# On the capitulation problem of some pure metacyclic fields
of degree 20

###### Abstract

Let $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$ be a pure quintic field, where $n$ is
a positive integer $5^{th}$ power-free, $k_{0}\,=\,\mathbb{Q}(\zeta_{5})$ be
the cyclotomic field containing a primitive $5^{th}$ root of unity
$\zeta_{5}$, and $k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_{5})$ the normal closure
of $\Gamma$. Let $k_{5}^{(1)}$ be the Hilbert $5$-class field of $k$,
$C_{k,5}$ the $5$-ideal classes group of $k$, and $C_{k,5}^{(\sigma)}$ the
group of ambiguous classes under the action of $Gal(k/k_{0})$ =
$\langle\sigma\rangle$. When $C_{k,5}$ is of type $(5,5)$ and rank
$C_{k,5}^{(\sigma)}\,=\,1$, we study the capitulation problem of the $5$-ideal
classes of $C_{k,5}$ in the six intermediate extensions of $k_{5}^{(1)}/k$.

Fouad ELMOUHIB
Department of Mathematics and Computer Sciences,
Mohammed first University, Oujda, Morocco,
Correspondence<EMAIL_ADDRESS>
Mohamed TALBI
Regional Center of Professions of Education and Training,
<EMAIL_ADDRESS>

Abdelmalek AZIZI
Department of Mathematics and Computer Sciences,
Mohammed first University, Oujda, Morocco,
<EMAIL_ADDRESS>
Key words: pure metacyclic fields, 5-class groups, Hilbert 5-class field,
Capitulation.
AMS Mathematics Subject Classification: 11R04, 11R18, 11R29, 11R37

##  1 Introduction

Let $K$ be a number field of finite degree over $\mathbb{Q}$. Let $F$ be an
unramified extension of $K$ of finite degree and let $\mathcal{O}_{F}$ be its
ring of integers. We say that an ideal $\mathcal{A}$ ( or the ideal class of
$\mathcal{A}$) of $K$ capitulates in $F$, if it becomes principal in $F$, i.e
if $\mathcal{A}\mathcal{O}_{F}$ is principal in $F$. Let $p$ a prime number
and $C_{K,p}$ be the $p$-ideal class group of $K$ such that $C_{K,p}$ is of
type $(p,p)$. Let $K_{p}^{(1)}$ be the Hilbert $p$-class field of $K$, i.e the
maximal abelian unramified extension of $K$. The most important result on
capitulation is the "Principal Ideal Theorem" conjectured by D. Hilbert in
References and proved by His student P.
Furtw$\displaystyle\stackrel{{\scriptstyle\text{..}}}{{a}}$ngler in
References. The theorem asserts that the class group of a given number field
capitulates completely in its Hilbert class field. Having established the
"Principal Ideal Theorem", the question remains which ideal classes of K
capitulate in a field which lies between $K$ and its Hilbert class field. A
number of researchers have studied this question, Taussky and Scholz in
References were the first who treated the capitulation in unramified cyclic
degree-3-extensions of several imaginary quadratic fields. Tannaka and Terada
proved that for a cyclic extension $K/K_{0}$ with Galois group $G$, the
$G$-invariant ideal classes in $K$ capitulate in the genus field of $K/K_{0}$,
for more details see References. Hilbert in his celebrated Zahlbericht, he
proved his Theorem 94 which states that in an unramfied cyclic extension
$L/K$, there are non-principal ideals which capitulates in $L$. For more
several results on capitulation we refer the reader to References, References,
References, References, References References…
In this paper we consider $p=5$ and $k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_{5})$,
$(\zeta_{5}\,=\,e^{\frac{2i\pi}{5}})$, be the normal closure of the pure
quintic field $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$, where $n$ is a positive
integer $5^{th}$ power free. We suppose that the $5$-class group $C_{k,5}$ of
$k$ is of type $(5,5)$, then the Hilbert $5$-class field $k_{5}^{(1)}$ of $k$
has degree $25$ over $k$, and the extension $k_{5}^{(1)}/k$ contains six
intermediate extensions. By $k_{0}\,=\,\mathbb{Q}(\zeta_{5})$ we denote the
cyclotomic field containing a primitive $5^{th}$ root of unity $\zeta_{5}$.
Let $C_{k,5}^{(\sigma)}$ be the group of ambiguous ideal classes of $k$ under
the action of $Gal(k/k_{0})\,=\,\langle\sigma\rangle$. According to
References, if $C_{k,5}$ is of type $(5,5)$ and rank
$C_{k,5}^{(\sigma)}\,=\,1$ we have three forms of the radicand $n$ as follows

  * -

$n\,=\,5^{e}p\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$e\in\\{1,2,3,4\\}$ and p prime such that
$p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$.

  * -

$n\,=\,p^{e}q\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$e\in\\{1,2,3,4\\}$, p and q are primes such that
$p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$, $q\,\not\equiv\,\pm
7\,(\mathrm{mod}\,25)$.

  * -

$n\,=\,p^{e}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with $e\in\\{1,2,3,4\\}$
and p prime such that $p\,\equiv\,-1\,(\mathrm{mod}\,25)$.

We will study the capitulation of $C_{k,5}$ in the six intermediate extensions
of $k_{5}^{(1)}/k$ In this cases.

Notations.

Throughout this paper, we use the following notations:

  * •

$\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$: a pure quintic field, where $n\neq 1$ is
a $5^{th}$ power-free positive integer.

  * •

$k_{0}\,=\,\mathbb{Q}(\zeta_{5})$, the cyclotomic field, where
$\zeta_{5}\,=\,e^{2i\pi/5}$ a primitive $5^{th}$ root of unity.

  * •

$k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_{5})$: the normal closure of $\Gamma$, a
quintic Kummer extension of $k_{0}$.

  * •

$\Gamma^{{}^{\prime}},\,\Gamma^{{}^{\prime\prime}},\,\Gamma^{{}^{\prime\prime\prime}},\,\Gamma^{{}^{\prime\prime\prime\prime}},\,$
the four conjugates quintic fields of $\Gamma$, contained in $k$.

  * •

$\langle\tau\rangle\,=\,\operatorname{Gal}(k/\Gamma)$ such that
$\tau^{4}\,=\,id,\,\tau^{3}(\zeta_{5})\,=\,\zeta_{5}^{3},\,\tau^{2}(\zeta_{5})\,=\,\zeta_{5}^{4},\,\tau(\zeta_{5})\,=\,\zeta_{5}^{2}$
and $\tau(\sqrt[5]{n})\,=\,\sqrt[5]{n}$.

  * •

$\langle\sigma\rangle\,=\,\operatorname{Gal}(k/k_{0})$ such that
$\sigma^{5}\,=\,id,\ \sigma(\zeta_{5})\,=\,\zeta_{5}$ and
$\sigma(\sqrt[5]{n})\,=\,\zeta_{5}\sqrt[5]{n},\,\sigma^{2}(\sqrt[5]{n})\,=\,\zeta_{5}^{2}\sqrt[5]{n},\,\\\
\\\
\sigma^{3}(\sqrt[5]{n})\,=\,\zeta_{5}^{3}\sqrt[5]{n},\,\sigma^{4}(\sqrt[5]{n})\,=\,\zeta_{5}^{4}\sqrt[5]{n}$.

