# Zero Shot Audio to Audio Emotion Transfer With Speaker Disentanglement

###### Abstract

The problem of audio-to-audio (A2A) style transfer involves replacing the
style features of the source audio with those from the target audio while
preserving the content related attributes of the source audio. In this paper,
we propose an efficient approach, termed as Zero-shot Emotion Style Transfer
(ZEST), that allows the transfer of emotional content present in the given
source audio with the one embedded in the target audio while retaining the
speaker and speech content from the source. The proposed system builds upon
decomposing speech into semantic tokens, speaker representations and emotion
embeddings. Using these factors, we propose a framework to reconstruct the
pitch contour of the given speech signal and train a decoder that reconstructs
the speech signal. The model is trained using a self-supervision based
reconstruction loss. During conversion, the emotion embedding is alone derived
from the target audio, while rest of the factors are derived from the source
audio. In our experiments, we show that, even without using parallel training
data or labels from the source or target audio, we illustrate zero shot
emotion transfer capabilities of the proposed ZEST model using objective and
subjective quality evaluations.

Index Terms: Speech Emotion modeling, Style Transfer, Disentangled
Representation Learning.

##  1 Introduction

Artificial emotional intelligence [1] encompasses methods that enable machines
to understand and interact with human expressions of emotions. The style
transfer approach to manipulating emotion, given a source and target data
sample, is the task of converting emotion in the source sample to match the
emotional style of the target sample while retaining rest of the attributes of
the source. While the task has shown promising results in image domain [2],
the applications in audio domain is more challenging [3, 4]. In this paper, we
explore the task of emotion style transfer in speech data.

Voice conversion of speech primarily explored converting the speaker identity
of a voice [5]. However, speech also contains information about the underlying
emotional trait of the speaker in varying levels [6]. The initial frameworks
using Gaussian mixture model (GMM) [7], hidden Markov model [8] and deep
learning [9] based conversion approaches have recently been advanced with
generative adversarial networks (GAN) [10] and sequence-to-sequence auto
encoding models [11].

In many of the prior emotion conversion approaches, the emotion targets are
treated as discrete labels. However, emotion is a fine-grained attribute which
has varying levels of granularities [6]. Forcing the emotion attribute to a
small number of discrete target labels may not allow the models to capture the
wide range of diverse and heterogeneous sentiments elicited in human speech.
Hence, we argue that the most natural form of emotion conversion is to
transfer the emotion expressed in a target audio to the source audio, a.k.a
A2A emotion style transfer. This motivation is also echoed in a recent work on
A2A style transfer [12]. In spite of these efforts, audio-to-audio (A2A) style
transfer in zero shot setting (unseen speakers and emotion classes) is
challenging.

On a separate front, representation learning of speech has shown remarkable
progress in the recent years. The wav2vec [13] models have been improved with
masked language modeling (MLM) objectives (for example, HuBERT [14]) in self-
supervised learning (SSL) settings. The derivation of speaker representations
have mostly been pursued with a supervised model [15]. Further, reconstructing
speech from factored representations of speaker, content and pitch contours
[16] has shown that models like Tacotron [17], AudioLM [18] and HiFi-GAN [19]
allow good quality speech generation.

In this paper, we propose a framework called, zero shot emotion transfer -
ZEST, which leverages the advances made in representation learning and speech
reconstruction. The proposed framework decomposes the given audio into
semantic tokens (using HuBERT model [14]), speaker representations (x-vectors
[15]), and emotion embeddings (derived from a pre-trained emotion classifier).
Inspired by speaker disentangling proposed for speech synthesis [20], we also
perform a single step of speaker and emotion disentanglement in the
embeddings. Since pitch (F0) is also a component that embeds content, speaker
and emotion, we investigate a cross-attention based model for predicting the
F0 contour of a given utterance. Using the three representations (speech,
speaker and emotion) along with the predicted F0 contours, the proposed ZEST
framework utilizes the HiFi-GAN [19] decoder model for reconstructing the
speech. During emotion conversion, the proposed ZEST approach does not use
text or parallel training and simply imports the emotion embedding from the
target audio for style transfer.

The experiments are performed on emotion speech dataset (ESD) [21]. We also
explore a zero shot setting, where an unseen emotion from a different dataset
is used as the reference audio. Further, a setting where the source speech is
derived from an unseen speaker is also investigated. We perform several
objective and subjective quality evaluations and compare with benchmark
methods to highlight the style transfer capability of the proposed framework.
The key contributions from this work can be summarized as follows,

  * •

Proposing a novel framework for predicting the pitch contour of a given audio
file using the semantic tokens from HuBERT, speaker embeddings and emotion
embeddings.

  * •

Enabling speaker-emotion disentanglement using adversarial training.

  * •

Illustrating zero shot emotion transfer capabilities from unseen emotion
categories, novel speakers and content.

##  2 Related Prior Works

A2A EST Using World Vocoder: One of the earliest attempts for EST involved
using the world vocoder, as proposed by Gao et al. [22]. This work used the
statistics of F0 and spectral components from the target speaker before
reconstruction using the decoder. Our work uses recent advances in speaker,
emotion and content embeddings for emotion style transfer. Further, we aim to
transfer the emotion style from the reference speech to the source speech
signal rather than just modifying the emotion category.

Expressive text-to-speech synthesis: The work by Li et al. [20] explored using
speaker disentanglement for generating emotional speech from text. Similarly,
Emovox proposed by Zhou et al. [6] used phonetic transcription for emotional
voice conversion. However, our work explores EST without using any linguistic
or phonetic transcriptions of the source or target speech.

Non-parallel and unseen emotion conversion: Recent work by Chen et al. [12]
explored using attention models for performing EST. However, this work forced
the source and target speech to be from the same speaker, limiting the utility
of the EST applications.

##  3 Method

Fig. 1: Training of the F0 contour predictor. EASE - Emotion Agnostic Speaker
Encoder. SACE - Speaker Adversarial Classifier of Emotions. The blue blocks
are kept frozen during training.

###  3.1 Content encoder

The content encoder used in the proposed framework is the HuBERT SSL model
[14]. The HuBERT model gives continuous valued vector representations for each
speech segment, which is subsequently converted into discrete tokens with a
k-means clustering.

###  3.2 Emotion Agnostic Speaker Encoder (EASE)

The speaker embeddings for each audio file are extracted using a enhanced
channel attention-time delay neural network (ECAPA-TDNN) [15] model. This
model is pre-trained on $2794$ hours and $7363$ speakers from the VoxCeleb
dataset [23], for the task of speaker classification. The model involves an
utterance level pooling of the frame-level embeddings, called x-vectors. The
x-vectors have been shown to encode emotion information [24, 25]. In order to
suppress the emotion information, inspired by the disentanglement approach
proposed in Li et al. [20], we add two fully connected layers to the x-vector
model and further train the model with an emotion adversarial loss [26]. We
refer to these vectors as the Emotion Agnostic Speaker Encoder (EASE) vectors.
The loss function is given by

$\mathcal{L}_{tot-
spkr}=\mathcal{L}_{ce}^{spkr}-\lambda_{adv}^{emo}\mathcal{L}_{ce}^{emo}$ (1)

###  3.3 Speaker Adversarial Classifier of Emotions (SACE)

A speaker adversarial classifier of emotions (SACE) classifier is designed
based on the wav2vec2.0 representations [13], similar to the one proposed by
Pepino et al. [27]. The wav2vec model, pre-trained on $300$ hours ($543$
speakers) of switchboard corpus [28], is used for extracting features from the
raw speech signal [29]. The convolutional feature extractors are kept frozen
while the transformer layers along with two position wise feed forward layers
are trained for the task of emotion recognition on the Emotional Speech
Dataset (ESD) [21]. The model is trained with speaker adversarial loss (the
emotion classifier equivalent of Eq. 1). The representations averaged over the
entire utterance are used as the emotion embedding.

###  3.4 Pitch Contour Predictor

The framework for the pitch (F0) predictor is shown in Figure 1. The HuBERT
tokens for the speech signal are converted to an sequence of vectors by means
of an embedding layer. This sequence, denoted by
$\mathbf{\mathcal{Q}}=\\{q_{1},..,q_{T}\\}_{t=1}^{T}$, is used as the query
sequence for cross-attention. The frame-level SACE embeddings are added with
speaker embedding (EASE) to form the key-value pair for the cross attention
module [30]. This is followed by a 1D-CNN network to predict the F0 contour.
The target pitch contour is the one derived using the YAAPT algorithm [31]. We
use the $\mathcal{L}_{1}$ loss between the predicted and target F0 contour.

Fig. 2: (a) During training, ZEST is learned to reconstruct the speech signal.
(b) During emotion conversion, the components that are derived from target
speech are coded in orange color. The yellow blocks are the learnable parts of
the model using an auto-encoding objective, while the blue blocks indicate
frozen components.

###  3.5 Speech reconstruction

The speech reconstruction framework is shown in Fig. 2(a). For reconstructing
the speech signal, the HuBERT tokens, SACE embedding, EASE vector and the
predicted F0 contour are used. The HuBERT tokens are converted to a sequence
of real-valued vectors with a learnable embedding layer. In order to add
contextual information during the speech reconstruction phase, the tokens and
$F0_{pred}$ are passed through two separate networks consisting of CNN layers
and bidirectional long-short term memory (BLSTM) layers. Finally, all the
components are passed through a HiFi-GAN vocoder [19] to reconstruct the
speech signal. More details of the HiFi GAN model are provided in Polyak et
al. [16].

###  3.6 Emotion conversion

The ZEST framework for emotion conversion is shown in Figure 2(b). The HuBERT
tokens and the speaker vector are extracted from the source speech while the
emotion embeddings are derived from the target speech. The emotion embedding
sequence, being extracted from the reference speech, may differ in length from
the source speech. However, as the query sequence in the cross-attention
(Figure 1) is driven by the HuBERT tokens of the source signal, the F0 contour
generated during conversion will match the length of the source signal. The
HuBERT tokens, speaker vector, predicted F0 contour and the emotion embedding
are then used to generate the converted speech through the pre-trained HiFi-
GAN model. The conversion phase does not involve any model training steps.

##  4 Experiments and Results

###  4.1 Datasets and Pretraining

The HuBERT model is the pre-trained $12$-layer base model described in Hsu et
al. [14]. This model is pre-trained on $960$ hours of Librispeech dataset. The
speaker encoder is initialized with the ECAPA-TDNN model [15] pre-trained on
the VoxCeleb dataset. Further, the emotion agnostic training of the EASE model
is performed on the Emotional Speech Database (ESD) [21]. We also train the F0
predictor as well as the HiFiGAN based reconstruction module on this dataset.
We use the gender balanced English partition of this dataset in our
experiments with $10$ speakers. The training, validation and test splits are
used as suggested in the dataset where, $300$ utterances from each of the $10$
speakers, for the five labelled emotions - neutral, angry, happy, sad and
surprise, are used for training. Thus, the training data consists of $15000$
utterances, while $2500$ ($50$ utterances per speaker per emotion) unseen
utterances are used for validation and testing.

We further evaluate the ZEST method on unseen emotions and speakers. For
unseen emotion targets, we use $2542$ utterances from the CREMA-D dataset [32]
belonging to two new emotion classes (fear and disgust). For unseen source
speakers, we investigate the use of the TIMIT database [33]. We choose $100$
utterances distributed across $10$ speakers selected at random from the TIMIT
dataset for these experiments. Note that, the TIMIT and the CREMA-D datasets
are only used in the conversion setting (Figure 2(b)). These evaluations
reflect the zero-shot capability of the model.

###  4.2 Implementation

The HuBERT representations for the encoder are taken from $9$-th layer of the
pre-trained model and clustered with a k-means algorithm ($K=100$) [16]. The
EASE model is trained for a total of $10$ epochs with $\lambda_{adv}^{emo}$
set to $10$ and batch size set to $32$. The SACE setup is trained with a
speaker adversarial loss ($\lambda=1000$) along with the pitch predictor
(Section 3.4) with a batch size of $24$ and a learning rate of $1e-4$, while
the speech reconstruction model is trained with a batch size of $32$ and a
learning rate of $2e-4$. The pitch predictor is trained for $50$ epochs and
the HiFi-GAN model is trained for $100K$ steps. The values of
$\lambda_{adv}^{emo}$ and $\lambda$ are set based on the validation set
performance of EASE and the pitch predictor respectively. Our code, converted
audio samples and further ablations are available at this
link111https://github.com/iiscleap/ZEST.

