commit c1a5a3de06e21f616d1e5a5d99579d9437f2b12c
parent ec512abcc914e42a0a8fabb51b4106aec90773c5
Author: Sebastiano Tronto <sebastiano.tronto@gmail.com>
Date: Fri, 20 Sep 2019 10:11:55 +0200
New documentation
Diffstat:
12 files changed, 0 insertions(+), 2310 deletions(-)
diff --git a/docs/README b/docs/README
@@ -1,7 +0,0 @@
-Here we collect some incomplete and work-in-progress documentation.
-
-In compute_degrees we prove that our way to compute the degrees gives the
-correct result.
-
-In af_code we break down the code for the "adelic failure" part of the script
-and we check that it computes the degrees as described in compute_degrees.
diff --git a/docs/af_code.aux b/docs/af_code.aux
@@ -1,15 +0,0 @@
-\relax
-\citation{DebryPerucca}
-\@writefile{toc}{\contentsline {section}{\numberline {1}The SageMath Code}{1}}
-\@writefile{toc}{\contentsline {section}{\numberline {2}The Pseudocode}{3}}
-\@writefile{loa}{\contentsline {algorithm}{\numberline {1}{\ignorespaces Compute the adelic failure}}{3}}
-\@writefile{toc}{\contentsline {section}{\numberline {3}Pseudocode, the sub-cases}{4}}
-\@writefile{toc}{\contentsline {subsection}{\numberline {3.1}Case $G\leq \mathbb {Q}_+^\times $}{4}}
-\@writefile{loa}{\contentsline {algorithm}{\numberline {2}{\ignorespaces Adelic failure, case $G\leq \mathbb {Q}^\times $}}{4}}
-\@writefile{toc}{\contentsline {subsection}{\numberline {3.2}Case $d\not =-1$, $n\leq d$}{5}}
-\@writefile{loa}{\contentsline {algorithm}{\numberline {3}{\ignorespaces Adelic failure, case $d\not =-1$, $n\leq d$}}{5}}
-\@writefile{toc}{\contentsline {subsection}{\numberline {3.3}Case $d\not =-1$, $n\geq d+2$}{6}}
-\@writefile{loa}{\contentsline {algorithm}{\numberline {4}{\ignorespaces Adelic failure, case $d\not =-1$, $n\geq d+2$}}{6}}
-\bibcite{DebryPerucca}{1}
-\@writefile{toc}{\contentsline {subsection}{\numberline {3.4}Case $d\not =-1$, $n= d+1$}{7}}
-\@writefile{loa}{\contentsline {algorithm}{\numberline {5}{\ignorespaces Adelic failure, case $d\not =-1$, $n= d+1$}}{7}}
diff --git a/docs/af_code.log b/docs/af_code.log
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-
-\author{Sebastiano Tronto}
-
-
-\begin{document}
-
-We begin by giving the code for the function that computes the adelic failure, both in SageMath and in pseudocode. Then we procede to breaking it down into different subcases, in order to check that it computes the correct values.
-
-\section{The SageMath Code}
-
-The function \texttt{adelic\_failure\_gb} takes two parameters as input: a list $B=\{B_0,\dots, B_t\}$ and an integer $d$. Each $B_i$ is itself a list of elements of $G$, and we require the following:
-\begin{itemize}
-\item Each element of $B_i=\{B_{i,0},\dots,B_{i,t_i}\}$ has $2$-divisibility $i$, using the terminology of \cite{DebryPerucca}.
-\item $\mathcal{B}=\bigcup_{i=1}^t B_i$ is a $2$-maximal basis for $G$.
-\item The integer $d$ is either $-1$ or $1\leq d\leq t$. For $i\in\{1,\dots,t\}\setminus\{d\}$ we have $B_i\subseteq \mathbb{Q}_+$. If $d\neq -1$ we have $B_{d,0}<0$ and $B_{d,j}>0$ for $j\neq 0$.
-\end{itemize}
-The output is a list $A=\{A_1,\dots,A_{N_0}\}$, where each $A_n=\{A_{n,0},\dots,A_{n,r_n}\}$ is a list of pairs $A_{n,i}=(d_{n,i},f_{n,i})$. We have $N_0=\max(3,t+1)$ if $d=t$, while $N_0=\max(3,t)$ otherwise. For each $1\leq n\leq N_0$ and each $i\leq r_n$, the integer $d_{n,i}$ is a divisor of $M_0=d_{N_0,r_{N_0}}$ and a multiple of $2^i$, and $f_{n,i}$ is the ``adelic failure'' (old definition), i.e.:
-\begin{align*}
-f_{n,i}=\left[\mathbb{Q}_{2^i}\left(G^{1/2^i}\right)\cap \mathbb{Q}_{d_{i,n}}:\mathbb{Q}_{2^i}\right].
-\end{align*}
-
-\lstset{language=Python}
-\begin{lstlisting}
-def adelic_failure_gb( B, d ):
-
- ad_fail = [] # The table to be returned at the end.
