1 = J_1, \dots, s_r = J_r$ and it will then appear with exponent $1$ as $|\llbracket t, r \rrbracket| = |\llbracket 1, r \rrbracket|_{\mathbb{C}}$ is even.

Also, note the other terms of the form $1 - \zeta_n^{v_1 \cdots v_{t-1}}$ that appear in the decomposition of $c$ satisfy $v_1 \cdots v_{t-1} = u$ modulo $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{K})$. At the end, we can see $1 - \zeta_n^u$ appears with exponent $2$ in the decomposition of $c$ in the basis of theorem 7 so that $\phi(c) = 1 - \zeta_n^u$ works and this term is not involved in the decomposition of the other elements of $C_{\llbracket 1,r \rrbracket}(\mathbb{K})$. Let us call this paragraph key fact number two.

We are then done with decomposing the elements of $C(\mathbb{K})$. We can now construct the matrix we talked about earlier.
To this aim, define the lexicographic order $\leq_{lexP}$ on the powerset of $\llbracket 1, r \rrbracket$ as follows:

$$ \forall \Omega_1, \Omega_2 \subset \llbracket 1, r \rrbracket, \quad \Omega_1 \leq_{lexP} \Omega_2 \iff \left\{ \begin{array}{l} |\Omega_1| < |\Omega_2| \quad \text{or} \\ |\Omega_1| = |\Omega_2| \quad \text{and } \exists i \in \Omega_1 \setminus \Omega_2 : \\ \qquad \Omega_1 \cap \llbracket 1, i \rrbracket = \Omega_2 \cap \llbracket 1, i \rrbracket. \end{array} \right. $$

Now, compare those elements $u_1 \cdots u_k$ from the $Y_\Omega$'s (or, equivalently, the elements of the $C'_\Omega s$) with the following binary relation:

$$ (\Omega_1, u_1 \cdots u_{k_1}) \leq (\Omega_2, v_1 \cdots v_{k_2}) \iff \left\{ \begin{array}{l} |\Omega_1| >_{lexP} |\Omega_2| \quad \text{or} \\ (\Omega_1 = \Omega_2 \text{ and } k_1 \leq k_2) \end{array} \right. $$

In particular, note this binary relation is not an order. For example, let $\Omega = \llbracket 1, r \rrbracket$ and let $u_1 \cdots u_r, v_1 \cdots v_r \in Y_\Omega(\mathbb{K})$ be two distinct elements (assume $p_r$ is big enough so that this situation actually occurs). We have $(\Omega, u_1 \cdots u_r) \leq (\Omega, v_1 \cdots v_r)$ and $(\Omega, v_1 \cdots v_r) \leq (\Omega, u_1 \cdots u_r)$.

Now, to create the matrix we mentioned before, place the elements $c$ from $C(\mathbb{K})$ (and place them in columns) from left to right by sorting the tuples associated to those $\phi(c)$ in an increasing order (there are many ways to do that but it does not matter). More precisely, we mean that we list all the elements of $C(\mathbb{K})$, say $c_1, \dots, c_N$, so that we have $\phi(c_1) \leq \phi(c_2) \leq \dots \leq \phi(c_N)$ (with $\leq$ denoting the binary relation we just introduced) and for any $i \in \llbracket 1, N \rrbracket$, the $i$-th column will be the vector made of the components of $c_i$ in the basis $C$ (we will order $C$ - that is will order the rows - just after that we are done with the columns).

Then, after those elements from $C(\mathbb{K})$, place the terms of $C \setminus \{\phi(c) : c \in C(\mathbb{K})\}$ from left to right in any order (that is for any $i > N$, the $i$-th column is the vector made of the components of some $c \in C \setminus \{\phi(c) : c \in C(\mathbb{K})\}$ in the basis $C$).

