\documentclass{article}
%\usepackage{mypack}
\usepackage{amsmath,amssymb,amsthm}
\usepackage[OT2,T1]{fontenc}
\DeclareSymbolFont{cyrletters}{OT2}{wncyr}{m}{n}
\DeclareMathSymbol{\Sha}{\mathalpha}{cyrletters}{"58}
\setlength{\textheight}{22cm} \headsep=15pt
%%change hight page
\setlength{\textwidth}{15.5cm}
%%change width page
\setlength{\oddsidemargin}{0.5cm} \setlength{\topmargin}{-.5cm}
\setlength{\evensidemargin}{\oddsidemargin}
\newcommand{\ttilde}{\lower4.5pt\hbox{$\widetilde{\ \ \ }$}}
\newcommand{\ETQP}{\E_\mathrm{tors}(\bbbq_p)}
\def\Z{{\mathbb Z}}
\def\Q{{\mathbb Q}}
\def\F{{\mathbb F}}
\def\a{{\alpha}}
\def\cE{{\mathcal E}}
\def\cD{{\mathcal D}}
\def\rTo{{\longrightarrow}}
\def\rInto{{\hookrightarrow}}
\def\bbbq{{\mathbb Q}}
\def\bbba{{\mathbb A}}
\def\adeles{{\bbba}_K}
\def\adelesQ{{\bbba}_\bbbq}
\def\bbbc{{\mathbb C}}
\newcommand{\TE}{{\widetilde{\mathcal{E}}}}
\def\bbbz{{\mathbb Z}}
\def\bbbn{{\mathbb N}}
\def\bbbr{{\mathbb R}}
\def\bbbf{{\mathbb F}}
\newcommand{\E}{\mathcal{E}}
\title{A non-abelian conjecture of Birch and Swinnerton-Dyer type}
\author{Minhyong Kim}
\date{Pohang, September, 2013}
\def\ms{\medskip}
\def\ft{\frametitle}
\def\fr{\frame}
\def\O{{\cal O}}
\def\piqp{\pi_1^{et, \Q_p}(\bX, b)}
\def\ms{\medskip}
\def\M{{\mathcal M}}
\def\dd{\hbox{d}}
\def\o{\underline{\bf o}}
\def\cD{{\mathcal D}}
\def\hphi{\hat{\phi}}
\def\hq{\hat{q}}
\begin{document}
\begin{flushleft}


1. (a) State and prove Hensel's Lemma.
\ms

\ms

Let~$K$ be a field, complete with respect to a non-Archimedean
valuation~$|\ \, |$, with valuation ring~$R = \{ x\in K : |x| \leqslant 1\}$.
If $f(x) \in R[x]$ and $a_0\in R$ satisfies
$| f(a_0) | < | f'(a_0) |^2$, then there exists a unique $a\in R$ such that
$f(a) = 0$ and $| a - a_0 | \leqslant | f(a_0) |/ | f'(a_0) |$.