  * •

For a number field $L$, denote by:

    * –

$\mathcal{O}_{L}$: the ring of integers of $L$;

    * –

$C_{L}$, $h_{L}$, $C_{L,5}$: the class group, class number, and $5$-class
group of $L$.

    * –

$L_{5}^{(1)},L^{*}$: the Hilbert $5$-class field of $L$, and the absolute
genus field of $L$.

$\mathbf{k_{5}^{(1)}}$$\mathbf{K_{4}}$$\mathbf{K_{5}}$$\mathbf{K_{6}}$$\mathbf{K_{3}}$$\mathbf{K_{2}}$$\mathbf{K_{1}}$$\mathbf{k}$$\mathbf{\Gamma}$
$\mathbf{\Gamma^{\prime}}$ $\mathbf{\Gamma^{\prime\prime}}$
$\mathbf{\Gamma^{\prime\prime\prime}}$
$\mathbf{\Gamma^{\prime\prime\prime\prime}}$ $\mathbf{k_{0}}$$\mathbb{Q}$
Figure 1

##  2 Construction of intermediate extensions

###  2.1 Absolut genus field

In chapter $7$ of [References], Ishida has explicitly given the genus field of
any pure field. For the pure quintic field $\Gamma=\mathbb{Q}(\sqrt[5]{n})$
where $n$ is $5^{th}$ power-free naturel number, we have the following
theorem.

######  Theorem 2.1.

Let $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$ be a pure quintic field, where $n\geq
2$ is $5^{th}$ power-free naturel number, and let $p_{1}\,,,,,p_{r}$ be all
primes divisors of $n$ such that $p_{i}$ congruent 1 $(\mathrm{mod}\,5)$ for
each $i\in\\{1,,,,,r\\}$. Let $\Gamma^{*}$ be the absolute genus field of
$\Gamma$, then

$\Gamma^{*}\,=\,\prod_{i=1}^{r}M(p_{i}).\Gamma$

where $M(p_{i})$ denotes the unique subfield of degree 5 of the cyclotomic
field $\mathbb{Q}(\zeta_{p_{i}})$. The genus number of $\Gamma$ is given by,
$g_{\Gamma}\,=\,[\Gamma^{*}:\Gamma]\,=\,5^{r}$.

######  Remark 2.1.

\- If no prime $p\,\equiv\,1\,(\mathrm{mod}\,5)$ divides $n$, i.e $r=0$, then
$\Gamma^{*}\,=\,\Gamma$.

\- For any $r\geq 0$, $\Gamma^{*}$ is contained in the Hilbert $5$-class field
$\Gamma_{5}^{(1)}$ of $\Gamma$.

######  Corollary 2.1.

(1) If $5$ divides exactly the class number $h_{\Gamma}$, then there is at
most one prime $p\,\equiv\,1\,(\mathrm{mod}\,5)$ divides the integer $n$.

(2) If $5$ divides exactly $h_{\Gamma}$, and if a prime
$p\,\equiv\,1\,(\mathrm{mod}\,5)$ divides $n$, then
$\Gamma^{*}\,=\,\Gamma_{5}^{(1)}$;
$(\Gamma^{{}^{\prime}})^{*}\,=\,(\Gamma^{{}^{\prime}})_{5}^{(1)}$;
$(\Gamma^{{}^{\prime\prime}})^{*}\,=\,(\Gamma^{{}^{\prime\prime}})_{5}^{(1)}$;
$(\Gamma^{{}^{\prime\prime\prime}})^{*}\,=\,(\Gamma^{{}^{\prime\prime\prime}})_{5}^{(1)}$;
$(\Gamma^{{}^{\prime\prime\prime\prime}})^{*}\,=\,(\Gamma^{{}^{\prime\prime\prime\prime}})_{5}^{(1)}$.
Furthermore,
$k.\Gamma_{5}^{(1)}\,=\,k.(\Gamma^{{}^{\prime}})_{5}^{(1)}\,=\,k.(\Gamma^{{}^{\prime\prime}})_{5}^{(1)}\,=\,k.(\Gamma^{{}^{\prime\prime\prime}})_{5}^{(1)}\,=\,k.(\Gamma^{{}^{\prime\prime\prime\prime}})_{5}^{(1)}$

###### Proof.

(1) If $p_{1},,,,p_{r}$ are all prime congruent to $1\,(\mathrm{mod\,}5)$ and
divides $n$, then $5^{r}|h_{\Gamma}$ therfore if $5$ divides exactly
$h_{\Gamma}$ then $r=1$, so one prime $p\,\equiv\,1\,(\mathrm{mod}\,5)$
divides $n$ exist, or no one.

(2) If $h_{\Gamma}$ is exactly divisible by $5$ and
$p\,\equiv\,1\,(\mathrm{mod}\,5)$ divides $n$, then $g_{\Gamma}=5$ so,
$\Gamma^{*}\,=\,\Gamma_{5}^{(1)}\,=\,\Gamma.M(p)$, we have $g_{\Gamma}=5$ so,
$(\Gamma^{{}^{\prime}})^{*}\,=\,\Gamma^{{}^{\prime}}\prod_{i=1}^{r}.M(p_{i})$
where $p_{1},,,,p_{r}$ are prime congruent to $1$ $(\mathrm{mod}\,5)$ and
divides $n$, if $5$ divides exactly $h_{\Gamma}$ then $5$ divides exactly also
$h_{\Gamma^{{}^{\prime}}}$, because $\Gamma$ and $\Gamma^{{}^{\prime}}$ are
isomorphic, so we obtaint
$(\Gamma^{{}^{\prime}})^{*}\,=\,(\Gamma^{{}^{\prime}})_{5}^{(1)}$. For the
others equality we use the same reasoning. Moreover,

$\left\\{\begin{array}[]{ll}k.\Gamma_{5}^{(1)}\,=\,k.\Gamma.M(p)\,=\,k.M(p)\\\
k.(\Gamma^{{}^{\prime}})_{5}^{(1)}\,=\,k.\Gamma^{{}^{\prime}}.M(p)\,=\,k.M(p)\\\
k.(\Gamma^{{}^{\prime\prime}})_{5}^{(1)}\,=\,k.\Gamma^{{}^{\prime\prime}}.M(p)\,=\,k.M(p)\\\
k.(\Gamma^{{}^{\prime\prime\prime}})_{5}^{(1)}\,=\,k.\Gamma^{{}^{\prime\prime\prime}}.M(p)\,=\,k.M(p)\\\
k.(\Gamma^{{}^{\prime\prime\prime\prime}})_{5}^{(1)}\,=\,k.\Gamma^{{}^{\prime\prime\prime\prime}}.M(p)\,=\,k.M(p)\\\
\end{array}\right.$ (1)

Hence,
$k.\Gamma_{5}^{(1)}\,=\,k.(\Gamma_{5}^{{}^{\prime}})^{(1)}\,=\,k.(\Gamma_{5}^{{}^{\prime\prime}})^{(1)}\,=\,k.(\Gamma_{5}^{{}^{\prime\prime\prime}})^{(1)}\,=\,k.(\Gamma_{5}^{{}^{\prime\prime\prime\prime}})^{(1)}$
∎

######  Remark 2.2.

If $h_{\Gamma}$ is exactly divisible by $5$, and the radicand $n$ is not
divisible by any prime $p\,\equiv\,1\,(\mathrm{mod}\,5)$, then the fields
$k.\Gamma_{5}^{(1)}$, $k.(\Gamma_{5}^{{}^{\prime}})^{(1)}$,
$k.(\Gamma_{5}^{{}^{\prime\prime}})^{(1)}$,
$k.(\Gamma_{5}^{{}^{\prime\prime\prime}})^{(1)}$ and
$k.(\Gamma_{5}^{{}^{\prime\prime\prime\prime}})^{(1)}$ are distincts.