-
- if d == len(B)-1:
- N = max(3,len(B)+1)
- else:
- N = max(3,len(B))
-
- # The shortlist grows at each step, so we build it incrementally.
- shortlist = []
- # The "special element" is (n,b) = \zeta_{2^n}\sqrt{b}.
- special_element = (1,1)
-
- M = 1 # M also grows with n.
-
- for n in range( 1, N+1 ): # Read as: 1 \leq n \leq N
-
- # We add the new elements to the shortlist, modifying M if needed.
- # This is not done in case we are in the extra "fake" level.
- if n-1 < len(B):
- for g in B[n-1]:
- if g < 0 and n > 1:
- special_element = ( n+1, abs(g)^(1/(2^(n-1))) )
- M = lcm( M, special_embed( special_element ) )
- else:
- b = g^(1/(2^(n-1))) # b is 2-indivisible
- shortlist.append( b )
- M = lcm( M, cyc_embed(b) )
-
- # We add a root of an even power of the negative generator, as soon as
- # we are beyond its level.
- if d != -1 and n == d+2:
- b = abs(B[d][0])^(1/2^d)
- shortlist.append( b )
- M = lcm( M, cyc_embed(b) )
-
- M = lcm(M,2^n)
-
- if n <= d:
- M = lcm( M, 2^(n+1) )
-
- if n == 1 and d >= 1:
- shortlist.append(-1)
- if n > 1 and -1 in shortlist:
- shortlist.remove(-1)
-
- aux = [] # Next line of ad_fail table
-
- for dM in divisors( M ):
- if dM % (2^n) != 0:
- continue
-
- S = [ product(s) for s in subsets( shortlist ) ]
- H = [ cyc_embed( s ) for s in S ]
- r = len( [ b for b in H if dM % b == 0 ] )
-
- if n <= d and dM % (2^(n+1)) == 0 and n > 1:
- r *= 2
-
- if 8 in H and dM % 8 == 0 and (n >= 3 or (n == 2 and n <= d)):
- r = r/2
-
- if special_element != (1,1) and special_element[0] == n+1:
- nothing_to_do = False
- intersecting_QdM = False
- for s in S:
- new_special = ( n+1, special_element[1] * s )
- m = special_embed( new_special )
- if n == 2 and m == 4: # \zeta_8 times 2 times square
- nothing_to_do = True
- if dM % m == 0:
- intersecting_QdM = True
- if intersecting_QdM and not nothing_to_do:
- r *= 2
-
- aux.append( (dM,r) )
-
- ad_fail.append(aux)
-
- return ad_fail
-\end{lstlisting}
-
-We have used the following auxiliary functions:
-
-\begin{lstlisting}
-# Computes the minimal cyclotomic field containing \sqrt(b)
-def cyc_embed( b ):
- m = squarefree_part(b)
- if m%4 != 1:
- m *= 4
- return abs(m)
-
-# Computes the minimal cyclotomic field containing \zeta_{2^n}\sqrt(b)
-def special_embed( (n,b) ):
- m = squarefree_part(b)
- if n == 3 and m % 2 == 0:
- return 4 * cyc_embed(m/2)
- else:
- return lcm( 2^n, cyc_embed(b) )
-\end{lstlisting}
-
-\section{The Pseudocode}
-We translate the SageMath code into pseudocode for ease of readability.
-
-\begin{algorithm}
-\caption{Compute the adelic failure}
-\begin{algorithmic}
-\State Let $B$, $t$, $d$ and $N$ as described in the previous section
-\State Let $M\leftarrow1$, $\texttt{special\_element}\leftarrow1$ and $\texttt{shortlist}\leftarrow[\,]$
-
-\State
-
-\For {$n=1$ to $N$}
-\If{$n-1<t$}
-\For{$g\in B_{n-1}$}
-\If{$g<0$ and $n>1$}
-\State $\texttt{special\_element}\leftarrow(n+1,\sqrt[2^{n-1}]{|g|})$
-\State $M\leftarrow\lcm(M,\texttt{special\_embed}(\texttt{special\_element}))$
-\Else
-\State Add $\sqrt[2^{n-1}]{g}$ to \texttt{shortlist}
-\State $M\leftarrow\lcm(M,\texttt{cyc\_embed}(g))$
-\EndIf
-\EndFor
-\EndIf
-
-\State
-
-\If{$n=d+2$ and $d\neq -1$}
-\State Add $\sqrt[2^{d}]{|B_{d,0}|}$ to \texttt{shortlist}
-\State $M\leftarrow\lcm(M,\texttt{cyc\_embed}(|B_{d,0}|))$
-\EndIf
-
-\State
-\If{$n\leq d$}
-\State $M\leftarrow\lcm(M,2^{n+1})$
-\Else
-\State $M\leftarrow\lcm(M,2^n)$
-\EndIf
-\State
-
-\If{$n=1$ and $d\geq 1$}
-\State Add $-1$ to \texttt{shortlist}
-\EndIf
-\State
-\If{$n>1$}
-\State Remove $-1$ from \texttt{shortlist} (if present)
