compute_degree.tex (15872B)
1 \documentclass[10pt,a4paper]{article} 2 \usepackage[utf8]{inputenc} 3 \usepackage{amsmath} 4 \usepackage{amsthm} 5 \usepackage[all]{xy} 6 \usepackage{amsfonts} 7 \usepackage{color} 8 \usepackage{amssymb} 9 \usepackage{float} 10 \usepackage[a4paper, top=3cm, bottom=3cm, left=2.5cm, right=2.5cm]{geometry} 11 12 \DeclareMathOperator{\alg}{alg} 13 \DeclareMathOperator{\obj}{Obj} 14 \DeclareMathOperator{\Hom}{Hom} 15 \DeclareMathOperator{\End}{End} 16 \DeclareMathOperator{\hol}{Hol} 17 \DeclareMathOperator{\aut}{Aut} 18 \DeclareMathOperator{\gal}{Gal} 19 \DeclareMathOperator{\id}{id} 20 \DeclareMathOperator{\res}{res} 21 \DeclareMathOperator{\im}{Im} 22 \DeclareMathOperator{\Id}{Id} 23 \DeclareMathOperator{\fib}{Fib} 24 \DeclareMathOperator{\spec}{Spec} 25 \DeclareMathOperator{\proj}{Proj} 26 \DeclareMathOperator{\trdeg}{trdeg} 27 \DeclareMathOperator{\car}{char} 28 \DeclareMathOperator{\Frac}{Frac} 29 \DeclareMathOperator{\reduced}{red} 30 \DeclareMathOperator{\real}{Re} 31 \DeclareMathOperator{\imag}{Im} 32 \DeclareMathOperator{\vol}{vol} 33 \DeclareMathOperator{\den}{den} 34 \DeclareMathOperator{\rank}{rank} 35 \DeclareMathOperator{\lcm}{lcm} 36 \DeclareMathOperator{\rad}{rad} 37 \DeclareMathOperator{\ord}{ord} 38 \DeclareMathOperator{\Br}{Br} 39 \DeclareMathOperator{\inv}{inv} 40 \DeclareMathOperator{\Nm}{Nm} 41 \DeclareMathOperator{\Tr}{Tr} 42 \DeclareMathOperator{\an}{an} 43 \DeclareMathOperator{\op}{op} 44 \DeclareMathOperator{\sep}{sep} 45 \DeclareMathOperator{\unr}{unr} 46 \DeclareMathOperator{\et}{\acute et} 47 \DeclareMathOperator{\ev}{ev} 48 \DeclareMathOperator{\gl}{GL} 49 \DeclareMathOperator{\SL}{SL} 50 \DeclareMathOperator{\mat}{Mat} 51 \DeclareMathOperator{\ab}{ab} 52 \DeclareMathOperator{\tors}{tors} 53 \DeclareMathOperator{\ed}{ed} 54 55 \newcommand{\grp}{\textsc{Grp}} 56 \newcommand{\set}{\textsc{Set}} 57 \newcommand{\x}{\mathbf{x}} 58 \newcommand{\naturalto}{\overset{.}{\to}} 59 \newcommand{\qbar}{\overline{\mathbb{Q}}} 60 \newcommand{\zbar}{\overline{\mathbb{Z}}} 61 62 \newcommand{\pro}{\mathbb{P}} 63 \newcommand{\aff}{\mathbb{A}} 64 \newcommand{\quat}{\mathbb{H}} 65 \newcommand{\rea}{\mathbb{R}} 66 \newcommand{\kiu}{\mathbb{Q}} 67 \newcommand{\F}{\mathbb{F}} 68 \newcommand{\zee}{\mathbb{Z}} 69 \newcommand{\ow}{\mathcal{O}} 70 \newcommand{\mcx}{\mathcal{X}} 71 \newcommand{\mcy}{\mathcal{Y}} 72 \newcommand{\mcs}{\mathcal{S}} 73 \newcommand{\mca}{\mathcal{A}} 74 \newcommand{\mcb}{\mathcal{B}} 75 \newcommand{\mcf}{\mathcal{F}} 76 \newcommand{\mcg}{\mathcal{G}} 77 \newcommand{\mct}{\mathcal{T}} 78 \newcommand{\mcq}{\mathcal{Q}} 79 \newcommand{\mcr}{\mathcal{R}} 80 \newcommand{\adl}{\mathbf{A}} 81 \newcommand{\mbk}{\mathbf{k}} 82 \newcommand{\m}{\mathfrak{m}} 83 \newcommand{\p}{\mathfrak{p}} 84 85 \newcommand{\kbar}{\overline{K}} 86 87 \newtheorem{lemma}{Lemma} 88 \newtheorem{proposition}[lemma]{Proposition} 89 \newtheorem{conjecture}[lemma]{Conjecture} 90 \newtheorem{corollary}[lemma]{Corollary} 91 \newtheorem{definition}[lemma]{Definition} 92 \newtheorem{theorem}[lemma]{Theorem} 93 \newtheorem{cond-thm}[lemma]{Conditional Theorem} 94 \theoremstyle{definition} 95 \newtheorem{remark}[lemma]{Remark} 96 97 \author{Sebastiano Tronto} 98 99 100 \begin{document} 101 102 \begin{lemma} 103 \label{lemma_zero} 104 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 105 \begin{align*} 106 \left[\mathbb{Q}_{2^m}\left(\sqrt{H}\right):\mathbb{Q}_{2^m}\right]=\begin{cases} 107 \#\overline H/2 & \text{ if }m\geq 3\text{ and }\exists b\in H\text{ with }b\equiv\pm2\pmod{\mathbb{Q}^{\times 2}},\\ 108 \#\overline H&\text{ otherwise}. 109 \end{cases} 110 \end{align*} 111 where $\overline{H}$ is the image of $H\cdot \mathbb{Q}^{\times 2}$ in $\mathbb{Q}^\times/\mathbb{Q}^{\times 2}$. 112 \begin{proof} 113 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, $g_1=\pm 2$. 