###  2.2 Relative genus field $(k/k_{0})^{*}$ of $k$ over $k_{0}$

Let $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$ be a pure quintic field,
$k_{0}\,=\,\mathbb{Q}(\zeta_{5})$ the $5^{th}$-cyclotomic field and
$k\,=\,\Gamma(\zeta_{5})$ be the normal closure of $\Gamma$. The relative
genus field $(k/k_{0})^{*}$ of $k$ over $k_{0}$ is the maximal abelian
extension of $k_{0}$ which is contained in the Hilbert $5$-class field
$k_{5}^{(1)}$ of $k$. A class $\mathcal{A}\in C_{k,5}$ is called ambiguous
class relativly to $k/k_{0}$ if $\mathcal{A}^{\sigma}\,=\,\mathcal{A}$ such
that $Gal(k/k_{0})\,=\,\langle\sigma\rangle$. We note the group of ambiguous
classes by $C_{k,5}^{(\sigma)}$.
The following proposition summarize some importants results of References.
Since $k$ is quintic Kummer extension of $k_{0}$, we can write
$k\,=\,k_{0}(\sqrt[5]{n})$ such that
$n\,=\,\mu\lambda^{e_{\lambda}}\pi_{1}^{e_{1}}....\pi_{g}^{e_{g}}$, where
$\mu$ is unity of $\mathcal{O}_{k_{0}}$, $\lambda\,=\,1-\zeta_{5}$ the unique
prime above $5$ in $k_{0}$, $\pi_{i}\,(1\leq i\leq g)$ are prime elements of
$k_{0}$ such that $\pi_{i}\,\equiv\,a\,(\mathrm{mod}\,5\mathcal{O}_{k_{0}})$
with $a\in\\{1,2,3,4\\}$ and $e_{\lambda},e_{i}\in\\{0,1,2,3,4\\}$. We note by
$d$ the number of ramified prime in $k/k_{0}$.

######  Proposition 2.1.

$C_{k}^{(\sigma)}\,=\,\\{\mathcal{A}\in
C_{k}\,|\,\mathcal{A}^{\sigma}=\mathcal{A}\\}$ the subgroup of ambiguous
classes coincide with the $5$-class group
$C_{k,5}^{(\sigma)}\,=\,\\{\mathcal{A}\in
C_{k,5}\,|\,\mathcal{A}^{\sigma}=\mathcal{A}\\}$.

rank $C_{k,5}^{(\sigma)}\,=\,d-3+q^{*}$.

$q^{*}$ = $\begin{cases}2&\text{if }\,\zeta_{5},\zeta_{5}+1\text{are norme of
element in}\,k^{*}.\\\ 1&\text{if
some}\,\zeta_{5}^{i}(\zeta_{5}+1)^{j}\text{are norme of element in}\,k^{*}.\\\
0&\text{if none}\,\zeta_{5}^{i}(\zeta_{5}+1)^{j}\text{are norme of element
in}\,k^{*}.\\\ \end{cases}$

Let $k\,=\,k_{0}(\sqrt[5]{n})$, writing $n$ =
$\mu\lambda^{e_{\lambda}}\pi_{1}^{e_{1}}....\pi_{f}^{e_{f}}\pi_{f+1}^{e_{f+1}}.....\pi_{g}^{e_{g}}$,
where each $\pi_{i}\,\equiv\,\pm 1,\pm 7\,(\mathrm{mod}\,\lambda^{5})$ for
$1\leq i\leq f$ and $\pi_{j}\,\not\equiv\,\pm 1,\pm
7\,(\mathrm{mod}\,\lambda^{5})$ for $f+1\leq j\leq g$. Then we have:

  * $(i)$

There exists $h_{i}\in\\{1,..,4\\}$ such that
$\pi_{f+1}\pi_{i}^{h_{i}}\,\equiv\,\pm 1,\pm 7\,(\mathrm{mod}\,\lambda^{5})$,
for $f+2\leq i\leq g$.

  * $(ii)$

if $n\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,\lambda^{5})$ and $q^{*}\,=\,1$,
then the genus field $(k/k_{0})^{*}$ is given as:

$(k/k_{0})^{*}$ =
$k(\sqrt[5]{\pi_{1}},....\sqrt[5]{\pi_{f}},\sqrt[5]{\pi_{f+1}\pi_{f+2}^{h_{f+2}}},....\sqrt[5]{\pi_{f+1}\pi_{g}^{h_{g}}})$

with $h_{i}$ as in (i).

  * $(iii)$

In the other cases of $q^{*}$ and the congruence of $n$, the genus field
$(k/k_{0})^{*}$ is given by deleting an appropriate number of $5^{th}$ root
from the right side of (ii).

For more details see References.
If $C_{k,5}$ is of type $(5,5)$ and rank $C_{k,5}^{(\sigma)}\,=\,1$, according
to References we have three possible forms of the radicand $n$, and the genus
field $(k/k_{0})^{*}$ is given explicitly as follows

######  Lemma 2.1.

Let $p$ and $q$ prime numbers such that $p\,\equiv\,-1\,(\mathrm{mod}\,5)$ and
$q\,\equiv\,\pm 2\,(\mathrm{mod}\,5)$. Let $\pi_{1}$ and $\pi_{2}$ primes of
$k_{0}$ such that $p\,=\,\pi_{1}\pi_{2}$. If $C_{k,5}$ is of type $(5,5)$ and
rank $C_{k,5}^{(\sigma)}\,=\,1$, then we have:

$n\,=\,5^{e}p\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$, where
$e\in\\{1,2,3,4\\}$ and $p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$. Then
$(k/k_{0})^{*}\,=\,k(\sqrt[5]{\lambda^{\alpha}\pi_{1}^{\alpha_{1}}\pi_{2}^{\alpha_{2}}})$,
with $\alpha,\alpha_{1},\alpha_{2}\in\\{1,2,3,4\\}$.

$n\,=\,p^{e}q\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$, where
$e\in\\{1,2,3,4\\}$, $p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$ and
$q\,\not\equiv\,\pm 7\,(\mathrm{mod}\,25)$. Then
$(k/k_{0})^{*}\,=\,k(\sqrt[5]{q\pi_{i}^{\alpha_{i}}})$, with $i=1,2$ and
$\alpha\in\\{1,2,3,4\\}$.

