\documentclass[10pt,a4paper]{article} \usepackage[utf8]{inputenc} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{amssymb} \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} \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} \author{Sebastiano Tronto} \title{On the Second Condition of Theorem 5.2 ($\ell=2$)} \begin{document} \maketitle \textbf{Question:} is Lemma 1 of Lang's \emph{Elliptic Curves Diophantine Analysis}, Chapter 5 Section 5, page 117 (References/Kummer theory/LANG-book-EC.pdf) consistent with Jones-Rouse's Theorem 5.2, i.e. can we deduce that $F(\beta_1)$ is partially contained in $F(A[2])$ already? I believe the answer is no. Lang is working over a base field that contains all the $\ell^\infty$ torsion of $A$, so it doesn't apply in our case (J\&R assume surjectivity of the torsion part). A counterexample is actually given by J\&R right after the proof of the theorem (``Remark''). They don't say explicitly that $\kiu(\beta_1)\cap \kiu(A[2])=\kiu$ in this case, but I have tested this with sage (see the file \texttt{2-division-counterex.sage}): $\kiu(\beta_1)$ has degree $24$ over $\kiu$ and $\kiu(A[2])$ has degree $2$, which together imply that $[\kiu(A[2],\beta_1):\kiu(A[2])]\geq 4$, so it is maximal. The file \texttt{2-division.sage} contains some code that looks for other counterexamples (varying the parameters of a short Weierstrass equation), but it is quite slow (about 3 minutes on my pc for each elliptic curve of which it computes the 4-torsion field). \end{document}