114 115 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 $\mathbb{Q}_4$: 116 \begin{align*} 117 d_1=\begin{cases} 118 0&\text{ if }g_1\neq\pm2\\ 119 1&\text{ if }g_1=\pm2 120 \end{cases}, 121 && 122 d_i=0 123 \quad \text{ for $i=2,\dots, r$},\\ 124 h_1=\begin{cases} 125 0&\text{ if } 0\leq g_1\neq2\\ 126 1&\text{ if } -2\neq g_1<0\\ 127 2&\text{ if } g_1=\pm 2 128 \end{cases}, && 129 h_i=\begin{cases} 130 0&\text{ if }g_i>0\\ 131 1&\text{ if }g_i<0 132 \end{cases} 133 \quad \text{ for $i=2,\dots, r$}. 134 \end{align*} 135 Thus, keeping the notation of the aformentioned Theorem, we get 136 \begin{align*} 137 n_1=\min(1,d_1)=\begin{cases} 138 0&\text{ if }g_1\neq\pm2\\ 139 1&\text{ if }g_1=\pm2 140 \end{cases},&& n_i=0\quad \text{ for $i=2,\dots, r$}. 141 \end{align*} 142 Thus we get 143 \begin{align*} 144 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=\\ 145 &=\begin{cases} 146 \max(3,m)-m+r-\sum_{i=1}^rn_i&\text{ if }\pm2\in H\\ 147 r-\sum_{i=1}^rn_i&\text{ if }\pm2\not \in H 148 \end{cases}\\ 149 &=\begin{cases} 150 1+r-1&\text{ if }m=2\text{ and }\pm2\in H\\ 151 r-1&\text{ if }m\geq3\text{ and }\pm2\in H\\ 152 r&\text{ if }\pm2\not\in H 153 \end{cases} 154 \end{align*} 155 which is what we want. 156 157 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. 158 \end{proof} 159 \end{lemma} 160 161 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 162 \begin{align} 163 \label{degree} 164 \left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]. 165 \end{align} 166 167 We will use the same notation as that of Remark 17 of Pietro's file. 168 169 \section{Case $G\leq \mathbb{Q}_+^\times$} 170 171 Assume that $G\leq \mathbb{Q}_+^\times$. In this case, by Remark 17, we have that 172 \begin{align*} 173 \mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M =\mathbb{Q}_{2^n}\left(\sqrt{H}\right). 174 \end{align*} 175 176 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 177 \begin{align*} 178 \left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]= 179 \begin{cases} 180 \#\overline H/2 & \text{if }n\geq 3\text{ and }2\in H,\\ 181 \#\overline H&\text{ otherwise}. 182 \end{cases} 183 \end{align*} 184 185 \section{General case} 186 187 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 $\mathcal{B}$. 188 189 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 $g_0$. 190 191 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: 192 \begin{align*} 193 \mathcal{B}_1,\dots,\mathcal{B}_{d-1},\mathcal{B}_{d+1},\dots,\mathcal{B}_L\subseteq \mathbb{Q}_+,\\ 194 g_0<0 \text{ and }\mathcal{B}_d\setminus \{g_0\}\subseteq \mathbb{Q}_+,\\ 195 d\geq 1. 196 \end{align*} 197 198 We also let 199 \begin{align*} 200 N=\begin{cases} 201 \max(3,L)&\text{if }d\neq L,\\ 202 \max(3,L+1)&\text{if }d=L. 203 \end{cases} 204 \end{align*} 205 206 \subsection{General case, $n=1(\leq d)$} 207 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 208 \begin{align*} 209 \left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]= 210 \#\overline{H'}, 211 \end{align*} 212 where $\#\overline{H'}$ is the image of $H'$ in $\mathbb{Q}^\times/\mathbb{Q}^{\times 2}$. 213 214 \subsection{General case, $n=2\leq d$} 215 We consider two cases: 216 \begin{itemize} 217 \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}. 218 \item If $v_2(M)\geq 3$ we have 219 \begin{align*} 220 \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], 221 \end{align*} 222 which, by Lemma \ref{lemma_zero}, is given by $\#\overline{H}$ if $2 \in H$ and by $2\#\overline{H}$ otherwise. 223 \end{itemize} 224 225 \subsection{General case, $3\leq n\leq d$} 226 We consider two cases: 227 \begin{itemize} 228 \item If $v_2(M)=3$, by lemma \ref{lemma_zero} we have 229 \begin{align*} 230 \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} 231 \#\overline H/2 & \text{ if }\pm 2\in H,\\ 232 \#\overline H&\text{ otherwise}. 