-\EndIf
-\State
-\algstore{alg1}
-\end{algorithmic}
-\end{algorithm}
-\pagebreak
-
-\begin{algorithm}
-\begin{algorithmic}
-\algrestore{alg1}
-\ForAll{$d_M\in \texttt{divisors}(M)$ such that $2^n\,|\,M$}
-\State $S\leftarrow\left\{\prod_{x\in T}x\,|\,T\subseteq\texttt{shortlist}\right\}$
-\State $H\leftarrow\left\{\min\left\{x\in\mathbb{Z}_{>0}\,|\sqrt{s}\in \mathbb{Q}_x\right\}\,|\,s\in S\right\}$
-\State $r\leftarrow\# \left\{s\in S\,|\, \sqrt{s}\in\mathbb{Q}_{d_M}\right\}$
-\State
-\If{$q<n\leq d$ and $2^{n+1}\,|\,{d_M}$}
-\State $r\leftarrow 2r$
-\EndIf
-\State
-\If{$8\in H$ and $8\,|\,d_M$ and (either $n\geq 3$ or $n=2\leq d$)}
-\State $r\leftarrow r/2$
-\EndIf
-\State
-
-\If{$\texttt{special\_element}=\zeta_{2^{n+1}}\sqrt{b}$ for some $b\in\mathbb{Q}$}
-\State $\texttt{specials}\leftarrow\{\zeta_{2^{n+1}}\sqrt{bs}\,|\,s\in S\}$
-\If{$\exists x\in \texttt{specials}$ such that $x\in\mathbb{Q}_{d_M}$ and $\texttt{special\_embed}(s)\neq 4\,\forall s\in\texttt{specials}$}
-\State $r\leftarrow 2r$
-\EndIf
-\EndIf
-\State
-\State Declare $\left[\mathbb{Q}_{2^n}\left(\sqrt[2^n]{G}\right)\cap \mathbb{Q}_{d_M}:\mathbb{Q}_{2^n}\right]=r$.
-
-\EndFor
-\EndFor
-\end{algorithmic}
-\end{algorithm}
-
-\section{Pseudocode, the sub-cases}
-
-We divide the pseudocode in sub-cases.
-
-\subsection{Case $G\leq \mathbb{Q}_+^\times$}
-
-\begin{algorithm}
-\caption{Adelic failure, case $G\leq \mathbb{Q}^\times$}
-
-\begin{algorithmic}
-\For {$n=1$ to $N$}
-\For{$g\in B_{n-1}$}
-\State Add $\sqrt[2^{n-1}]{g}$ to \texttt{shortlist}
-\State $M\leftarrow\lcm(M,\texttt{cyc\_embed}(g))$
-\EndFor
-\State
-\State $M\leftarrow\lcm(M,2^n)$
-\State
-\ForAll{$d_M\in \texttt{divisors}(M)$ such that $2^n\,|\,M$}
-\State $S\leftarrow\left\{\prod_{x\in T}x\,|\,T\subseteq\texttt{shortlist}\right\}$
-\State $H\leftarrow\left\{\min\left\{x\in\mathbb{Z}_{>0}\,|\sqrt{s}\in \mathbb{Q}_x\right\}\,|\,s\in S\right\}$
-\State $r\leftarrow\# \left\{s\in S\,|\, \sqrt{s}\in\mathbb{Q}_{d_M}\right\}$
-%\State
-\State Declare $\left[\mathbb{Q}_{2^n}\left(\sqrt[2^n]{G}\right)\cap \mathbb{Q}_{d_M}:\mathbb{Q}_{2^n}\right]=\begin{cases}
-r/2&\text{ if }8\in H\text{ and }n\geq 3,\\
-r&\text{ otherwise}.
-\end{cases}$
-\EndFor
-\EndFor
-\end{algorithmic}
-
-\end{algorithm}
-\pagebreak
-\subsection{Case $d\neq -1$, $n\leq d$}
-For this and the following cases, we assume we are already inside the main \texttt{for} cycle, since we have particular assumptions on $n$.
-\begin{algorithm}
-\caption{Adelic failure, case $d\neq -1$, $n\leq d$}
-\begin{algorithmic}
-\For{$g\in B_{n-1}$}
-\State Add $\sqrt[2^{n-1}]{g}$ to \texttt{shortlist}
-\State $M\leftarrow\lcm(M,\texttt{cyc\_embed}(g))$
-\EndFor
-\State
-\State $M\leftarrow\lcm(M,2^{n+1})$
-\State
-\If{$n=1$ and $d\geq 1$}
-\State Add $-1$ to \texttt{shortlist}
-\EndIf
-\State
-\If{$n>1$}
-\State Remove $-1$ from \texttt{shortlist} (if present)
-\EndIf
-\State
-\ForAll{$d_M\in \texttt{divisors}(M)$ such that $2^n\,|\,M$}
-\State $S\leftarrow\left\{\prod_{x\in T}x\,|\,T\subseteq\texttt{shortlist}\right\}$
-\State $H\leftarrow\left\{\min\left\{x\in\mathbb{Z}_{>0}\,|\sqrt{s}\in \mathbb{Q}_x\right\}\,|\,s\in S\right\}$
-\State $r\leftarrow\# \left\{s\in S\,|\, \sqrt{s}\in\mathbb{Q}_{d_M}\right\}$
-\State
-\If{$n>1$ and $2^{n+1}\,|\,d_M$}
-\State $r\leftarrow 2r$
-\EndIf
-\State Declare $\left[\mathbb{Q}_{2^n}\left(\sqrt[2^n]{G}\right)\cap \mathbb{Q}_{d_M}:\mathbb{Q}_{2^n}\right]=\begin{cases}
-r/2&\text{ if }8\in H\text{ and }n\geq 3,\\
-r/2&\text{ if }8\in H\text{ and }n=2\text{ and }8\,|\,d_M\\
-r&\text{ otherwise}.