\ms

\begin{proof}

[{\it From lectures}] Define $f_j(x)$ by:
\par $f(x + y) = f_0(x) + f_1(x) y + f_2(x) y^2 + \ldots,$
\par\noindent so that $f_0(x) = f(x), f_1(x) = f'(x)$. Define
$b_0 = -f(a_0)/f'(a_0)$. By $(*)$, $|b_0| < 1$.
\par Define $a_1 = a_0 + b_0 = a_0 - f(a_0)/f'(a_0)$. Then:
\par $| f'(a_1) - f'(a_0) | = | f'(a_0+b_0) - f'(a_0) |
= | (\hbox{poly in }a_0) b_0 + (\hbox{poly in }a_0) b_0^2 + \ldots |$
\par \ \ \ \ \ \ \ \ \ \ \ $\leqslant | b_0 | < | f'(a_0) |$
\ \ (by $(*)$),
\par\noindent  so that $| f'(a_1) | = | f'(a_0) |$.
\par Also,
$| f(a_1) | = | f(a_0 + b_0) | =
| f_0(a_0) + f_1(a_0)b_0 + f_2(a_0)b_0^2 + \ldots |$
\par\ \ \ $= | f_2(a_0)b_0^2 + \ldots |$\ \
[since $f_0(a_0) + f_1(a_0)b_0 = 0$]
\par\ \ \ $\leqslant \hbox{max}_{j\geqslant 2} |f_j(a_0)| |b_0|^j
\leqslant | b_0 |^2 = \frac{ |f(a_0 )|^2}{|f'(a_0 )|^2}
= \rho | f(a_0) | < | f(a_0) |$,
where $\rho = \frac{ |f(a_0 )|}{|f'(a_0 )|^2} < 1$.
\par Summarising: $ |f'(a_1) | = |f'(a_0)|$ and
$| f(a_1) | \leqslant \rho | f(a_0) | < | f(a_0) |$, where
$\rho = \frac{ |f(a_0 )|}{|f'(a_0 )|^2} < 1$.
\par\noindent For all~$n$, given $a_n\in R$, define
$b_n = -f(a_n)/f'(a_n)$ and $a_{n+1} = a_n + b_n = a_n -f(a_n)/f'(a_n)$.
\par\noindent Assume, as induction hypothesis, that:
\par $|f'(a_n)| = \ldots = |f'(a_1)| = |f'(a_0)|$
and $ | f(a_n) | \leqslant \rho |f(a_{n-1})| \leqslant \ldots
\leqslant \rho^n |f(a_0)|$.\ \ \ \ \ \ (1)
\par\noindent Then, as above: $|f'(a_{n+1})| = \ldots = |f'(a_1)| = |f'(a_0)|$.
\par\noindent Then $| f(a_{n+1}) | \leqslant |b_n|^2$ \ \ [justified as
for the case~$n=0$ above]
\par \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
$= \frac{ | f(a_n) |^2 }{ | f'(a_n) |^2 }
= \frac{ | f(a_n) |^2 }{ | f'(a_0) |^2 }$\ \ [by~(1), the induction hypothesis]
\par \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
$\leqslant \frac{ | f(a_0) | }{ | f'(a_0) |^2 } |f(a_n)|$
\ \ [since $|f(a_n)| \leqslant |f(a_0)|$ by~(1), the induction hypothesis]
\par \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
$= \rho | f(a_n) | \leqslant \rho^{n+1} | f(a_0) |$
\ \ [by~(1), the induction hypothesis].
\par\noindent By induction, $\forall n$, $|f'(a_n)| = |f'(a_0)|$ and
$ | f(a_{n}) | \leqslant \rho^n | f(a_0) |$ which $\rightarrow 0$
as $n\rightarrow \infty$. \ \ \ \ \ \ (2)
\par\noindent Now, $| b_n | = | f(a_n) |/|f'(a_n)|
= | f(a_n) |/|f'(a_0)| \rightarrow 0$, so
[by the theorem from lectures that a series converges
in a non-Archimedean field iff its terms converge to~$0$]:
\par $a_n = a_0 + b_0 + b_1 + \ldots + b_n$
converges to~$a$, say.
\par\noindent By continuity of polynomials,
$f(a) = \lim f(a_n) = 0$\ \ [by~(2)].
Furthermore: 
\par\noindent $| a - a_0 | = | \sum b_n | \leqslant
\max | b_n | = \max \frac{| f(a_n) |}{|f'(a_n)|}
= \max \frac{| f(a_n) |}{|f'(a_0)|}
= \frac{| f(a_0) |}{|f'(a_0)|}$ [by~(2)], as required.
\par\noindent For uniqueness, imagine that $\hat a \not= a$ also satisfied
$f(\hat a) = 0$ and $| \hat a - a_0 | \leqslant | f(a_0) |/ | f'(a_0) |$.
Let $\hat b = \hat a - a \not= 0$.
\par Then $0 = f(\hat a) - f(a) = f(a + \hat b) - f(a)
= {\hat b} f_1(a) + {\hat b}^2 f_2(a) + \ldots$ \ \ \ \ \ \ (3)
\par\noindent But $| \hat b | = | \hat a - a_0 + a_0 - a |
\leqslant \max( | \hat a - a_0 |, | a - a_0 | )
\leqslant | f(a_0) |/ | f'(a_0) | $
\par \ \ \ $< | f'(a_0) \hbox{ [by (*)] }|
= | f_1(a_0) | = | f_1(a) |$ \ \ [by~(1) and continuity of $|f'(x)|$].
\par\noindent This gives $|{\hat b}^j f_j(a)| \leqslant
|{\hat b}^j| \leqslant |{\hat b}^2| < |{\hat b} f_1(a)|$
(since $|{\hat b}| \not= 0$ \& $|{\hat b}| < |f_1(a)|$)
for~$j\geqslant 2$, so that the leading term of the sum in~(3)
has valuation strictly greater than the valuations of the other terms,
which is inconsistent with the sum being~$0$. Hence~$a$
is unique.
\end{proof}
\hfill {\bf [10~marks]}
\ms