$n\,=\,p^{e}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$, where
$e\in\\{1,2,3,4\\}$, $p\,\equiv\,-1\,(\mathrm{mod}\,25)$. Then
$(k/k_{0})^{*}\,=\,k(\sqrt[5]{\pi_{1}^{\alpha_{1}}\pi_{2}^{\alpha_{2}}})$,
with $\alpha_{1},\alpha_{2}\in\\{1,2,3,4\\}$.

###### Proof.

For the three forms of $n$, we refer the reader to References. For the
relative genus field $(k/k_{0})^{*}$ of each case, its follows from $(4)$ of
proposition 2.1 and [References, theorem 5.16]. ∎

###  2.3 intermediate extensions of $k_{5}^{(1)}/k$

Let $k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_{5})$ be the normal closure of the
pure quintic field $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$, $C_{k,5}$ be the
$5$-class group of $k$. When $C_{k,5}$ is of type $(5,5)$, it has $6$
subgroups of order $5$, denoted by $H_{i}$, $1\leq i\leq 6$. Let $K_{i}$ be
the intermediate extension of $k_{5}^{(1)}/k$, corresponding by class field
theory to $H_{i}$. In References we proved that
$C_{k,5}\,\cong\,C_{k,5}^{+}\times C_{k,5}^{-}$ such that
$C_{k,5}^{+}\,=\,\\{\mathcal{A}\in
C_{k,5}\,|\,\mathcal{A}^{\tau^{2}}=\mathcal{A}\\}$ and
$C_{k,5}^{-}\,=\,\\{\mathcal{X}\in
C_{k,5}\,|\,\mathcal{X}^{\tau^{2}}=\mathcal{X}^{-1}\\}$ with
$Gal(k/\Gamma)\,=\,\langle\tau\rangle$. The following theorem allow us to
determine the six intermediate extensions of $k_{5}^{(1)}/k$, using the action
of $Gal(k/\mathbb{Q})$ on $C_{k,5}$.

######  Theorem 2.2.

rank $C_{k,5}^{(\sigma)}\,\geq\,1$.

If rank $C_{k,5}^{(\sigma)}\,=\,1$ then
$C_{k,5}^{(\sigma)}\,=\,C_{k,5}^{+}\,=\,C_{k,5}^{1-\sigma}$ with
$C_{k,5}^{1-\sigma}$ the principal genus.

Let
$\Gamma_{5}^{(1)},(\Gamma^{\prime})_{5}^{(1)},(\Gamma^{{}^{\prime\prime}})_{5}^{(1)},(\Gamma^{{}^{\prime\prime\prime}})_{5}^{(1)}$
and $(\Gamma^{{}^{\prime\prime\prime\prime}})_{5}^{(1)}$ be respectivly the
Hilbert $5$-class fields of
$\Gamma^{{}^{\prime}},\,\Gamma^{{}^{\prime\prime}},\,\Gamma^{{}^{\prime\prime\prime}}$
and $\Gamma^{{}^{\prime\prime\prime\prime}}$. If rank
$C_{k,5}^{(\sigma)}\,=\,1$, then
$k\Gamma_{5}^{(1)},k(\Gamma^{\prime})_{5}^{(1)},k(\Gamma^{{}^{\prime\prime}})_{5}^{(1)},k(\Gamma^{{}^{\prime\prime\prime}})_{5}^{(1)},k(\Gamma^{{}^{\prime\prime\prime\prime}})_{5}^{(1)}$
and $(k/k_{0})^{*}$ are the six intermediate extensions of $k_{5}^{(1)}/k$.
Furthermore $\tau^{2}$ permutes the fields $k.\Gamma_{5}^{(1)}$,
$k.(\Gamma_{5}^{{}^{\prime}})^{(1)}$,
$k.(\Gamma_{5}^{{}^{\prime\prime}})^{(1)}$,
$k.(\Gamma_{5}^{{}^{\prime\prime\prime}})^{(1)}$ and
$k.(\Gamma_{5}^{{}^{\prime\prime\prime\prime}})^{(1)}$

###### Proof.

Let $\mathcal{A}$ be a class that generate $C_{k,5}^{+}$. According to
[References, lemma 5.2], $C_{k,5}^{+}\,\cong\,C_{\Gamma,5}$, then
$\mathcal{A}$ can be identified with a class that generate $C_{\Gamma,5}$.

If $\mathcal{A}\in C_{k,5}^{(\sigma)}$ then $C_{k,5}^{+}\subset
C_{k,5}^{(\sigma)}$ and rank $C_{k,5}^{(\sigma)}\geq 1$.

If $\mathcal{A}\notin C_{k,5}^{(\sigma)}$ then
$\mathcal{A}^{\sigma}\neq\mathcal{A}$. Furthermore
$\mathcal{A}^{\sigma}\neq\mathcal{A}^{2}$, otherwise
$\mathcal{A}^{\sigma}=\mathcal{A}^{2}\Rightarrow\mathcal{A}^{\sigma^{2}}=(\mathcal{A}^{\sigma})^{2}=\mathcal{A}^{4}\Rightarrow\mathcal{A}^{\sigma^{3}}=(\mathcal{A}^{\sigma})^{4}=\mathcal{A}^{8}=\mathcal{A}^{3}\Rightarrow\mathcal{A}^{\sigma^{4}}=(\mathcal{A}^{\sigma})^{3}=\mathcal{A}^{6}=\mathcal{A}\Rightarrow\mathcal{A}^{\sigma^{5}}=\mathcal{A}^{\sigma}=\mathcal{A}$,
which is impossible since $\mathcal{A}^{\sigma}\neq\mathcal{A}$. By the same
reasoning we show that $\mathcal{A}^{\sigma}\neq\mathcal{A}^{3}$,
$\mathcal{A}^{\sigma}\neq\mathcal{A}^{4}$. Ainsi $\mathcal{A}$ et
$\mathcal{A}^{\sigma}$ generate $C_{k,5}$. Since $\mathcal{A}^{5}=1$ and
$\mathcal{A}^{1+\sigma+\sigma^{2}+\sigma^{3}+\sigma^{4}}=1$ then
$\mathcal{A}^{\sigma^{4}}=\mathcal{A}^{4+4\sigma+4\sigma^{2}+4\sigma^{3}}$,
this equality allow us to prove that
$\mathcal{A}^{\sigma^{3}+3\sigma^{2}+2\sigma^{2}-1}$ is ambiguous class. Thus
rank $C_{k,5}^{(\sigma)}\geq 1$