-\end{cases}$
-\EndFor
-\end{algorithmic}
-
-\end{algorithm}
-
-
-\pagebreak
-\subsection{Case $d\neq -1$, $n\geq d+2$}
-
-\begin{algorithm}
-\caption{Adelic failure, case $d\neq -1$, $n\geq d+2$}
-\begin{algorithmic}
-\If{$n-1<t$}
-\For{$g\in B_{n-1}$}
-\State Add $\sqrt[2^{n-1}]{g}$ to \texttt{shortlist}
-\State $M\leftarrow\lcm(M,\texttt{cyc\_embed}(g))$
-\EndFor
-\EndIf
-\State
-
-\If{$n=d+2$}
-\State Add $\sqrt[2^{d}]{|B_{d,0}|}$ to \texttt{shortlist}
-\State $M\leftarrow\lcm(M,\texttt{cyc\_embed}(|B_{d,0}|))$
-\EndIf
-
-\State
-\State $M\leftarrow\lcm(M,2^{n})$
-\State
-\ForAll{$d_M\in \texttt{divisors}(M)$ such that $2^n\,|\,M$}
-\State $S\leftarrow\left\{\prod_{x\in T}x\,|\,T\subseteq\texttt{shortlist}\right\}$
-\State $H\leftarrow\left\{\min\left\{x\in\mathbb{Z}_{>0}\,|\sqrt{s}\in \mathbb{Q}_x\right\}\,|\,s\in S\right\}$
-\State $r\leftarrow\# \left\{s\in S\,|\, \sqrt{s}\in\mathbb{Q}_{d_M}\right\}$
-\State Declare $\left[\mathbb{Q}_{2^n}\left(\sqrt[2^n]{G}\right)\cap \mathbb{Q}_{d_M}:\mathbb{Q}_{2^n}\right]=\begin{cases}
-r/2&\text{ if }8\in H,\\
-r&\text{ otherwise}.
-\end{cases}$
-\EndFor
-\end{algorithmic}
-
-\end{algorithm}
-
-\pagebreak
-
-\subsection{Case $d\neq -1$, $n= d+1$}
-
-\begin{algorithm}
-\caption{Adelic failure, case $d\neq -1$, $n= d+1$}
-\begin{algorithmic}
-\For{$g\in B_{n-1}$}
-\If{$g<0$}
-\State $\texttt{special\_element}\leftarrow(n+1,\sqrt[2^{n-1}]{|g|})$
-\State $M\leftarrow\lcm(M,\texttt{special\_embed}(\texttt{special\_element}))$
-\Else
-\State Add $\sqrt[2^{n-1}]{g}$ to \texttt{shortlist}
-\State $M\leftarrow\lcm(M,\texttt{cyc\_embed}(g))$
-\EndIf
-\EndFor
-\State
-\State $M\leftarrow\lcm(M,2^{n})$
-\State
-\State Remove $-1$ from \texttt{shortlist} (if present)
-
-
-\State
-\ForAll{$d_M\in \texttt{divisors}(M)$ such that $2^n\,|\,M$}
-\State $S\leftarrow\left\{\prod_{x\in T}x\,|\,T\subseteq\texttt{shortlist}\right\}$
-\State $H\leftarrow\left\{\min\left\{x\in\mathbb{Z}_{>0}\,|\sqrt{s}\in \mathbb{Q}_x\right\}\,|\,s\in S\right\}$
-\State $r\leftarrow\# \left\{s\in S\,|\, \sqrt{s}\in\mathbb{Q}_{d_M}\right\}$
-\State
-%\State $\texttt{specials}\leftarrow\{\zeta_{2^{n+1}}\sqrt{bs}\,|\,s\in S\}$
-\If{$\exists x\in \{\zeta_{2^{n+1}}\sqrt{bs}\,|\,s\in S\}\cap\mathbb{Q}_{d_M}$ and $\texttt{special\_embed}(s)\neq 4\,\forall s\in\texttt{specials}$}
-\State $r\leftarrow 2r$
-\EndIf
-\State Declare $\left[\mathbb{Q}_{2^n}\left(\sqrt[2^n]{G}\right)\cap \mathbb{Q}_{d_M}:\mathbb{Q}_{2^n}\right]=\begin{cases}
-r/2&\text{ if }8\in H\text{ and }n\geq 3,\\
-r&\text{ otherwise}.