\ms

(b)

(i) Which numbers $a\in \Q_2$ have cube roots  in $\Q_2$?
\ms

\ms
[\textit{Similar seen in problem sheet}]

Obviously, $a=0$ has a cube root. If $a\neq 0$, then
we will have $a=b^3$ if and only if $a=2^{3n}u$ where $u$ is a unit in $\Z_2$ that has a cube root.
But $u\equiv 1 \mod 2$, so
If we put $f(x)=x^3-u$, then $x=1$ will be a root of $f$ $\mod 2$, and $f'(1)=3$. Hence,
$|f(1)|_2<1$, while $|f'(1)|_2=1$. Thus, $|f(1)|_2<|f'(1)|^2$, and Hensel's Lemma applies to give us
a root $\a$ of $f(x)$ in $\Z_2$. That is, all units in $\Z_2$ will have cube roots.
Therefore, $a\in \Q_2^{\times}$ has a cube root if and only if $a=2^{3n}u$ whre $u$ is any unit.

\hfill {\bf [2~marks]}
\ms

\ms
(ii)


Which numbers $a\in \Q_3$ have cube roots  in $\Q_3$?
\ms

\ms

[\textit{Similar seen in problem sheet}]


Obviously, $a=0$ has a cube root. If $a\neq 0$, then $a$ has a cube root if and only if
$a=3^{3n}u$ where $u\in \Z_3$ is a unit that has a cube root. If $u$ has a cube root, then
$u\mod 3^m$ has a cube root for every $m$, and hence, $u\mod 27$ has a cube root. Conversely, suppose $u \mod 27 $ has a cube root. So there is a $b\in \Z_3$ such that $b^3\equiv u \mod 27.$
If we define $f(x)=x^3-u$, then $|f(b)|_3\leq 3^{-3}$. Now, since $u$ is a unit, $b$ is also a unit.
Hence, since $f'(b)=3b$, we $|f'(b)|_3=3^{-1}$. Therefore, $|f(b)|_3< |f'(b)|^2.$
By Hensel's lemma, $f$ would then have a root in $\Z_3$, which would then be a cube root of $u$.
To sum, $a\in \Q_3^{\times}$ has a cube root if and only if $a=3^{3n}u$ where $u$ is a unit in
$\Z_3$ such that $u\mod 27 $ has a cube root. Which units of $\Z/27$ have cube roots?
To find this out, first compute the cubes in $(\Z/9)^{\times}$. We  quickly see that the only possibilities are
$\pm 1\mod 9$. Thus, the cubes in $(\Z/27)^{\times}$ must be among
$1, 8, 10,-10,-8, -1.$ Clearly, all are cubes except possibly $\pm 10$. But $4^3\equiv 10 \mod 27$
and $(-4)^3\equiv -10 \mod 27$. Therefore, $a\in \Q_3^{\times}$ has a cube root if and only if
$a=3^{3n}u$ where $u\in \Z_3^{\times}$ is congruent to $\pm 1 \mod 9$.