We use the VAWGAN model proposed by Zhou et al. [34] as the baseline setup for
bench-marking the proposed ZEST framework. As described in the paper, we
supply the target emotion classes instead of the reference audio file for
conversion. We also explore the reconstruction model proposed in Polyak et al.
[16] for benchmarking. Here, the F0 contour is derived from the target audio,
while the speech and speaker contents are derived from the source audio.

###  4.3 Evaluation settings

  * •

Same-Speaker-Same-Text (SSST): The source and reference speech are from the
same speaker and with the same textual content. Further, the source speech is
always in neutral emotion, while the reference speech can be from any other
emotional category. This test set has a total of $1200$ samples ($30$ neutral
samples from the $10$ speakers to be converted to $4$ other emotion
categories)

  * •

Same-Speaker-Different-Text (SSDT): The source and reference speech are from
the same speaker but with different textual content. This test set is created
by randomly choosing $10$ neutral utterances (one per speaker) as the source
speech signals, and the remaining $29$ test set utterances for conversion into
each of the four target emotions. This leads to a total of $1160$ test
utterances.

  * •

Different-Speaker-Same-Text (DSST): The source and reference speech are
derived from different speakers using the same textual content. Each source
speech signal (in neutral emotion) has $36$ reference signals (the other $9$
speakers having the same utterance in each of the $4$ emotions). As each
speaker has $30$ test utterances, this test set has a total of $10800$
utterances.

  * •

Different-Speaker-Different-Text (DSDT): This is the most generic setting,
where the source and reference speech do not share speaker or textual content.
In this case, $10$ neutral utterances are chosen as the source speech signal
set (one per speaker). For each of these utterances, the remaining $29$
utterances are selected as the reference. This results in a test set of
$10400$ samples.

  * •

Unseen Target Emotions (UTE): The $2542$ utterances from CREMA-D are
considered as target speech signals for each of the $10$ randomly chosen
neutral source utterances from ESD dataset, resulting in a total of $25420$
evaluation utterances.

  * •

Unseen Source Speakers (USS): In this setting, the $100$ utterances from TIMIT
are chosen to be the source speech signals and $8$ randomly chosen emotional
utterances from ESD dataset ($2$ per emotion class barring neutral) are used
as the reference speech signals. Each of the $8$ utterances from ESD is from a
different speaker. This results in a total of $800$ evaluation utterances.

Table 1: Objective evaluation results (%). Here, Emo. Acc.- Emotional
Accuracy, CER - Character Error Rate, Spk. Acc. - Speaker accuracy. The
different test settings are described in Sec. 4.3.

| Method | SSST | SSDT | DSST | DSDT | UTE | USS
---|---|---|---|---|---|---|---
CER $\downarrow$ |  VAWGAN [34] | $7.9$ | $6.6$ | $7.9$ | $6.6$ | - | -
Polyak [16] | $\mathbf{7.0}$ | $5.7$ | $\mathbf{7.0}$ | $5.7$ | $6.1$ | $\mathbf{14.6}$
ZEST-no-adv. | $7.9$ | $\mathbf{5.6}$ | $7.6$ | $5.7$ | $6.2$ | $14.9$
ZEST-no-F0-pred. | $8.6$ | $6.3$ | $8.5$ | $6.3$ | $7.1$ | $14.9$
ZEST | $7.2$ | $\mathbf{5.6}$ | $7.2$ | $\mathbf{5.4}$ | $\mathbf{5.9}$ | $14.8$
Emo. Acc. $\uparrow$ |  VAWGAN [34] | $41.4$ | $44.5$ | $41.4$ | $44.5$ | - | -
Polyak [16] | $36.4$ | $33.3$ | $28.3$ | $25.9$ | - | $26.6$
ZEST-no-adv. | $56.1$ | $42.4$ | $46.8$ | $38.6$ | - | $36.4$
ZEST-no-F0-pred. | $55.8$ | $39.9$ | $45.3$ | $36.6$ | - | $28.4$
ZEST | $\mathbf{69.0}$ | $\mathbf{63.8}$ | $\mathbf{60.0}$ | $\mathbf{55.6}$ | - | $\mathbf{72.5}$
Spk. Acc. $\uparrow$ |  VAWGAN [34] | $62.4$ | $76.0$ | $62.4$ | $76.0$ | - | -
Polyak [16] | $\mathbf{99.6}$ | $\mathbf{100}$ | $\mathbf{99.7}$ | $\mathbf{100}$ | $\mathbf{99.1}$ | -
ZEST-no-adv. | $98.3$ | $\mathbf{100}$ | $97.5$ | $99.8$ | $98.7$ | -
ZEST-no-F0-pred. | $96.8$ | $99.4$ | $93.4$ | $97.9$ | $97.6$ | -
ZEST | $99.4$ | $99.8$ | $97.8$ | $99.7$ | $98.6$ | -

####  4.3.1 Objective evaluation

We use three objective metrics on the converted speech signals.

  * •

Emotion conversion accuracy: We evaluate the target emotion class accuracy on
the converted audio using the trained emotion classifier (Sec. 3.3)

  * •

Textual content preservation: The converted speech is passed through an
automatic speech recognition (ASR) system. This ASR system used the HuBERT
encoder as a pre-training module [14] and was trained on the supervised $960$h
audio in the Librispeech [35] dataset using the CTC loss. The character error
rate (CER) is computed for the converted audio with respect to the ground
truth transcription of the source speech signal.

  * •

Speaker preservation: We train a speaker classification model using the source
speech and the ECAPA-TDNN embeddings to classify among the $10$ training
speakers. A two-layer feed-forward model is employed as backend for this task.
The speaker recognition accuracy of the converted audio, with the source
speaker as target, is used as measure of the speaker preservation property.

The results for these objective tests are shown in Table 1. The following are
the insights drawn from the objective evaluations.

  * •

The emotion transfer accuracy for ZEST is seen to be signficantly improved
over the baseline systems compared here. Even in challenging conditions, like
different speakers and different text content present in the target audio, and
for unseen source speakers, the ZEST is seen to perform emotion transfer
effectively.

  * •

The CER results show that both the baseline system (Polyak et al. [16] and the
ZEST provide similar ASR performance for all the conditions. Even on the
unseen target emotions (UTE), the ZEST is seen to preserve the CER. On the USS
condition, both models show a degradation in the CER, potentially due to the
mismatch in the training/test domain of HiFi-GAN model.

  * •

The speaker accuracy of the VAWGAN is found to be inferior (SSST setting). The
baseline model of Polyak et al. [16] allows near perfect speaker
classification on all settings, while the proposed ZEST also matches this
performance on all settings except the DSST. However, the model based on
Polyak et al. [16] is not seen to be effective in transfer of emotions.

  * •

We show two ablations of ZEST in Table 1, where either the emotion adversarial
loss in the EASE module or the pitch predictor module is absent. The
adversarial training allows disentangled representations of speaker and
emotion, which is shown to improve the objective quality results. Further, the
utility of EASE is seen from the improvement in the emotion accuracy across
all the $6$ test settings. The pitch prediction module also aids the system to
achieve a better ASR performance.

Fig. 3: Subjective evaluation on the different test settings. Abbreviations
used: MOS- Mean Opinion Score, SMOS - Similarity Mean Opinion Score. The
definition of the different test settings is given in Sec. 4.3. ★ indicates
that the difference in scores from the SSST test setting is statistically
significant (p $<$ 0.05)

###  4.4 Subjective tests

We conduct listening tests in order to judge the effectiveness of our method.
We used Prolific222https://www.prolific.co tool to setup the subjective
listening tests. We recruited $20$ participants to perform the subjective
evaluation. We choose $52$ recordings, with $8$ recordings from each of the
$4$ test settings (SSST, SSDT, DSST, DSDT) and $10$ recordings each from UTE,
and USS settings. The recordings were presented in a random order. The
participants were also provided with training examples to illustrate the
objective of the test.

All the participants in the survey were asked to give their opinion score on
the audio files (range of $1$-$5$) based on three criteria - i) Emotion
transfer between the converted and the reference signal, ii) Quality of the
converted speech and iii) Speaker similarity between the converted and the
source signal. The subjective evaluation results (in terms of mean opinion
score (MOS)) are reported in Figure 3. As seen in these results, the emotion
transfer and reconstruction speech quality MOS values are the best for SSST
condition. This is expected as the transfer of emotion from a reference speech
of the same speaker with same textual content is the easiest test setting
among all. However, the MOS results for the other challenging conditions also
compare well with the SSST setting (except the speaker MOS on the unseen
source speaker (USS) condition).

We also measure a statistical significance (unpaired t-test) between the
scores obtained for the SSST test setting with each of the other $5$
conditions. The emotion MOS for SSST is statistically significantly higher
than all the other conditions. However on the speech quality, all the
conditions, except UTE and DSDT, generate statistically similar results
compared to SSST setting. For the speaker MOS scores, the unseen emotion
targets and unseen speaker sources generate statistically different results
compared to SSST. Part of the reason for this behavior may be attributed to
the limited ($10$ speakers with same speech content) training employed for the
reconstruction model (HiFi-GAN). In future, we plan to train the
reconstruction model on a larger resource of neutral speech along with the
emotional speech to improve the generalization to novel speakers. Further,
leveraging larger emotional speech datasets for training the emotion embedding
extractor and the F0 predictor may improve the quality of the emotion transfer
to unseen emotions.

##  5 Summary

We have presented an approach for zero shot emotion style transfer (ZEST) for
audio-to-audio emotion conversion. The proposed ZEST method leverages pre-
trained representations of speech content, speaker embeddings and emotion
embeddings. Further, a pitch predictor model is designed in a self-supervised
setting in order to learn the mapping from the representations to the pitch
contour. A reconstruction module based on the HiFi-GAN allows the re-
composition of the factored representations to generate the speech signal. For
emotion conversion, we only derive the emotion embeddings from the target
speech and perform the reconstruction. Various objective quality evaluation
experiments with different levels of source/target mis-match, transfer to
unseen emotions and from unseen source speakers elicit the zero shot emotion
transfer capability of the proposed model. The subjective listening tests also
validate the benefits of the proposed setting.