Place the rows from top to bottom, that is place the elements from $C$, according to the same order (say the $i$-th column represents $c \in C(\mathbb{K})$ then the $i$-th row represents $\phi(c)$ and if the $i$-th column represents an element of $C \setminus \{\phi(c) : c \in C(\mathbb{K})\}$ then the $i$-th row also represents this same element). Observe we have a triangular matrix that is just as expected. Now, apply lemma 13 to
conclude. Let us now explain why this matrix is triangular. First, as we ordered the first $N$ columns in an increasing order according to the binary relation we introduced, the matrix we constructed is of the following form

$$ \left( \begin{array}{c|c} M_{\llbracket 1, r \rrbracket} & 0 \\ \hline * & M_{\{r\}} \\ \hline & I \end{array} \right) $$

where $I$ denotes the identity matrix with size $Card(C \setminus \{\varphi(c) : c \in C(\mathbb{K})\})$ which represents the terms of $C \setminus \{\varphi(c) : c \in C(\mathbb{K})\}$. Each matrix $M_{\Omega}$ represents partially the $C_{\Omega}$-part of the decomposition of the elements of $C_{\Omega}(\mathbb{K})$ in the basis $C$. Indeed, the increasing order we chose is so that the first $N$ columns $c_1, \dots, c_N$ are first gathered according to their conductor so that we have blocs $M_{\Omega}$ appearing as we said. The reason why the identity matrix appears is clear. Next, the zeros appear as any term from $C_{\Omega}(\mathbb{K})$ decomposes in $C$ with terms that have lower or equal level to $\Omega$ as explained after theorem 7.

Then, let $\Omega$ be a non empty subset of $\llbracket 1, r \rrbracket$. If $\Omega$ has cardinality 1, then $M_{\Omega}$ is the identity matrix as seen above. From now, suppose $|\Omega| \ge 2$. If $\Omega_{\mathbb{C}} = \emptyset$, then $M_{\Omega}$ is the identity matrix because, as seen above, in this case, the decompositions are pairwise disjoint and $\phi(c)$ appears with exponent 1 for any $c \in C_{\Omega}(\mathbb{K})$. If $\Omega_{\mathbb{R}} = \emptyset$, then $M_{\Omega}$ is the identity matrix because of the paragraphs we made on the case $\Omega_{\mathbb{R}} = \emptyset$. Now, suppose we have $\Omega_{\mathbb{R}} \ne \emptyset$ and $\Omega_{\mathbb{C}} \ne \emptyset$. The matrix $M_{\Omega}$ is the identity matrix if $|\Omega|$ is odd and is of the following form if $|\Omega|$ is even

$$ \left( \begin{array}{c|c} 2I & 0 \\ \hline * & I \end{array} \right) $$

where the upper scaling matrix corresponds to the problematic terms and the lower identity matrix corresponds to the other terms of $C_{\Omega}(\mathbb{K})$. To observe this, first keep in mind that the columns of $M_{\Omega}$ represent all the terms of $C_{\Omega}(\mathbb{K})$, the rows represent their associated tuples and both columns and rows are sorted with the binary relation we introduced. Then, note that the last condition that defines the binary relation that we used makes the problematic terms appear first. Also, the key fact number one explains why $M_{\Omega}$ is the identity matrix when $|\Omega|$ is odd as, in
this case, if we consider the decomposition of any term $c \in C_{\Omega}(K)$, it decomposes with terms that have lower or equal level to $\Omega$ and the terms from $C_{\Omega}$ that appear in this decomposition are not involved in the decomposition of any other term of $C_{\Omega}(K)$ (observe any two terms from $Y_{\Omega}(K)$ are never equal modulo $\text{Gal}(\mathbb{Q}(\zeta_{\Omega})/K_{\Omega}))$.

Now, if $|\Omega|$ is even, the lower identity matrix appears for the same reason and it also explains the zero matrix on the upper right side (that is the decomposition of any non problematic term of $C_{\Omega}(K)$ does not involve any problematic term with level $\Omega$). Next, note the upper scaling matrix appears because of the key fact number two. $\square$

**Remark 24.** Equation (4) from lemma 23 shows Gold and Kim's basis has the cardinality it should have to be a basis.

**Corollary 25.** Suppose $K$ is totally deployed and keep the same notation as in theorem 22. Suppose $K_r$ is imaginary and $K_1, \dots, K_{r-1}$ are real. Then $C(K)$ is a $\mathbb{Z}$-basis of $\text{Was}(K)$ and

$$ \text{Was}(K) = \mathbb{Z}(K) \text{Was}^{+}(K^{+}). $$

*Proof.* Observe the proof of theorem 22 shows that, for any $c \in C(K)$ (resp. $c \in C(K^{+})$), the element $\phi(c)$ appears with exponent 1 in the decomposition of $c$ in this case so that, at the end, the matrix we considered is invertible in $\mathbb{Z}$. Hence our theorem 22 can be stated with no tensor with $\mathbb{Z}[1/2]$ for $K$ (resp. $K^{+}$).