\hfill {\bf [3~marks]}
\ms

\ms

(iii)

How many solutions are there to
$$x^{37}-x=37$$
in $\Q_{37}$?
\ms

\ms
[\textit{unseen}]

Let $f(x)=x^{37}-x-37$.
For $a=0, 1, \ldots, 36$, we have
$f(a)\equiv 0 \mod 37$.
On the other hand,
$f'(a)=37a^{36}-1\equiv -1 \mod 37.$
Therefore,
for each such $a$, we have $|f(a)|_{37}<1=|f'(a)|_{37}^2$.
Therefore, for each such $a$, there exists an $\a$ that is a root of  $f(x)$ such that
$|\a-a|_{37}\leq |f(a)|_{37}/|f'(a)|_{37}<1.$ That is, each $\a$ is congruent $\mod 37 $ to $a$.
In particular, all these roots are distinct. Therefore, $f(x)$ has at least 37 distinct roots.
Obviously, it can't have more, since $\Q_{37}$ is a field. Therefore, the equation has exactly 37 solutions.

\hfill {\bf [3~marks]}

\ms

\ms

(c)

Consider the equation
$$y^2=x^3+6.$$
Find all $p$ for which the equation has a solution in $\Z_p$.

[\textit{Warning: You are not allowed to use the point at `infinity'.}]

[\textit{You may use standard facts from the lectures as long as they are stated clearly.}]

\ms

\ms

[\textit{Similar seen in lecture}]

If we take $x=3$, then we get the equation $y^2=33$, which gives $y^2\equiv 1 \mod 8$.
Thus, 33 has a square root mod 8, and hence, a square root in $\Z_2$. Therefore, there is a solution in $\Z_2$.

If we take $x=1$, then we get $y^2=7$, which is $y^2\equiv 1 \mod 3$. Thus, 7 has a square root mod 3, and hence, a square root in $\Z_3$. So the  equation has a solution in $\Z_3$. Taking $x=0$, we get
$y^2=6$, which becomes $y^2\equiv 1 \mod 5$. Hence, $6$ has a square root mod 5, and therefore, a square root in $\Z_5$.

If we take $y=0$, then we get the equation $0=x^3+6$ or
$x^3-1\equiv 0 \mod 7$. This clearly has the root 1 mod 7, and hence, has a root in $\Z_7$.
Therefore, the equation has a solution in $\Z_7$.

Consider now the situation of $\Z_p$ for $p\geq 11$. The discriminant of the equation
is $27\times 6 $ so the corresponding elliptic curve only has bad reduction at 2 and 3. Thus, for
all prime $p\geq 11$, we will get an elliptic curve upon reduction mod $p$, which
has numbers of point $\geq p+1-2\sqrt{p}$. For $p\geq 11$ this is always greater than 4. Hence, the elliptic curve will have some point $(x_0, y_0)$ in $\F_p$ such that $y_0\neq 0$. Now let $(x,y)$ be a pair in $\Z_p$ that lifts the pair $(x_0, y_0)$ and consider the polynomial
$h(t)=t^2-(x^3+6)$. Clearly, $|h(y)|_p<1$ and $h'(y)=2y$ is non-zero mod $p$, and hence,
$|h'(y)|_p=1$. Therefore, by Hensel's Lemma, $h(t)$ has a root $y_1$ in $\Z_p$, upon which
$(x,y_1)$ will be a solution of our original equation in $\Z_p$.

To conclude, the equation has solutions in $\Z_p$ for all primes $p$.