According to References or [References, Lemma 3.3], since rank
$C_{k,5}^{(\sigma)}\,=\,1$ then the radicand $n$ is not divisible by any prime
$p$ such that $p\,\equiv\,1\,(\mathrm{mod}\,5)$. By [References, Theorem 3.2]
all elements of the group of strong ambiguous ideal classes, denoted
$C_{k,s}^{(\sigma)}$ are fixed by $\tau^{2}$ and according to References
$C_{k,5}^{(\sigma)}/C_{k,s}^{(\sigma)}$ is generated by a class modulo
$C_{k,s}^{(\sigma)}$ of element choosen in $(C_{k,5}^{(\sigma)})^{+}$. Thus
all elements of $C_{k,5}^{(\sigma)}$ are fixed by $\tau$ and $\tau^{2}$, then
$C_{k,5}^{(\sigma)}\subset C_{k,5}^{+}$. Therefor
$C_{k,5}^{(\sigma)}\,=\,C_{k,5}^{+}$ because they are groups of order $5$. Let
$\mathcal{A}\in C_{k,5}^{+}$, since
$(\mathcal{A}^{1-\sigma})^{\tau^{2}}\,=\,\mathcal{A}^{\tau^{2}-\tau^{2}\sigma}\,=\,\mathcal{A}^{\tau^{2}-\sigma\tau^{2}}\,=\,(\mathcal{A}^{\tau^{2}})^{1-\sigma}\,=\,\mathcal{A}^{1-\sigma}$
then $(C_{k,5}^{+})^{1-\sigma}\,\subset\,C_{k,5}^{+}$. By the same reasoning
for $\mathcal{X}\in C_{k,5}^{-}$ we can prove that
$(C_{k,5}^{-})^{1-\sigma}\,\subset\,C_{k,5}^{+}$ because $\mathcal{X}$ and
$\mathcal{X}^{-1}$ are in $C_{k,5}^{-}$. Since
$C_{k,5}\,=\,\langle\mathcal{A},\mathcal{X}\rangle$ we get that
$C_{k,5}^{1-\sigma}\,\subset\,C_{k,5}^{+}$. Thus
$C_{k,5}^{(\sigma)}\,=\,C_{k,5}^{+}$ because they are groups of order $5$.

Its sufficient to see that the six extensions are quintic cyclic and
unramified extensions of $k$, then they are all contained in $k_{5}^{(1)}$.
Precisly, since $Gal(k/\Gamma)\,=\,\langle\tau\rangle$, and by class field
theory, $k\Gamma_{5}^{(1)}$ correspond to the subgroup $C_{k,5}^{-}$. Its easy
to see that $C_{k,5}^{-}\,=\,C_{k,5}^{1-\tau^{2}}$. The relative genus field
$(k/k_{0})^{*}$ correspond to the principal genus $C_{k,5}^{1-\sigma}$ and by
the previous point $(k/k_{0})^{*}$ correspond to $C_{k,5}^{(\sigma)}$. We
denote by $K_{1}\,=\,(k/k_{0})^{*}$, $K_{2}\,=\,k\Gamma_{5}^{(1)}$,
$K_{3}\,=\,k(\Gamma^{\prime})_{5}^{(1)}$,
$K_{4}\,=\,k(\Gamma^{\prime\prime})_{5}^{(1)}$,
$K_{5}\,=\,k(\Gamma^{\prime\prime\prime})_{5}^{(1)}$ and
$K_{6}\,=\,k(\Gamma^{\prime\prime\prime\prime})_{5}^{(1)}$. Since
$K_{2}^{\tau^{2}}\,=\,K_{4}$, $K_{4}^{\tau^{2}}\,=\,K_{6}$,
$K_{6}^{\tau^{2}}\,=\,K_{3}$, $K_{3}^{\tau^{2}}\,=\,K_{5}$ and
$K_{5}^{\tau^{2}}\,=\,K_{2}$, then $\tau^{2}$ permutes the fields
$K_{2},K_{3},K_{4},K_{5}$ and $K_{6}$. According to [References, proposition
5.1], Since $C_{k,5}$ is of type $(5,5)$, we have $5$ divides exactly
$h_{\Gamma}$, and by remark 2.2, the six fields
$K_{1},K_{2},K_{3},K_{4},K_{5}$ and $K_{6}$ are distincts ∎

##  3 Capitulation problem

Let $k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_{5})$ be the normal closure of the
pure quintic field $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$, $C_{k,5}$ be the
$5$-class group of $k$. $C_{k,5}$ is elementary abelian bicyclic of type
$(5,5)$. We begin with illuminating the generators of $C_{k,5}$ whenever rank
$C_{k,5}^{(\sigma)}\,=\,1$.

###  3.1 Generators of $C_{k,5}$

According to References, if $C_{k,5}$ is of type $(5,5)$ and rank
$C_{k,5}^{(\sigma)}\,=\,1$, we have an explicit study of generators of
$C_{k,5}$, by means of the relation $C_{k,5}\,\cong\,C_{k,5}^{+}\times
C_{k,5}^{-}$ and the three forms of the radicand $n$ proved in References, and
already stated in the introduction. we have the following:

######  Theorem 3.1.

Let $k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_{5})$, where $n$ is a positive integer
$5^{th}$ power-free, be the normal closure of the pure quintic field
$\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$ and
$Gal(k/\Gamma)\,=\,\langle\tau\rangle$. Let $p$,$q$ and $l$ primes such that,
$p\,\equiv\,-1\,(\mathrm{mod}\,5)$, $q\,\equiv\,\pm 2\,(\mathrm{mod}\,5)$ and
$l\,\neq\,p$, $l\,\neq\,q$ . Assume that $C_{k,5}$ is of type $(5,5)$ and
rank$(C_{k,5}^{(\sigma)})\,=\,1$, then we have:

  * (1)

If $n\,=\,5^{e}p\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$, with
$e\in\\{1,2,3,4\\}$ and $p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$. The prime $p$
decompose in $k$ as
$p\mathcal{O}_{k}\,=\,\mathcal{P}^{5}_{1}\mathcal{P}^{5}_{2}$, where
$\mathcal{P}_{1}$ and $\mathcal{P}_{2}$ are prime ideals of $k$. Let
$\mathcal{L}$ a prime ideal of $k$ above $l$. If $5$ and $l$ are not a quintic
residue modulo $p$, then the $5$-class group $C_{k,5}$ is generated by classes
$[\mathcal{P}_{1}]$ and $[\mathcal{L}]^{1-\tau^{2}}$ and we have:

$C_{k,5}$ =
$\langle[\mathcal{P}_{1}]\rangle\times\langle[\mathcal{L}]^{1-\tau^{2}}\rangle$
= $\langle[\mathcal{P}_{1}],[\mathcal{L}]^{1-\tau^{2}}\rangle$

  * (2)

If $n\,=\,p^{e}q\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$, with
$e\in\\{1,2,3,4\\}$ and $p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$,
$q\,\not\equiv\,\pm 7\,(\mathrm{mod}\,25)$. The prime $p$ decompose in $k$ as
$p\mathcal{O}_{k}\,=\,\mathcal{P}^{5}_{1}\mathcal{P}^{5}_{2}$, where
$\mathcal{P}_{1}$ and $\mathcal{P}_{2}$ are prime ideals of $k$. Let
$\mathcal{L}$ a prime ideal of $k$ above $l$. If $q$ and $l$ are not a quintic
residue modulo $p$, then the $5$-class group $C_{k,5}$ is generated by classes
$[\mathcal{P}_{1}]$ and $[\mathcal{L}]^{1-\tau^{2}}$ and we have:

$C_{k,5}$ =
$\langle[\mathcal{P}_{1}]\rangle\times\langle[\mathcal{L}]^{1-\tau^{2}}\rangle$
= $\langle[\mathcal{P}_{1}],[\mathcal{L}]^{1-\tau^{2}}\rangle$

  * (3)

