\documentclass{oxmathexam}
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\school{Honour School of Mathematics Part C:  Paper C3.7\\
	Master of Science in Mathematical Sciences: Paper C3.7}
%Please put the name of the examination in the brackets above, e.g. Honour Moderations, Final Honour School of Mathematics Part B, etc.
\title{Elliptic Curves\\
	James Newton\\
	Draft solutions}
%Please put your course title, your name and the name of your checker in the brackets above.
\date{08/03/2024}
%Please be sure to mark each draft submitted with the correct date.

\begin{document}

\makecoverpage

\begin{questions}
	
\question 
\begin{parts}
	
	\part[6]
	\begin{subparts}\subpart State Hensel's lemma.
		\subpart Let $a \in \Q_7^\times$. Show that $a$ has a cube root in $\Q_7$ if and only if $a = 7^nu$ with $n \in \Z$ divisible by $3$, and $u \in \Z_7$ with $u\equiv \pm1 \mod{7}$.
		%	\subpart Consider the elliptic curve over $\Q_7$ with equation $\mathcal{E}: y^2 = x^3 + 7x - 1$.
		%	How many $2$-torsion points are there in $\mathcal{E}(\Q_7)$?
		
		
	\end{subparts}
	\part[19] Consider the elliptic curve with (projective) equation \[\mathcal{C}: X^3 + Y^3 = 2Z^3\] and distinguished identity point $(1:-1:0) \in \mathcal{C}(\Q)$. 
	\begin{subparts}
		\subpart For a point $P = (x_1:y_1:1)$ of $\mathcal{C}$, find a formula for the doubled point $2P$ in terms of rational functions in $x_1$ and $y_1$.
		
		[\textit{You may find it useful to note that the tangent line to $\mathcal{C}$ at $(1:-1:0)$ has equation $X+Y = 0$.}]  
		\subpart What are the $2$-torsion points of $\mathcal{C}(\Q_7)$?
		
		\subpart By considering your formula for $2P$, or otherwise, show that $\mathcal{C}(\Q_7)$ contains no point $P$ of order $4$. 
		
		[\textit{You may assume without proof that the points in $\mathcal{C}(\Q_7) \cap \{ Z = 0\}$ do not have order $4$.}]
	\end{subparts}
\end{parts}

\textbf{Solution 1.}
\begin{itemize}
	\item[(a)(i)] [Bookwork, 2 marks] Let~$K$ be a field, complete with respect to a non-Archimedean valuation~$|~|$, with valuation ring~$R = \{ x\in K : |x| \leqslant 1\}$.	
	
	Let $f(x) \in R[x]$ and let $a_0\in R$ satisfy
$| f(a_0) | < | f'(a_0) |^2$. Then there exists a unique $a\in R$ such that
	$f(a) = 0$ and $| a - a_0 | < |f'(a_0)|$. This solution moreover satisfies $|a - a_0|\leqslant | f(a_0) |/ | f'(a_0) |$. 
	
	[\textit{Also acceptable to drop the `moreover' part, or state that there exists a unique solution $a$ with $|a - a_0|\leqslant | f(a_0) |/ | f'(a_0) |$, or state the special case with $|f'(a_0)|=1$.}]
	
	\item[(a)(ii)] [Easy seen similar, 4 marks] Suppose we have $\alpha^3 = a$. First we consider the $7$-adic valuation of both sides. This tells us that $|a|_7 = 7^{-n}$ for $n \in \Z$ divisible by $3$. Setting $u = 7^{-n}a$, we compute $|u|_7 = 1$ and $a = 7^nu$ as required. Now it remains to show that $u \in \Z_7^\times$ is a cube if and only if $u\equiv \pm1 \mod{7}$. First The cubes in $\mathbb{F}_7^\times$ are $\pm 1$ so that tells us that $u \equiv \pm1 \mod{7}$ is necessary. Conversely, if $u\equiv \pm1 \mod{7}$ we can apply Hensel's lemma to the polynomial $x^3 - (\pm1)$, with approximate root $a_0 = \pm 1$. The derivative $f'(a_0) = 3a_0^2$ has $7$-adic valuation $1$ in both cases, and $|f(a_0)|_7 < 1$. So we deduce that $u$ does have a cube root in $\Q_7$.
	
%	\item[(a)(iii)] [Seen similar applications of Hensel's lemma. 6 marks] The $2$-torsion points correspond to points $(\alpha,0)$ with $\alpha$ a root of the cubic $x^3 + 7x-1$, together with the point at infinity. So we must determine how many roots the cubic has in $\Q_7$. There are $3$ roots in $\mathbb{F}_7$: the three cube roots of unity, $1, 2, 4 \mod{7}$. We can apply Hensel's lemma with $a_0 = 1, 2, 4$ respectively (in each case, $f'(a_0) = 3a_0^2 + 7$ has $7$-adic valuation $1$). That gives us three distinct roots in $\Z_7$. So we conclude that there are $4$ $2$-torsion points in $\mathcal{E}(\Q_7)$. 
	
	\item[(b)(i)] [Similar to bookwork, but all calculations in lectures/notes have been for cubics in Weierstrass form. 8 marks]  First we will need to compute the inverse map for this curve. Since $O = (1:-1:0)$ is an inflection point, $-P$ is the third point on the intersection of $\mathcal{C}$ with the line through $O$ and $P$. This line has equation $X + Y - (x_1+y_1)Z = 0$. By symmetry (or a computation of the intersection), the third intersection point is $(y_1:x_1:1)$, so we have $-(x_1:y_1:1) = (y_1:x_1:1)$. 
	