\hfill {\bf [7~marks]}

\ms

2. Let~$R$ be any ring (commutative, with~1), and let~$F,G$ be
formal groups over~$R$.
\ms

\ms

(a) 

Show that there exists a unique
normalised invariant differential for~$F$, which is given by
$\omega = F_X(0, T)^{-1}\hbox{d} T \in R[[T]] \hbox{d} T$,
and that every invariant differential for~$F$ is of the form $a\omega$
for some $a\in R$.
%\hfill {\bf [6~marks]}

\ms

\begin{proof}
 [{\it From lectures}]
Let $P(T) = F_X(0, T)^{-1}$, so that $\omega(T) = P(T)\dd T$.
Note that $F_X(0, T) = 1 + \ldots$ is invertible, so
that $P(T)$ is indeed a member of~$R[[T]]$.
Furthermore, $P(0) = 1$, so that it is normalised.
\par We need to show that $\omega$ is an invariant differential;
that is, $\omega \circ F(T,S) = \omega(T)$, which is equivalent to:
$P\bigl( F(T,S) \bigr) F_X (T,S) = P(T)$ so, in our case,
it is sufficient to show:
$$ F_X\bigl( 0, F(T,S) \bigr)^{-1} F_X(T,S) = F_X(0,T)^{-1}, $$ 
which is true iff:
$$ F_X\bigl( 0, F(T,S) \bigr) = F_X(T,S) F_X(0,T). $$
But this last statement is immediate from differentiating
$F\bigl( U, F(T,S) \bigr) = F\bigl( F(U,T), S \bigr)$
[associativity]
with respect to~$U$ to get: 
$F_X\bigl( U, F(T,S) \bigr) = F_X\bigl( F(U,T), S\bigr) F_X(U,T)$
and setting~$U = 0$. Hence~$\omega$ is an invariant differential. 
\par Suppose that $\hat \omega = Q(T)\dd T \in R[[T]] \dd T$ is also an 
invariant differential, so that $Q(T)$ satisfies
$Q\bigl( F(T,S) \bigr) F_X(T,S) = Q(T)$.
Substituting~$T=0$ gives $Q(S) F_X(0,S) = Q(0)$, so
that $Q(S) = Q(0) F_X(0,S)^{-1}$. It follows that
$\hat \omega = a \omega$, where $a = Q(0)$.

\par\noindent
\end{proof}

\hfill {\bf [8~marks]}
\ms

\ms

(b)
Let~$f$ be a homomorphism over~$R$ from~$F$
to~$G$. Let~$\omega_F, \omega_G$ be
the normalised invariant differentials on~$F,G$, respectively.
Show that $\omega_G \circ f = f'(0)\ \omega_F$.
Deduce that, for any prime~$p$, there
exist $f,g\in R[[T]]$
such that $[p](T) = p f(T) + g(T^p)$ [where~$[p]$ represents
the multiplication-by-$p$ map on~$F$].
%\hfill {\bf [6~marks]}
\ms

\ms
\begin{proof}
[{\it From lectures}]
First, note that $\omega_G \circ f\bigl( F(T,S) \bigr)
= \omega_G \bigl( G( f(T), f(S) \bigr) = \omega_G \circ f (T)$,
so that $\omega_G \circ f$ is an inviariant differential
on~$G$. From part~(a), it follows
that $\omega_G \circ f = a\ \omega_F$, for some~$a\in R$.
Since~$\omega_F,\omega_G$ are normalised,
$(1 + \ldots ) \dd f(T)  = a (1 + \ldots )\dd T$,
and so $(1 + \ldots ) f'(T) \dd T  = a (1 + \ldots )\dd T$; 
equating constant terms gives $a = f'(0)$, as required.
\par Let~$\omega$ be the normalised invariant differential on~$F$.
Since $[p](T) = p T + \dots$, it satisfies $[p]'(0) = p$.
Applying the previous result to~$[p]$, a homomorphism from~$F$
to itself, gives: $\omega \circ [p] = [p]'(0) \omega = p \omega$,
and so
$$ p \omega(T) = \omega \circ [p](T)
= (1 + \ldots ) \dd ( [p](T) ) = (1 + \ldots ) [p]'(T) \dd T.$$
Hence $[p]'(T) \in p\, R_T$. Each term $a_n T^n$ in $[p](T)$
must then satisfy $ p | n a_n$ and so $p | n$ or $p | a_n$,
as required.
\end{proof}