If $n\,=\,p^{e}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$, with
$e\in\\{1,2,3,4\\}$ and $p\,\equiv\,-1\,(\mathrm{mod}\,25)$. $5$ decompose in
$k$ as
$5\mathcal{O}_{k}\,=\,\mathcal{B}^{4}_{1}\mathcal{B}^{4}_{2}\mathcal{B}^{4}_{3}\mathcal{B}^{4}_{4}\mathcal{B}^{4}_{5}$,
where $\mathcal{B}_{i}$ are prime ideals of $k$. If $5$ is not a quintic
residue modulo $p$, then the $5$-class group $C_{k,5}$ is generated by classes
$[\mathcal{B}_{i}]$ and $[\mathcal{B}_{j}]$, $i\,\neq\,j\in\\{1,2,3,4,5\\}$
and we have:

$C_{k,5}$ =
$\langle[\mathcal{B}_{i}]\rangle\times\langle[\mathcal{B}_{j}]\rangle$ =
$\langle[\mathcal{B}_{i}],[\mathcal{B}_{j}]\rangle$

###  3.2 Study of capitulation

Let $k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_{5})$ be the normal closure of the
pure quintic field $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$, $C_{k,5}$ be the
$5$-class group of $k$. When $C_{k,5}$ is of type $(5,5)$, it has $6$
subgroups of order $5$, denoted by $H_{i}$, $1\leq i\leq 6$. Let $K_{i}$ be
the intermediate extension of $k_{5}^{(1)}/k$, corresponding by class field
theory to $H_{i}$. As each $K_{i}$ is cyclic of order $5$ over $k$, there is
at least one subgroup of order $5$ of $C_{k,5}$, i.e., at least one $H_{l}$
for some $l\in\\{1,2,3,4,5,6\\}$, which capitulates in $K_{i}$ (by Hilbert’s
theorem $94$).

######  Definition 3.1.

Let $\mathcal{S}_{j}$ be a generator of $H_{j}$ $(1\leq j\leq 6)$
corresponding to $K_{j}$. For $1\leq j\leq 6$, let
$i_{j}\in\\{0,1,2,3,4,5,6\\}$. We say that the capitulation is of type
$(i_{1},i_{2},i_{3},i_{4},i_{5},i_{6})$ to mean the following:

when $i_{j}\in\\{1,2,3,4,5,6\\}$, then only the class $\mathcal{S}_{i_{j}}$
and its powers capitulate in $K_{j}$;

when $i_{j}\,=\,0$, then all the $5$-classes capitulate in $K_{j}$.

By theorem 2.2,we have that $C_{k,5}^{(\sigma)}\,=\,C_{k,5}^{+}$. Let
$C_{k,5}^{+}\,=\,\langle\mathcal{A}\rangle$ and
$C_{k,5}^{-}\,=\,\langle\mathcal{X}\rangle$. According to [References, Lemma
5.2], $C_{k,5}^{+}\,\cong\,C_{\Gamma,5}$, we can consider $C_{\Gamma,5}$ as
subgroup of $C_{k,5}$, and we may choose the class $\mathcal{A}$ as
$\mathcal{A}\,=\,[j_{k/\Gamma}(\mathcal{B})]$, when $\mathcal{B}$ is ideal of
$\Gamma$ such that its class generates $C_{\Gamma,5}$.
We order the six intermediat fields $K_{i}$ of $k_{5}^{(1)}/k\,(1\leq i\leq
6)$ as follows

  * -

$K_{1}\,=\,(k/k_{0})^{*}$ correspond to the subgroup
$H_{1}\,=\,C_{k,5}^{(\sigma)}\,=\,C_{k,5}^{+}\,=\,\langle\mathcal{A}\rangle$.

  * -

$K_{2}\,=\,k\Gamma_{5}^{(1)}$ correspond to the subgroup
$H_{2}\,=\,C_{k,5}^{1-\tau^{2}}\,=\,C_{k,5}^{-}\,=\,\langle\mathcal{X}\rangle$.

  * -

$K_{3}\,=\,k(\Gamma^{{}^{\prime}})_{5}^{(1)}$,
$K_{4}\,=\,k(\Gamma^{{}^{\prime\prime}})_{5}^{(1)}$,
$K_{5}\,=\,k(\Gamma^{{}^{\prime\prime\prime}})_{5}^{(1)}$ and
$K_{6}\,=\,k(\Gamma^{{}^{\prime\prime\prime\prime}})_{5}^{(1)}$.

Our principal result on capitulation can be stated as follows:

######  Theorem 3.2.

Using the same notation as above.

The class $\mathcal{A}$ capitulates in the six quintic extensions
$K_{i},\,i=1,2,3,4,5,6.$

The same number of classes capitulate in $K_{2},K_{3},K_{4},K_{5}$ and
$K_{6}$. More precisely, the possible capitulation types are $(0,0,0,0,0,0)$,
$(1,0,0,0,0,0)$, $(0,1,1,1,1,1)$ or $(1,1,1,1,1,1)$.

###### Proof.

Let $C_{k,5}^{(\sigma)}\,=\,C_{k,5}^{+}\,=\,\langle\mathcal{A}\rangle$, where
$\mathcal{A}\,=\,[j_{k/\Gamma}(\mathcal{B})]$ such that $\mathcal{B}$ is ideal
of $\Gamma$. Since $\Gamma_{5}^{(1)}$ is Hilbert $5$-class field of $\Gamma$,
then $\mathcal{B}$ becames pricipal in $\Gamma_{5}^{(1)}$. Thus when
$\mathcal{B}$ considred as ideal of $k\Gamma_{5}^{(1)}$, $\mathcal{B}$ becames
principal in $k\Gamma_{5}^{(1)}$. Then we get that $\mathcal{A}$ capitulates
in $k\Gamma_{5}^{(1)}\,=\,K_{2}$. As $K_{4}\,=\,K_{2}^{\tau^{2}}$,
$K_{6}\,=\,K_{4}^{\tau^{2}}$, $K_{3}\,=\,K_{6}^{\tau^{2}}$,
$K_{5}\,=\,K_{3}^{\tau^{2}}$ and $\mathcal{A}^{\tau^{2}}\,=\,\mathcal{A}$, we
deduce that $\mathcal{A}$ capitulates in $K_{2},K_{3},K_{4},K_{5}$ and
$K_{6}$. To prove that $\mathcal{A}$ capitulates in $K_{1}$, we use Tanaka-
Therada’s theorem. Since $K_{1}\,=\,(k/k_{0})^{*}$ is the genus field of
$k/k_{0}$, and $\mathcal{A}$ is ambiguous class, then $\mathcal{A}$
capitulates in $K_{1}$.