## References

  * [1] Dagmar Schuller and Björn W Schuller,  “The age of artificial emotional intelligence,”  Computer, vol. 51, no. 9, pp. 38–46, 2018.
  * [2] Yongcheng Jing et al.,  “Neural style transfer: A review,”  IEEE transactions on visualization and computer graphics, vol. 26, no. 11, pp. 3365–3385, 2019.
  * [3] Kaizhi Qian et al.,  “Autovc: Zero-shot voice style transfer with only autoencoder loss,”  in ICML. PMLR, 2019, pp. 5210–5219.
  * [4] Shrutina Agarwal, Sriram Ganapathy, and Naoya Takahashi,  “Leveraging symmetrical convolutional transformer networks for speech to singing voice style transfer,”  arXiv preprint arXiv:2208.12410, 2022.
  * [5] Berrak Sisman et al.,  “An overview of voice conversion and its challenges: From statistical modeling to deep learning,”  IEEE/ACM Transactions on Audio, Speech, and Language Processing, vol. 29, pp. 132–157, 2020.
  * [6] Kun Zhou et al.,  “Emotion intensity and its control for emotional voice conversion,”  IEEE Transactions on Affective Computing, vol. 14, no. 1, pp. 31–48, 2022.
  * [7] Ryo Aihara et al.,  “GMM-based emotional voice conversion using spectrum and prosody features,”  American Journal of Signal Processing, vol. 2, no. 5, pp. 134–138, 2012.
  * [8] Zeynep Inanoglu and Steve Young,  “Data-driven emotion conversion in spoken english,”  Speech Communication, vol. 51, no. 3, pp. 268–283, 2009.
  * [9] Huaiping Ming et al.,  “Deep Bidirectional LSTM Modeling of Timbre and Prosody for Emotional Voice Conversion,”  in INTERSPEECH 2016, 2016, pp. 2453–2457.
  * [10] Georgios Rizos et al.,  “StarGAN for emotional speech conversion: Validated by data augmentation of end-to-end emotion recognition,”  in ICASSP. IEEE, 2020, pp. 3502–3506.
  * [11] Ravi Shankar et al.,  “Multi-speaker emotion conversion via latent variable regularization and a chained encoder-decoder-predictor network,”  arXiv preprint arXiv:2007.12937, 2020.
  * [12] Yun Chen et al.,  “Attention-based Interactive Disentangling Network for Instance-level Emotional Voice Conversion,”  in INTERSPEECH 2023, 2023, pp. 2068–2072.
  * [13] Alexei Baevski et al.,  “wav2vec 2.0: A framework for self-supervised learning of speech representations,”  NeurIPS, vol. 33, pp. 12449–12460, 2020.
  * [14] Wei-Ning Hsu et al.,  “HuBERT: Self-supervised speech representation learning by masked prediction of hidden units,”  IEEE/ACM Transactions on Audio, Speech, and Language Processing, vol. 29, pp. 3451–3460, 2021.
  * [15] Brecht Desplanques, Jenthe Thienpondt, and Kris Demuynck,  “Ecapa-tdnn: Emphasized channel attention, propagation and aggregation in tdnn based speaker verification,”  arXiv preprint arXiv:2005.07143, 2020.
  * [16] Adam Polyak et al.,  “Speech resynthesis from discrete disentangled self-supervised representations,”  arXiv preprint arXiv:2104.00355, 2021.
  * [17] Jonathan Shen et al.,  “Natural tts synthesis by conditioning wavenet on mel spectrogram predictions,”  in ICASSP. IEEE, 2018, pp. 4779–4783.
  * [18] Zalán Borsos et al.,  “Audiolm: a language modeling approach to audio generation,”  IEEE/ACM Transactions on Audio, Speech, and Language Processing, 2023.
  * [19] Jungil Kong, Jaehyeon Kim, and Jaekyoung Bae,  “Hifi-gan: Generative adversarial networks for efficient and high fidelity speech synthesis,”  NeurIPS, vol. 33, pp. 17022–17033, 2020.
  * [20] Tao Li et al.,  “Cross-speaker emotion disentangling and transfer for end-to-end speech synthesis,”  IEEE/ACM Transactions on Audio, Speech, and Language Processing, vol. 30, pp. 1448–1460, 2022.
  * [21] Kun Zhou et al.,  “Seen and unseen emotional style transfer for voice conversion with a new emotional speech dataset,”  in ICASSP. IEEE, 2021, pp. 920–924.
  * [22] Jian Gao et al.,  “Nonparallel emotional speech conversion,”  arXiv preprint arXiv:1811.01174, 2018.
  * [23] Arsha Nagrani et al.,  “Voxceleb: Large-scale speaker verification in the wild,”  Computer Speech & Language, vol. 60, pp. 101027, 2020.
  * [24] Raghavendra Pappagari et al.,  “x-vectors meet emotions: A study on dependencies between emotion and speaker recognition,”  in ICASSP. IEEE, 2020, pp. 7169–7173.
  * [25] Zein Shaheen et al.,  “Exploiting Emotion Information in Speaker Embeddings for Expressive Text-to-Speech,”  in INTERSPEECH 2023, 2023, pp. 2038–2042.
  * [26] Yaroslav Ganin et al.,  “Domain-adversarial training of neural networks,”  The journal of machine learning research, vol. 17, no. 1, pp. 2096–2030, 2016.
  * [27] Leonardo Pepino, Pablo Riera, and Luciana Ferrer,  “Emotion recognition from speech using wav2vec 2.0 embeddings,”  arXiv preprint arXiv:2104.03502, 2021.
  * [28] John J Godfrey, Edward C Holliman, and Jane McDaniel,  “Switchboard: Telephone speech corpus for research and development,”  in ICASSP. IEEE Computer Society, 1992, vol. 1, pp. 517–520.
  * [29] Soumya Dutta and Sriram Ganapathy,  “HCAM–Hierarchical Cross Attention Model for Multi-modal Emotion Recognition,”  arXiv preprint arXiv:2304.06910, 2023.
  * [30] Ashish Vaswani et al.,  “Attention is all you need,”  NeurIPS, vol. 30, 2017.
  * [31] Kavita Kasi and Stephen A Zahorian,  “Yet another algorithm for pitch tracking,”  in ICASSP. IEEE, 2002, vol. 1, pp. I–361.
  * [32] Houwei Cao et al.,  “CREMA-D: Crowd-sourced emotional multimodal actors dataset,”  IEEE transactions on affective computing, vol. 5, no. 4, pp. 377–390, 2014.
  * [33] John S Garofolo et al.,  “DARPA TIMIT acoustic-phonetic continous speech corpus CD-ROM. NIST speech disc 1-1.1,”  NASA STI/Recon technical report n, vol. 93, pp. 27403, 1993.
  * [34] Kun Zhou, Berrak Sisman, and Haizhou Li,  “Vaw-gan for disentanglement and recomposition of emotional elements in speech,”  in IEEE SLT. IEEE, 2021, pp. 415–422.
  * [35] Vassil Panayotov et al.,  “Librispeech: an asr corpus based on public domain audio books,”  in ICASSP. IEEE, 2015, pp. 5206–5210.


# On the Hilbert $2$-class field towers of some cyclotomic
$\mathbb{Z}_{2}$-extensions

Mohamed Mahmoud Chems-Eddin  Mohamed Mahmoud CHEMS-EDDIN: Mohammed First
University, Mathematics Department, Sciences Faculty, Oujda, Morocco
<EMAIL_ADDRESS>,  Abdelkader Zekhnini  Abdelkader Zekhnini: Mohammed
First University, Mathematics Department, Pluridisciplinary faculty, Nador,
Morocco<EMAIL_ADDRESS>and  Abdelmalek Azizi  Abdelmalek Azizi: Mohammed
First University, Mathematics Department, Sciences Faculty, Oujda, Morocco
<EMAIL_ADDRESS>
###### Abstract.

In this paper, we study the length of the $2$-class field towers and the
structure of the Galois groups $\mathrm{Gal}(\mathcal{L}(K_{n})/K_{n})$ of the
maximal unramified $2$-extensions of the layers $K_{n}$ of the cyclotomic
$\mathbb{Z}_{2}$-extension of some special Dirichlet fields. The capitulation
problem is investigated too.

###### Key words and phrases:

Hilbert $2$-class field towers, Cyclotomic $\mathbb{Z}_{2}$-extensions,
Multiquadratic number fields, unit groups, $2$-class groups.

###### 2010 Mathematics Subject Classification:

11R11, 11R23, 11R27, 11R29, 11R37.

##  1\. Introduction

Let $K$ be a number field and $\mathrm{Cl}_{2}(K)$ its $2$-class group, that
is the $2$-Sylow subgroup of its class group in the wide sense. Let $K^{(1)}$
denote the Hilbert $2$-class field of $K$, that is the maximal unramified
(including the infinite primes) abelian extension of $K$ whose degree is a
power of $2$. For any positive integer $i\geq 1$, define inductively $K^{(i)}$
as $K^{(0)}=K$ and $K^{(i)}=(K^{(i-1)})^{(1)}$. Then the sequence of fields

$K=K^{(0)}\subset K^{(1)}\subset K^{(2)}\subset\cdots\subset
K^{(i)}\subset\cdots$

is called the $2$-class field tower of $K$. If for all $i\geq 1$, $K^{(i)}\neq
K^{(i-1)}$, the tower is said to be infinite, otherwise the tower is said to
be finite, and the minimal integer $i$ such that $K^{(i)}=K^{(i-1)}$ is called
the length of the tower. The determination of the length of the $2$-class
field tower of $K$ and the structure of the Galois groups of the tower are
until nowadays two classical and difficult open problems of class field
theory. Actually, there is no known decision procedure to determine whether or
not the $2$-class field of an arbitrary number field $K$ is finite. However,
it is known that if the rank of $\mathrm{Cl}_{2}(K^{(1)})\leq 2$, then by
group theoretic means, the tower is finite and its length is at most $3$ (cf.
[11, Theorem 1]). If the $2$-class group of $K$ is of type $(2,2)$ (here
$(a_{1},...,a_{r})$ denotes the direct sum of cyclic groups of order $a_{i}$,
for $i=1,...,r$), then the Hilbert class field tower of such field terminates
in at most two steps and the Galois groups of the tower are characterized by
means of capitulation problem in the three unramified quadratic extension of
$K$ (cf. [27]). If the $2$-class group of $K$ is of type $(2,2^{n})$, with
$n\geq 2$, the situation becomes more difficult, and many authors treated it
using the capitulation problem in the unramified extensions of $K$ for fields
of low degree (e.g. $2$ or $4$, cf. [4, 6, 7, 9, 10]). Our contribution in
this paper is to determine the length of the $2$-class field tower, the
structure of its Galois group and to study the capitulation problem for some
families of number fields of large degree over $\mathbb{Q}$ whose $2$-class
groups are of type $(2,2^{n})$, with $n\geq 2$. The novelty in our procedure
to deal with this problem is the combination of some tools from Iwasawa theory
and the rank of some unramified extensions.

Let $d$ be an odd positive square-free integer. Consider the biquadratic
number field $K=\mathbb{Q}({\sqrt{d},i})$. Denote by $K_{\infty}$ the
cyclotomic $\mathbb{Z}_{2}$-extension of $K$, that is a Galois extension of
$K$ whose Galois group is topologically isomorphic to the $2$-adic ring
$\mathbb{Z}_{2}$. It is well known that for any integer $n\geq 0$, the field
$K_{\infty}$ contains a unique cyclic subfield $K_{n}$ of degree $2^{n}$ over
$K$ called the $n$-th layer of the $\mathbb{Z}_{2}$-extension of $K$. Denote
by $A(K_{n})$ (or simply $A_{n}$) the $2$-part of the ideal class group of
$K_{n}$ and let $\mathcal{L}(K_{n})$ be the maximal unramified $2$-extension
of $K_{n}$. The cyclotomic $\mathbb{Z}_{2}$-extension of $K$ is given by the
sequence $K_{n}=K_{n,d}=\mathbb{Q}({\zeta_{2^{n+2}},\sqrt{d}})$ such that

$K_{1,d}=\mathbb{Q}({\zeta_{2^{3}},\sqrt{d}})\subset
K_{2,d}=\mathbb{Q}({\zeta_{2^{4}},\sqrt{d}})\subset\cdots\subset
K_{n,d}=\mathbb{Q}({\zeta_{2^{n+2}},\sqrt{d}})\cdots$

In this article, we are interested in giving, for $n\geq 1$, the length of the
$2$-class field tower of $K_{n}$, the structure of the Galois group
$\mathrm{Gal}(\mathcal{L}(K_{n})/K_{n})$ and the capitulation of the $2$-ideal
classes of $K_{n}$ in its three unramified quadratic extensions for some
fields $K$. Our main theorem is the following.

######  Theorem 1.1.

Let $d$ be a square-free integer and $n\geq 1$ a positive integer. Let $m$ be
such that $2^{m}=h_{2}(-2d)$ and
$K_{n}=\mathbb{Q}({\zeta_{2^{n+2}},\sqrt{d}})$. Let $\mathcal{L}(K_{n})$ be
the maximal unramified $2$-extension of $K_{n}$ and put
$G_{n}=\mathrm{Gal}(\mathcal{L}(K_{n})/K_{n})$. Then we have

  1. 1.

If $d=q_{1}q_{2}$, where $q_{i}\equiv 3\pmod{8}$ are two distinct primes, then
$\mathcal{L}(K_{n})=K_{n}^{(1)}$ and $G_{n}$ is abelian isomorphic to
$\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2^{n+m-2}\mathbb{Z}$.

  2. 2.

If $d=p$, where $p\equiv 9\pmod{16}$ is a prime such that
$(\frac{2}{p})_{4}=1$, then $G_{n}$ is abelian isomorphic to
$\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2^{n+m-2}\mathbb{Z}$ or modular, and
$\mathcal{L}(K_{n})=K_{n}^{(1)}$ or $K_{n}^{(2)}$. In particular for $n=1$,
$G_{1}=Gal(K_{1,d}^{(1)}/K_{1,d})\simeq\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2^{m-1}\mathbb{Z}$
is abelian.