\hfill {\bf [7~marks]}

\ms

\ms

(c) 
Let $\cE$ be given by an equation
$$\cE: y^2=x^3+(3m+1)x+9n^2,$$
where the polynomial $x^3+(3m+1)x+9n^2$ has no rational root.
Prove that $\cE(\Q)$ is infinite.
\ms

\ms


[\textit{unseen}] 

Reducing mod 3, we get the equation
$$y^2=x^3+x$$
which defines an elliptic curve over $\F_3$. This curve has the four points
$\o, (0,0), (2, \pm 1)$. Since
$\cE_{tor}(\Q)$ admits an injective homomorphism to $\cE(\F_3)$,
we see that $\cE_{tor}(\Q)$ has order at most four. Hence, the only possible points of finite order must have order 2 or 4. But $\cE(\Q)$ clearly has no points of order 2, since $x^3+(3m+1)x+9n^2$ is assumed to have no rational roots.  Hence, $\cE_{tor}(\Q)=\o$.
On the other hand, the curve has the obvious point $(0, 3n)$, which must then be a point of infinite order.

\hfill {\bf [5~marks]}



\ms

\ms
(d)  
Let $\cE$ be given by an equation
$$\cE: y^2=x^3+ax,$$
where $a$ is a non-zero square-free integer. Show that $\cE(\Q)$ has no element of order 4.
\ms

\ms

[\textit{Similar seen in lecture}]

First note: the question should also read that $a \not= \pm 1$.
However, even just taking the question as read, one can just
deal separately with $y^2 = x^3 - x$ and reduce mod~$3$ to
get $C_2 \times C_2$ and 
so the torsion group over~$\Q$ is $C_2 \times C_2$; also
for $y^2 = x^3 + x$, reduce mod~$5$ to get $C_2 \times C_2$
and so the torsion group over~$\Q$ is $C_2$; in both cases,
no point of order~$4$.
If $P$ were a point of order 4, we would have $2P=(0,0)$, since $(0,0)$ is the only point of
order 2. This is saying that the tangent line to $\cE$ at $P$ must go through $(0,0)$.
Hence, we must consider the non-vertical 
lines through $(0,0)$ and see which ones are tangent to $\cE$ at the other point of intersection.
Theses lines have the equation $y=mx$. Thus, we get
$$m^2x^2=x^3+ax,$$
and the other points of intersection will have $x$-coordinates that are the roots of
$x^2-m^2x+a$. This polynomial will have multiple roots if and only if
$m^4-4a=0$. Hence, we get $m=\pm \sqrt{2}a^{1/4}, \pm i\sqrt{2}a^{1/4}$.
The $x$-values at the intersection points in these cases are $x=m^2/2=\pm \sqrt{a}.$
These values will never be rational. Hence, $\cE(\Q)$ has no point of order 4.

\hfill {\bf [5~marks]}

\smallskip
\par\noindent

\ms

\ms

\ms
3.

Let $l$ be a prime and ${\mathcal E}_l$ be the elliptic curve
$$y^2=x^3-l^2x.$$

\ms

\ms

(a) Find the torsion subgroup ${\mathcal E}_{l, \mbox{tor}}(\Q)$ of the Mordell-Weil group.


\ms

\ms
[\textit{Similar seen in lecture}]

Note the four points $\o$, $(0,0)$, $(\pm l,0)$ of order two. Consider first the case $l\neq 3$.

