From c1a5a3de06e21f616d1e5a5d99579d9437f2b12c Mon Sep 17 00:00:00 2001 From: Sebastiano Tronto Date: Fri, 20 Sep 2019 10:11:55 +0200 Subject: New documentation --- docs/compute_degree.tex | 320 ------------------------------------------------ 1 file changed, 320 deletions(-) delete mode 100644 docs/compute_degree.tex (limited to 'docs/compute_degree.tex') diff --git a/docs/compute_degree.tex b/docs/compute_degree.tex deleted file mode 100644 index 836a229..0000000 --- a/docs/compute_degree.tex +++ /dev/null @@ -1,320 +0,0 @@ -\documentclass[10pt,a4paper]{article} -\usepackage[utf8]{inputenc} -\usepackage{amsmath} -\usepackage{amsthm} -\usepackage[all]{xy} -\usepackage{amsfonts} -\usepackage{color} -\usepackage{amssymb} -\usepackage{float} -\usepackage[a4paper, top=3cm, bottom=3cm, left=2.5cm, right=2.5cm]{geometry} - -\DeclareMathOperator{\alg}{alg} -\DeclareMathOperator{\obj}{Obj} -\DeclareMathOperator{\Hom}{Hom} -\DeclareMathOperator{\End}{End} -\DeclareMathOperator{\hol}{Hol} -\DeclareMathOperator{\aut}{Aut} -\DeclareMathOperator{\gal}{Gal} -\DeclareMathOperator{\id}{id} -\DeclareMathOperator{\res}{res} -\DeclareMathOperator{\im}{Im} -\DeclareMathOperator{\Id}{Id} -\DeclareMathOperator{\fib}{Fib} -\DeclareMathOperator{\spec}{Spec} -\DeclareMathOperator{\proj}{Proj} -\DeclareMathOperator{\trdeg}{trdeg} -\DeclareMathOperator{\car}{char} -\DeclareMathOperator{\Frac}{Frac} -\DeclareMathOperator{\reduced}{red} -\DeclareMathOperator{\real}{Re} -\DeclareMathOperator{\imag}{Im} -\DeclareMathOperator{\vol}{vol} -\DeclareMathOperator{\den}{den} -\DeclareMathOperator{\rank}{rank} -\DeclareMathOperator{\lcm}{lcm} -\DeclareMathOperator{\rad}{rad} -\DeclareMathOperator{\ord}{ord} -\DeclareMathOperator{\Br}{Br} -\DeclareMathOperator{\inv}{inv} -\DeclareMathOperator{\Nm}{Nm} -\DeclareMathOperator{\Tr}{Tr} -\DeclareMathOperator{\an}{an} -\DeclareMathOperator{\op}{op} -\DeclareMathOperator{\sep}{sep} -\DeclareMathOperator{\unr}{unr} -\DeclareMathOperator{\et}{\acute et} -\DeclareMathOperator{\ev}{ev} -\DeclareMathOperator{\gl}{GL} -\DeclareMathOperator{\SL}{SL} -\DeclareMathOperator{\mat}{Mat} -\DeclareMathOperator{\ab}{ab} -\DeclareMathOperator{\tors}{tors} -\DeclareMathOperator{\ed}{ed} - -\newcommand{\grp}{\textsc{Grp}} -\newcommand{\set}{\textsc{Set}} -\newcommand{\x}{\mathbf{x}} -\newcommand{\naturalto}{\overset{.}{\to}} -\newcommand{\qbar}{\overline{\mathbb{Q}}} -\newcommand{\zbar}{\overline{\mathbb{Z}}} - -\newcommand{\pro}{\mathbb{P}} -\newcommand{\aff}{\mathbb{A}} -\newcommand{\quat}{\mathbb{H}} -\newcommand{\rea}{\mathbb{R}} -\newcommand{\kiu}{\mathbb{Q}} -\newcommand{\F}{\mathbb{F}} -\newcommand{\zee}{\mathbb{Z}} -\newcommand{\ow}{\mathcal{O}} -\newcommand{\mcx}{\mathcal{X}} -\newcommand{\mcy}{\mathcal{Y}} -\newcommand{\mcs}{\mathcal{S}} -\newcommand{\mca}{\mathcal{A}} -\newcommand{\mcb}{\mathcal{B}} -\newcommand{\mcf}{\mathcal{F}} -\newcommand{\mcg}{\mathcal{G}} -\newcommand{\mct}{\mathcal{T}} -\newcommand{\mcq}{\mathcal{Q}} -\newcommand{\mcr}{\mathcal{R}} -\newcommand{\adl}{\mathbf{A}} -\newcommand{\mbk}{\mathbf{k}} -\newcommand{\m}{\mathfrak{m}} -\newcommand{\p}{\mathfrak{p}} - -\newcommand{\kbar}{\overline{K}} - -\newtheorem{lemma}{Lemma} -\newtheorem{proposition}[lemma]{Proposition} -\newtheorem{conjecture}[lemma]{Conjecture} -\newtheorem{corollary}[lemma]{Corollary} -\newtheorem{definition}[lemma]{Definition} -\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} \ No newline at end of file -- cgit v1.3