Now, the last part of the corollary results from the fact that we have

$$ \mathbb{Z}(K) \text{Was}^{+}(K) \subset \text{Was}(K) $$

and $\text{Was}^{+}(K)$ is a direct factor of $\text{Was}(\mathbb{Q}(\zeta_n))$ as

$$ \text{Was}(K^{+}) = \text{Was}(K_{1} \cdots K_{r-1} K_{r}^{+}) $$

(see lemma 13). $\square$

**Remark 26.** We can also observe $\text{Was}(\mathbb{Q}(\zeta_n)^{+})$ is not a direct factor of $\text{Was}(\mathbb{Q}(\zeta_n))$ because, if it was a direct factor, we would get

$$ \text{Was}(\mathbb{Q}(\zeta_n)) = \mathbb{Z}(\mathbb{Q}(\zeta_n)) \text{Was}^{+}(\mathbb{Q}(\zeta_n)) $$
by lemma 13 but this equality is not true in general as shows theorem 11.

**Corollary 27.** Suppose $K$ is totally deployed. The quotient group $\mathbf{Was}(K)/\mathbf{Sin}(K)$ is a 2-group.

*Proof.* Indeed, any generator that is (mentioned in the previous theorem 22 and) associated to some $\Omega \subset [1, r]$ such that $|\Omega_C| \ge 1$ and $|\Omega| \ge 2$ is already an element of $\mathbf{Sin}(K_\Omega)$. All the other generators have order 1 or 2 in the quotient group $\mathbf{Was}(K_\Omega)/\mathbf{Sin}(K_\Omega)$ (see [17], equation 11 and corollary 3). Hence, the quotient group $(\mathbf{Was}(K)/\mathbf{Sin}(K)) \otimes \mathbb{Z}[1/2]$ is trivial. $\square$

**Remark 28.** Werl Milàn stated and proved in a special case (see [17] remark 4) this quotient group is an elementary 2-group with rank $[K : \mathbb{Q}] - 1$ if $K_1, \dots, K_r$ are real.

We may also observe if $K_1, \dots, K_r$ are imaginary, we have $\mathbf{Was}(K) = \mathbf{Sin}(K)$ so that the previous theorem gives a $\mathbb{Z}$-basis of $\mathbf{Sin}(K)$ (again, in this case, in the proof of the previous theorem, the term $\phi(c)$ appears with exponent 1 in the decomposition of any $c \in C(K)$). The same $\mathbb{Z}$-basis of $\mathbf{Sin}(K)$ has been given in [10], [6] and the author also proved $\mathbf{Sin}(K) = \mathbf{Was}(K)$ in this same case (see [6], proposition that follows theorem 2).

**Corollary 29.** Suppose $K$ is totally deployed. Let $M(K)$ denote the abelian group generated by $C(K)$ and $\mathbb{Z}(K)$. We have

$$[\mathbf{Was}(K) : M(K)] \le 2^{\alpha}$$

with

$$\alpha = (2^{r-t} - 1) ([K_1 \cdots K_{t-1} : \mathbb{Q}] - 1).$$

*Proof.* Quotient by the roots of unity and keep the same notation for $M(K)$ and $\mathbf{Was}$.

Indeed, through the proof of theorem 22, we see that we have a subgroup $T \le \mathbf{Was}(\mathbb{Q}(\zeta_n))$ such that $T \cap M(K) = \{1\}$ and

$$[\mathbf{Was}(\mathbb{Q}(\zeta_n)) : T \oplus M(K)] = 2^{\alpha}$$
with

$$ \alpha = \sum_{\substack{\Omega \subset \llbracket 1, r \rrbracket \\ |\Omega_C| \in 2\mathbb{N}^* \\ \Omega_{\mathbb{R}} \neq \emptyset}} \prod_{i \in \Omega_{\mathbb{R}}} (d_i - 1). $$