Then the equation defines an elliptic curve over $\F_3$. In fact, for any $l\neq 3$, the equation over $\F_3$ is
$$y^2=x^3-x.$$
This has the four points $\o, (0,0), (2,0), (3,0).$  Since the reduction map defines an injection
$${\mathcal E}_{l, \mbox{tor}}(\Q)\hookrightarrow {\mathcal E}(\F_3),$$
we get that
$${\mathcal E}_{l, \mbox{tor}}(\Q)={\mathcal E}(\Q)[2]\simeq C_2\times C_2.$$

Now suppose $l=3$ and consider  the equation
$$y^2=x^3-3^2x=x^3+x$$
over $\F_5$ (where it has good reduction). It has the points
$\o, (0,0), (1,0), (4,0)$. Since we still have an injection
$${\mathcal E}_{3, \mbox{tor}}(\Q)\hookrightarrow {\mathcal E}_3(\F_5),$$
we again get
$${\mathcal E}_{3, \mbox{tor}}(\Q)={\mathcal E}_3(\Q)[2]\simeq C_2\times C_2.$$

\hfill {\bf [5~marks]}
\ms

\ms

(b) If $l$ is a prime, show that the rank of
$${\mathcal E}_l: y^2=x^3-l^2x$$
is at most two.

\ms

\ms
[\textit{Similar seen in lecture}]

We have the standard isogeny
$$\phi: \cE_l \rTo \cD$$
where the latter is defined by
$$y^2=x^3+4l^2x,$$
and the dual isogeny
$$\hphi: \cD\rTo \cE_l.$$
The maps are given by
$$\phi(x,y)=(y^2/x^2, y+25y/x^2)$$
and
$$\phi(x,y)=((1/4)(y/x)^2, (1/8)(y-100y/x^2)).$$
There is the map
$$q:\cD(\Q)/\phi(\cE_l(\Q))\rInto \Q^*/(\Q^*)^2$$
that takes $(u,v)$ to $u$ if $(u,v)\neq (0,0)$ and $q(0,0)=4l^2=1$.
Also,
$$\hq: \cE_l(\Q)/\hphi(\cD(\Q))\rInto \Q^*/(\Q^*)^2$$
takes $(u,v)$ to $u$ for $(u,v)\neq (0,0)$ and  $\hq(0,0)=-l^2=-1.$
Now $Im(\hq)$ is given by square-free integers that divide $l$, and hence,
$\pm 1, \pm l$. Also, $Im(q)$ is given by square-free integers dividing
$2l$, that is, $\pm 1, \pm 2, \pm l, \pm 2l$.
We have $r\in Im (\hq) $ if and only if
$$r^2s^4-l^2m^4=rn^2$$
admits a solution $(s,m,n)$ such that $gcd(s,m)=1$. But for $r=1,-1, l, -l$, we will have the corresponding solutions
$(1,0,1)$, $(0,1,l)$, $(1,1,0)$ and $(1,1,0)$. Therefore, $Im(\hq)=\{\pm 1, \pm l\}$.
On the other hand, $r\in Im(q)$ if and only if
$$r^2s^4+4l^2m^4=rn^2$$
has a solution $(s,m,n)$ such that $gcd(s,m)=1$. Hence, we must have $r>0$.
That is, $Im(q)\subset \{ 1,2,l, 2l\}$.

We conclude that $\cE_l(\Q)/\hphi(\cD(\Q))$ has rank 2 and $\cD(\Q)/\phi(\cE_l(\Q))$ has rank at most 2.
Therefore, $\cE_l(\Q)/2\cE_l(\Q)$ has rank at most 4. Then, since $\cE_l(\Q)[2]=C_2\times C_2$, we get that
the Mordell-Weil rank is at most 2.

