\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\g{{\cal G}}
\def\h{{\cal H}}

\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[11]
State and prove {\it Hensel's Lemma}.
\part[6]
Let $p$ be a prime and $a$ an integer. 
\begin{subparts}
\subpart
Show that if $p$ divides $a$ then $x^p - x + a$ splits into linear factors
over $\Z_p$, and otherwise $x^p - x + a$ has no roots in $\Z_p$.
\subpart
Compute an integer $x$ such that $x \equiv 1 \bmod{3}$ and $x^3 - x + 3 = 0 \bmod{3^{4}}$.
\subpart
Hence, or otherwise, find integers $x$ and $y$ such that $x^3 - x = 3^y - 3$ and $x > 3$.
\end{subparts}
\part[8]
Let $p$ be a prime. Show that $x^2 + 7 y^3 + 5 z^5 = 0$ has a solution $x,y,z \in \Z_p$ with
$(x,y,z) \ne (0,0,0)$. 

[{\it Hint: It may help to observe $\frac{1}{2} + \frac{1}{3} + \frac{1}{5} > 1$.}]
\end{parts}

\question
\begin{parts}
\part[6]
Compute the torsion subgroup of the elliptic curve $y^2 = x^3 + 5x^2 + 3x + 7$.
\part[6]
Let $n$ be an integer with $n > 2$. Show that the equation $y^2 = x^3 + x + (n^2 - 2)$ has
infinitely many rational solutions.
\part[13]
Let $R$ be a field of characteristic zero and let
\[ F(X,Y) := \frac{X + Y}{1 + XY} \in R[[X,Y]].\]
\begin{subparts}
\subpart
Prove that $F$ defines a formal group over $R$.
\subpart
Compute explicitly the formal logarithm $\log_F(T)$.
\subpart 
Hence find an explicit isomorphism $f(T) \in T R[[T]]$ from $F$ to the formal
multiplicative group
\[ \widehat{G}_m(X,Y) = X + Y + XY\]
with $f(T)$ a rational function in $T$.
\end{subparts}
\end{parts}

\question
\begin{parts}
\part[14] 
Let ${\mathcal C} : Y^2 = X(X^2 + aX + b)$, where $a,b\in \Z, b \not= 0,
a^2 - 4b \not= 0$, and let ${\mathcal D} : V^2 = U(U^2 + a_1 U + b_1)$,
where $a_1 = -2a$ and $b_1 = a^2 - 4b$.  Let the map $q$ be defined by
\[ q: {\mathcal D}(\Q) \rightarrow \Q^* / (\Q^*)^2, \,\,\, (u,v) \mapsto u \mbox{ when } u \ne 0,\]
\[ q: (0,0) \mapsto b_1,\,\,\, q: \underline{\bf o} \mapsto 1.\]
\begin{subparts}
\subpart
Show that $q(P + Q) = q(P) q(Q)$ in the case when none of $P,Q,P + Q$ are 
$(0,0)$ or $\underline{\bf o}$.
\subpart
Let $P = (u,v) \in {\mathcal D}(\Q)$ with $P \ne (0,0)$.
Explicitly compute the sum $P + (0,0)$,
%% Prove that 
%% \[ P + (0,0) = \left( \frac{b_1}{u}, - \frac{b_1 v}{u^2} \right).\]
and deduce that $q(P + (0,0)) = q(P)q((0,0))$. 
\subpart
By considering the remaining cases, prove that $q$ is a homomorphism.

\end{subparts}
\part[11]
Compute the rank of the elliptic curve $y^2 = x(x^2 - 3 x + 10)$.

%%  [{\it You may use any results from lectures provided they are clearly stated.}]
\end{parts}

\newpage

{\bf Solutions:}

1. (a) [Bookwork]
(Hensel's Lemma).
Let~$K$ be a field, complete with respect to a non-Archimedean
valuation $| \cdot |$, with valuation ring~$R = \{ x\in K : |x| \leqslant 1\}$.
\par
Let $f(x) \in R[x]$ and let $a_0\in R$ satisfy:
\ \ \ \ $| f(a_0) | < | f'(a_0) |^2. \ \ \ \ \ \ \ (*)$ 
\par\noindent Then there exists a unique $a\in R$ such that
$f(a) = 0$ and $| a - a_0 | \leqslant | f(a_0) |/ | f'(a_0) |$. [2 marks]\\