-\end{cases}$
-\EndFor
-\end{algorithmic}
-
-\end{algorithm}
-
-\begin{thebibliography}{10} \expandafter\ifx\csname url\endcsname\relax \def\url#1{\texttt{#1}}\fi \expandafter\ifx\csname urlprefix\endcsname\relax\def\urlprefix{URL }\fi
-
-\bibitem{DebryPerucca}
-\textsc{Debry, C. - Perucca, A.}: \emph{Reductions of algebraic integers}, J. Number Theory, {\bf 167} (2016), 259--283.
-
-%\bibitem{PeruccaSgobba}
-%\textsc{Perucca, A. - Sgobba, P.}: \emph{Kummer Theory for Number Fields}, preprint.
-
-\end{thebibliography}
-
-\end{document}
-\ No newline at end of file
diff --git a/docs/compute_degree.aux b/docs/compute_degree.aux
@@ -1,14 +0,0 @@
-\relax
-\citation{DebryPerucca}
-\citation{DebryPerucca}
-\newlabel{lemma_zero}{{1}{1}}
-\newlabel{degree}{{1}{1}}
-\@writefile{toc}{\contentsline {section}{\numberline {1}Case $G\leq \mathbb {Q}_+^\times $}{1}}
-\@writefile{toc}{\contentsline {section}{\numberline {2}General case}{2}}
-\@writefile{toc}{\contentsline {subsection}{\numberline {2.1}General case, $n=1(\leq d)$}{2}}
-\@writefile{toc}{\contentsline {subsection}{\numberline {2.2}General case, $n=2\leq d$}{2}}
-\@writefile{toc}{\contentsline {subsection}{\numberline {2.3}General case, $3\leq n\leq d$}{2}}
-\@writefile{toc}{\contentsline {subsection}{\numberline {2.4}General case, $n\geq d+2$}{3}}
-\@writefile{toc}{\contentsline {subsection}{\numberline {2.5}General case, $n=d+1$}{3}}
-\bibcite{DebryPerucca}{1}
-\bibcite{PeruccaSgobba}{2}
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@@ -1,320 +0,0 @@
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-\author{Sebastiano Tronto}
-
-
-\begin{document}
-
-\begin{lemma}
-\label{lemma_zero}
-Let $H\leq \mathbb{Q}^\times$ be a finitely generated subgroup. Assume that $H$ does not contain minus a square of $\mathbb{Q}^\times$ or that $m=1$. Then we have
-\begin{align*}
-\left[\mathbb{Q}_{2^m}\left(\sqrt{H}\right):\mathbb{Q}_{2^m}\right]=\begin{cases}
-\#\overline H/2 & \text{ if }m\geq 3\text{ and }\exists b\in H\text{ with }b\equiv\pm2\pmod{\mathbb{Q}^{\times 2}},\\
-\#\overline H&\text{ otherwise}.
-\end{cases}
-\end{align*}
-where $\overline{H}$ is the image of $H\cdot \mathbb{Q}^{\times 2}$ in $\mathbb{Q}^\times/\mathbb{Q}^{\times 2}$.
-\begin{proof}
-Clearly we may assume that $H$ is generated by suqarefree integers $\{g_1,\dots, g_r\}$, where $r=\#\overline H$. In fact, we have that $\mathbb{Q}_{2^m}(\sqrt{H})=\mathbb{Q}_{2^m}(\sqrt{H'})$ for any $H'$ such that $(H\cdot \mathbb{Q}^{\times 2})/\mathbb{Q}^{\times 2}=(H'\cdot \mathbb{Q}^{\times 2})/\mathbb{Q}^{\times 2}$. Recall moreover that by {\color{red}Lemma 13} if there is $\pm2$ times a square in $H$ we can assume that, say, $g_1=\pm 2$.
-
-Assume first that $m\geq 2$, so that $-1\not\in H$ by assumption. In this case we can work over $\mathbb Q_4$ and use Theorem 18 of \cite{DebryPerucca}. We just need to compute the divisibility parameters over $\mathbb{Q}_4$:
-\begin{align*}
-d_1=\begin{cases}
-0&\text{ if }g_1\neq\pm2\\
-1&\text{ if }g_1=\pm2
-\end{cases},
-&&
-d_i=0
-\quad \text{ for $i=2,\dots, r$},\\
-h_1=\begin{cases}
-0&\text{ if } 0\leq g_1\neq2\\
-1&\text{ if } -2\neq g_1<0\\
-2&\text{ if } g_1=\pm 2
-\end{cases}, &&
-h_i=\begin{cases}
-0&\text{ if }g_i>0\\
-1&\text{ if }g_i<0
-\end{cases}
-\quad \text{ for $i=2,\dots, r$}.
