\documentclass{oxmathexam}




\input{standardmacros.sty}

\newcommand{\F}{{\mathbb F}}
\renewcommand{\ss}{s} %% the prime p - 2 in solution

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% USER DEFINED MACROS - IF YOU MUST USE THEM, INSERT THEM HERE

\def\m{{m}}
\def\a{{\bf a}}
\def\b{{\bf b}}
\def\c{{\bf c}}
\def\d{{\bf d}}
\def\e{{\bf e}}
\def\f{{\bf f}}
\def\u{{\bf u}}
\def\v{{\bf v}}
\def\w{{\bf w}}
\def\k{{\bf k}}

\def\o{\underline{\bf o}}

\renewcommand{\C}{C}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



%% \school{Part C}
%% \title{Modular Forms}
%% \date{Version December 7, 2016. Checker: Victor Flynn}


\begin{document}

\begin{questions}


\question
\begin{parts}
\part[15]\hfill
\begin{subparts}
\subpart
Let ${\mathcal C}$ be a non-singular projective cubic curve defined over a field $K$, with a $K$-rational
point $\underline{\bf o}$. Describe the standard law for adding two points on ${\mathcal C}$, and prove that
it is associative. 

[{\it Standard results from geometry may be assumed provided they are clearly stated. For associativity, you may
assume points are in a suitably ``general position''. You do not need to describe the inverse of a point, verify other
group axioms, or check that ${\mathcal C}(K)$ is a subgroup.}]

\subpart
Now let ${\mathcal C}$ be the non-singular projective curve $X^3 + Y^3 + Z^3 = 0$ over the field $\F_5$ and
$\underline{\bf o} = (1,2,1)$. Compute the sum $\a+ \b$ where $\a = (2,1,1)$ and $\b = (3,3,1)$ are points in
 ${\mathcal C}(\F_5)$.
 
[{\it It may simplify your calculations to first list all the points in ${\mathcal C}(\F_5)$.}]

\end{subparts}

\part[10]
%% More complicated application. (See Buzzard's thing about the Selmer equation. Basically
%% e.g. one of $3,5,15,45$ must be a cube modulo a prime (not $3$ or $5$) and would be nice
%% to find a question using this idea.

Let $p \ne 3, 5$ be a prime and $f(x,y) = 3 x^3 + 4 y^3 - 5$.
\begin{subparts}
\subpart
Show that at least one element of the set $\{3,5,15,45\}$ must be
a cube modulo $p$.\newline [{\it It may help to consider the quotient group of $(\Z/p \Z)^*$ modulo cubes.}]
\subpart
If $3$ is a cube modulo $p$ then show that there exists $x \in \Q_p$ with
$f(x,1) = 0$.
\subpart
If $5$ is a cube modulo $p$ then show that there exists $x \in \Q_p$ with
$f(x,-x) = 0$.
\subpart
If $15$ is a cube modulo $p$ then show that there exists $x \in \Q_p$ with
$f(x,\frac{5}{7}) = 0$. %% (Note that $5 \cdot 7^3 - 4 \cdot 5^3 = 3^5 \cdot 5$.)
\subpart
If $45$ is a cube modulo $p$ then show that there exists $x \in \Q_p$ with
$f(x,0) = 0$.
\end{subparts}
[{\it Hensel's lemma may be used without proof provided it is clearly stated.}]
\end{parts}
\vspace{1cm}

{\bf Solution} 
1. 

(a) (i) [Bookwork: 10 marks] (See page 1-2 of the course notes.)

Let $\C$ be a nonsingular projective cubic curve, with a $K$-rational
point, which we shall denote~$\o$.
For any two points~$\a,\b$ on~$\C$, let~$\ell_{\a,\b}$
denote the line which meets~$\C$ at~$\a,\b$ [if~$\a,\b$ are distinct
then~$\ell_{\a,\b}$ is the unique line through~$\a,\b$;
if~$\a=\b$ then $\ell_{\a,\b}$ is the line tangent to~$\C$ at~$\a=\b$].
\bigskip