Put $G=\mathrm{Gal}(\mathcal{L}(K_{\infty})/K_{\infty})$. For the abelian case
we have $G\simeq X_{\infty}\simeq\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}_{2}$,
otherwise $G/G^{\prime}\simeq
X_{\infty}\simeq\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}_{2}$, where
$G^{\prime}$ is the commutator subgroup of $G$.

The proof of this theorem is detailed in the next sections. Therein one can
find results on unit groups, $2$-class groups, computations of Iwasawa
$\lambda$-invariant and some corollaries.

## Notations

Let $k$ be a number field. Throughout this paper, we adopt the following
notations.

  1. $\star$

$q_{1}$, $q_{2}$: Two primes congruent to $3\pmod{8}$,

  2. $\star$

$k^{+}$: The maximal real subfield of $k$,

  3. $\star$

$n$: An integer $\geq 1$,

  4. $\star$

$k_{n}$: The $n$-th layer of the $\mathbb{Z}_{2}$-extension of $k$,

  5. $\star$

$\lambda(k)$: The Iwasawa $\lambda_{2}$-invariant of $k$,

  6. $\star$

$\lambda(k)^{-}$: The Iwasawa $\lambda_{2}^{-}$-invariant of $k$,

  7. $\star$

$\mathrm{Cl}_{2}(k)$: The $2$-class group of $k$,

  8. $\star$

$rank(\mathrm{Cl}_{2}(k))$: The rank of the $2$-class group of $k$,

  9. $\star$

$K_{n,d}$: $\mathbb{Q}({\zeta_{2^{n+2}},\sqrt{d}})$,

  10. $\star$

$F_{n}$: $\mathbb{Q}(\zeta_{2^{n+2}},\sqrt{q_{1}},\sqrt{q_{2}})$,

  11. $\star$

$h(k)$: The class number of the field $k$,

  12. $\star$

$h_{2}(d)$: The $2$-class number of the quadratic field
$\mathbb{Q}(\sqrt{d})$,

  13. $\star$

$\varepsilon_{d}$: The fundamental unit of the quadratic field
$\mathbb{Q}(\sqrt{d})$,

  14. $\star$

$E_{k}$: The unit group of $k$,

  15. $\star$

$q(k):=[E_{k}:\prod_{i}E_{k_{i}}]$, with $k_{i}$ are the quadratic subfields
of $k$,

  16. $\star$

$Q_{k}$: The Hasse’s unit index, that is $[E_{k}:W_{k}k^{+}]$, if $k/k^{+}$ is
CM,

  17. $\star$

FSU: A fundamental system of units,

  18. $\star$

$N_{K/k}$: The norm map of an extension $K/k$,

  19. $\star$

$e$: Defined by $[E_{k}:E_{k}\cap N_{K/k}(K^{*})]=2^{e}$, for a $2$-extension
$K/k$.

##  2\. Units and $2$-class numbers of some multiquadratic fields

Let us first recall the result stated by Wada [31] to determine a fundamental
system of units of a multiquadratic number field. Let $K_{0}$ be a
multiquadratic number field. Denote by $\sigma$ and $\tau$ two different
generators of the group $\mathrm{Gal}(K_{0}/\mathbb{Q})$, let then $K_{1}$,
$K_{2}$ and $K_{3}$ be respectively the invariant subfields of $K_{0}$ by
$\sigma$, $\tau$ and $\sigma\tau$, and $E_{K_{i}}$ the unit group of $K_{i}$.
Then the unit group $E_{K_{0}}$ of $K_{0}$ is generated by the elements of
each $E_{i}$ and the square roots of elements of the product
$E_{K_{1}}E_{K_{2}}E_{K_{3}}$ that are perfect squares in $K_{0}$.

Put $K=K_{0}(i)$, then to determine a fundamental system of units of $K$, we
will use the following result (cf. [2, p. 18]) that the third author has
deduced from a theorem of Hasse [20, §21, Satz 15 ].

######  Lemma 2.1.

Let $K_{0}$ be a real number field, $K=K_{0}(i)$ a quadratic extension of
$K_{0}$, $n\geq 2$ be an integer and $\xi_{n}$ a $2^{n}$-th primitive root of
unity, then $\xi_{n}=\frac{1}{2}(\mu_{n}+\lambda_{n}i)$, where
$\mu_{n}=\sqrt{2+\mu_{n}}$, $\lambda_{n}=\sqrt{2-\mu_{n-1}}$, $\mu_{2}=0$,
$\lambda_{0}=2$ and $\mu_{3}=\lambda_{3}=\sqrt{2}$. Let $n_{0}$ be the
greatest integer such that $\xi_{n_{0}}$ is contained in $K$,
$\\{\varepsilon_{1},...,\varepsilon_{r}\\}$ a fundamental system of units of
$K_{0}$ and $\varepsilon$ a unit of $K_{0}$ such that
$(2+\mu_{n_{0}})\varepsilon$ is a square in $K_{0}\ ($if it exists$)$. Then a
fundamental system of units of $K$ is one of the following systems:

  1. 1.

$\\{\varepsilon_{1},...,\varepsilon_{r-1},\sqrt{\xi_{n_{0}}\varepsilon}\\}$ if
$\varepsilon$ exists, in this case
$\varepsilon=\varepsilon_{1}^{j_{1}}...\varepsilon_{r-1}^{j_{1}}\varepsilon_{r}$,
where $j_{i}\in\\{0,1\\}$.

  2. 2.

$\\{\varepsilon_{1},...,\varepsilon_{r}\\}$ elsewhere.

We will also need the following lemmas about units.

######  Lemma 2.2 ([1], Lemma 5).

Let $d>1$ be a square-free integer and $\varepsilon_{d}=x+y\sqrt{d}$, where
$x$, $y$ are integers or semi-integers. If $N(\varepsilon_{d})=1$, then
$2(x+1)$, $2(x-1)$, $2d(x+1)$ and $2d(x-1)$ are not squares in $\mathbb{Q}$.

######  Lemma 2.3 ([33]).

Let $d\equiv 1\pmod{4}$ be a positive square free integer and
$\varepsilon_{d}=x+y\sqrt{d}$ be the fundamental unit of
$\mathbb{Q}(\sqrt{d})$. Assume $N(\varepsilon_{d})=1$, then

  1. 1.

$x+1$ and $x-1$ are not squares in $\mathbb{N}$, i.e., $2\varepsilon_{d}$ is
not a square in $\mathbb{Q}(\sqrt{d})$.

  2. 2.

For all prime $p$ dividing $d$, $p(x+1)$ and $p(x-1)$ are not squares in
$\mathbb{N}$.

######  Lemma 2.4.

Let $q_{1}\equiv q_{2}\equiv 3\pmod{8}$ be two primes such that
$\left(\dfrac{q_{1}}{q_{2}}\right)=1$.

  1. 1.

Let $x$ and $y$ be two integers such that
$\varepsilon_{2q_{1}q_{2}}=x+y\sqrt{2q_{1}q_{2}}$. Then

    1. a.

$(x-1)$ is a square in $\mathbb{N}$,

    2. b.

$\sqrt{2\varepsilon_{2q_{1}q_{2}}}=y_{1}+y_{2}\sqrt{2q_{1}q_{2}}$ and
$2=-y_{1}^{2}+2q_{1}q_{2}y_{2}^{2}$, for some integers $y_{1}$ and $y_{2}$
such that $y=y_{1}y_{2}$.

  2. 2.

There are two integers $a$ and $b$ such that
$\varepsilon_{q_{1}q_{2}}=a+b\sqrt{q_{1}q_{2}}$ and we have

    1. a.

$2q_{1}(a+1)$ is a square in $\mathbb{N}$,

    2. b.

$b$ is even,
$\sqrt{\varepsilon_{q_{1}q_{2}}}=b_{1}\sqrt{q_{1}}+b_{2}\sqrt{q_{2}}$ and
$1=q_{1}b_{1}^{2}-q_{2}b_{2}^{2}$ for some integers $b_{1}$ and $b_{2}$ such
that $b=2b_{1}b_{2}$.

  3. 3.

Let $c$ and $d$ be two integers such that
$\varepsilon_{2q_{i}}=c+d\sqrt{2q_{i}}$. Then we have

    1. a.

$c-1$ is a square in $\mathbb{N}$,

    2. b.

$\sqrt{2\varepsilon_{2q_{i}}}=d_{1}+d_{2}\sqrt{2q_{i}}$ and
$2=-d_{1}^{2}+2q_{i}d_{2}^{2}$, for some integers $d_{1}$ and $d_{2}$ such
that $d=d_{1}d_{2}$.

  4. 4.

Let $\alpha$ and $\beta$ be two integers such that
$\varepsilon_{q_{i}}=\alpha+\beta\sqrt{q_{i}}$. Then we have

    1. a.

$\alpha-1$ is a square in $\mathbb{N}$,

    2. b.

$\sqrt{2\varepsilon_{q_{i}}}=\beta_{1}+\beta_{2}\sqrt{q_{i}}$ and
$2=-\beta_{1}^{2}+q_{i}\beta_{2}^{2}$, for some integers $\beta_{1}$ and
$\beta_{2}$ such that $\beta=\beta_{1}\beta_{2}$.

###### Proof.

  1. 1.

It is known that $N(\varepsilon_{2q_{1}q_{2}})=1$, so
$x^{2}-1=y^{2}2q_{1}q_{2}$, thus by the unique prime factorization theorem in
$\mathbb{Z}$ and Lemma 2.2 we have

$(1):\ \left\\{\begin{array}[]{ll}x\pm 1=2q_{1}y_{1}^{2}\\\ x\mp
1=q_{2}y_{2}^{2},\end{array}\right.\quad(2):\ \left\\{\begin{array}[]{ll}x\pm
1=q_{1}y_{1}^{2}\\\ x\mp 1=2q_{2}y_{2}^{2},\end{array}\right.\quad\text{ or
}\quad(3):\ \left\\{\begin{array}[]{ll}x\pm 1=y_{1}^{2}\\\ x\mp
1=2q_{1}q_{2}y_{2}^{2},\end{array}\right.$

For some integers $y_{1}$ and $y_{2}$ such that $y=y_{1}y_{2}$.

    1. $*$

System $(1)$ can not occur. Indeed,

      1. $\bullet$

if $2q_{1}(x+1)$ is a square, then
$-1=\left(\frac{q_{2}}{q_{1}}\right)=\left(\frac{x-1}{q_{1}}\right)=\left(\frac{x+1-2}{q_{1}}\right)=\left(\frac{-2}{q_{1}}\right)=1.$
Which is absurd.

      2. $\bullet$

similarly, if $2q_{1}(x-1)$ is a square, then
$-1=\left(\frac{2q_{1}}{q_{2}}\right)=\left(\frac{x-1}{q_{2}}\right)=\left(\frac{x+1-2}{q_{2}}\right)=\left(\frac{-2}{q_{2}}\right)=1.$
Which is absurd too.

    2. $*$

We similarly eliminate the cases of system $(2)$ and the system
$\left\\{\begin{array}[]{ll}x+1=y_{1}^{2}\\\
x-1=2q_{1}q_{2}y_{2}^{2}.\end{array}\right.$

Thus, the only possible case is $\left\\{\begin{array}[]{ll}x-1=y_{1}^{2}\\\
x+1=2q_{1}q_{2}y_{2}^{2}.\end{array}\right.$
Which implies that
$\sqrt{2\varepsilon_{2q_{1}q_{2}}}=y_{1}+y_{2}\sqrt{2q_{1}q_{2}}$ and
$2=-y_{1}^{2}+2q_{1}q_{2}y_{2}^{2}$.

  2. 2.

We have $q_{1}q_{2}\equiv 1\pmod{8}$, then there are two integers
$\varepsilon_{q_{1}q_{2}}=({x+y\sqrt{d}})/{2}$. As
$N(\varepsilon_{q_{1}q_{2}})=1$, so $x^{2}-4=y^{2}d$, hence $x^{2}-4\equiv
y^{2}\pmod{8}$. On the other hand, if we suppose that $x$ and $y$ are odd,
then $x^{2}\equiv y^{2}\equiv 1\pmod{8}$, but this implies the contradiction
$-3\equiv 1\pmod{8}$. Thus $x$ and $y$ are even. It follows that there are two
integers $a$ and $b$ such that
$\varepsilon_{q_{1}q_{2}}=a+b\sqrt{q_{1}q_{2}}$.