-\end{align*}
-Thus, keeping the notation of the aformentioned Theorem, we get
-\begin{align*}
-n_1=\min(1,d_1)=\begin{cases}
-0&\text{ if }g_1\neq\pm2\\
-1&\text{ if }g_1=\pm2
-\end{cases},&& n_i=0\quad \text{ for $i=2,\dots, r$}.
-\end{align*}
-Thus we get
-\begin{align*}
-v_2\left[\mathbb{Q}_{2^m}(\sqrt{H}):\mathbb Q_{2^m}\right]&=\max(h_1+n_1,\dots, h_r+n_r,m)-m+r-\sum_{i=1}^rn_i=\\
-&=\begin{cases}
-\max(3,m)-m+r-\sum_{i=1}^rn_i&\text{ if }\pm2\in H\\
-r-\sum_{i=1}^rn_i&\text{ if }\pm2\not \in H
-\end{cases}\\
-&=\begin{cases}
-1+r-1&\text{ if }m=2\text{ and }\pm2\in H\\
-r-1&\text{ if }m\geq3\text{ and }\pm2\in H\\
-r&\text{ if }\pm2\not\in H
-\end{cases}
-\end{align*}
-which is what we want.
-
-Assume now that $m=1$. If $-1\not\in H$, we get the desired result directly from Lemma 19 of \cite{DebryPerucca} applied with $G=H$, using the computations that we did in the previous case. In case $-1\in H$, let $H'$ be any subgroup of $H$ such that $H=H'\oplus\langle-1\rangle$. Notice that we have $\#\overline {H'}=r-1$, so that Lemma 19 with $G=H'$ again gives our result, and the Proposition is proved.
-\end{proof}
-\end{lemma}
-
-Let $G\leq \mathbb{Q}^\times$ be a finitely generated torsion-free subgroup of rank $r$ and let $M$ and $n$ be integers such that $2^n\,|\,M$. We want to compute the degree
-\begin{align}
-\label{degree}
-\left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right].
-\end{align}
-
-We will use the same notation as that of Remark 17 of Pietro's file.
-
-\section{Case $G\leq \mathbb{Q}_+^\times$}
-
-Assume that $G\leq \mathbb{Q}_+^\times$. In this case, by Remark 17, we have that
-\begin{align*}
-\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M =\mathbb{Q}_{2^n}\left(\sqrt{H}\right).
-\end{align*}
-
-Let $\overline{H}$ be the image of $H$ in $\mathbb{Q^\times}/\mathbb{Q}^{\times 2}$. By Remark 17 and Lemma \ref{lemma_zero} above, the degree (\ref{degree}) is given by
-\begin{align*}
-\left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]=
-\begin{cases}
-\#\overline H/2 & \text{if }n\geq 3\text{ and }2\in H,\\
-\#\overline H&\text{ otherwise}.
-\end{cases}
-\end{align*}
-
-\section{General case}
-
-Let $\mathcal{B}$ be a basis for $G$ and let $\mathcal{B}_i\subseteq \mathcal{B}$ be the subset of basis elements of $2$-divisibility $i$. Call also $L=\max d_i$ the largest $2$-divisiblity parameter. In this way $\mathcal{B}_0,\dots,\mathcal{B}_L$ is a partition of $\mathcal{B}$.
-
-As explained in ({\color{red}ref}) we may assume that there is at most one negative basis element. Since we have dealt with the $G\subseteq \mathbb{Q}_+$ case in the previous section, we assume that such an element exists and that it has $2$-divisibility $d$. We call this element $g_0$.
-
-It is (or will be?) clear ({\color{red}but we should explain it}) that it actually does not matter if we have negative elements of divisibility $0$: that case is treated exactly as the case $G\subseteq \mathbb{Q}_+$. In conclusion, we assume that:
-\begin{align*}
-\mathcal{B}_1,\dots,\mathcal{B}_{d-1},\mathcal{B}_{d+1},\dots,\mathcal{B}_L\subseteq \mathbb{Q}_+,\\
-g_0<0 \text{ and }\mathcal{B}_d\setminus \{g_0\}\subseteq \mathbb{Q}_+,\\
-d\geq 1.
-\end{align*}
-
-We also let
-\begin{align*}
-N=\begin{cases}
-\max(3,L)&\text{if }d\neq L,\\
-\max(3,L+1)&\text{if }d=L.
-\end{cases}
-\end{align*}
-
-\subsection{General case, $n=1(\leq d)$}
-This case can be treated as follows: let $\mathcal{S}'=\mathcal{S}\cup \{-1\}$ and let $H'$ be constructed from $\mathcal{S}'$ in the exact same way as $H$ is constructed from $\mathcal{S}$. Then it's easy to check ({\color{red}it follows from the ``torsion case'' for $G$, it is for sure in some other file}) that $\mathbb{Q}_{2^n}\left(\sqrt{H'}\right)=\mathbb{Q}_{2^{w'}}\left(\sqrt{H}\right)$, where $w'=\min(v_2(M),n+1)$ (as in Remark 17). Then we can again use Lemma \ref{lemma_zero} and conclude that
-\begin{align*}
-\left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]=
-\#\overline{H'},
-\end{align*}
-where $\#\overline{H'}$ is the image of $H'$ in $\mathbb{Q}^\times/\mathbb{Q}^{\times 2}$.