We have $\tau^{2}$ permute the fields $K_{6},K_{5},K_{4},K_{3},$ and $K_{2}$,
and $C_{k,5}^{\tau^{2}}\,=\,C_{k,5}$. Since each element of
$C_{k,5}^{(\sigma)}\,=\,\langle\mathcal{A}\rangle$ capitulates in
$K_{6},K_{5},K_{4},K_{3},$ and $K_{2}$, then we have two possibilities: Only
$\mathcal{A}$ and its powers capitulate in the extensions
$K_{6},K_{5},K_{4},K_{3},$ and $K_{2}$, or all the $5$-classes capitulate in
$K_{6},K_{5},K_{4},K_{3},$ and $K_{2}$. This give us as type of capitulation,
$(i_{1},0,0,0,0,0)$ or $(i_{1},1,1,1,1,1)$ with $i_{1}\in\\{0,1,2,3,4,5,6\\}$.
Since $\mathcal{A}$ capitulates in $K_{1}$, then $i_{1}\,\neq\,0$, so $i_{1}$
is necessary $1$. Thus we prove that the types of capitulation possibles are
$(0,0,0,0,0,0)$, $(0,1,1,1,1,1)$, $(1,0,0,0,0,0)$ or $(1,1,1,1,1,1)$. ∎

## References

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# Latest results from Daya Bay using the full dataset

Zhiyuan Chen
Institute of High Energy Physics, CAS, Beijing 100049, China
On behalf of the Daya Bay collaboration

(17-21 July 2023)

###### Abstract

The Daya Bay Reactor Neutrino Experiment was designed with the primary goal of
precisely measuring the neutrino mixing parameter, $\theta_{13}$. Eight
identically-designed gadolinium-doped liquid scintillator detectors installed
in three underground experimental halls measure the reactor antineutrinos from
six nuclear reactors at different distances. Until its shutdown at the end of
2020, Daya Bay experiment has acquired nearly 6 million inverse beta decay
candidates with neutron captured on gadolinium. In this talk, the latest
neutrino oscillation analysis results based on full data will be presented.
The resulting oscillation parameters are $\sin^{2}2\theta_{13}=0.0851\pm
0.0024$, $\Delta m^{2}_{32}=(2.466\pm 0.060)\times
10^{-3}$\mathrm{e}\mathrm{V}$^{2}$ for the normal mass ordering or $\Delta
m^{2}_{32}=-(2.571\pm 0.060)\times 10^{-3}$\mathrm{e}\mathrm{V}$^{2}$ for the
inverted mass ordering, which are the most precise measurement of
$\theta_{13}$ and $\Delta m^{2}_{32}$ so far. Moreover, latest results on
other topics such as the search of high energy reactor neutrino is included as
well.

##  1 The Daya Bay Reactor Neutrino Experiment

The Daya Bay Reactor Neutrino Experiment was designed to measure the mixing
angle $\theta_{13}$ via the investigation of reactor antineutrino
disappearance at a about 2 km baseline resulting from neutrino oscillation. It
began data taking in late 2011 and finished operation at the end of 2020.

To reduce systematic issues, Daya Bay performed relative measurement with
Far/Near ratio. As shown in Figure 1 (left), Daya Bay used eight antineutrino
detectors (ADs) to detect $\overline{\nu}_{e}$s emitted from the reactors at
the Daya Bay-Ling Ao nuclear power facility in China. The ADs were installed
in three underground experimental halls and were submerged in water pools to
reduce ambient radiation, shown in Fig. 1 (right). Each pool was optically
divided into inner (IWS) and outer (OWS) water Cherenkov detectors for
detecting cosmic-ray muons as muon veto systems. Four layers of resistive
plate chambers (RPCs) cover the top of each water pool to provide another
independent muon detector. As an neutrino target, 20 tonnes of liquid
scintillator doped with 0.1% gadolinium by weight (GdLS) was contained in a
3-m-diameter acrylic cylinder enclosed inside a 4-m-diameter acrylic cylinder
filled with 22 tonnes of undoped liquid scintillator (LS) in each AD. 192
photomultiplier tubes (PMTs) were installed on the barrel surface of the AD to
detect optical photons generated in the scintillator.

Reactor antineutrinos are detected via the inverse beta decay (IBD) reaction:
$\overline{\nu}_{e}+p\to e^{+}+n$. Positron deposits the energy quickly and
forms the prompt signal. Neutron is captured on a nucleus and becomes the
delayed signal. With the coincidence of prompt and delayed signals, we are
able to suppress the backgrounds remarkably.

Figure 1: (Left) Layout of the Daya Bay experiment. Two near experimental
halls, EH1 and EH2, monitor reactor neutrino flux and spectrum, while the far
hall EH3 observes the oscillation driven by $\theta_{13}$ mixing angle.
(Right) Sketch of the detectors in one of the near halls. The ADs are
installed in a water pool and covered with RPCs.

##  2 Neutrino Oscillation Results

In this section, we report a new measurement of $\sin^{2}2\theta_{13}$ and
$\Delta m^{2}_{32}$ using a final sample of $5.55\times 10^{6}$ IBD candidates
with the final-state neutron captured on gadolinium (nGd) acquired in 3158
days of operation [1]. Details of the analysis process and techniques can be
found in Refs. [2, 3]. In this Proceeding, we focus on the improvements to the
analysis techniques.

In terms of the energy response of the detector, a correction for the
nonlinear response of the electronics was applied to each channel. This
correction was derived from the waveform output from a flash-ADC readout
system running in parallel with the default ADC system of EH1-AD1 in 2016 [4].

A new source of PMT flashers was observed in the 7-AD operation period that
were not rejected by the previous criteria. By utilizing the characteristic
charge pattern and temporal distribution of these new flashers, we proposed
additional selection criteria that removed over 99% of this instrumental
background with a negligible IBD selection inefficiency.

The largest correlated background is $\beta$-n decay of cosmogenic
radioisotopes 9Li/8He. In order to determine this background, we paired muons
with all IBD candidates within $\pm$2 s. To improve discrimination of 9Li/8He
from other processes, candidate events were separated into several samples
based on the visible energy deposited by the muon in the AD and the distance
between the prompt and delayed signals, $\Delta r$. The rates and energy
spectra of the dominant cosmogenic radioisotopes were extracted with a
simultaneous fit to 12 two-dimensional histograms defined by the different
muon samples in the three experimental halls for the two $\Delta r$ regions.
We simply measured the sum of these two radioisotopes, in consideration of the
comparable lifetimes of 9Li and 8He. This method provides higher statistics
and a better determination of the low-energy part of the $\beta$ spectrum of
9Li/8He than the previous determination while reducing the rate uncertainty to
less than 25%.

Due to the gradual loss of functional PMTs near the top of the water pools,
the muon detection efficiency of water pools dropped with time, particularly
in the 7-AD period. As a consequence, a new muon-induced background, named as
“muon-x” became significant. The muon-x background was caused by low energy
muons that passed through the IWS undetected. These events typically consisted
of the muon as the prompt signal and a Michel electron from muon decay a
product of muon capture or a spallation neutron as the delayed signal. Muon-x
background was efficiently suppressed by rejecting events with a delayed
signal less than 410 $\mu$s after a muon identified with a more stringent IWS
PMT-hit multiplicity requirement of 6 <nHit $\leq$ 12 which led to a <0.1%
loss in livetime. To determine the rate of these two backgrounds, the prompt-
energy spectra of the IBD-candidate sample were extended to 250 MeV and were
fitted to the spectra of the previously described fast-neutron sample and the
muon-x sample with IWS nHit = 7. Through extrapolation, we obtained their
rates in the range of 0.7 MeV <$E_{p}$ <12 MeV.