\hskip 120 pt \vrule width 1 pt depth 2 pt height 130 pt
\nopagebreak
\vskip-150pt
\hskip 37 pt {\lower 6.8pt \hbox{\a}} \hskip 28 pt 
{\lower 6.8pt \hbox{\b}}
\par\noindent $\ell_{\a,\b}$ \hskip 30 pt {\lower 3pt\hbox{$|$}} 
\hskip 30 pt {\lower 3pt\hbox{$|$}}
\hskip 40 pt $\d$
\hrule width 150 pt depth 1 pt height 1 pt
\vskip 30 pt \hskip 117 pt --- \c
\vskip 20 pt \hskip 117 pt --- \o
\vskip 20 pt \hskip 123 pt $\ell_{\o,\d}$
\nopagebreak
\vskip-125pt
\par\noindent
\hskip 150 pt Let $\ell_{\a,\b}$ denote the line which meets~$\C$ at~$\a,\b$.
\par\noindent
\hskip 150 pt Then~$\ell_{\a,\b}$ and~$\C$ have~$3$ points of intersection
(B\'ezout).
\par\noindent
\hskip 150 pt Let~$\d$ be the third point of intersection between~$\C$
and~$\ell_{\a,\b}$.
\par\noindent
\hskip 150 pt Now, let~$\ell_{\o,\d}$ denote the line which meets~$\C$ at~$\o$
and~$\d$.
\par\noindent
\hskip 150 pt Let~$\c$ be the third point of intersection between~$\C$
and~$\ell_{\o,\d}$.
\par\noindent
\hskip 150 pt Define~$\a + \b = \c$.
\vskip 45 pt


For associativity we need the
following technical lemma.

Lemma: Let~$P_1,\ldots , P_8$ be such that no~$4$ points lie on
a line and no~$7$ points lie on a conic. Then there exists
a unique point~$P_9$ which is
a 9th point of intersection of any two cubics passing
through~$P_1,\ldots ,P_8$.

%% \newpage
In order
to prove associativity, consider the following diagram.
\par
\medskip
\hskip 50 pt \vrule width 1 pt depth 2 pt height 240 pt \hskip 60 pt
\vrule width 1 pt depth 2 pt height 240 pt \hskip 60 pt
\vrule width 1 pt depth 2 pt height 240 pt 
\nopagebreak
\vskip-230pt \hskip 127 pt {\lower 5.5pt \hbox{\bf w}} \vskip -5 pt
\hskip 53 pt {\bf a} \hskip 67 pt {\lower 3pt\hbox{$|$}} 
\hskip 36 pt {\bf v} \hskip 60 pt $r$ \vskip -2.5 pt
\hrule width 240 pt depth 1 pt height 1 pt \vskip 15 pt
\hskip 98 pt {\bf f} \hskip -2 pt --- \vskip 28 pt
\hskip 52.5 pt {\bf b} \hskip 50 pt {\bf c} \hskip 50 pt {\bf u}
\hskip 60 pt $s$ \vskip 2pt
\hrule width 240 pt depth 1 pt height 1 pt \vskip 65 pt
\hskip 52.5 pt {\bf d} \hskip 50 pt {\bf e} \hskip 51 pt \o
\hskip 62 pt $t$ \vskip 1pt
\hrule width 240 pt depth 1 pt height 1 pt \vskip 50 pt
\hskip 49 pt $\ell$ \hskip 51 pt $m$ \hskip 48 pt $n$
\medskip
\par
Here, $r,s,t,\ell,m,n$ are lines. On each line, the labelled points
are the points of intersection between~$\C$ and that line.
From the definition of addition:
\par
\centerline{$\a + \b = \e$,}
\par\noindent and so:
\par
\centerline{$(\a + \b) + \c =$ 3rd point of intersection on~$\ell_{\o,\f}$.}
Similarly:
\par
\centerline{$\b + \c = \v$,}
\par
\centerline{$\a + (\b + \c) =$ 3rd point of intersection on~$\ell_{\o,\w}$.}
\par\noindent To show $(\a + \b) + \c = \a + (\b + \c)$, it
is sufficient to show that~$\f = \w$. Let $F_1 = \ell m n$
and $F_2 = rst$, both of which are cubic curves.
\par $\C$ and~$F_1$ have~$8$ common points: $\a,\b,\c,\d,\e,\u,\v,\o$.
\par $\C$ and~$F_2$ also have these~$8$ 
common points: $\a,\b,\c,\d,\e,\u,\v,\o$.
\par\noindent From the Lemma, the 9th point of
intersection of~$\C$ and~$F_1$ must be the same as the 9th
point of intersection of~$\C$ and~$F_2$; that is, $\f = \w$,
as required. 
\vspace{1cm}

(ii) [New: 5 marks. {\it They have not actually applied the point addition formula above except for an elliptic
curve in standard form}.]