-
-\subsection{General case, $n=2\leq d$}
-We consider two cases:
-\begin{itemize}
-\item If $v_2(M)=2$ we have $\left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]=\left[\mathbb{Q}_4\left(\sqrt{H}\right):\mathbb{Q}_4\right]=\#\overline{H}$ by Lemma \ref{lemma_zero}.
-\item If $v_2(M)\geq 3$ we have
-\begin{align*}
-\left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]&=\left[\mathbb{Q}_8\left(\sqrt{H}\right):\mathbb{Q}_4\right]=\left[\mathbb{Q}_8\left(\sqrt{H}\right):\mathbb{Q}_8\right]\cdot \left[\mathbb{Q}_8:\mathbb{Q}_4\right]=\\&=2\left[\mathbb{Q}_8\left(\sqrt{H}\right):\mathbb{Q}_8\right],
-\end{align*}
-which, by Lemma \ref{lemma_zero}, is given by $\#\overline{H}$ if $2 \in H$ and by $2\#\overline{H}$ otherwise.
-\end{itemize}
-
-\subsection{General case, $3\leq n\leq d$}
-We consider two cases:
-\begin{itemize}
-\item If $v_2(M)=3$, by lemma \ref{lemma_zero} we have
-\begin{align*}
-\left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]=\left[\mathbb{Q}_8\left(\sqrt{H}\right):\mathbb{Q}_8\right]=\begin{cases}
-\#\overline H/2 & \text{ if }\pm 2\in H,\\
-\#\overline H&\text{ otherwise}.
-\end{cases}
-\end{align*}
-\item If $v_2(M)\geq 4$ we have
-\begin{align*}
-\left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]&=\left[\mathbb{Q}_{16}\left(\sqrt{H}\right):\mathbb{Q}_8\right]=\left[\mathbb{Q}_{16}\left(\sqrt{H}\right):\mathbb{Q}_{16}\right]\cdot \left[\mathbb{Q}_{16}:\mathbb{Q}_8\right]=\\&=2\left[\mathbb{Q}_{16}\left(\sqrt{H}\right):\mathbb{Q}_{16}\right],
-\end{align*}
-which, by Lemma \ref{lemma_zero}, is given by $\#\overline{H}$ if $2 \in H$ and by $2\#\overline{H}$ otherwise.
-\end{itemize}
-
-\subsection{General case, $n\geq d+2$}
-By the corresponding case in Remark 17, we simply have
-\begin{align*}
-\left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]=\begin{cases}
-\#\overline {H'}/2 & \text{ if }\pm 2\in H,\\
-\#\overline {H'}&\text{ otherwise}.
-\end{cases}
-\end{align*}
-where $H'$ is constructed from $\mathcal{S}'=\mathcal{S}\cup\{B_0\}$ and $\overline{H'}$ is the image of $H'$ in $\mathbb{Q}^\times/\mathbb{Q}^{\times 2}$.
-
-\subsection{General case, $n=d+1$}
-We distinguish between some cases.
-\begin{itemize}
-\item Assume $n=2$ (thus $d=3$) and $v_2(g_0)=2$ (i.e. $2$ divides the square-free part of $B_0$, where $g_0=-B_0^{2^d}$). Then we write the square-free part of $B_0$ as $2s$ for some odd square-free $s\in\mathbb{Z}$. Then letting $\mathcal{S}':=\mathcal{S}\cup \{s\}$ and construct $H'$ from $\mathcal{S}'$ in the usual way. By Remark 17 we have
-\begin{align*}
-\left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]=\left[\mathbb{Q}_{2^n}\left(\sqrt{H'}\right):\mathbb{Q}_{2^n}\right]=\#\overline{H'}.
-\end{align*}
-But we can be more precise and say that
-\begin{align*}
-\#\overline{H'}=\begin{cases}
-2\#\overline{H}&\text{if }\sqrt{xs}\in\mathbb{Q}_M\text{ for some }x\in\mathcal{S}\text{ and }s\not\in \mathcal{S},\\
-\#\overline{H}&\text{otherwise}.
-\end{cases}
-\end{align*}
-%\item Assume $n=2$, $2^{n+1}\nmid M$ and either $v_2(g_0)>2$ or $g_0$ is odd. Then $\left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]=\#\overline H$.
-%\item Assume $n=2$, $2^{n+1}\,|\,M$ and either $v_2(g_0)>2$ or $g_0$ is odd. ({\color{red}TODO})
-\item Assume $n\geq 2$ and $2^{n+1}\nmid M$. Then
-\begin{align*}
-\left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]=\begin{cases}
-\#\overline H/2 & \text{ if }\pm 2\in H\text{ and }n\geq 3,\\
-\#\overline H&\text{ otherwise}.