% N.B. I might verbally add: note that this is in contrast
% with Newton-Raphson in R, where we need to know something
% about f(x) for all x in an interval about the initial approximation.
Proof: Define $f_j(x)$ by:
\ \ \ \ $f(x + y) = f_0(x) + f_1(x) y + f_2(x) y^2 + \ldots,$
\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 $f_0(a_0) + f_1(a_0)b_0 = 0$]
% N.B. I might omit the following line:
\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
$\rho = \frac{ |f(a_0 )|}{|f'(a_0 )|^2} < 1$. [3 marks]

\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 As above: $|f'(a_{n+1})| = \ldots |f'(a_1)| = |f'(a_0)|$.
%\ \ \ \ \ \ (1)
\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 a standard theorem in
non-archimedean analysis)
\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. [4 marks]
\par\noindent For uniqueness, imagine $\hat a \not= a$ 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~(2) and continuity of $|f'(x)|$].  
\par\noindent This gives $|{\hat b}^j f_j(a)| \leqslant
|{\hat b}^j| \leqslant |{\hat b}^2| < |{\hat b} f_1(a)|$
% N.B. I might omit the next line:
(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. [2 marks]\\

(b) 
(i) 
[2 marks] Let $f(x) = x^p - x + a$. If $p|a$ then $f(x) = x^p - x = \prod_{\alpha = 0}^{p-1} (x - \alpha) \bmod{p}$. Also
for any $\alpha \in \Z$, $f^\prime(\alpha) = p \alpha^{p-1} - 1 + a \equiv  - 1 \not = 0 \bmod{p}$. So we may apply
Hensel's lemma to each $0 \leq \alpha \leq p-1$ to get a root $\hat{\alpha}$ and factorisation
$f(x) =  \prod_{\alpha = 0}^{p-1} (x - \hat{\alpha})$.

[1 mark] Suppose now $a \bmod{p} \ne 0$. Then for any $\alpha \in \Z$ we have $f(\alpha) = \alpha^p - \alpha + a \equiv a \ne 0 \bmod{p}$,
so there are no roots modulo $p$ and thus none in $\Z_p$.

(ii) [2 marks]
Let $a_0 = 1$ and $f(x) = x^3 - x + 3$, so $f^\prime(x) = 3 x^2 - 1$. Then iterating Newton we find
\[ a_1 = 1 -  \frac{3}{2} = - \frac{1}{2},\, a_2 = - \frac{1}{2} - \frac{-\frac{1}{8}+ \frac{1}{2} + 3}{\frac{3}{4} - 1} = 13.\]
We know as the lifting is in fact quadratically convergent this should be a solution. And indeed
$13^3 - 13 + 3 = 2187 = 3^7$.

(iii) [1 mark: {\it this also gives a hint for the calculation in (ii)}]  $x = 13$ and $y = 7$.

(c) [{\it This is new, but a similar idea is used in question 1 of the 2017 paper.}]

Let $p$ be a prime and consider the sets
\[ A_2:= \{ x^2 : x \in \F_p^\star\},\, A_3 :=\{ 7 y^3 : y \in \F_p^\star\},\, A_5 :=\{ 5 z^5: z \in \F_p^\star\}.\]
Assume now $p \ne 5, 7$ (to make sure none of these sets are just $\{0\}$).
Since always $x \mapsto x^m$ is at most $m$ to $1$ on $\F_p^\star$ these sets have size
at least $(p-1)/2$, $(p-1)/3$ and $(p-1)/5$ respectively. As
\[ \frac{1}{2} + \frac{1}{3} + \frac{1}{5}  > 1\]
they must overlap. [1 mark]

If $A_2 \cap A_3 \ne \emptyset$ we can find non-zero $x,y\in \F_p$ so that $x^2 = 7 y^3 \bmod{p}$ and then
$x^2 + 7 (-y)^3 + 5 \cdot 0^5 \equiv 0 \mod{p}$. We can now apply Hensel, using for $p \ne 2$ the polynomial
$f(X) = X^2 + 7 (-y)^3$ with starting value $X = x$, to get a solution $(\hat{x},-y,0)$. And using for
$p = 2$ the polynomial $f(Y) = 7 Y^3 + x^2$ with starting value $Y = -y$, to get a solution $(x,\hat{y},0)$. [2 marks]