As $N(\varepsilon_{q_{1}q_{2}})=1$, then, by Lemma 2.3 and the unique prime
factorization theorem, one deduces

$(1):\ \left\\{\begin{array}[]{ll}a+1=2q_{1}b_{1}^{2}\\\
a-1=2q_{2}b_{2}^{2},\end{array}\right.\quad\text{ or }\quad(2):\
\left\\{\begin{array}[]{ll}a-1=2q_{1}b_{1}^{2}\\\
a+1=2q_{2}b_{2}^{2},\end{array}\right.$

For some integers $b_{1}$ and $b_{2}$ such that $b=2b_{1}b_{2}$. As above we
show that $\left\\{\begin{array}[]{ll}a+1=2q_{1}b_{1}^{2}\\\
a-1=2q_{2}b_{2}^{2}.\end{array}\right.$ Thus,
$\sqrt{\varepsilon_{q_{1}q_{2}}}=b_{1}\sqrt{q_{1}}+b_{2}\sqrt{q_{2}}$ and
$1=q_{1}b_{1}^{2}-q_{2}b_{2}^{2}$.

The items $3$ and $4$ are proved similarly. ∎

Let us now recall the following class number formula for a multiquadratic
number field which is usually attributed to Kuroda [28], but it goes back to
Herglotz [21] (cf. [8, p. 27]).

######  Lemma 2.5.

Let $K$ be a multiquadratic number field of degree $2^{n}$, $n\in\mathds{N}$,
and $k_{i}$ the $s=2^{n}-1$ quadratic subfields of $K$. Then

$h(K)=\frac{1}{2^{v}}[E_{K}:\prod_{i=1}^{s}E_{k_{i}}]\prod_{i=1}^{s}h(k_{i}),$

with

$v=\left\\{\begin{array}[]{cl}n(2^{n-1}-1);&\text{ if }K\text{ is real, }\\\
(n-1)(2^{n-2}-1)+2^{n-1}-1&\text{ if }K\text{ is
imaginary.}\end{array}\right.$

To use the above lemma we need to recall the values of the class numbers of
some quadratic fields.

######  Lemma 2.6.

Let $q=q^{\prime}\equiv 3\pmod{4}$ be two distinct primes. Then

  1. 1.

By [15, Corollary 18.4], we have
$h_{2}(q)=h_{2}(-q)=h_{2}(2q)=h_{2}(2)=h_{2}(-2)=h_{2}(-1)=h_{2}(qq^{\prime})=1$.

  2. 2.

If $q\equiv 3\pmod{8}$, then by [15, Corollary 19.6], $h_{2}(-2q)=2$.

  3. 3.

If $q$ or $q^{\prime}\equiv 3\pmod{8}$, then by [25, p. 345],
$h_{2}(2qq^{\prime})=2$.

  4. 4.

If $q^{\prime}\equiv 3\pmod{8}$, with $\left(\frac{q}{q^{\prime}}\right)=1$,
then by [25, p. 354], $h_{2}(-qq^{\prime})=4$.

Now we can state the important result of this section which gives a FSU of a
triquadratic number field.

######  Proposition 2.7.

Let $q_{1}\equiv q_{2}\equiv 3\pmod{8}$ be two different prime integers and
$\mathbb{L}=\mathbb{Q}(\sqrt{2},\sqrt{q_{1}},\sqrt{q_{2}})$. Then the unit
group of $\mathbb{L}$ is

$E_{\mathbb{L}}=\left\langle-1,\varepsilon_{2},\sqrt{\varepsilon_{q_{1}}},\sqrt{\varepsilon_{2q_{1}}},\sqrt{\varepsilon_{q_{2}}},\sqrt{\varepsilon_{q_{1}q_{2}}},\sqrt{\sqrt{\varepsilon_{q_{1}}}\sqrt{\varepsilon_{q_{2}}}\sqrt{\varepsilon_{2q_{1}q_{2}}}},\sqrt{\sqrt{\varepsilon_{2q_{1}}}\sqrt{\varepsilon_{2q_{2}}}\sqrt{\varepsilon_{2q_{1}q_{2}}}}\right\rangle,$

and its class number is odd.

###### Proof.

Without loss of generality we may suppose that
$\left(\dfrac{q_{1}}{q_{2}}\right)=1$. Consider the following diagram (Figure
1 below):

---
$\textstyle{\mathbb{L}=\mathbb{Q}(\sqrt{2},\sqrt{q_{1}},\sqrt{q_{2}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{L_{1}=\mathbb{Q}(\sqrt{2},\sqrt{q_{1}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{L_{2}=\mathbb{Q}(\sqrt{2},\sqrt{q_{2}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{L_{3}=\mathbb{Q}(\sqrt{2},\sqrt{q_{1}q_{2}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{Q}(\sqrt{2})}$
Figure 1. Subfields of $\mathbb{L}/\mathbb{Q}(\sqrt{2})$

Put
$\mathrm{Gal}(\mathbb{L}/\mathbb{Q})=\langle\sigma_{1},\sigma_{2},\sigma_{3}\rangle$,
where

$\sigma_{1}(\sqrt{2})=-\sqrt{2}$, $\sigma_{2}(\sqrt{q_{1}})=-\sqrt{q_{1}}$,
$\sigma_{3}(\sqrt{q_{2}})=-\sqrt{q_{2}}$,

$\sigma_{i}(\sqrt{2})=\sqrt{2}$ for $i\in\\{2,3\\}$,
$\sigma_{i}(\sqrt{q_{1}})=\sqrt{q_{1}}$ for $i\in\\{1,3\\}$ and

$\sigma_{i}(\sqrt{q_{2}})=\sqrt{q_{2}}$ for $i\in\\{1,2\\}$.

Hence $L_{1}$, $L_{2}$ and $L_{3}$ are the fixed fields of the subgroups of
$\mathrm{Gal}(\mathbb{L}/\mathbb{Q})$ generated respectively by $\sigma_{3}$,
$\sigma_{2}$ and $\sigma_{2}\sigma_{3}$. By Lemma 2.4, we easily infer that a
FSU of $L_{1}$ is
$\\{\varepsilon_{2},\sqrt{\varepsilon_{q_{1}}},\sqrt{\varepsilon_{2q_{1}}}\\}$,
a FSU of $L_{2}$ is
$\\{\varepsilon_{2},\sqrt{\varepsilon_{q_{2}}},\sqrt{\varepsilon_{2q_{2}}}\\}$
and a FSU of $L_{3}$ is
$\\{\varepsilon_{2},\varepsilon_{q_{1}q_{2}},\sqrt{\varepsilon_{2q_{1}q_{2}}}\\}$.
Thus

$\displaystyle
E_{L_{1}}E_{L_{2}}E_{L_{3}}=\langle-1,\varepsilon_{2},\varepsilon_{q_{1}q_{2}},\sqrt{\varepsilon_{q_{1}}},\sqrt{\varepsilon_{2q_{1}}},\sqrt{\varepsilon_{q_{2}}},\sqrt{\varepsilon_{2q_{2}}},\sqrt{\varepsilon_{2q_{1}q_{2}}}\rangle.$
(1)

To find a fundamental system of units of $\mathbb{L}$, it suffices, as we have
said in the beginning of the section, to find elements $\xi$ of
$E_{L_{1}}E_{L_{2}}E_{L_{3}}$ which are squares in $\mathbb{L}$ (cf. page 2).
According to Lemma 2.4, $\varepsilon_{q_{1}q_{2}}$ is a square in
$\mathbb{L}$, then it suffices to look for the units $\xi$ such that

$\displaystyle\xi^{2}=\varepsilon_{2}^{a}\sqrt{\varepsilon_{q_{1}}}^{b}\sqrt{\varepsilon_{2q_{1}}}^{c}\sqrt{\varepsilon_{q_{2}}}^{d}\sqrt{\varepsilon_{2q_{2}}}^{e}\sqrt{\varepsilon_{2q_{1}q_{2}}}^{f}.$
(2)

where $a,b,c,d,e$ and $f\in\\{0,1\\}$. Let us eliminate some forms of
$\xi^{2}$ such that $\xi$ can not be in $\mathbb{L}$. Consider
$L_{4}=\mathbb{Q}(\sqrt{q_{1}},\sqrt{q_{2}})$, we will apply the norm
$N_{\mathbb{L}/L_{4}}=1+\sigma_{1}$ to the equation (2). For this, by using
Lemma 2.4 we have the following table:

$\varepsilon$ | $\varepsilon_{2}$ | $\sqrt{\varepsilon_{q_{1}}}$ | $\sqrt{\varepsilon_{2q_{1}}}$ | $\sqrt{\varepsilon_{q_{2}}}$ | $\sqrt{\varepsilon_{2q_{2}}}$ | $\sqrt{\varepsilon_{2q_{1}q_{2}}}$
---|---|---|---|---|---|---
$\varepsilon^{1+\sigma_{1}}$ | $-1$ | ${-\varepsilon_{q_{1}}}$ | $1$ | ${-\varepsilon_{q_{2}}}$ | $1$ | $1$

From which we deduce that

$\displaystyle N_{\mathbb{L}/L_{4}}(\xi^{2})=N_{\mathbb{L}/L_{4}}(\xi)^{2}$
$\displaystyle=$
$\displaystyle(-1)^{a}\cdot(-1)^{b}\cdot\varepsilon_{q_{1}}^{b}\cdot
1\cdot(-1)^{d}\cdot\varepsilon_{q_{2}}^{d}\cdot 1\cdot 1,$ $\displaystyle=$
$\displaystyle(-1)^{a+b+d}\cdot\varepsilon_{q_{1}}^{b}\cdot\varepsilon_{q_{2}}^{d}>0.$

Thus, $a+b+d=0\pmod{2}$, and since by Lemma 2.4, $\varepsilon_{q_{1}}$ and
$\varepsilon_{q_{2}}$ are not squares in $L_{4}$ whereas their product is, one
deduces that $b=d$ and $a=0$. Therefore

$\displaystyle\xi^{2}=\sqrt{\varepsilon_{q_{1}}}^{b}\sqrt{\varepsilon_{2q_{1}}}^{c}\sqrt{\varepsilon_{q_{2}}}^{b}\sqrt{\varepsilon_{2q_{2}}}^{e}\sqrt{\varepsilon_{2q_{1}q_{2}}}^{f}.$

Similarly, we will consider $L_{3}=\mathbb{Q}(\sqrt{2},\sqrt{q_{1}q_{2}})$ and
apply the norm $N_{\mathbb{L}/L_{3}}=1+\sigma_{2}\sigma_{3}$. Firstly, we need
the following computations:

$\varepsilon$ | $\varepsilon_{2}$ | $\sqrt{\varepsilon_{q_{1}}}$ | $\sqrt{\varepsilon_{2q_{1}}}$ | $\sqrt{\varepsilon_{q_{2}}}$ | $\sqrt{\varepsilon_{2q_{2}}}$ | $\sqrt{\varepsilon_{2q_{1}q_{2}}}$
---|---|---|---|---|---|---
$\varepsilon^{1+\sigma_{2}\sigma_{3}}$ | $\varepsilon_{2}^{2}$ | $-1$ | $-1$ | $-1$ | $-1$ | $\varepsilon_{2q_{1}q_{2}}$

So

$\displaystyle N_{\mathbb{L}/L_{3}}(\xi^{2})$ $\displaystyle=$
$\displaystyle(-1)^{b}\cdot(-1)^{c}\cdot(-1)^{b}\cdot(-1)^{e}\cdot\varepsilon_{2q_{1}q_{2}}^{f},$
$\displaystyle=$
$\displaystyle(-1)^{b+c+b+e}\cdot\varepsilon_{2q_{1}q_{2}}^{f}>0.$