233 \end{cases} 234 \end{align*} 235 \item If $v_2(M)\geq 4$ we have 236 \begin{align*} 237 \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], 238 \end{align*} 239 which, by Lemma \ref{lemma_zero}, is given by $\#\overline{H}$ if $2 \in H$ and by $2\#\overline{H}$ otherwise. 240 \end{itemize} 241 242 \subsection{General case, $n\geq d+2$} 243 By the corresponding case in Remark 17, we simply have 244 \begin{align*} 245 \left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]=\begin{cases} 246 \#\overline {H'}/2 & \text{ if }\pm 2\in H,\\ 247 \#\overline {H'}&\text{ otherwise}. 248 \end{cases} 249 \end{align*} 250 where $H'$ is constructed from $\mathcal{S}'=\mathcal{S}\cup\{B_0\}$ and $\overline{H'}$ is the image of $H'$ in $\mathbb{Q}^\times/\mathbb{Q}^{\times 2}$. 251 252 \subsection{General case, $n=d+1$} 253 We distinguish between some cases. 254 \begin{itemize} 255 \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 256 \begin{align*} 257 \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'}. 258 \end{align*} 259 But we can be more precise and say that 260 \begin{align*} 261 \#\overline{H'}=\begin{cases} 262 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},\\ 263 \#\overline{H}&\text{otherwise}. 264 \end{cases} 265 \end{align*} 266 %\item Assume $n=2$, $2^{n+1}\nmid M$ and either $v_2(g_0)>2$ or $g_0$ is odd. Then $\left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]=\#\overline H$. 267 %\item Assume $n=2$, $2^{n+1}\,|\,M$ and either $v_2(g_0)>2$ or $g_0$ is odd. ({\color{red}TODO}) 268 \item Assume $n\geq 2$ and $2^{n+1}\nmid M$. Then 269 \begin{align*} 270 \left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]=\begin{cases} 271 \#\overline H/2 & \text{ if }\pm 2\in H\text{ and }n\geq 3,\\ 272 \#\overline H&\text{ otherwise}. 273 \end{cases} 274 \end{align*} 275 \item Assume $n\geq 2$ and $2^{n+1}\,|\,M$. Following the notation of Remark 17, we have 276 \begin{align*} 277 \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) 278 \end{align*} 279 hence 280 \begin{align*} 281 \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]=\\ 282 &=\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]. 283 \end{align*} 284 We claim that 285 \begin{align*} 286 \left[\mathbb{Q}_{2^n}\left(\sqrt{\langle H, H'\rangle}\right):\mathbb{Q}_{2^n}\left(\sqrt{H}\right)\right]=\begin{cases} 287 1&\text{ if }H'=\emptyset\text{ or }H'=\{2\zeta_4\},\\ 288 2&\text{ otherwise}. 289 \end{cases} 290 \end{align*} 291 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?}). 292 293 We conclude that 294 \begin{align*} 295 \left[\mathbb{Q}_{2^n}\left(G^{1/2^n}\right)\cap \mathbb{Q}_M:\mathbb{Q}_{2^n}\right]=\begin{cases} 296 \#\overline{H}/2&\text{ if } n\geq 3,\,\pm2 \in H\text{ and }H'\subseteq\{2\zeta_4\},\\ 297 \#\overline{H}&\text{ if }(n<3\text{ or }\pm2\not\in H)\text{ and }H'\subseteq \{2\zeta_4\},\\ 298 \#\overline{H}&\text{ if } n\geq 3,\,\pm2 \in H\text{ and }H'\not\subseteq\{2\zeta_4\},\\ 299 2\cdot \#\overline{H}&\text{ if }(n<3\text{ or }\pm2\not\in H)\text{ and }H'\not\subseteq \{2\zeta_4\}. 300 \end{cases} 301 \end{align*} 302 %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 303 %\begin{align*} 304 %\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]=\\ 305 %=&\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]. 306 %\end{align*} 307 %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 $1$ 308 \end{itemize} 309 310 \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 311 312 \bibitem{DebryPerucca} 313 \textsc{Debry, C. - Perucca, A.}: \emph{Reductions of algebraic integers}, J. Number Theory, {\bf 167} (2016), 259--283. 314 315 \bibitem{PeruccaSgobba} 316 \textsc{Perucca, A. - Sgobba, P.}: \emph{Kummer Theory for Number Fields}, preprint. 317 318 \end{thebibliography} 319 320 \end{document}