If $A_2 \cap A_5 \ne 0$ we can find non-zero $x,z\in \F_p$ so that $x^2 = 5 z^5 \bmod{p}$ and then
$x^2 +  7 \cdot 0^3 + 5 (-z)^5 \equiv 0 \mod{p}$. Now apply Hensel, using $f(X) = X^2 + 5 (-z)^5$ for $p \ne 2$
and $f(Z) = 5 Z^5 + x^2$ for $p = 2$. [2 marks]

If $A_3 \cap A_5 \ne 0$ we can find non-zero $y,z\in \F_p$ so that $7 y^3 = 5 z^5 \bmod{p}$ and then
$0^2 + 7 (-y)^3 + 5 z^5 \equiv 0 \mod{p}$. Now apply Hensel, using $f(Y) = 7 Y^3 +  5 (-z)^5 $ for $p \ne 3$
and $f(Z) = 5 Z^5 + 2(-y)^3$ for $p= 3$. [2 marks]

Finally when $p = 5$ as $x \mapsto x^3$ is a bijection on $\F_p$ we have $A_2 \cap A_3 \ne \emptyset$, and
the argument before works. And for $p = 7$, since $x \mapsto x^5$ is a bijection on $\F_p$ we have $A_2 \cap A_5 \ne 0$,
and we find a solution as before. [1 mark]
 
\newpage

2 (a) [Similar: I think this is a little tricky though, as it is somewhat different from the other computations
of this sort they have done.]

Reducing modulo $3$ we get the equation $y^2 = x^3 + 2x^2 + 7$. Now the polynomial on the RHS has
no solutions in $\F_3$, and thus must be squarefree (any repeated root must be defined over the ground field).
So the reduction is smooth, and once checks the points over $\F_3$ are
\[ \{ \o, (0,1),(0,2),(1,1),(1,2) \}.\]
So by the results in lectures the torsion subgroup has order dividing $5$. [2 marks]

Now there is the obvious point $P = (1,4)$ on the curve. So all we need to do is check whether this order $5$ (rather than
infinite order). The tangent at a point $(x,y)$ has gradient
\[ \frac{3x^2 + 10 x + 3}{2 y}.\]
So the tangent at $P$ is $y = 2x + 2$. Putting into the curve we get
\[ 4 x^2 + \cdots = x^3 + 5 x^2 + \cdots \]
\[ x^3 + x^2 + \cdots = (x - x(2P))(x-1)(x-1).\]
So $x(2P) = -3$, and $2P = (-3,4)$.
Doubling again we see that the tangent at $2P$ has gradient
\[ \frac{3 \cdot 9 + 10 \cdot (-3) + 3}{8} = 0.\]
So this tangent is $y = 4$. Putting in we get
\[ (x^3 + 5 x^2 + 3x + 7) -  16 = (x + 3)(x + 3)(x - x(4P)).\]
So looking at constants we get $x(4P) = 1$, and so
$4P = (1,-4)$. Without further computation one now sees
$P + 4P = \o$. [4 marks]

(b) [Similar / New: they have seen that a non-rational solution implies positive rank.]
This equation defines an elliptic curve exactly when $\Delta_n := 4 + 27 (n^2 - 2)^2 \ne 0$. So 
let us first assume $\Delta_n \ne 0$. [1 mark]

The curve has the obvious point $P = (1,n)$. The strategy is to double it and observe $2P$ is not
integral, when $n \ne \pm 2$, so $2P$ has infinite order (by an important result from lectures) giving our infinite set of solutions. [1 mark]

The tangent at a point $(x,y)$ has gradient
\[ \frac{3 x^2 + 1}{2 y} \]
so at $P$ this is $\frac{4}{2n} = \frac{2}{n}$. Hence the tangent line is $y = \frac{2}{n} x + n - \frac{2}{n}$.
Thus
\[ \left( \frac{2}{n} x + \cdots\right)^2 = x^3 + 0 x^2 + \cdots\]
and
\[ x^2 - \frac{4}{n^2} x + \cdots = (x - 1)(x - 1)(x - x(2P)).\]
Thus we find
\[ x(2P) = \frac{4}{n^2} - 2. \]
This is not integral provided $n \ne  \pm 1, \pm 2$. [3 marks]

On the discriminant, if we suppose $\Delta_n= 0$ then reducing mod $5$ we get $-1 + 2(n^2 - 2)^2  \equiv 0 \bmod{5}$
and so $(n^2 - 2)^2 = \frac{1}{2} \equiv 3 \bmod{5}$, which is a contradiction. Or alternatively, note 
$\Delta_n$ is in fact always positive. [1 mark]