In affine coordinates the equations is $x^3 + y^3 + 1 = 0$, that is $y^3 = -x^3 - 1$. Cubing is a bijection on $\F_5$ and so
the affine points are easily found to be
\[ (1,2),(2,1),(3,3),(4,0),(0,4).\]
We have $\a = (2,1), \b = (3,3)$. The line
joining $\a$ and $\b$ is $y = 2x - 3$. By inspection the other point on this line is $\d = (4,0)$. The line
joining $\d = (4,0)$ and $\o = (1,2)$ is $y = x + 1$. By inspection no other affine point satisfies this equation.
Now the point at infinity is $(-1,1,0)$ and this does not lie on the projective line $Y = X + Z$. So our line
must be tangent to the curve at one of the two points. By calculus the gradient at a point $(x,y)$ is
$\frac{d y}{d x } = - \frac{x^2}{y^2}$ and indeed the gradient at $\o$ is $- \frac{1}{4} \equiv 1$. So
the third point of intersection is $\o$. So overall $\a + \b = \o$.


(b) [New, but solving different equations using Hensel's lemma is in the problem sheets]\\

(i) [4 marks] (This argument is due to Kevin Buzzard.) The quotient group of $(\Z / (p \Z))^\star$ modulo cubes either has order
$3$ or $1$, depending upon whether $p \equiv 1$ or $p \equiv 2$ mod $3$. In the latter case
all non-zero residues are cubes. In the former case, let $\alpha$ and $\beta$ respectively be the images of
$3$ and $5$ in this quotient. So, written additively, $\alpha,\beta \in \{0,1,2\}$. If either of them are
zero then one of $3$ or $5$ is a cube. Otherwise we $(\alpha,\beta) = (1,2),(2,1),(1,1),(2,2)$. In the first
two cases $\alpha + \beta = 0$ and so $3 \cdot 5$ is a cube. In the second two cases $2\alpha + \beta = 0$
and so $3^2 \cdot 5$ is a cube.\\

(ii) [1 mark] In this case the equation becomes $3x^3 = 1 \bmod{p}$ and since $3$ is a cube so is $1/3$. Now apply Hensel's
lemma to get a solution in $\Q_p$.\\

(iii) [1 mark] Here the equation is $(-x)^3 = 5 \bmod{p}$ and since $5$ is a cube we can solve this as before.\\

(iv) [3 marks] Here the equation becomes $3(7x)^3 + 4 \cdot 5^3 = 5 \cdot 7^3$ and so $3 (7x)^3 = 5\cdot 7^3 - 4 \cdot 5^3 = 3^5 \cdot
5$. (Note $7^3 - 4 \cdot 5^2 = 343 - 100 = 243 = 3^5$.)
Multiplying by $3^2$ we get $(21 x)^3 = 3^6 \cdot 15 \bmod{p}$, and since $15$ is a cube we can solve
this as before.

(v) [1 mark] Here we can write the equation as $3^2 \cdot 3x^3 = 3^2 \cdot 5 \bmod{p}$, that is
$(3x)^3 = 45 \bmod{p}$, which we can solve.


\vspace{1cm}


\newpage


\question
\begin{parts}
\part[10]
Let $R$ be a commutative ring with an identity. Let $F$ and $G$ be formal groups defined over $R$ and
let $f(T) \in T R[[T]]$ be a homomorphism from $F$ to $G$.
\begin{subparts}
\subpart
Assume $g(T) \in T R[[T]]$ is such that
$f(g(T)) = g(f(T)) = T$. Prove that $g(T)$ is a homomorphism from $G$ to $F$.
\subpart
Let $\omega_F$ and $\omega_G$ be the normalised invariant differentials on $F$ and $G$, respectively.
Show that $\omega_G \circ f = f^\prime(0) \omega_F$.