Thus all what we can deduce is that $c=e$. Hence

$\displaystyle\xi^{2}=\sqrt{\varepsilon_{q_{1}}}^{b}\sqrt{\varepsilon_{2q_{1}}}^{c}\sqrt{\varepsilon_{q_{2}}}^{b}\sqrt{\varepsilon_{2q_{2}}}^{c}\sqrt{\varepsilon_{2q_{1}q_{2}}}^{f}.$

Similarly, we consider $L_{5}=\mathbb{Q}(\sqrt{q_{1}},\sqrt{2q_{2}})$ and we
apply $N_{\mathbb{L}/L_{5}}=1+\sigma_{1}\sigma_{3}$. By Lemma 2.4 we have

$\varepsilon$ | $\varepsilon_{2}$ | $\sqrt{\varepsilon_{q_{1}}}$ | $\sqrt{\varepsilon_{2q_{1}}}$ | $\sqrt{\varepsilon_{q_{2}}}$ | $\sqrt{\varepsilon_{2q_{2}}}$ | $\sqrt{\varepsilon_{2q_{1}q_{2}}}$
---|---|---|---|---|---|---
$1+\sigma_{1}\sigma_{3}$ | $-1$ | $-\varepsilon_{q_{1}}$ | $1$ | $1$ | $-\varepsilon_{2q_{2}}$ | $-\varepsilon_{2q_{1}q_{2}}$

Then

$\displaystyle N_{\mathbb{L}/L_{5}}(\xi^{2})$ $\displaystyle=$
$\displaystyle(-1)^{b}\cdot\varepsilon_{q_{1}}^{b}\cdot 1\cdot
1\cdot(-1)^{c}\cdot\varepsilon_{2q_{2}}^{c}\cdot(-1)^{f}\cdot\varepsilon_{2q_{1}q_{2}}^{f},$
$\displaystyle=$
$\displaystyle(-1)^{b+c+f}\cdot\varepsilon_{q_{1}}^{b}\cdot\varepsilon_{2q_{2}}^{c}\cdot\varepsilon_{2q_{1}q_{2}}^{f}>0.$

Thus all what we can deduce is that $b+c+f\equiv 0\pmod{2}$. Applying the norm
maps on all other subextensions of $\mathbb{L}$ we deduce no more information
on $b,c,e$ and $f$. On the other hand, by Lemmas 2.5 and 2.6 we have

$\displaystyle h_{2}(\mathbb{L})$ $\displaystyle=$
$\displaystyle\frac{1}{2^{9}}q(\mathbb{L})h_{2}(2)h_{2}(q_{1})h_{2}(2q_{1})h_{2}(q_{2})h_{2}(2q_{2})h(q_{1}q_{2})h_{2}(2q_{1}q_{2})$
(3) $\displaystyle=$ $\displaystyle\frac{1}{2^{9}}\cdot q(\mathbb{L})\cdot
1\cdot 1\cdot 1\cdot 1\cdot 1\cdot 1\cdot 2=\frac{1}{2^{8}}\cdot
q(\mathbb{L}).$

If we suppose that $b=c=f=0$, then there is no element of
$E_{L_{1}}E_{L_{2}}E_{L_{3}}$ which is a square in $\mathbb{L}$ except
${\varepsilon_{q_{1}q_{2}}}$. So $q(\mathbb{L})=2^{6}$. Hence formula 3,
implies that $h_{2}(\mathbb{L})=\frac{1}{2^{8}}\cdot 2^{6}=\frac{1}{2^{4}}$.
Which is absurd. Thus the case $b=c=f=0$ is impossible. Hence $\xi^{2}$ can
take one of the following forms:

$\displaystyle\xi_{1}^{2}=\sqrt{\varepsilon_{q_{1}}}\sqrt{\varepsilon_{q_{2}}}\sqrt{\varepsilon_{2q_{1}q_{2}}}\quad(\text{i.e.,
}c=0\text{ and }b=f=1),$ (4)

$\displaystyle\xi_{2}^{2}=\sqrt{\varepsilon_{2q_{1}}}\sqrt{\varepsilon_{2q_{2}}}\sqrt{\varepsilon_{2q_{1}q_{2}}}\quad(\text{i.e.,
}b=0\text{ and }c=f=1),$ (5)

$\displaystyle\xi_{3}^{2}=\sqrt{\varepsilon_{q_{1}}}\sqrt{\varepsilon_{2q_{1}}}\sqrt{\varepsilon_{q_{2}}}\sqrt{\varepsilon_{2q_{2}}}\quad(\text{i.e.,
}f=0\text{ and }c=b=1).$ (6)

Remark that by the equalities (1) and (3) at least two equations from (4), (5)
and (6) are verified for some $\xi\in\mathbb{L}$ (elsewhere, we get a
contradiction as for the case $b=c=f=0$). Remark also that if two of the forms
in the equations (4), (5) and (6), are verified for some $\xi_{i}$ and
$\xi_{j}\in\mathbb{L}$ ($i\not=j$), then by their product we may deduce that
the third form is also verified for some $\xi_{k}\in\mathbb{L}$. So the
result. ∎

We close this section by the following result on class numbers of some number
fields.

######  Lemma 2.8.

Let $q_{1}\equiv q_{2}\equiv 3\pmod{8}$ be two distinct primes. Then

  1. 1.

The class numbers of $F=\mathbb{Q}(\sqrt{q_{1}},\sqrt{q_{2}},i)$ and
$F^{+}=\mathbb{Q}(\sqrt{q_{1}},\sqrt{q_{2}})$ are odd.

  2. 2.

The $2$-class number of $K=\mathbb{Q}(\sqrt{-q_{1}},\sqrt{q_{2}},\sqrt{2})$
equals $h_{2}(-2q_{1}q_{2})$.

###### Proof.

  1. 1.

Note first that by Lemma 2.4, we have $\sqrt{2\varepsilon_{q_{1}}}$,
$\sqrt{2\varepsilon_{q_{2}}}\in F^{+}$, so
$\sqrt{\varepsilon_{q_{1}}\varepsilon_{q_{2}}}\in F^{+}$ and
$\sqrt{\varepsilon_{q_{i}}}\not\in F^{+}$. Thus a fundamental system of unit
of $F^{+}$ is
$\\{\varepsilon_{q_{1}},\sqrt{\varepsilon_{q_{1}}\varepsilon_{q_{2}}},\sqrt{\varepsilon_{q_{1}q_{2}}}\\}$
and that of $F$ is
$\\{\sqrt{\varepsilon_{q_{1}}\varepsilon_{q_{2}}},\sqrt{\varepsilon_{q_{1}q_{2}}},\sqrt{i\varepsilon_{q_{1}}}\\}$
(see Lemma 2.1). Therefore, $q(F)=2^{3}$. Thus by Lemmas 2.5 and 2.6 we obtain

$\displaystyle h_{2}(F)$ $\displaystyle=$
$\displaystyle\frac{1}{2^{5}}q(F)h_{2}(q_{1})h_{2}(q_{2})h_{2}(-q_{1})h_{2}(-q_{2})h_{2}(q_{1}q_{2})h_{2}(-q_{1}q_{2})h_{2}(-1),$
$\displaystyle=$ $\displaystyle\frac{1}{2^{5}}q(F)\cdot 4=1.$

We similarly prove that the $2$-class number of $F^{+}$, is odd. Hence the
first assertion of the lemma.

  2. 2.

For $K=\mathbb{Q}(\sqrt{-q_{1}},\sqrt{q_{2}},\sqrt{2})$, by Lemma 2.4, we have
$\\{\varepsilon_{2},\sqrt{\varepsilon_{q_{2}}},\sqrt{\varepsilon_{2q_{2}}}\\}$
is a FSU of $K^{+}=\mathbb{Q}(\sqrt{q_{2}},\sqrt{2})$. One can check easily
that $q_{1}\varepsilon$ is not a square in $K^{+}$, for all
$\varepsilon\in\\{\varepsilon_{2}^{i}\sqrt{\varepsilon_{q_{1}}}^{j}\sqrt{\varepsilon_{2q_{2}}}^{k},\text{
with }i,j\text{ and }k\in\\{0,1\\}\\}$. It follows by [2, Proposition 3], that
$Q_{K}=1$. Thus $q(K)=4$. So using Lemmas 2.5 and 2.6 we get

$\begin{array}[]{ll}h_{2}(K)&=\frac{1}{2^{5}}q(K)h_{2}(-q_{1})h_{2}(q_{2})h_{2}(-2q_{1})h_{2}(2q_{2})h_{2}(-q_{1}q_{2})h_{2}(-2q_{1}q_{2})h_{2}(2)\\\
&=\;\frac{1}{2^{5}}\cdot 4\cdot 1\cdot 1\cdot 2\cdot 1\cdot 4\cdot
h_{2}(-2q_{1}q_{2})\cdot 1,\\\ &=h_{2}(-2q_{1}q_{2}).\end{array}$

Which completes the proof. ∎

##  3\. Some Iwasawa theory results

Let $K$ be a number field and $p$ a prime number. Let $K_{\infty}$ be a
$\mathbb{Z}_{p}$-extension of $K$. Denote by $K_{n}$, with $n\geq 0$ an
integer, the unique cyclic subfield of $K_{\infty}$ of degree $p^{n}$ over
$K$, i.e. the $n$-th layer of the $\mathbb{Z}_{p}$-extension of $K$. In [23],
Iwasawa proved that for $n$ sufficiently large, the highest power of $p$
dividing the class number of $K_{n}$ is
$p^{\lambda_{p}n+\mu_{p}p^{n}+\nu_{p}}$ for some integers $\lambda_{p}$,
$\mu_{p}\geq 0$ and $\nu_{p}$, all independent of $n$, called the Iwasawa
invariants of $K_{\infty}$. Later, for abelian number fields, Ferrero and
Washington proved that $\mu_{p}=0$ (cf. [16]). Denote by $A_{p}(K_{n})$ (or
simply $A_{n}$) the $p$-Sylow subgroup of the ideal class group of $K_{n}$,
and by $L(K_{\infty})$ (resp. $L(K_{n})$) the Hilbert $p$-class field of
$K_{\infty}$ (resp. $K_{n}$), then
$X=\mathrm{Gal}(L(K_{\infty})/K_{\infty})=\displaystyle\lim_{\longleftarrow}A_{p}(K_{n})$
is a module over the formal power series ring $\Lambda=\mathbb{Z}_{p}[[T]]$
(here the action of $1+T$ is defined by an inner automorphism induced from a
fixed topological generator of $\mathrm{G}(K_{\infty}/K)$). For $m\geq n\geq
0$, put $\omega_{n}=(1+T)^{p^{n}}-1\in\Lambda$, and
$\nu_{m,n}=\omega_{m}/\omega_{n}\in\Lambda$.

For $p=2$, the tower $K\subset K_{1}=K(\sqrt{2})\subset
K_{2}=K(\sqrt{2+\sqrt{2}})\subset\cdots$ defines a $\mathbb{Z}_{2}$-extension
called the cyclotomic $\mathbb{Z}_{2}$-extension of $K$ of layers $K_{i}$. We
need the following lemmas.

######  Lemma 3.1 ([13]).

Let $n\geq 1$ and $q$ a prime such that $q\equiv 3\pmod{8}$. Then $q$
decomposes into the product of $2$ prime ideals of
$K_{n}=\mathbb{Q}{(\zeta_{2^{n+2}})}$ while it is inert in $K_{n}^{+}$.

######  Lemma 3.2 ([13]).

Let $q$ be prime such that $q\equiv 3\pmod{8}$. Then the class number of
$\mathbb{Q}{(\zeta_{2^{n+2}},\sqrt{q})}$ is odd.

Let $rank_{2}(G)$ denote the $2$-rank of a group $G$.

######  Lemma 3.3 ([17]).

Let $K/k$ be a $\mathbb{Z}_{2}$-extension, $k_{n}$ its n-th layer and $n_{0}$
an integer such that any prime of $K$ which is ramified in $K/k$ is totally
ramified in $K/k_{n_{0}}$. If there exists an integer $n\geq n_{0}$ such that
$rank_{2}(\mathrm{Cl}(k_{n}))=rank_{2}(\mathrm{Cl}(k_{n+1}))$, then
$rank_{2}(\mathrm{Cl}(k_{m}))=rank_{2}(\mathrm{Cl}(k_{n}))$ for all $m\geq n$.