It is easily seen that this definition of $\alpha$ matches the value given in the statement of our corollary 29. Then, we have

$$ [\mathbf{Was}(\mathbb{K}) \oplus T : M(\mathbb{K}) \oplus T] \leq 2^{\alpha}. $$

Now, observe the natural map $\mathbf{Was}(\mathbb{K})/M(\mathbb{K}) \to \mathbf{Was}(\mathbb{K}) \oplus T/M(\mathbb{K}) \oplus T$ is injective. Indeed, let $x \in \mathbf{Was}(\mathbb{K})$ be such that $x = yz$ with $y \in M(\mathbb{K})$ and $z \in T$. As $M(\mathbb{K})$ has finite index in $\mathbf{Was}(\mathbb{K})$, there is $k \in \mathbb{N}$ such that $x^k \in M(\mathbb{K})$, then we have $z^k \in T \cap M(\mathbb{K})$ so that $z^k = 1 = z$ since $T$ is torsion-free and $x = y \in M(\mathbb{K})$. $\square$

**Corollary 30.** Let $Q$ denote the Hasse's unit index of $\mathbb{K}$. Assuming $\mathbb{K}$ is a totally deployed abelian number field with $\mathbb{K}_r$ being imaginary, we have

$$ [\mathbf{E}(\mathbb{K}) : \mathbf{Was}(\mathbb{K})] = h^+(\mathbb{K})Q2^x $$

for some $x \in \mathbb{Z}$ satisfying

$$ -\nu \leq x \leq \alpha - \mu $$

with $\nu$ being the number of integers $i$ such that $\mathbb{K}_i/\mathbb{Q}$ has even degree and $\mu = r-t+1$ being the number of integers $i$ such that $\mathbb{K}_i$ is imaginary.

*Proof.* This results of the previous corollary and the formula Sinnott has given for the index of $\mathbf{Sin}(\mathbb{K})$ in $\mathbf{E}(\mathbb{K})$ (see [14] proposition 4.1, theorem 4.1 and theorem 5.4). $\square$

**Remark 31.** If $\mathbb{K}_r$ is real, then we have $[\mathbf{E}(\mathbb{K}) : \mathbf{Was}(\mathbb{K})] = h(\mathbb{K})$, see [17] remark 4.

**Corollary 32.** Suppose $\mathbb{K} = \mathbb{K}_1 \cdots \mathbb{K}_r$ is totally deployed. Let $A_1, \dots, A_k$ be disjoint subsets of $\llbracket 1, r \rrbracket$. We have a canonical injective map

$$ \prod_{j=1}^{k} \mathbf{E}(\mathbb{K}_{A_j}) / \mathbf{Was}(\mathbb{K}_{A_j}) \otimes \mathbb{Z}[1/2] \hookrightarrow \mathbf{E}(\mathbb{K}) / \mathbf{Was}(\mathbb{K}) \otimes \mathbb{Z}[1/2]. $$
In particular, if we let $h_p^+(K)$ denote the $p$-part of the class number of $K^+$, we have for any odd prime $p$

$$ \prod_{j=1}^{k} h_{p}^{+}(\mathbb{K}_{A_{j}}) \mid h_{p}^{+}(\mathbb{K}). $$

*Proof.* Let $x = x_1 \cdots x_k \in \textbf{Was}_2(\mathbb{K})$ with $x_j \in \mathbf{E}(\mathbb{K}_{A_j}) \otimes \mathbb{Z}[1/2]$. We have to show $x_j \in \textbf{Was}_2(\mathbb{K}_{A_j})$. There is an integer $N$ such that $x_j^N \in \textbf{Was}_2(\mathbb{K}_{A_j})$. Modulo roots of unity of $\mathbb{K}$, we have

$$ x = \prod_{c \in C(\mathbb{K})} c^{x_c} \\ x_j^N = \prod_{c \in C(\mathbb{K}_{A_j})} c^{x_{j,c}} $$

that is the decomposition of $x$ and the $x_j^N$'s in the $\mathbb{Z}[1/2]$-basis we gave in the previous theorem 22. This theorem shows the following module is a direct factor of $\textbf{Was}_2(\mathbb{K})$

$$ \prod_{j=1}^{k} \textbf{Was}_2(\mathbb{K}_{A_j}). $$

Now, we may identify the exponents of $x^N$ so that we get

$$ \forall j \in [1, k], \forall c \in C(\mathbb{K}_{A_j}), \quad N x_c = x_{j,c} $$

then we have

$$ x_j^N = \left( \prod_{c \in C(\mathbb{K}_{A_j})} c^{x_j} \right)^N $$

hence $x_j \in \textbf{Was}_2(\mathbb{K}_{A_j})$.