[{\it You may assume that every invariant differential on $F$ is of the form $a \omega_F$ for some $a \in R$}.]
\subpart
Deduce that for any prime $p$ there exists  $a,b \in R[[T]]$ such that $[p](T) = p a(T) + b(T^p)$.
(Here $[p](T)$ denotes the multiplication by $p$ map in the formal group $F$.)

\end{subparts}

\part[9]
Now let $R$ be a commutative ring with an identity and consider the formal multiplicative group
\[ \hat{G}_m(X,Y) = X + Y + XY.\]
\begin{subparts}
\subpart
Let $p$ be prime and $[p](T)$ be the multiplication by $p$ map on the formal group $\hat{G}_m(X,Y)$.
Find explicitly $a(T),b(T) \in R[[T]]$ such that $[p](T) = p a(T) + b(T^p)$. 
\subpart
With $[p](T)$ as in part (b)(i), assume now that $R$ is a field of characteristic zero. Write down explicitly a series $g(T) \in T R[[T]]$ with
$[p](g(T)) = g([p](T)) = T$, justifying your answer.
\subpart
Let $n \geq 1$ be an integer. Using your construction in part (b)(ii), show that the rational number
\[ \frac{\frac{1}{p} \left(\frac{1}{p} - 1\right) \left(\frac{1}{p} - 2 \right) \cdots \left(\frac{1}{p} - n + 1 \right)}{n!} \]
when written in reduced form has denominator a power of $p$.

\end{subparts}
\part[6]
Let $P$ be the point $(1,-1)$ on the elliptic curve $y^2 = x^3 + 5 x - 5$. Using the Elliptic Curve Method, factorise $2491$ by attempting to calculate $4P$.

\end{parts}

\vspace{1cm}
{\bf Solution:}
(a)
(i) [3 marks. {\it This argument is mentioned in lectures but is not in the actual lecture notes.}]

We have $f(F(X,Y)) = G(f(X),f(Y))$, since $f$ is a homomorphism. Evaluating $g(T)$ on both sides and using that
$g(f(T)) = T$ we get that $F(X,Y) = g(G(f(X),f(Y))$. Now put $X = g(X_1)$ and $Y = g(Y_1)$ and use that
$f(g(T)) = T$ to get $F(g(X_1),g(Y_1) = g(G(X_1,Y_1))$.

(ii) [4 marks. Bookwork.]

First, note that $\omega_G \circ f\bigl( F(T,S) \bigr)
= \omega_G \bigl( G( f(T), f(S) ) \bigr) = \omega_G \circ f (T)$,
% N.B. I might mention verbally that this last step is
% due to to the invariant differential property of \omega_G.
so that $\omega_G \circ f$ is an invariant differential
on~$F$. From the italicised hint, it follows
that $\omega_G \circ f = a\ \omega_F$, for some~$a\in R$.
Since~$\omega_F,\omega_G$ are normalised,
$(1 + \cdots ) {\rm d}  f(T)  = a (1 + \cdots ) {\rm d} T$,
and so $(1 + \cdots ) f'(T) {\rm d} T  = a (1 + \cdots ) {\rm d} T$; 
% N.B. the fact that \omega_G \circ f [of T] becomes (1 + \ldots ) \dd f(T)
% (above) is due both to the fact that \omega_G is normalised
% and that f has no constant term.
equating constant terms gives $a = f'(0)$, as required.


(iii) [3 marks. Bookwork.]

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
$$ p \omega(T) = \omega \circ [p](T)
= (1 + \ldots ) {\rm d} ( [p](T) ) = (1 + \ldots ) [p]'(T) {\rm d} T.$$
Hence $[p]'(T) \in p\, R_T$. 
% N.B. This last step uses the fact that 1 + \ldots is invertible in R_T.
Each term $a_n T^n$ in $[p](T)$
must then satisfy 
$p | n a_n$ in $R$, and so $p | n$ in~$\Z$ or $p | a_n$ in~$R$,
as required. 


