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-\newtheorem{theorem}[lemma]{Theorem}
-\newtheorem{cond-thm}[lemma]{Conditional Theorem}
-\theoremstyle{definition}
-\newtheorem{remark}[lemma]{Remark}
-
-\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}
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