<figure><img src="image_2.png" /></figure>

It turns out we can prove this last result on class numbers through class field theory. We have found no reference related to the proof of this next proposition so far and that is why we prove it next.
**Proposition 33.** Suppose

$$K = K_1 \cdots K_r$$

with $K_i \subset \mathbb{Q}(\zeta_{q_i})$. Let $A_1, \dots, A_k$ be a partition of $\lfloor 1; r \rfloor$. For any odd prime number $p$, we have

$$\prod_{j=1}^{k} h_{p}^{+}(K_{A_{j}}) \mid h_{p}^{+}(K).$$

*Proof.* For any number field $\mathbb{L}$, let $\mathbf{H}_{\mathbb{L}}$ denote the Hilbert class field of $\mathbb{L}$. Let $A = A_2 \cup \dots \cup A_k$. By restriction, we have

$$\mathrm{Gal}(\mathbf{H}_{\mathbb{K}^{+}}/\mathbb{K}^{+}) \to \mathrm{Gal}(\mathbb{K}^{+}\mathbf{H}_{\mathbb{K}_{A_1}^{+}}\mathbf{H}_{\mathbb{K}_{A}^{+}}/\mathbb{K}^{+}).$$

For any finite abelian group $G$ let $G_2$ denote the product of the $p$-Sylow's of $G$ for $p$ running over the set of odd prime numbers. Then, we get

$$\mathrm{Gal}(\mathbf{H}_{\mathbb{K}^{+}}/\mathbb{K}^{+})_2 \to \mathrm{Gal}(\mathbb{K}^{+}\mathbf{H}_{\mathbb{K}_{A_1}^{+}}\mathbf{H}_{\mathbb{K}_{A}^{+}}/\mathbb{K}^{+})_2$$

and note that this last group is also $\mathrm{Gal}(\mathbf{H}_{\mathbb{K}_{A_1}^{+}}\mathbf{H}_{\mathbb{K}_{A}^{+}}/\mathbf{H}_{\mathbb{K}_{A_1}^{+}}\mathbf{H}_{\mathbb{K}_{A}^{+}} \cap \mathbb{K}^{+})_2$. Now, observe that we have

$$\mathbb{K}_{A_1}^{+} \mathbb{K}_{A}^{+} \subset \mathbf{H}_{\mathbb{K}_{A_1}^{+}} \mathbf{H}_{\mathbb{K}_{A}^{+}} \cap \mathbb{K}^{+} \subset \mathbb{K}_{A_1 \cup A}^{+}$$

and these extensions have at most degree 2 so that we get

$$|\mathrm{Gal}(\mathbf{H}_{\mathbb{K}_{A_1}^{+}}\mathbf{H}_{\mathbb{K}_{A}^{+}}/\mathbb{K}_{A_1}^{+}\mathbb{K}_{A}^{+})_2| \mid |\mathrm{Gal}(\mathbf{H}_{\mathbb{K}^{+}}/\mathbb{K}^{+})_2|.$$

To conclude, observe that we have $\mathbf{H}_{\mathbb{K}_{A_1}^{+}} \cap \mathbf{H}_{\mathbb{K}_{A}^{+}} = \mathbb{Q}$. Indeed, each prime number ramifies in $\mathbf{H}_{\mathbb{K}_{A_1}}$ (resp. $\mathbf{H}_{\mathbb{K}_{A}}$) if and only if it ramifies in $\mathbb{K}_{A_1}$ (resp. $\mathbf{H}_{\mathbb{K}_{A}}$) and $\mathbb{Q}$ has no unramified extension (see [11] theorem 2.18). Hence, we have

$$|\mathrm{Gal}(\mathbf{H}_{\mathbb{K}_{A_1}^{+}}\mathbf{H}_{\mathbb{K}_{A}^{+}}/\mathbb{K}_{A_1}^{+}\mathbb{K}_{A}^{+})| = |\mathrm{Gal}(\mathbf{H}_{\mathbb{K}_{A_1}^{+}}/\mathbb{K}_{A_1}^{+})| |\mathrm{Gal}(\mathbf{H}_{\mathbb{K}_{A}^{+}}/\mathbb{K}_{A}^{+})|$$

which allows us to make the same procedure on $\mathbb{K}_A$ so that we get our result by induction on $r$. $\square$