(b)
(i) [3 marks. {\it New: this is an illustration of Corollary 4.13 (part (a)(ii)) in the notes but looking at the proof will not help much.}]

The group operation is just $(1 + X)(1 + Y) - 1$ (key observation) and so $[p](T) = (1 + T)^p - 1$.
Now we know $(1 + T)^p - 1 \equiv 1 + T^p - 1 \equiv T^p \bmod{p}$. So we can take
$b(T) = T$ (so $b(T^p) = T^p$) and 
\[ a(T) = \frac{(1 + T)^p - (1 + T^p)}{p}.\]

(ii) [3 marks: {\it New.}]

Intuitively one sees that $g(T) = (1 + T)^{1/p} - 1$ should work. For indeed it can be expanded as a series
\[ \sum_{n = 1}^\infty \binom{1/p}{n} T^n \in T R[[T]].\]
Note all the binomial coefficients lie in $R$ since it must contain a copy of $\Q$.
%% ({\it That $p \in R^*$ should be enough, but I didn't want to get into proving the binomial coefficients
%% then lie in $R^*$.})
Moreover
\[ [p](g(T)) = (1 + (1 +T)^{1/p} - 1)^p - 1 = T,\, g([p](T)) = (1 + (1 + T)^p - 1)^{1/p} - 1 = T.\]

(iii) [3 marks: {\it New.}]\\
 The rational number is our binomial coefficient from (ii). Now if in (ii) we take instead $R = \Z[1/p]$ so that
$p \in R^*$ then Lemma 4.7 in the lecture notes tells us a unique $g(T) \in T R[[T]]$ exists with $[p](g(T)) = T$. Now such
$g(T)$ must also be the unique functional inverse taking $R$ to be the field of fractions $\Q$ of $\Z[1/p]$. Thus
our $g(T)$ must be the power series written above, and so all the coefficients of this series lie in $\Z[1/p]$.

(c) [6 marks: {\it Similar}.]

The gradient of the tangent at a point $(x,y)$ on the curve is $\frac{d y}{dx} = \frac{3 x^2 + 5}{2 y}$. So at $P = (1,-1)$
we get gradient $-4$. Hence the tangent line is $y = -4 x + 3$. Putting into the equation of the curve gives
\[ (-4 x + 3)^2 = x^3 + 5 x - 5 \]
and so
\[ x^3 - (-4 x)^2 + \cdots = 0.\]
Looking at the coefficient of $x^2$ we get $1 + 1 + x_3 = 16$ where $(x_3,y_3)$ is the third point of intersection of the tangent
line with the curve. So this point is $(14,-4 \cdot 14 + 3) = (14,-53)$. Hence $2P = (14,53)$. (Note this point has order $2$ modulo
$53$ which tells us already that $4 P$ is the identity modulo $53$ and the doubling will break down modulo any multiple of $53$.)

The gradient of the tangent at $2 P$ is
\[ \frac{ 3 \cdot 14^2 + 5}{2 \cdot 53}.\]
So we need to attempt to compute the inverse of $53$ modulo $2491$and some long division then reveals $2491 = 53 \cdot 47$.

\newpage 

\question
\begin{parts}
\part[9] Compute the Mordell-Weil group of the curve $y^2 = x (x^2 + x  - 1)$.
%% The different curve x (x^2 - x - 1) is done in 2017, asking for the rank only for 7 marks. My curve
%% has a 3-torsion point, so probably worth 10 marks. Idea is that for an open book exam it can be
%% used in place of the unused bookwork for 2020. Then for part (b) do something more difficult.

\part[16] Let $p$ be a prime number such that $s = p - 2$ is also prime. Consider the elliptic curve
\[ {\mathcal E}_p: y^2 = x (x - 2)(x - p).\]
\begin{subparts}
\subpart
Show that the rank of ${\mathcal E}_p$ is at most $2$.
\subpart
When $p \equiv 3 \bmod{8}$ show that the rank of ${\mathcal E}_p$ is at most $1$.
[{\it It may help to observe that the curve $v^2 = u(u^2 + 42 u + 289)$ 
has the point $(17/16,1207/64)$}.]
\subpart
What is the rank of ${\mathcal E}_{19}$? Briefly justify your answer.

\end{subparts}
\end{parts}

\vspace{1cm}

{\bf Solution:}

(a) [{\it Routine and similar to questions in lectures and notes.}]
Following notation in the lecture notes, we have $a = 1$, $b = -1$, $a_1 = -2a = -2$, $b_1 = a^2 - 4b = 5$.
Thus ${\mathcal D}: v^2 = u(u^2 - 2 u  + 5)$.