We extracted the oscillation parameters using the survival probability of
three-flavor oscillation given by

$\displaystyle P=$ $\displaystyle
1-\cos^{4}\theta_{13}\sin^{2}2\theta_{12}\sin^{2}\Delta_{21}$ (1)
$\displaystyle-\sin^{2}2\theta_{13}(\cos^{2}\theta_{12}\sin^{2}\Delta_{31}+\sin^{2}\theta_{12}\sin^{2}\Delta_{32}),$

where $\Delta_{ij}=\Delta m^{2}_{ij}L/(4\hbar cE)$ with $\Delta m^{2}_{ij}$ in
eV2, L is the baseline in meters between an AD and a reactor core and E is the
energy of the $\overline{\nu}_{e}$ in MeV. Alternatively, for short baselines
of a few kilometers, the survival probability can be parametrized as

$\displaystyle
P=1-\cos^{4}\theta_{13}\sin^{2}2\theta_{12}\sin^{2}\Delta_{21}-\sin^{2}2\theta_{13}\sin^{2}\Delta_{ee}.$
(2)

Here, the effective mass-squared difference $\Delta m^{2}_{ee}$ is related to
the wavelength of the oscillation observed at Daya Bay.

Figure 2: Error ellipses in the $\Delta m^{2}_{ee}-\sin^{2}2\theta_{13}$ space
with the bestfit point indicated. The error bars display the one-dimensional
1-standard-deviation confidence intervals. The colored contours correspond to
1, 2, and 3 standard deviations.

Fitting method B desribed in Ref. [2] was used in this work. Figure 2 shows
the covariance contours in the $\Delta m^{2}_{ee}-\sin^{2}2\theta_{13}$ space.
The best-fit point with $\chi^{2}$/ndf = 559/517 yields
$\sin^{2}2\theta_{13}=0.0851\pm 0.0024$, $\Delta m^{2}_{32}=(2.466\pm
0.060)\times 10^{-3}$\mathrm{e}\mathrm{V}$^{2}$ for the normal mass ordering
or $\Delta m^{2}_{32}=-(2.571\pm 0.060)\times
10^{-3}$\mathrm{e}\mathrm{V}$^{2}$ for the inverted mass ordering [1].

As shown in Fig. 3, the best-fit prompt-energy distribution is in excellent
agreement with the observed spectra in each experimental hall.

Figure 3: The measured prompt-energy spectra of EH1, EH2, and EH3 with the
best-fit and no-oscillation curves superimposed in the upper panels. The shape
of the backgrounds are apparent in the spectra with a logarithmic ordinate
shown in the insets.

The normalized signal rate of the three halls as a function of $L_{\rm
eff}/\langle E_{\overline{\nu}_{e}}\rangle$ with the best-fit curve
superimposed is ploted in Fig. 4, where $L_{\rm eff}$ and $\langle
E_{\overline{\nu}_{e}}\rangle$ are the effective baseline and average
$\overline{\nu}_{e}$ energy, respectively. The oscillation pattern related to
$\theta_{13}$ is unambiguous.

Figure 4: Measured disappearance probability as a function of the ratio of the
effective baseline $L_{\rm eff}$ to the mean antineutrino energy $\langle
E_{\overline{\nu}_{e}}\rangle$.

##  3 First Evidence of High Energy Reactor Antineutrinos

High-energy reactor antineutrinos are likely generated by only a handful of
short-lived $\beta$-decay nuclei with high end-point energies, such as 88,90Br
and 94,96,98Rb. Nevertheless, for reactor antineutrino experiments, all
previous measurements have focused on $E_{p}<$ 8 MeV, due to rare signals in
the higher-energy region and large contamination with cosmogenic backgrounds.
In spite of the rarity, high-energy reactor antineutrinos serve as a
significant background for future measurements, such as the diffuse supernova
neutrino background expected to permeate the Universe. Moreover, direct
measurements of high-energy antineutrinos can provide a valuable new
perspective for nuclear data validations relevant well beyond the bounds of
neutrino physics.

In this section, we report the first measurement of high-energy reactor
antineutrinos at Daya Bay, with nearly 9000 inverse beta decay candidates in
the prompt energy region of 8–12 MeV from 1958 days of data collection [5].

The main backgrounds with $E_{p}$ in 8–12 MeV are from muon decays, cosmogenic
fast neutrons, and cosmogenic isotope decays. A multivariate analysis is
performed to distinguish 2500 signal events from background statistically. The
hypothesis of no reactor antineutrinos with neutrino energy above 10 MeV is
rejected with a significance of 6.2 standard deviations. We observed a 29%
antineutrino flux deficit in the prompt energy region of 8–11 MeV compared to
a recent model prediction. Additionally, our work provide the unfolded
antineutrino spectrum above 7 MeV as a data-based reference for other
experiments, as shown in Fig. 5.

Figure 5: (Left) Measured prompt energy spectrum compared with the prediction
from the SM2018 model. (Right) Reactor antineutrino energy spectrum above 7
MeV and its comparison with the prediction from SM2018.

##  4 Summary

To sum up, we have made a new measurement of $\sin^{2}2\theta_{13}$ with a
precision of 2.8% and the mass-squared differences reaching a precision of
about 2.4%. The reported $\sin^{2}2\theta_{13}$ will likely remain the most
precise measurement of $\theta_{13}$ in the foreseeable future and be the
crucial input for next-generation neutrino experiments studying the mass
hierarchy and CP violation. Moreover, the hypothesis of no reactor
antineutrinos with energy above 10 MeV is rejected with a significance of
6.2$\sigma$. More importantly, the energy region of reactor antineutrinos
above 10 MeV is extended by direct measurement for the first time. Other
latest results such as joint analysis of reactor antineutrino spectra with
PROSPECT, and the measurement of the evolution of the flux and spectrum are
not covered here, which can be seen in Ref. [6, 7].

## References

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Oscillation at Kilometer-Scale Baselines by Daya Bay_, Phys. Rev. Lett. 130,
161802 (2023).

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based on 1230 days of operation of the Daya Bay experiment, Phys. Rev. D 95,
072006 (2017).

  * [3] D. Adey _et al_. (Daya Bay Collaboration), _Measurement of the Electron Antineutrino
Oscillation with 1958 Days of Operation at Daya Bay_, Phys. Rev. Lett. 121,
241805 (2018).

  * [4] Y. B. Huang et al., The Flash ADC system and PMT waveform reconstruction for the
Daya Bay experiment, Nucl. Instrum. Methods Phys. Res., Sect. A 895, 48
(2018).

  * [5] F. P. An _et al_. (Daya Bay Collaboration), _First Measurement of High-Energy Reactor
Antineutrinos at Daya Bay_, Phys. Rev. Lett. 129, 041801 (2022).

  * [6] F. P. An _et al_. (Daya Bay Collaboration and PROSPECT Collaboration), Joint Determination
of Reactor Antineutrino Spectra from 235U and 239Pu Fission by Daya Bay and
PROSPECT,
Phys. Rev. Lett. 128, 081801 (2022).

  * [7] F. P. An _et al_. (Daya Bay Collaboration), Improved measurement of the evolution of the
reactor antineutrino flux and spectrum at Daya Bay, Phys. Rev. Lett. 130,
211801 (2023).