Next lemma 34 allows us to state corollary 35 as an equivalent of corollary 30 with $\mathbb{K}^{+}$ replacing $\mathbb{K}$.
**Lemma 34.** Let $K$ be an abelian number field. We have

$$[\mathbf{E}^{+}(K) : \mathbf{Was}^{+}(K)] = [\mathbf{E}(K) : \mathbf{Was}(K)] \times 2^{\epsilon}$$

with $\epsilon \in \{0, -1\}$ and

$$2^{\epsilon} = \frac{[\mathbf{Was}(K)\mathbf{E}^{+}(K) : \mathbf{Z}(K)\mathbf{E}^{+}(K)]}{[\mathbf{E}(K) : \mathbf{Z}(K)\mathbf{E}^{+}(K)]}.$$

*Proof.* As the index is multiplicative, we get these two equalities (independently)

$$ \begin{aligned} [\mathbf{E}(K) : \mathbf{Was}^{+}(K)] &= [\mathbf{E}(K) : \mathbf{Was}(K)][\mathbf{Was}(K) : \mathbf{Z}(K) \mathbf{Was}^{+}(K)] \\ & \quad [\mathbf{Z}(K) \mathbf{Was}^{+}(K) : \mathbf{Was}^{+}(K)] \\ &= [\mathbf{E}(K) : \mathbf{Z}(K)\mathbf{E}^{+}(K)][\mathbf{Z}(K)\mathbf{E}^{+}(K) : \mathbf{E}^{+}(K)] \\ & \quad [\mathbf{E}^{+}(K) : \mathbf{Was}^{+}(K)]. \end{aligned} $$

It remains to see the second isomorphism theorem gives

$$[\mathbf{Z}(K)\mathbf{E}^{+}(K) : \mathbf{E}^{+}(K)] = \frac{|\mathbf{Z}(K)|}{2} = [\mathbf{Z}(K) \mathbf{Was}^{+}(K) : \mathbf{Was}^{+}(K)]$$

and

$$\mathbf{Was}(K)\mathbf{E}^{+}(K)/\mathbf{Z}(K)\mathbf{E}^{+}(K) \simeq \mathbf{Was}(K)/\mathbf{Z}(K)\mathbf{Was}^{+}(K).$$

□

**Corollary 35.** Assuming $K$ is a totally deployed number field such that $K_r$ is imaginary, we have

$$[\mathbf{E}(K^{+}) : \mathbf{Was}(K^{+})] = h^{+}y2^{x}$$

for some $x \in \mathbb{Z}$ satisfying

$$-\nu \le x \le \alpha - \mu$$

and

$$y = [\mathbf{Was}(K)\mathbf{E}^{+}(K) : \mathbf{Z}(K)\mathbf{E}^{+}(K)].$$

Moreover, if $n$ is odd, we have

$$y = \begin{cases} 2 & \text{if } |\mathbb{[}1, r\mathbb{]}_{\mathbb{C}}| \ge 2 \\ 1 & \text{if } |\mathbb{[}1, r\mathbb{]}_{\mathbb{C}}| = 1 \end{cases}$$
*Proof.* The first formula results from corollary 30 and the previous lemma 34. The value of $y$ is given by the following observation. If $||[1, r]_c| < 2$, then we have $\mathbf{Was}(K) = \mathbf{Was}(K^+)$ (see corollary 25) so that $y = 1$. If $||[1, r]_c| \ge 2$, note that we have $y \le 2$ because of proposition 1. Then

$$N_{\mathbb{Q}(\zeta_{q_{r-1}q_r})/\mathbb{K}_{r-1}\mathbb{K}_r}(1 - \zeta_{q_{r-1}q_r}) \in \mathbf{Was}(K)\mathbf{E}^+(K)/\mathbf{Z}(K)\mathbf{E}^+(K)$$

has order 2 as a consequence of Eq. (3). □

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