\documentclass[11pt]{article}
\usepackage{amssymb,amsmath}
\def\ms{\medskip}
\def\Q{\mathbb{Q}}
\def\cE{{\cal E}}

\def\Z{\mathbb{Z}}
\begin{document}
\begin{flushleft}

Question 1.
\ms

(a) 

To see this, note that if $|y|>1$, then from the equation
$$|y|^2=|x^3+Ax+B|>1,$$
we must have $|x|>1$. On the other hand, if $|x|>1$, then
$|x^3+Ax+B|=|x^3|=|y|^2$, so $|y|>1$. That is, we have
$|x|>1$ and $|y|>1$ if and only if $|x|>1$ or $|y|>1$.

Now if $(x,y)=(x:y:1)$ reduces to $(0:1:0)$ modulo $p$, then either $|x|>1$ or $|y|>1$, since otherwise, the reduction would be $(\tilde{x}:\tilde{y}:1)$.  
Now suppose that $|x|>1$ or $|y|>1$. Then from
$|y|^2=|x^3+Ax+B|=|x|^3$, we see that $|y|>1$, $|x|>1$, and $|y|>|x|$. So the reduction is
$(\widetilde{x/y}:1:\widetilde{1/y})=(0:1:0).$

[7 marks. Similar to bookwork.]
\ms

(b)

If $a\in \Q^*$ has a square root in $\Q_p^*$ for all $p$ then $a$ has a square root in $\Q$.

\ms

{\em Proof.} Write $a=\pm\prod_p p^{n_p}$ for distinct primes $p$ and integers $n_p$. For each $p$, since
$a=x^2$ in $\Q_p$, we have that $n_p$ is even. Thus, we can write
$a=\pm \prod_p p^{2m_p}$ for integers $m_p$. But then, since $\prod_pp^{-n_p}$
has a square root $\prod_pp^{-m_p}$ in $\Q$, $a\prod_pp^{-n_p}=\pm 1$ would have a square root in each $\Q_p$. But $-1$ does not have a square root in, say, $\Q_3$, since any square root would be in $\Z_3$, giving rise to a square root of  $-1$ in $\Z/3$, which doesn't exist.
Thus, we must have $a\prod_pp^{-n_p}= 1$, and hence,
$a=\prod_pp^{2m_p}$, which has the square root $\prod_pp^{m_p}$ in $\Q$.

[7 marks. Unseen, but similar seen.]
\ms

(c)

Hensel: Let $K$ be a field, complete w.r.t. a non-archimedean
valuation $|\cdot|$, with valuation ring $R$.  Let $f(x)\in R[x]$, and let
$a\in R$ satisfy 
\[|f(a)|<|f'(a)|^2.\]
Then there is a unique $a^*\in R$ with $f(a^*)=0$ and 
\[|a^{*}-a|\le |f(a)|/|f'(a)|.\]

[2 marks. Bookwork]
\ms

(d) Let $p\neq 2$. 
Suppose $a$ has a square root. Then clearly $n=2m$, so that $u=ap^{-2m}$ has a square root.
Conversely, suppose $a=p^nu$ with $n$ even  such that $u\in \Z_p^*$ has a square root modulo $p$. Then
clearly, $p^n$ has a square root, so it suffices to show that $u$ has a square root. For this, consider $f(x)=x^2-u$ and let $a\in \Z_p$ such that $f(a)\equiv 0 \mod p$. Then
$a\in \Z_p^*$, so  $f'(a)=2a\in \Z_p^*$. Hence,
$|f'(a)|^2=1>|f(a)|$. Hence, by Hensel's Lemma, there is a $b\in \Z_ps$ such that $b\equiv a\mod p$ and
$f(b)=0$, which then is a square root of $u$.  

When $p=2$, $a\in \Q_2^*$ has a square root if and only if $a=2^nu$ where
$n$ is even and where $u\in \Z_2^* $has a square root modulo 8. That the condition is necessary is clear exactly as in the $p\neq 2$ case. To show that it is sufficient, note again that it suffices to show that $u\in \Z_2^*$ has a square root. Again, put $f(x)=x^2-u$ and let $a$ be a square root of $u$ modulo 8. Then $a\in \Z_s^*$ and
$f'(a)=2a$. So $|f'(a)|^2=2^{-2}> |f(a)|=2^{-3}.$ Therefore, Hensel's Lemma applies again to give us a $b\in \Z_2$ such that $f(b)=0$.