######  Lemma 3.4.

[26] Let $K$ and $F$ be $CM$-fields and $F/K$ a finite $2$-extension. Assume
that $\mu^{-}(K)=0$, then $\mu^{-}(F)=0$ and

$\lambda^{-}(F)-\delta(F)=[F_{\infty}:K_{\infty}]\left(\lambda^{-}(K)-\delta(K)\right)+\sum(e_{\beta}-1)-\sum(e_{\beta^{+}}-1),$

where $\delta(k)=1$ or $0$ according to whether $K_{\infty}$ contains the
fourth roots of unity or not, $e_{\beta}$ is the ramification index of a prime
$\beta$ in $F_{\infty}/K_{\infty}$ coprime to $2$ and $e_{\beta^{+}}$ is the
ramification index of a prime coprime to $2$ in
$F_{\infty}^{+}/K_{\infty}^{+}$.

Let us close this section with the following important lemma.

######  Lemma 3.5.

Let $q_{1}\equiv q_{2}\equiv 3\pmod{8}$ be two primes and $n\geq 1$. Let
$F_{n}=\mathbb{Q}(\zeta_{2^{n+2}},\sqrt{q_{1}},\sqrt{q_{2}})$. Then we have

  1. 1.

$\lambda(F)=1$, where $F=\mathbb{Q}(\sqrt{q_{1}},\sqrt{q_{2}},i)$.

  2. 2.

The class numbers of the layers of the $\mathbb{Z}_{2}$-extension of
$F^{+}=\mathbb{Q}(\sqrt{q_{1}},\sqrt{q_{2}})$ are odd.

  3. 3.

For all $n\geq 1$, the $2$-class group of $F_{n}$ is cyclic non trivial.

###### Proof.

Without loss of generality we may assume that $q_{1}$ and $q_{2}$ satisfy the
condition $\left(\dfrac{q_{1}}{q_{2}}\right)=1$. Set
$k=\mathbb{Q}(\sqrt{q_{1}},i)$. Let us firstly prove that the $2$-class group
of $F_{1}$ is cyclic non trivial. It is easy to see that there are exactly two
primes of $F$ that ramify in $F_{1}$. Since, by Lemma 2.8, the class number of
$F$ is odd, so by ambiguous class number formula (cf. [18])
$rank(\mathrm{Cl}_{2}(F_{1}))=2-1-e=1-e$, where $e$ is defined by
$[E_{F}:E_{F}\cap N_{F_{1}/F}(F_{1})]=2^{e}$. Then the $2$-class group of
$F_{1}$ is trivial or cyclic non-trivial.

Since $\left(\dfrac{q_{1}}{q_{2}}\right)=1$, then $q_{2}$ decomposes into the
product of $2$ primes of $\mathbb{Q}(\sqrt{q_{1}})$. So by Lemma 3.1, $q_{2}$
decomposes into the product of $4$ primes in $k_{n}(\sqrt{q_{1}})$ while it
decomposes into $2$ primes in $k_{n}^{+}(\sqrt{q_{1}})$. Since
$[F_{\infty}:k_{\infty}]=2$, then by Lemma 3.4, we have

$\lambda^{-}(F)-1=2\cdot\left(0-1\right)+4-2,$ and thus $\lambda^{-}(F)=1$.

Assume that the $2$-class group of $F_{1}$ is trivial. Note that the layers of
the cyclotomic $\mathbb{Z}_{2}$-extension of $F$ are given by
$F_{n}=\mathbb{Q}(\zeta_{2^{n+2}},\sqrt{q_{1}},\sqrt{q_{2}})$. Then by Lemma
3.3, the $2$-class group of $F_{n}$ is trivial for all $n\geq 1$. Thus
$\lambda(F)=0$, which contradicts the fact that $\lambda^{-}(F)=1$. Thus the
$2$-class group of $F_{1}$ is cyclic non-trivial. By Lemma 2.8 and Proposition
2.7, the class numbers of $F^{+}$ and $F_{1}^{+}$ are odd, so by Lemma 3.3 the
class number of $F_{n}^{+}$ is odd for all $n\geq 1$. Therefore
$\lambda(F)^{+}=0$ and $\lambda(F)=\lambda(F)^{-}=1$. We adopt the definitions
of $A_{n}^{+}$ and $A_{n}^{-}$ given in [30]. Note that there is no finite
submodule in $A_{\infty}^{-}$ (cf. [30, Theorem 2.5]).

Since $h(F_{n}^{+})$ is odd for all $n$, then by the definition of $A_{n}^{+}$
is trivial or contains only the primes above $2$ that are ramified in
$F_{n}/F_{n}^{+}$. Note that index of ramification of $2$ in $F_{1}$ (resp.
$F_{1}^{+}$) equals $4$. Therefore the primes above $2$ are unramified in
$F_{1}/F_{1}^{+}$ and so they are also unramified in $F_{n}/F_{n}^{+}$, for
all $n$. Then $A_{n}^{+}$ is trivial. Hence there is no finite part in
$A_{\infty}$. Thus $A_{n}$ is cyclic for all $n\geq 1$.

∎

##  4\. Proof of the main Theorem

Let us quote the last ingredient needed to prove the main theorem.

######  Lemma 4.1 ([22]).

Let $K/k$ be a quadratic extension and $\mu\in k$ prime to $2$ such that
$K=k(\sqrt{\mu})$. The extension $K/k$ is unramified at finite primes if and
only if $\mu$ verify the following properties:

  1. 1.

The principal ideal generated by $\mu$ is a square of a fractional ideal of
$k$.

  2. 2.

There exists $\xi\in k$ such that $\mu\equiv\xi^{2}\pmod{4}$.

###### Proof of Theorem 1.1 .

We know, by [14], that the $2$-class group of $K_{n}=K_{n,d}$ is isomorphic to
$\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2^{n+m-2}\mathbb{Z}$, whenever $d$
takes one of the forms appearing in Theorem 1.1. Let us start by proving the
assertions of the first item.

  1. 1.

Note that $F_{n}$ is the genus field of $K_{n,d}$, $[F_{n}:K_{n,d}]=2$ and its
$2$-class group is cyclic, then by class field theory the first Hilbert
$2$-class field $F_{n}^{(1)}$ of $F_{n}$ coincides with the second Hilbert
$2$-class field $K_{n,d}^{(2)}$ of $K_{n,d}$.

To this end, we claim that $K_{n,d}^{(1)}=F_{n}^{(1)}=K_{n,d}^{(2)}$. To prove
this it suffices then to verify that
$h_{2}(F_{n})=\frac{h_{2}(K_{n,d})}{2}=2^{n+m-2}.$

Note that by Lemma 3.5, $A_{n}(F_{n})$, the $2$-class group of $F_{n}$ is
cyclic and that $\lambda=1$. Therefore, the $\Lambda$-module
$A_{\infty}=\displaystyle\lim_{\longleftarrow}(A_{n}(F_{n}))$ is isomorphic to
its elementary module. Since every prime which ramify in $F_{\infty}/F$ is
totally ramified, then according to [32, pp. 282-283], one gets
$\nu_{n,0}A_{\infty}=2\nu_{n-1,0}A_{\infty}$ for all $n\geq 1$. Hence,

$|A_{n}|=|A_{\infty}/\nu_{n,0}A_{\infty}|=|A_{\infty}/2^{n-1}A_{\infty}||A_{\infty}/\nu_{1,0}A_{\infty}|=2^{n+s},$

for $n\geq 1$ and some constant $s\geq 0$. Hence the Iwasawa formula holds for
all $n\geq 1$, from which we infer that $h_{2}(F_{1})=2^{1+s}$. It follows, by
Lemmas 2.5 and 2.6 that
$\begin{array}[]{ll}h_{2}(F_{1})&=\frac{1}{2^{16}}q(F_{1})h_{2}(q_{1})h_{2}(-q_{1})h_{2}(q_{2})h_{2}(-q_{2})h_{2}(2q_{1})h_{2}(-2q_{1})h_{2}(2q_{2})\\\
&\qquad
h_{2}(-2q_{2})h_{2}(q_{1}q_{2})h_{2}(-q_{1}q_{2})h_{2}(2q_{1}q_{2})h_{2}(-2q_{1}q_{2})h_{2}(2)h_{2}(-2)h_{2}(-1)\\\
&=\;\frac{1}{2^{16}}\cdot q(F_{1})\cdot 1\cdot 1\cdot 1\cdot 1\cdot 1\cdot
2\cdot 1\cdot 2\cdot 1\cdot 4\cdot 2\cdot h_{2}(-2q_{1}q_{2})\cdot 1\cdot
1\cdot 1,\\\ &=\frac{1}{2^{11}}q(F_{1})h_{2}(-2q_{1}q_{2}).\end{array}$
We have to compute $q(F_{1})$, i.e., $Q_{F_{1}}$. Note that $Q_{F_{1}}$ must
be $2$, for if $Q_{F_{1}}=1$, then by Proposition 2.7, $q(F_{1})=2^{9}$. So
$h_{2}(F_{1})=\frac{1}{4}h_{2}(-2q_{1}q_{2})$. Considering the field
$K=\mathbb{Q}(\sqrt{-q_{1}},\sqrt{q_{2}},\sqrt{2})$, since $F_{1}/K$ is an
unramified quadratic extension, then $h_{2}(F_{1})\geq\frac{1}{2}h_{2}(K)$,
and thus by Lemma 2.8, we get
$h_{2}(F_{1})\geq\frac{1}{2}h_{2}(-2q_{1}q_{2})$. So we have a contradiction.
Hence $Q_{F_{1}}=2$, and so by Lemma 2.1 and Proposition 2.7,
$q(F_{1})=2^{10}$. Therefore,
$h_{2}(F_{1})=\frac{h_{2}(K_{1,d})}{2}=2^{1+m-2}=2^{1+s}$, this in turn
implies, using Iwasawa formula, that
$h_{2}(F_{n})=2^{n+m-2}=\frac{h_{2}(K_{n,d})}{2}$. Hence
$K_{n,d}^{(1)}=K_{n,d}^{(2)}$. Which gives the first item of the main theorem.

  2. 2.

Now assume that we are in the conditions of the second item of the main
theorem. Put $K_{n}^{\prime}=\mathbb{Q}{(\zeta_{2^{n+2}})}$,
$p=a^{2}+16b^{2}=e^{2}-32f^{2}$ and $\pi_{1}^{\prime}=a+4bi$,
$\pi_{1}^{\prime\prime}=a-4bi$, $\pi_{2}^{\prime}=e+4f\sqrt{2}$ and
$\pi_{2}^{\prime\prime}=e-4f\sqrt{2}$.

So $p=\pi_{r}^{\prime}\pi_{r}^{\prime\prime}$, for $r\in\\{1,2\\}$. Note that
by [12, Proposition 1], $p$ splits into $4$ ideals of $K_{n}^{\prime}$. Thus,
there are exactly $2$ primes of $K_{n}^{\prime}$ above $\pi_{r}^{\prime}$
(resp. $\pi_{r}^{\prime\prime}$). Note also that by [12, Lemma 6], we have the
following norm residue symbols
$\left(\frac{\zeta_{2^{n+2}},p}{\mathfrak{p}_{K_{n}^{\prime}}}\right)=-1=\left(\frac{\zeta_{2^{n+2}},\pi_{r}^{\prime}\pi_{r}^{\prime\prime}}{\mathfrak{p}_{K_{n}^{\prime}}}\right)$.
So choose $\pi_{r}\in\\{\pi_{r}^{\prime},\pi_{r}^{\prime\prime}\\}$ such that
$\left(\frac{\zeta_{2^{n+2}},\pi_{r}}{\mathfrak{p}_{K_{n}^{\prime}}}\right)=-1$,
where $\mathfrak{p}_{K_{n}^{\prime}}$ is a prime ideal of $K_{n}^{\prime}$
above $\pi_{r}$.