So for ${\mathcal H}/\phi({\mathcal G})$ we need to consider $r  \in \{\pm 1, \pm 5\}$, and
we have $q(0) = 1,\, q((0,0)) = b_1 = 5$. For $r = - 1$ we have $W_{-1}: - \ell^4 
- 2\ell^2 m^2  - 5 m^4 = n^2$. But
we see this has no non-trivial solutions since always $LHS \leq 0$ and $RHS \geq 0$.

Hence ${\mathcal H}/\phi({\mathcal G}) = \langle
(0,0) \rangle \cong C_2$.

For ${\mathcal G}/\hat{\phi}({\mathcal H})$ we need to consider $r \in \{\pm 1\}$, and
$\hat{q}(0) = 1,\, \hat{q}((0,0)) = b = -1$. 
So  ${\mathcal G}/\hat{\phi}({\mathcal H}) = \langle (0,0) \rangle$.

Overall ${\mathcal G}/2{\mathcal G}$  is generated by $(0,0)$ and $\hat{\phi}((0,0)) = \underline{o}$, 
so is $C_2$.
Since the two torsion of our curve is $C_2$ we get rank $1 - 1 = 0$.

For the torsion subgroup, the reduction is smooth mod $p$ provided $b_1 = 5 \ne 0 \mod{p}$. So we can look modulo
$3$ and find the points are $\{(0,1),(0,0), (1,1),(1,2),(2,1),(2,2)\}$. So the torsion has order $2$ or $6$. But aside from the
point $(0,0)$ on the curve there is the obvious point $(1,1)$. So the torsion subgroup must have order $6$. (Now we don't know whether $(1,1)$ has order $3$ or $6$ (it has order $3$) but this is not necessary to answer the question.)

Hence the Mordell-Weil group is isomorphic to $C_6$.\\

(b) [{\it Part (i) is fairly straightforward and similar to other questions. The method used for (ii) has been seen before in many other
exam questions, but the execution of this method is quite delicate. Part (iii) is an original argument not seen before. The notation
$\ss$ for $p-2$ is introduced in the question just to make marking it easier for me.}]\\

(i) [8 marks] We have $x(x - 2)(x - p) = x (x^2 - (2 + p) x + 2p)$ and so $a = -(p+2), b = 2p, a_1 = -2a = 2(p+2)$ and
$b_1 = a^2 - 4 b = (p+2)^2 - 8 p = (p-2)^2 = \ss^2$. Thus ${\mathcal D}: v^2 = u(u^2 + 2(p+2) u  + (p-2)^2)$.

So for ${\mathcal H}/\phi({\mathcal G})$ we need to consider $r|\ss^2$, that is $\{\pm 1, \pm \ss\}$, and
we have $q(0) = 1,\, q((0,0)) = b_1 = 1$. So we know that this quotient is $C_2^a$ where $0 \leq a \leq 2$.

For ${\mathcal G}/\hat{\phi}({\mathcal H})$ we need to consider $r|2p$ and so in $\{\pm 1, \pm 2, \pm p\}$, and
$\hat{q}(0) = 1,\, \hat{q}((0,0)) = b = 2p$. But notice that $a = - (p+2) < 0$ and $b = 2 p > 0$ so by the usual argument
we can rule out negative $r$ (since $LHS \leq 0$ and $RHS \geq 0$ etc). But also
we have that $q((2,0)) = 2$ and $q((p,0)) = p$ and so this quotient is $C_2^2$.

Overall ${\mathcal G}/2{\mathcal G}$  is generated by a $C_2^2$ and ``$\hat{\phi}(C_2^a)$'' the latter
having rank between $0$ and $2$.  (Note that though $\hat{\phi}((0,0)) = \underline{o}$ it was already trivial
in the first quotient.)
Thus we have at most a
$C_2^{a + 2 }$, that is at most a $C_2^4$. Now the two torsion of the curve is $C_2^2$ and so the rank is at most $2$.\\

(ii) [6 marks] To reduce the rank we need to consider ${\mathcal H}/\phi({\mathcal G})$.
We should try and prove that one of the following homogenous spaces has no suitable points:


%% p = 1 or 7 mod 8 get rank usually 0 but sometimes 2.
%% p = 3 or 5 mod 8 only saw rank 1. So here may be able to reduce rank.

%% When p = 5 get that -1 is in the image. When p = 13 get -1 in image
%% When p = 19 get that q is in the image.