[4 marks. Seen in example in lecture.]
\ms

(e)
First, since 11 has a square root in $\mathbb{R}$, it is clear that the polynomial has a  root in $\mathbb{R}$.  In $\Q_2$, we have $-7\equiv 1 \mod 8$, so that $-7$ has a square root in
$\Z/8$, and hence, in $\Z_2$ by Hensel's lemma. In $\Q_7$, $11\equiv 4=2^2\mod 7$, so 11 has a square root in $\Z_7$ by Hensel's lemma. Also, $-7\equiv 4\equiv 2^2 \mod 11$, so that
$-7$ has a square root in $\Z_{11}$.
Now, let $p$ be any prime other than 2, 7 and 11. Then $-7$ and 11 are units in $\Z_p$, and
$$(-77/p)=(-7/p)(11/p),$$
so that at least one of $-77$, $-7$ and 11 must have a square root mod $p$. Hence, by Hensel's lemma, one of them must have a square root in $\Z_p$.

[5 marks. Unseen, but similar seen.]
\bigskip

\bigskip

Question 2.
\ms

(a) A formal group over $R$ is a power series $F(X,Y)\in R[[X,Y]]$ such that

$$F(X,Y)=X+Y+\mbox{terms of degree $\geq 2$};$$
$$F(F(X,Y),Z)=F(X,F(Y,Z));$$
$$F(X,Y)=F(Y,X).$$
Given an elliptic curve $y^2=x^3+Ax+B$, there is a formal group
$F_{\cE}(X,Y)\in \Z[A,B][[X,Y]]$ such that
$p\Z_p$ with the group law induced by $F_{\cE}$ is isomorphic to $\cE_1(\Q_p)$ with the elliptic curve group law.

[2 marks.]
\ms

(b) Let $\omega=F_X(0,T)^{-1}dT$, where the subcript refers to the derivative with respect to the first variable. Then
$$\omega\circ F(T,S)=F_X(0, F(T,S))^{-1}F_X(T,S)dT.$$
So we must show
$$F_X(0,F(T,S))^{-1}F_X(T,S)=F_X(0,T)^{-1},$$
or
$$F_X(0,T)F_X(T,S)=F_X(0,F(T,S)).$$
But we have
$$F(U, F(T,S))=F(F(U,T),S).$$
Differentiation with respect to $U$ yields
$$F_X(U, F(T,S))=F_X(F(U,T),S)F_X(U,T).$$
Setting $U=0$, we get
$$F_X(0, F(T,S))=F_X(T,S)F_X(0,T),$$
which is what we wanted. Hence, $\omega$ is an invariant differential form. Since $F(X,Y)\equiv X+Y$ up to terms of higher order, it is clear $\omega $ is normalized, i.e., of the form
$$\omega=a(T)dT,$$
where $a(T)=1+Tb(T)$ and $b(T)\in \Z_p[T]$.
\ms

Let $\alpha =P(T)dT$ be an invariant differential form. Then
$$P(F(T,S))F_X(T,S)dT=P(T)dT.$$
So
$$P((F(T,0))F_X(T,0)=P(0),$$
or
$$P(T)=P(0)F_X(T,0)^{-1}.$$
That is $\alpha=P(0)\omega.$ Hence, if $\alpha $ is normalized, then $\alpha =\omega$.

[3 marks]

\ms
(c) We put $[0](T)=0$, $[1](T)=T$ and $[m](T)=F([m-1](T),T)$ for $m>1$. Also,
$[m](T)=F([m+1](T), i(T))$ for $m<1$, where $i(T)\in T\Z_p[[T]]$ is the unique power series such that $F(T,i(T))=0$. 