\hfill {\bf [5~marks]}

\ms

\ms


(c) Compute the rank of
$${\mathcal E}_5: y^2=x^3-25x.$$

\ms

\ms

[\textit{Similar seen in lecture}]

We have seen already that
$\cE_5(\Q)/\hphi(\cD(\Q))$  has rank 2. Consider $\cD(\Q)/\phi(\cE_5(\Q))$.
For each of $r\in \{2,5, 10\}$ 
we must consider the equation
$$r^2s^4+100m^4=rn^2.$$
When $r=2$, this becomes
$$4s^4+100m^4=2n^2.$$
After replacing $r$ by $2\times 2^2$ and dividing by 4, we get the equation
$$s^4+25m^4=2n^2.$$
If there were a solution, then we could find one of minimal length.
Now, suppose $5\mid n$. Then we would have $5|s$, after which we would get
$$m^4+25(s/5)^4=2(n/5)^2,$$
contradicting the minimality. Therefore, $5$ does not divide $n$. But then, reducing mod 5 would give us a square root of 2 in $\F_5$, a contradiction. Therefore, the equation has no solution, that is, $2\notin Im(q)$.

On the other hand, if we try $r=5$, we get
$$25s^4+100m^4=5n^2$$
which clearly has the solution $(1,1,5)$. Therefore, $Im(q)=\{1,5\}$ and
 $\cD(\Q)/\phi(\cE_5(\Q))$ has rank 1.
The point on $\cD$ corresponding to $r=5$  is $(5, 25)$ and we calculate quickly that
$\hphi(5,25)=(25/4, -75/8).$ This is not a torsion point and contributes a generator to
$\cE_5/2\cE_5$. Therefore, $\cE_5/2\cE_5$ has rank 3 and $\cE_5$ has rank one.


\hfill {\bf [10~marks]}
\ms

\ms



\ms

\ms

(d) For $l\equiv 3 \mod 8$, show that ${\mathcal E}_l$ has rank zero.

\ms

\ms

[\textit{unseen}]

The key again is to analyze $\cD(\Q)/\phi(\cE_l(\Q))$, that is, the equation


$$r^2s^4+4l^2m^4=rn^2,$$
for $r\in \{2, l, 2l\}$. First consider $r=2$. Then we get as above,
$$s^4+l^2m^4=2n^2.$$
Also as above, we get that when the solution is chosen to have minimal length, $l\nmid n$.
Then reducing mod $l$, $2$ would have  a square root in $\F_l$, which cannot happen since 
$l\equiv 3 \mod 8$. Therefore, $2\notin Im(q)$.

For $r=l$, the equation eventually reduces to
$$s^4+4m^4=ln^2.$$
Again, take a solution of minimal length. If $2|n$, then $2|s$, and we get the equation
$$m^4+4(s/2)^4=l(n/2)^2,$$
contradicting minimality. Therefore, $2\nmid n$. But then reducing mod 4 would give us a square root of 3 mod 4, a contradiction. Therefore, $l\notin Im(q)$.

Now suppose $2l\in Im(q)$. This gives a solution to the equation
$$4l^2s^4+4l^2m^4=2ln^2,$$
which then reduces to
$$s^4+m^4=2ln^2,$$
to which we can assume we have a minimal solution. Since, furthermore, $gcd(s,m)=1$, they cannot both be divisible by $l$. Hence, neither is divisible by $l$ (by the equation). Thus, we can reduce mod $l$ and get a non-trivial solution in $\F_l$ to the equation
$$s^4+m^4=0.$$
But then,  -1 would have a fourth root in $\F_l$ and, a fortiori, a square root. This is impossible for $l\equiv 3\mod 4.$ Therefore, $Im(q)=1$ and $\cD(\Q)/\phi(\cE_l(\Q))$ is trivial. We conclude that
$$\cE_l(\Q)/2\cE_l(\Q)$$
has rank 2, and hence, that the Mordell-Weil rank of $\cE_l$ is zero.


\hfill {\bf [5~marks]}
\end{flushleft}
\end{document}