$r = -1$: Here $\hat{W}_{-1}: - \ell^4 + 2 (\ss+4) \ell^2 m^2 - \ss^2 m^4 = n^2$.
%% Should be able to rule this out in the case p = 3 mod 8, i.e. q = 1 mod 8

$r = \ss$: Here $\hat{W}_{\ss}: \ss \ell^4 + 2 (\ss+4) \ell^2 m^2 + \ss m^4 = n^2$.
%% Should be able to rule this out when p = 5 mod 8, i.e. q = 2 mod 8.

But from the hint when $p = 19$ the equation $\hat{W}_{17}$ does have a solution.
So this indicates we should prove that $\hat{W}_{-1}$ has no suitable solution.

Note that when $\ell$ and $m$ are both even then $n$ is. So
at least one of $\ell$ and $m$ must be odd.

Assume $p \equiv 3 \bmod{8}$ and so $\ss \equiv 1 \bmod{8}$. Suppose first $\ell$ and $m$ are both odd.
Then $\ell^4,m^4 \equiv 1 \bmod{16}$ and $\ell^2 m^2 = 1,9 \bmod{16}$.
Also $\ss^2 \equiv 1 \bmod{16}$ and $2(\ss + 4) \equiv 10 \mbox{ or }26 \equiv 10 \bmod{16}$.
Note that always $2(\ss + 4) \ell^2 m^2 \equiv 10 \mbox{ or }90 \equiv 10 \bmod{16}$.
So we get
\[ \hat{W}_{-1}: - 1 + 10 - 1 = n^2 \bmod{16}\]
which is impossible as $8$ is not a square modulo $16$.

Next assume $\ell$ is odd and $m$ is even. Then $\ell^2 m^2 \equiv 0 \mbox{ or } 4 \bmod{16}$
and $m^4 \equiv 0 \bmod{16}$. Hence $2 (\ss+4) \ell^2 m^2 \equiv 0 \mbox{ or } 8 \bmod{16}$.
So we have
\[ \hat{W}_{-1}: - 1 + 0  -  0 = n^2 \bmod{16} \mbox{ or } - 1 + 8 - 0 \equiv n^2 \bmod{16} \]
which is impossible as the squares modulo $16$ are $0,1,4,9$.

Finally assume $\ell$ is even and $m$ is odd. Since $\ss^2 \equiv 1 \bmod{16}$ the equation $\hat{W}_{-1}$ modulo
$16$ is symmetrical in $\ell$ and $m$ and the preceding argument applies.

So we conclude that $\hat{W}_{-1}$ has no suitable solutions, and hence ${\mathcal H}/\phi({\mathcal G})$ has
$2$-rank either $0$ or $1$, which reduces our rank bound to $1$.
[Added: An alternative argument starts with $\ell^4 + 2 (2 + p) \ell^2 m^2 - (p - 2)^2 m^2 = n^2$ and reduces it modulo
$p$ to get $-(\ell -  2m^2) = n^2 \bmod{p}$. As $p \equiv 3 \bmod{8}$ we have $-1$ is not a square modulo $p$,
and hence find in particular that $\ell - 2 m^2 \equiv 0 \bmod{p}$. Then we see that unless $p|m$ we can write
$2 = (\ell / m)^2 \bmod{p}$ and this contradicts that $2$ is not a square modulo $p$ etc.]

(iii) [2 marks] The rank is $1$. We are told by the hint that $\hat{W}_{17}$ has a solution, which should prove the rank is $1$ except that
the theory in the lecture notes does not guarantee that $\hat{\phi}((17/16,1207/64))$ is independent of the other points
we already have in ${\mathcal G}/2{\mathcal G}$. However, if we let $Q = \hat{\phi}((17/16,1207/64))$ and suppose this
point has finite order $m$ then
\[ [m] [2] (17/16,1207/64) = [m] \phi (\hat{\phi}((17/16,1207/64))) = [m] \phi (Q) = \phi([m] Q) =\phi( \underline{o}) = \underline{o}\]
which shows $(17/16,1207/64)$ has order dividing $2m$. But this is impossible since it is not integral. [Added: In fact Sheet 4 Question 3 of the
revised sheets for 2021-22 guarantees independence.]

\end{questions}



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