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-rw-r--r--tex/old/compute_degree.aux14
-rw-r--r--tex/old/compute_degree.log706
-rw-r--r--tex/old/compute_degree.pdfbin0 -> 176578 bytes
-rw-r--r--tex/old/compute_degree.synctex.gzbin0 -> 44626 bytes
-rw-r--r--tex/old/compute_degree.tex320
5 files changed, 1040 insertions, 0 deletions
diff --git a/tex/old/compute_degree.aux b/tex/old/compute_degree.aux
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@@ -0,0 +1,14 @@
1\relax
2\citation{DebryPerucca}
3\citation{DebryPerucca}
4\newlabel{lemma_zero}{{1}{1}}
5\newlabel{degree}{{1}{1}}
6\@writefile{toc}{\contentsline {section}{\numberline {1}Case $G\leq \mathbb {Q}_+^\times $}{1}}
7\@writefile{toc}{\contentsline {section}{\numberline {2}General case}{2}}
8\@writefile{toc}{\contentsline {subsection}{\numberline {2.1}General case, $n=1(\leq d)$}{2}}
9\@writefile{toc}{\contentsline {subsection}{\numberline {2.2}General case, $n=2\leq d$}{2}}
10\@writefile{toc}{\contentsline {subsection}{\numberline {2.3}General case, $3\leq n\leq d$}{2}}
11\@writefile{toc}{\contentsline {subsection}{\numberline {2.4}General case, $n\geq d+2$}{3}}
12\@writefile{toc}{\contentsline {subsection}{\numberline {2.5}General case, $n=d+1$}{3}}
13\bibcite{DebryPerucca}{1}
14\bibcite{PeruccaSgobba}{2}
diff --git a/tex/old/compute_degree.log b/tex/old/compute_degree.log
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index 0000000..571b9ec
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@@ -0,0 +1,706 @@
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87\newtheorem{lemma}{Lemma}
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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}
104Let $H\leq \mathbb{Q}^\times$ 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*}
111where $\overline{H}$ is the image of $H\cdot \mathbb{Q}^{\times 2}$ in $\mathbb{Q}^\times/\mathbb{Q}^{\times 2}$.
112\begin{proof}
113Clearly 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
115Assume 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*}
117d_1=\begin{cases}
1180&\text{ if }g_1\neq\pm2\\
1191&\text{ if }g_1=\pm2
120\end{cases},
121&&
122d_i=0
123\quad \text{ for $i=2,\dots, r$},\\
124h_1=\begin{cases}
1250&\text{ if } 0\leq g_1\neq2\\
1261&\text{ if } -2\neq g_1<0\\
1272&\text{ if } g_1=\pm 2
128\end{cases}, &&
129h_i=\begin{cases}
1300&\text{ if }g_i>0\\
1311&\text{ if }g_i<0
132\end{cases}
133\quad \text{ for $i=2,\dots, r$}.
134\end{align*}
135Thus, keeping the notation of the aformentioned Theorem, we get
136\begin{align*}
137n_1=\min(1,d_1)=\begin{cases}
1380&\text{ if }g_1\neq\pm2\\
1391&\text{ if }g_1=\pm2
140\end{cases},&& n_i=0\quad \text{ for $i=2,\dots, r$}.
141\end{align*}
142Thus we get
143\begin{align*}
144v_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\\
147r-\sum_{i=1}^rn_i&\text{ if }\pm2\not \in H
148\end{cases}\\
149&=\begin{cases}
1501+r-1&\text{ if }m=2\text{ and }\pm2\in H\\
151r-1&\text{ if }m\geq3\text{ and }\pm2\in H\\
152r&\text{ if }\pm2\not\in H
153\end{cases}
154\end{align*}
155which is what we want.
156
157Assume 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
161Let $G\leq \mathbb{Q}^\times$ 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
167We will use the same notation as that of Remark 17 of Pietro's file.
168
169\section{Case $G\leq \mathbb{Q}_+^\times$}
170
171Assume 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
176Let $\overline{H}$ 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
187Let $\mathcal{B}$ 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
189As 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
191It 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}_+,\\
194g_0<0 \text{ and }\mathcal{B}_d\setminus \{g_0\}\subseteq \mathbb{Q}_+,\\
195d\geq 1.
196\end{align*}
197
198We also let
199\begin{align*}
200N=\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, $n=1(\leq d)$}
207This 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*}
212where $\#\overline{H'}$ is the image of $H'$ in $\mathbb{Q}^\times/\mathbb{Q}^{\times 2}$.
213
214\subsection{General case, $n=2\leq d$}
215We 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*}
222which, 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, $3\leq n\leq d$}
226We 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*}
239which, 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, $n\geq d+2$}
243By 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*}
250where $H'$ 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, $n=d+1$}
253We 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*}
259But we can be more precise and say that
260\begin{align*}
261\#\overline{H'}=\begin{cases}
2622\#\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*}
279hence
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*}
284We 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}
2871&\text{ if }H'=\emptyset\text{ or }H'=\{2\zeta_4\},\\
2882&\text{ otherwise}.
289\end{cases}
290\end{align*}
291To 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
293We 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\},\\
2992\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} \ No newline at end of file

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