Since the primes of $K_{n}^{\prime}$ that ramify in $K_{n,p}$ are exactly the
divisors of $p$ in $K_{n}^{\prime}$, then the ideals of $K_{n,p}$ generated by
$\pi_{1}$ and $\pi_{2}$ are squares of ideals of $K_{n,p}$. As $a\equiv
e\equiv\pm 1\equiv i^{2}\pmod{4}$, since they are odd. It follows that the
equations $\pi_{r}\equiv\xi^{2}\pmod{4}$, $r=1$ or $2$, have solutions.
Therefore, By Lemma 4.1, $L_{n,1}=K_{n,p}(\sqrt{\pi_{1}})$ and
$L_{n,2}=K_{n,p}(\sqrt{\pi_{2}})$ are two distinct unramified quadratic
extensions of $K_{n,p}$.

Let us now show that the $2$-class groups of $K_{1}$ and $K_{2}$ are cyclic.
Put $k_{n,r}=\mathbb{Q}(\zeta_{2^{n+2}},\sqrt{\pi_{r}})$. By the ambiguous
class number formula (cf. [18]) applied on the extension
$k_{n,r}/K_{n}^{\prime}$, we have $rank(\mathrm{Cl}_{2}(k_{n,r}))=2-1-e=1-e$.
Since
$\left(\frac{\zeta_{2^{n+2}},\pi_{r}}{\mathfrak{p}_{K_{n}^{\prime}}}\right)=-1$,
then $e\not=0$. Hence $rank(\mathrm{Cl}_{2}(k_{n,r}))=0$. So the class number
of $k_{n,r}$ is odd. Then again by the ambiguous class number formula (cf.
[18]) applied on the extension $L_{n,r}/k_{n,r}$, we have
$rank(\mathrm{Cl}_{2}(L_{n,i}))=2-1-e=1-e$. Thus $\mathrm{Cl}_{2}(L_{n,r})$,
is either trivial or cyclic non trivial, but since $\mathrm{Cl}_{2}(K_{n,p})$
is of type $(2,2^{2^{n+m-2}})$ and $L_{n,r}/K_{n,p}$ is an unramified
quadratic extension, we infer that $\mathrm{Cl}_{2}(L_{n,r})$, can not be
trivial. Hence $\mathrm{Cl}_{2}(L_{n,r})$ is cyclic. It follows that the
second $2$-class group of $K_{n,p}$ is abelian or modular (cf. [19, Theorem
12.5.1]).

Assume now that $n=1$, we use the fact that $K_{1,p}$ is the genus field of
$\mathbb{Q}(\sqrt{2p},i)$. As, by [4, Théorème 5], the second $2$-class group
of $\mathbb{Q}(\sqrt{2p},i)$ is abelian, then by class field theory the second
$2$-class group of $K_{1,p}$ is abelian too. Which completes the proof of the
main Theorem.

∎

######  Remark 4.2.

Keep the notations of the second part of the above proof. Then $L_{n,r}$, with
$r=1$ or $2$, are two unramified quadratic extensions of $K_{n,p}$, for which
the $2$-class groups are cyclic.

######  Remark 4.3.

Let $d$ be in one of the forms appearing in Theorem 1.1. Since by [29,
Proposition 2], the $2$-class group of $K_{0,d}:=K=\mathbb{Q}(\sqrt{d},i)$ is
cyclic, then the $2$-class field tower of $K_{0,d}$ terminates at the first
step.

Now we treat the capitulation of the $2$-classes of each $K_{n}$ and the
orders of the three unramified quadratic extensions of $K_{n}$. We need the
following lemma:

######  Lemma 4.4 ([9]).

Let $k$ be a number field such that $\mathrm{Cl}_{2}(k)\simeq(2^{m},2^{n})$
for some positive integers $m$ and $n$. If there is an unramified quadratic
extension of $k$ with $2$-class number $2^{m+n-1}$, then all three unramified
quadratic extensions of $k$ have $2$-class number $2^{m+n-1}$, and the
$2$-class field tower of $k$ terminates at $k^{(1)}$.

Conversely, if the $2$-class field tower of $k$ terminates at $k^{(1)}$, then
all three unramified quadratic extensions of $k$ have $2$-class number
$2^{m+n-1}$.

By the main Theorem, Lemma 4.4 and [10] (or [7]), one easily deduces the
following corollary.

######  Corollary 4.5.

Let $d$ be a positive square-free integer and $m$ such that
$h_{2}(-2d)=2^{m}$.

  1. 1.

If $d=q_{1}q_{1}$, where $q_{i}\equiv 3\pmod{8}$ are two distinct primes, then
for all $n\geq 1$ the class numbers of the three unramified quadratic
extensions of $K_{n,d}$ equal $h_{2}(F_{n})=2^{n+m-2}$. Furthermore, there are
exactly $4$ classes of the $2$-class group of $K_{n,d}$ capitulating in each
unramified quadratic extension of $K_{n,d}$.

  2. 2.

Assume $d=p\equiv 9\pmod{16}$ is a prime such that $(\frac{2}{p})_{4}=1$.

    1. a.

In the abelian case, for all $n\geq 1$, the class numbers of the three
unramified quadratic extensions of $K_{n,d}$ equal $2^{n+m-2}$, and there are
exactly $4$ classes of the $2$-class group of $K_{n,d}$ capitulating in each
unramified quadratic extension of $K_{n,d}$.

    2. b.

In the modular case and for $n\geq 2$, there are only $2$ classes of the
$2$-class group of $K_{n,d}$ capitulating in each unramified quadratic
extensions of $K_{n,d}$.

######  Corollary 4.6.

Let $q_{1}\equiv q_{2}\equiv 3\pmod{8}$ be two primes. Then we have

$\mathrm{Cl}_{2}(F_{n})\simeq\mathbb{Z}/2^{n+m-2}\mathbb{Z},$

for all $n\geq 1$.

######  Remark 4.7.

Let $d$ be a square-free integer such that the $2$-class group of
$K_{1,d}=\mathbb{Q}(\sqrt{d},\sqrt{2},\sqrt{-1})$ is of type $(2,4)$, for the
conditions that make the $2$-class group of $K_{1,d}$ of type $(2,4)$ see [3].
Then by Theorem 1.1 the class field tower of $K_{1,d}$ stops at the first step
$K_{1,d}^{(1)}$ and $G=\mathrm{Gal}(K_{1,d}^{(1)}/K_{1,d})$ is abelian.
Therefore, by [5, page 112], $4$ classes (resp. all the classes) of the
$2$-class group of $K_{n,d}$ capitulate in each unramified quadratic (resp.
biquadratic) extension of $K_{n,d}$.

## Acknowledgments

The authors would like to express their gratitude to Katharina Müller for her
comments on the previous versions of this paper. Particularly the first author
would like to thank her for explaining to him many things in Iwasawa Theory.

## References

  * [1] A. Azizi, Sur la capitulation des $2$-classes d’idéaux de $\mathds{k}=\mathbb{Q}(\sqrt{2pq},i)$, où $p\equiv-q\equiv 1\pmod{4}$,  Acta. Arith. 94 (2000), 383-399.
  * [2] A. Azizi, Unités de certains corps de nombres imaginaires et abéliens sur $\mathbb{Q}$,  Ann. Sci. Math. Québec, 23 (1999), 15-21.
  * [3] A. Azizi, M. M. Chems-Eddin and A. Zekhnini, On the $2$-class group of some imaginary triquadratic number fields, arXiv:2002.03094.
  * [4] A. Azizi et M. Taous, Capitulation des $2$-classes d’idéaux de $k=\mathbb{Q}(\sqrt{2p},i)$,  Acta Arith., 131 (2008), 103–123.
  * [5] A. Azizi, M. Taous and A. Zekhnini, On the $2$-groups whose abelianizations are of type $(2,4)$ and applications, Publ. Math. Debrecen 88/1-2 (2016), 93-117.
  * [6] E. Benjamin, On the second Hilbert $2$-class field of real quadratic number fields with $2$-class group isomorphic to $(2,2^{n})$, $n\geq 2$, Rocky Mt. J. Math., 29 (1999) 763-788.
  * [7] E. Benjamin, On the $2$-class field tower of some imaginary biquadratic number fields, Ramanujan J., 11 (2006) 103-110.
  * [8] E. Benjamin, F. Lemmermeyer and C. Snyder, On the unit group of some multiquadratic number felds, Paci. J. Math., 230 (2007), 27–40.
  * [9] E. Benjamin, F. Lemmermeyer and C. Sneyder, Real quadratic fields with abelian $2$-class field tower, J. Number Theory, 73 (1998) 182-194 (13).
  * [10] E. Benjamin and C. Snyder,  Number fields with $2$-class number isomorphic to $(2,2^{m})$, $m>1$, preprint, 1993.
  * [11] N. Blackburn On prime-power groups in which the derived group has two generators. Proceedings of the Cambridge Philosophical Society (1957) 53, 19-27.
  * [12] M. M. Chems-Eddin, The rank of the $2$-class group of some fields with large degree.Algebra Colloq. (To appear), arXiv:2001.00865v2
  * [13] M. M. Chems-Eddin, A. Azizi and A. Zekhnini, On the $2$-class group of some number fields with large degree, Arch. Math. (Brno), 57 (2021), 13-26.
  * [14] M. M. Chems-Eddin and K. Müller, $2$-Class groups of cyclotomic towers of imaginary biquadratic fields and applications, Int. J. Number Theory, https://doi.org/10.1142/S1793042121500627.
  * [15] P. E. Conner and J. Hurrelbrink, Class number parity,  Ser. Pure. Math. $8$, World Scientific, 1988.
  * [16] B. Ferrero and L. C. Washington, The Iwasawa invariant $\mu_{p}$ vanishes for abelian number fields, Annals. Math., Second Series, 109 (1979), 377–395.
  * [17] T. Fukuda, Remarks on $\mathbb{Z}_{p}$-extensions of number fields. Proc. Japan Acad. Ser. A Math. Sci. $70$ $(1994)$, $264$–$266$.
  * [18] G. Gras, Sur les l-classes d’idéaux dans les extensions cycliques relatives de degré premier l, Ann. Inst. Fourier (Grenoble) 23 (1973), 1–48.
  * [19] M. Hall, The theory of groups.  American Mathematical Soc., (1999).
  * [20] H. Hasse, Über die Klassenzahl abelscher Zahlkörper, Berlin, Akademie-Verlag, (1952).
  * [21] G. Herglotz, Über einen Dirichletschen Satz, Math. Z., 12 (1922), 255–261.
  * [22] D. Hilbert. Über die Theorie des relativquadratischen Zahlkörpers. Mathematische Annalen 51.1 (1898), 1-127.
  * [23] K. Iwasawa, On $\varGamma$-extensions of algebraic number fields, Bull. Amer. Math. Soc. 65 (1959), 183–226.
  * [24] M. Ishida, The genus fields of algebraic number fields, Lecture Notes in Math. 555, Springer, 1976.
  * [25] P. Kaplan, Sur le $2$-groupe des classes d’idéaux des corps quadratiques,  J. Reine. Angew. Math., 283/284 (1976), 313-363.
  * [26] Y. Kida, Cyclotomic $\mathbb{Z}_{2}$-extensions of J-fields, J. Number Theory, 14 (1982), 340-352.
  * [27] H. Kisilevsky, Number fields with class number congruent to $4\pmod{8}$ and Hilbert’s theorem $94$,  J. Number Theory, 8 (1976), 271-279.
  * [28] S. Kuroda, Über die Klassenzahlen algebraischer Zahlkörper, Nagoya Math. J., 1 (1950), 1–10.
  * [29] T. M. McCall, C. J. Parry and R. R. Ranalli, Imaginary bicyclic biquadratic fields with cyclic $2$-class group, J. Number Theory, 53 (1995), 88–99.
  * [30] K. Müller, Capitulation in the $\mathbb{Z}_{2}$ extension of CM number fields. Math. Proc. Camb. Phil. Soc., 166 (2019), 371-380.
  * [31] H. Wada, On the class number and the unit group of certain algebraic number fields. J. Fac. Sci. Univ. Tokyo, 13 (1966), 201–209.
  * [32] L. C. Washington, Introduction to cyclotomic fields. Graduate Texts in Mathematics, 83. Springer, 1997.
  * [33] A. Zekhnini, A. Azizi and M. Taous, On the generators of the $2$-class group of the field $\mathbb{Q}(\sqrt{q_{1}q_{2}p},i)$ Correction to Theorem $3$ of [5], IJPAM, Volume 103, No. 1 (2015), 99-107.