(c)

(i) 
[4 marks: similar]
We need to check the axioms. Commutativity is obvious, and also
\[ F(X,Y) = X + Y + \cdots \]
since $(1 + XY)^{-1} = 1 + (XY) + (XY)^2 + \cdots$.
For associativity we have
\[ F(X,F(Y,Z)) = \frac{X + \frac{Y + Z}{1 + YZ}}{1 + X\left(\frac{Y + Z}{1 + YZ}\right)}
\]
and 
\[ F(F(X,Y),Z) = \frac{\frac{X + Y}{1 + XY} + Z}{1 + \frac{X + Y}{1 + XY} Z} \]
and both are
\[
\frac{X + Y + Z + XYZ}{1 + YZ + XY + XZ}.
\]

(ii)
[4 marks: similar]

The canonical differential is $F_X(0,T)^{-1} dT$. One computes
\[ F_X(X,Y) = \frac{(1 + XY) - (X + Y)Y}{(1 + XY)^2} = \frac{1 - Y^2}{(1+XY)^2}.\]
Hence $F_X(0,T) = 1 - T^2$. So
\[ \omega = \frac{dT}{1 - T^2} \]
and the formal logarithm is
\[ \int \omega = \int \frac{dT}{(1 + T)(1 - T)} = \frac{1}{2} \log \left(\frac{1 + T}{1 - T}\right) \,\,\left( = \tanh^{-1}(T) \right).\]

(iii) [New]
Thus by the lectures 
\[ \log_F(T) =  \frac{1}{2} \log \left(\frac{1 + T}{1 - T}\right) \]
is an isomorphism from $F$ to the formal additive group $\widehat{G}_a$. [1 mark]

Now $\log(1 + T)$ is an isomorphism from $\widehat{G}_m$ to $\widehat{G}_a$, with
inverse $\exp(T) - 1$.  [1 mark]
Thus the composition
\[ \exp\left( \frac{1}{2} \log \left(\frac{1 + T}{1 - T}\right)  \right) - 1 \]
is an isomorphim. But this is not a rational function in $T$. [2 marks]

However, recall that multiplication by $2$ is an isomorphism from $\widehat{G}_a$ to
itself. This allows us to remove the factor $\frac{1}{2}$ and take 
\[ \exp\left(\log \left(\frac{1 + T}{1 - T}\right)  \right)  - 1 =\frac{2T}{1  - T}.\] 
[2 marks]
(And indeed one can check by a computation this is indeed a homomorphism, and the
series is certainly functionally invertible.)



\newpage



3 (a) (i) [Bookwork] [6 marks] Let~$(u_1,v_1),
(u_2,v_2),(u_3,v_3)$ be~3 points on~${\mathcal H} = {\mathcal D}(\Q)$ which sum to~$\o$,
[so that $(u_1,v_1) + 
(u_2,v_2)= (u_3,-v_3)$].
Then these are the~3 points of intersection between~${\mathcal D}$
and some line defined over~$\Q$: 
$V = \ell U + m$, say. Substituting~$V = \ell U + m$
into ${\mathcal D}$ gives: $U(U^2 + a_1 U + b_1) - (\ell U + m)^2$,
whose 3~roots must be~$u_1,u_2,u_3$.  
That is: $U(U^2 + a_1 U + b_1) - (\ell U + m)^2
= (U-u_1)(U-u_2)(U-u_3)$. Equating constant terms gives:
$u_1 u_2 u_3 = m^2 = 1$ in $ \Q^* / (\Q^*)^2$, and so $u_1 u_2 = 1/u_3 = u_3$
in $ \Q^* / (\Q^*)^2$. (Note $u_1 u_2 u_3 \ne 0$, by our assumption.)
Therefore, by the definition of~$q$ we have:
$q\bigl( (u_1,v_1) \bigr) q\bigl( (u_2,v_2) \bigr)
= q\bigl( (u_3,-v_3) \bigr)= q\bigl( (u_1,v_1) + (u_2,v_2) \bigr)$.\\

(ii) [The first part is in the lectures, but on the curve ${\mathcal C}$. The second part is new.] [5 marks]