We see by induction that $[m](T)=mT+h(T)$ where $T^2|h(T)$.
Now, from
$$F([m](X), [m](Y))=[m](F(X,Y)),$$
we see that
$$F_X([m](X), [m](Y))[m]'(X)=[m]'(F(X,Y))F_X(X,Y).$$
Setting $X=0$, we get
$$F_X(0, [m](Y))[m]'(0)=[m]'(Y)F_X(0,Y),$$
so that
$$[m]'(Y)F_X(0,[m](Y))^{-1}=[m]'(0)F_X(0,Y)^{-1}.$$
That is,
$$\omega\circ [m]=[m]'(0)\omega.$$
Applying this to
$[p]$, we get
$$\omega\circ [p]=p\omega.$$
If we put
$\omega=P(T)dT$, where $P(T)=1+Th(T)$, this implies $p|[p]'(T)$. Writing
$$[p](T)=pT+\sum_{n=2}^{\infty} a_nT^n,$$
we get $p|na_n$ for all $n$. Thus, we can group the terms for which $p|n$ into $g(T^p)$ to get the form we want.

[5 marks]

\ms

(d) We have the isomorphism
$$(p\Z_p, F_{\cE}(X,Y))\simeq (\cE, +_{\cE}).$$
So the torsion subgroups will be isomorphic. If $m$ is not divisible by $p$, then
$$[m](T)=mT+T^2g(T),$$
is an isomorphism (since $m$ is a unit in $\Z_p$), so no non-trivial element is killed by $[m]$. If $p|m$, and $p^a$ is the largest power of $p$ that divides $m$, then $m=p^am_0$ for $(m_0,p)=1$. Therefore,
for any point $x$ of order $m$, $m_0x$ has order $p^a$. So by induction on $a$, it suffices to show there is no $p$-torsion. If $p=2$, then the 2 torsion points  in $\cE(\Q_p)$ satisfy
$y=0$, so they cannot reduce mod $p$ to the origin. Hence, we assume $p>2$. Let $z\in p\Z_p$ correspond to a $p$-torsion point in $\cE_1(\Q_p)$.
If $[p](z)=0$, then we have
$$pz=z^pr(z)$$for some power series $r\in \Z_p[[T]]$. Thus,
$$p^{-1}|z|=|z|^p|r(z)|\leq |z|^p.$$
If $|z|\neq 0$, then we get $1/p^{1/(p-1)}\leq |z|\leq 1/p$, which is a contradiction for $p>2$. So we must have $z=0$.

[7 marks. Bookwork to here.]
\ms

(e) Note that $(1,2)$ is a rational point. We compute $2(1,2)=(a,b)$.
The slope of the tangent line is $y'=3x^2/2y=3/4$ at$ (1,2)$.
Thus, the tangent line is
$y=3x/4+5/4.$ To find the point of intersection, we substitute
$$x^3-(3x/4+5/4)^2+3.$$
The $x^2$ term has coefficient $-9/16$. Thus $1+1+a=9/16$. Therefore, $a=-23/16$ has non-integral coordinates. But then, $(a,b)\in \cE_1(\Q_2)$ and  $(a,b)$ is not torsion.
Hence, $\cE(\Q)$ is infinite.

[8 marks. Unseen.]
\ms

\ms




Question 3.

\bigskip

(a)

As coset representatives for $\mathbb{Q}^{\times}/(\mathbb{Q}^{\times})^2$
we may take non-zero square-free values $r\in\mathbb{Z}$. Such an $r$
arises from $q(x,y)$ if and only if $x=ru^2$ and $x^2+ax+b=rv^2$ for
some rational $u,v$.  If $u=l/m$ in lowest terms, then
$r^2l^4+ral^2m^2+bm^4=rn^2$ with integers $l,m$ for which $(l,m)=1$
and $m\neq 0$.  Since $rn^2$ is an integer and $r$ is square-free it
follows that $n$ is also an integer.  We claim that $r|b$.  Otherwise 
there would be a prime $p|r$ with $p\nmid b$. Then
$p|r(n^2-rl^4-al^2m^2)=bm^4$, whence $p|m$.  This entails
$p^2|r^2l^4+ral^2m^2+bm^4$, whence $p|rn^2$, from which we deduce that
$p|n$.  Finally we would have $p^3|rn^2-ral^2m^2-bm^4=r^2l^4$, which
imples that $p|l$.  This contradicts our assumption that $(l,m)=1$,
and therefore shows that no such $p$ can exist.  It therefore follows
that $r|b$, so that $q$ must have finite image.