-\end{cases}
-\end{align*}
-\item Assume $n\geq 2$ and $2^{n+1}\,|\,M$. Following the notation of Remark 17, we have
-\begin{align*}
-\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M=\mathbb{Q}_{2^n}\left(\sqrt{\langle H, H'\rangle}\right)
-\end{align*}
-hence
-\begin{align*}
-\left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]&=\left[\mathbb{Q}_{2^n}\left(\sqrt{\langle H, H'\rangle}\right):\mathbb{Q}_{2^n}\right]=\\
-&=\left[\mathbb{Q}_{2^n}\left(\sqrt{\langle H, H'\rangle}\right):\mathbb{Q}_{2^n}\left(\sqrt{H}\right)\right]\cdot \left[\mathbb{Q}_{2^n}\left(\sqrt{H}\right):\mathbb{Q}_{2^n}\right].
-\end{align*}
-We claim that
-\begin{align*}
-\left[\mathbb{Q}_{2^n}\left(\sqrt{\langle H, H'\rangle}\right):\mathbb{Q}_{2^n}\left(\sqrt{H}\right)\right]=\begin{cases}
-1&\text{ if }H'=\emptyset\text{ or }H'=\{2\zeta_4\},\\
-2&\text{ otherwise}.
-\end{cases}
-\end{align*}
-To see this, notice that $\sqrt{2\zeta_4}=\zeta_8\sqrt{2}\in\mathbb{Q}_4\subseteq\mathbb{Q}_{2^n}\left(\sqrt{H}\right)$, so the first case is settled. Assume now that there is $x=\zeta_{2^n}b\in H'$ with $x\neq 2\zeta_4$. If $y=\zeta_{2^n}c$ is any other element of $H'$, then we have $\sqrt{x/y}=\sqrt{b/c}$. So if $x,y\in \mathbb{Q}_{2^n}\left(\sqrt{\langle H, H'\rangle}\right)$ we have also $\sqrt{b/c}\in \mathbb{Q}_{2^n}\left(\sqrt{\langle H, H'\rangle}\right)$, which by Kummer theory implies $bc\in H$. But then $y\in \mathbb{Q}_{2^n}\left(\sqrt{H}\right)\left(x\right)$. So we have $\mathbb{Q}_{2^n}\left(\sqrt{\langle H, H'\rangle}\right)=\mathbb{Q}_{2^n}\left(\sqrt{H}\right)(x)$, and the sought degree is $\left[\mathbb{Q}_{2^n}\left(\sqrt{H}\right)(x):\mathbb{Q}_{2^n}\left(\sqrt{H}\right)\right]$, which is in fact $2$ ({\color{red}Do we need to explain this better?}).
-
-We conclude that
-\begin{align*}
-\left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]=\begin{cases}
-\#\overline{H}/2&\text{ if } n\geq 3,\,\pm2 \in H\text{ and }H'\subseteq\{2\zeta_4\},\\
-\#\overline{H}&\text{ if }(n<3\text{ or }\pm2\not\in H)\text{ and }H'\subseteq \{2\zeta_4\},\\
-\#\overline{H}&\text{ if } n\geq 3,\,\pm2 \in H\text{ and }H'\not\subseteq\{2\zeta_4\},\\
-2\cdot \#\overline{H}&\text{ if }(n<3\text{ or }\pm2\not\in H)\text{ and }H'\not\subseteq \{2\zeta_4\}.
-\end{cases}
-\end{align*}
-%Let $s$ as in the first subcase of this section and let $\mathcal{C}'$ and $H'$ be as in the last case of Remark 17. We have
-%\begin{align*}
-%\left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]=&\left[\mathbb{Q}_{2^n}\left(\sqrt{\langle H,\zeta_{2^n}H'\rangle}\right):\mathbb{Q}_{2^n}\right]=\\
-%=&\left[\mathbb{Q}_{2^n}\left(\sqrt{\langle H,\zeta_{2^n}H'\rangle}\right):\mathbb{Q}_{2^n}\left(\sqrt{H}\right)\right]\cdot \left[\mathbb{Q}_{2^n}\left(\sqrt{H}\right):\mathbb{Q}_{2^n}\right].
-%\end{align*}
-%Notice that, by construction of $H$ and $H'$, the degree $\left[\mathbb{Q}_{2^n}\left(\sqrt{\langle H,\zeta_{2^n}H'\rangle}\right):\mathbb{Q}_{2^n}\left(\sqrt{H}\right)\right]$ is either $1$
-\end{itemize}
-
-\begin{thebibliography}{10} \expandafter\ifx\csname url\endcsname\relax \def\url#1{\texttt{#1}}\fi \expandafter\ifx\csname urlprefix\endcsname\relax\def\urlprefix{URL }\fi
-
-\bibitem{DebryPerucca}
-\textsc{Debry, C. - Perucca, A.}: \emph{Reductions of algebraic integers}, J. Number Theory, {\bf 167} (2016), 259--283.
-
-\bibitem{PeruccaSgobba}
-\textsc{Perucca, A. - Sgobba, P.}: \emph{Kummer Theory for Number Fields}, preprint.
-
-\end{thebibliography}
-
-\end{document}
-\ No newline at end of file
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