When $u\not= 0$, we first find the line through $(0,0)$ and
$(u,v)$, which is: $V = \frac{v}{u} U$. Substituting this
into~${\mathcal D}$ gives:
\[
\Bigl( \frac{v}{u} \Bigr)^2 U^2 = U(U^2 + a_1 U + b_1)
\]
\[
v^2 U^2 = u^2 U^3 + a_1 u^2 U^2 + b_1 u^2 U\]
\[u(u^2 + a_1 u + b_1) U^2 = u^2 U^3 + a_1 u^2 U^2 + b_1 u^2 U
\mbox{ [since $(u,v)$ is on ${\mathcal D}$]}
\]
\[0 = u U^3 - (u^2 + b_1) U^2 + b_1 u U,
\mbox{ [since $u\not= 0$]}
\]
and so $U(U - u)(u U - b_1) = 0$. The roots of this cubic are:
$U = 0, U = u, U = b_1/u$. 
The line $V = \frac{v}{u} U$ and~${\mathcal D}$ intersect at:
$$ (0,0), (u,v)\hbox{ and } \Bigl( \frac{b_1}{u}, \frac{b_1 v}{u^2} \Bigr)
\hbox{ [since $U = \frac{b_1}{u}$ gives 
$V = \frac{v}{u}\frac{b_1}{u} = \frac{b_1 v}{u^2}$]}
$$
and so $(u,v) + (0,0) = \Bigl( \frac{b_1}{u}, -\frac{b_1 v}{u^2} \Bigr)$

Thus $q(P + (0,0)) = b_1/u \equiv b_1 \cdot u = q((0,0)) q(P)$.

(iii) [3 marks: new, and I think a little tricky.] If $P$ or $Q$ is $\underline{\bf o}$ the result is immediate. If $P + Q = \underline{\bf o}$ then
we have $Q = - P$ and then always $q(Q) = q(P)$ and so $q(P)q(Q) = q(P)^2 \equiv 1 = q( \underline{\bf o})$.
If $P = Q = (0,0)$ then $P + Q = \underline{\bf o}$ and $b_1 \cdot b_1 \equiv 1$. It remains then to consider the case $P + Q = (0,0)$.

Then $Q = -P + (0,0)$ so by (ii) we have $q(Q) = q(-P + (0,0)) = q(-P)q((0,0))$ which is
$q(P)q((0,0))$. Thus $q(P)q(Q) = q(P)^2 q((0,0)) \equiv q((0,0)) = q(P + Q)$.

(b) [Similar: this is a standard type of exam question.]
Using the notation from lectures we have $a = -3, b = 10, a_1 = 6, b_1 = -31$, and $C: y^2 = x(x^2 - 3x + 10)$, 
$D:v^2 = u(u^2 + 6 u - 31)$. [1 mark]

We first consider $\h / \phi(\g)$. Under $q$ this is isomorphic to a subgroup of the squarefree divisors
of $b_1 = -31$. So ${\rm im}(q) \leq \langle -1,-31 \rangle$. Taking $r = -1$ we easily spot the point $(-1,6) \in D(\Q)$
with $q((-1,6)) = -1$.  And $q((0,0)) = -31$.
%% For $r = 31$, this being quite a big number, we first examine the homogeneous space
%% \[ W_{31}: 31 \ell^4 + 6 \ell^2 m^2  - m^2 = n^2.\]
%% This has the obvious solution $(\ell,m,n) = (1,1,6)$. 
%% (This corresponds to the point $(31,6 \cdot  31)$: there
%% is a formula for this point in the notes given  $(\ell,m,n)$. Note finding this point by inspection would be much harder.)
So ${\rm im}(q) = \langle -1,31 \rangle \cong C_2^2$. [4 marks]

We next consider $\g / \hat{\phi}(\h)$. Under $\hat{q}$ this is isomorphic to a subgroup of the squarefree divisors
of $b = 10$. So ${\rm im}(\hat{q}) \leq \langle -1,2,5 \rangle$. Again we easily spot points $(2,4)$ and $(5,10)$ on $C(\Q)$
and so $2$ and $5$ are in the image. For $r = -1$ the homogeneous space is
\[ W_{-1} : - \ell^4 - 3 \ell^2 m^2 - 10 m^4 = n^2.\]
This has no non-trivial solutions in $\R$, since always LHS is $\leq 0$ and RHS is $\geq 0$, so none in $\Z$.
Hence ${\rm im}(\hat{q}) \leq \langle 2,5 \rangle> \cong C_2$. [5 marks]

So in conclusion using the formula from Sheet 4 Question 3, the rank is $2 + 2 - 2 = 2$. [1 mark]


\end{questions}

\end{document}

