From 5ea79c7ae0d44686f1df05c4a016652afbe58968 Mon Sep 17 00:00:00 2001
From: Sebastiano Tronto <sebastiano@tronto.net>
Date: Sun, 14 Jun 2026 09:58:21 +0200
Subject: Initial commit

---
 necessity_52b/necessity_52b.tex | 96 +++++++++++++++++++++++++++++++++++++++++
 1 file changed, 96 insertions(+)
 create mode 100755 necessity_52b/necessity_52b.tex

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+\documentclass[10pt,a4paper]{article}
+\usepackage[utf8]{inputenc}
+\usepackage{amsmath}
+\usepackage{amsthm}
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+
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+\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}
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