[7 marks]

\bigskip

(b)

The maps $\phi:\mathcal{C}\rightarrow\mathcal{D}$ and
$\hat{\phi}:\mathcal{D}\rightarrow\mathcal{C}$ are homomomorphisms with
$\phi\hat{\phi}$ and $\hat{\phi}\phi$ being $[2]$ on the respective
groups $\mathcal{D}$ and $\mathcal{C}$.  We are told that ${\rm ker}(q)=\hat{\phi}(\mathcal{D})$, and we have a similar map
$\hat{q}$ on $\mathcal{D}$ with ${\rm ker}(\hat{q})=\phi(\mathcal{C})$. We then obtain induced injective
homomorphisms from $\mathcal{C}/\hat{\phi}(\mathcal{D})$ and
$\mathcal{D}/\phi(\mathcal{C})$ into
$\mathbb{Q}^{\times}/(\mathbb{Q}^{\times})^2$, each with finite image,
as above.  Thus $\mathcal{C}/\hat{\phi}(\mathcal{D})$ and
$\mathcal{D}/\phi(\mathcal{C})$ are finite.



Let $\{g_1,\ldots,g_k\}$ be coset representatives for 
$\mathcal{C}/\hat{\phi}(\mathcal{D})$, and similarly $\{h_1,\ldots,h_l\}$ for 
$\mathcal{D}/\phi(\mathcal{C})$.  Given $g\in \mathcal{C}$ there is a $g_i$ and an $h\in\mathcal{D}$
such that $g=g_i+\hat{\phi}(h)$.  Similarly there is an $h_j$ and a $g'\in \mathcal{C}$ so that $h=h_j+\phi(g')$.  Since $\hat{\phi}$

is a homomorphism, and $\hat{\phi}\phi=[2]$ we have
\[g=g_i+\hat{\phi}(h)=g_i+\hat{\phi}(h_j+\phi(g'))
=g_i+\hat{\phi}(h_j)+\hat{\phi}\phi(g')=g_i+\hat{\phi}(h_j)+2g'.\]
Thus the cosets $g_i+\hat{\phi}(h_j)+2\mathcal{C}(\mathbb{Q})$ cover
$\mathcal{C}$, whence $\mathcal{C}/2\mathcal{C}$ is finite.



[7 marks --- Bookwork to here]

\bigskip


(c)

For $y^2=x(x^2-p^2)$ we have $s=2$.  The image of $q$ corresponds to
solvable equations $r^2l^4-p^2m^4=rn^2$, with $r|p$; and since $\mathcal{E}'$ is
given by $y^2=x(x^2+4p^2)$ the image of $\hat{q}$ corresponds to
solvable equations $r^2l^4+4p^2m^4=rn^2$, with $r|2p$.  For
$r^2l^4-p^2m^4=rn^2$ there are solutions $(1,0,1),(0,1,p), (1,1,0)$
and $(1,1,0)$ for $r=1,-1,p$ and $-p$ respectively.  For $r^2l^4+4p^2m^4=rn^2$
we must clearly have $r>0$.  Thus there are at most 4 possibilities
for $r$, namely $1,2,p$ and $2p$.  Hence, from the above,
\[2^{R+2}=\#\mathcal{C}/2\mathcal{C}\le
\big(\#\mathcal{C}/\hat{\phi}(\mathcal{D})\big)
\big(\#\mathcal{D}/\phi(\mathcal{C})\big)\le 4\times 4,\]
so that the rank is always at most 2.



[5 marks]
\bigskip

(d)


If $x^4+p^2y^4=2z^2$ has a non-trivial integral solution we may choose
a minimal one.  If $p|z$ we would have $p|x$ and so
$y^4+p^2(x/p)^4=2(z/p)^2$, contradicting minimality.  Hence $p\nmid
z$.  However $2z^2\equiv x^4\pmod{p}$, and therefore $(\tfrac{2}{p})=+1$.



[2 marks]

\bigskip

(e)

Thus if $r^2l^4+4p^2m^4=rn^2$ and $r=2$ then $p\equiv\pm 1\pmod{8}$.
Thus for $p\equiv 3$ or $5\pmod{8}$ the case $r=2$ cannot occur, giving
$\#\mathcal{D}/\phi(\mathcal{C})\le 3$, whence $R<2$.  



[2 marks]

\bigskip



For $p=5$ the curve $C$ has a point $(x,y)=(25/4,75/8)$, which is
non-integral and therefore of infinite order.  Thus $R\ge 1$, and
hence $R=1$.



[2 marks --- Unseen, but similar to previous exam questions]


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