	Now we compute the doubling map. The point $-2P$ is the third point on the intersection with $\mathcal{C}$ of the tangent to $\mathcal{C}$ at $P$. This tangent line has (affine) equation $y = \frac{-x_1^2 x + 2}{y_1^2}$. Intersecting with the cubic gives an equation 
	\[x^3 + \left(\frac{-x_1^2 x + 2}{y_1^2}\right)^3 = 2\] which rearranges to a cubic in $x$ with first two terms \[\left(\frac{y_1^6 - x_1^6}{y_1^6}\right) x^3 + \frac{6x_1^4}{y_1^6}x^2 + \cdots \] We deduce that the point $-2P$ has $x$-co-ordinate $x_2$, where \[2x_1 + x_2 = \frac{-6x_1^4}{y_1^6-x_1^6} = \frac{-6x_1^4}{(2-x_1^3)^2-x_1^6} = \frac{-6x_1^4}{4-4x_1^3} \] hence $x_2 = \frac{2x_1^4-8x_1}{4-4x_1^3} = \frac{x_1^4-4x_1}{2-2x_1^3}.$ Substituting in to the equation of the tangent line, or using the fact that $-2(-P) = 2P$, tells us that $-2P = (x_2, y_2)$ with $y_2 = \frac{y_1^4-4y_1}{2-2y_1^3}$. Taking the inverse gives \[2P = \left(\frac{y_1^4-4y_1}{2-2y_1^3},\frac{x_1^4-4x_1}{2-2x_1^3}\right) \]
	

	\item[(b)(ii)] [A simple computation with the tangent line at a point. 5 marks.] We are looking for points $P = (x_1:y_1:z_1)$ whose tangent line passes through $O = (1:-1:0)$. The tangent line has projective equation 
	\[x_1^2 X + y_1^2 Y - 2 z_1^2 Z = 0\] and substituting in $O$ gives us $x_1^2 - y_1^2 = 0$. We deduce that $y_1 = \pm x_1$. Substituting back in to the equation for the curve gives $x_1^3 \pm x_1^3 = 2z_1^3$. So $y_1 = -x_1$ gives us $z_1 = 0$ and this corresponds to the point $O$. On the other hand $y_1 = x_1$ gives us $x_1^3 = z_1^3$ and this corresponds to points $(1:1:\zeta)$ with $\zeta^3 = 1$ (or alternatively $(\zeta:\zeta:1)$). That gives us the 4 $2$-torsion points of $\mathcal{C}(\Q_7)$, since Hensel's lemma tells us (as in part (a)) that $\Q_7$ contains $3$ cube roots of $1$. 
	
	[\textit{Students may begin with the formula in the previous part, which shows that if $P$ satisfies $x_1^3 = y_1^3 = 1$, then it gets mapped to a point at $\infty$. However there are three points in $\mathcal{C}(\Q_7)$ with $Z = 0$: $(1:\zeta:0)$ with $\zeta^3 = -1$. An extra argument with the doubling formula is needed to show that $x_1 = y_1$. 2 marks for noticing that $x_1^3 = y_1^3 = 1$ is required for a 2-torsion point.}]
	
	\item[(b)(iii)] [New (or a more difficult version of examples they have seen, if they use the reduction map). 6 marks.] We are looking for a point $P \in \mathcal{C}(\Q_7)$ with $2P = (\zeta:\zeta:1)$, $\zeta^3 = 1$. Looking at the formula for $2P$, we need \[\frac{x^4 - 4x}{2-2x^3} = \zeta.\] Taking $7$-adic valuations gives $|x|_7|x^3-4| = |2-2x^3|_7$. We see from this that $|x|_7 \ge 1$. Also, if we had $|x|_7 > 1$, the valuation of the left hand side would be $|x|_7^4$, whereas the valuation of the right hand side would be $|x|_7^3$. We deduce that $|x|_7 = 1$. We know that $x^3 = \pm1 \mod 7$, so we deduce also from comparing valuations that $x^3 = -1 \mod 7$. Now we go back to the equation \[\frac{x^4 - 4x}{2-2x^3} = \zeta.\] Reducing mod $7$ gives us $x = 2\zeta \mod 7$. Cubing both sides gives $x^3 = 1 \mod 7$, a contradiction.  
	  
	An alternative is to use the reduction mod $7$ map. They will need to compute that $\widetilde{\mathcal{C}}(\mathbb{F}_7) \cong C_2\times C_2\times C_3$ so it contains no element of order $4$ (showing that the cardinality is $12$ is not sufficient). 
\end{itemize}


\newpage

\question Let $p$ be a prime with $p > 5$, and consider the elliptic curve over $\Q$ with equation ${\mathcal E} : y^2 = x^3 + p$.

[{\it Throughout this question, any results on formal groups from lectures may be assumed provided they are clearly stated.}]
\begin{parts}
\part[7] Let $q$ be a prime with $q \nmid 6p$. Show that the order $\#{\mathcal E}(\Q)_{{\rm tors}}$ of the torsion subgroup of
${\mathcal E}$ divides the order $\#\widetilde{\mathcal E}_{q}(\mathbb{F}_q)$ of the group of
points on the reduction $\widetilde{\mathcal E}_{q}$ of the curve modulo $q$.


\part[8]

\begin{subparts}
\subpart Compute the order $\#\widetilde{\mathcal E}_{5}(\mathbb{F}_5)$ of the group of
points on the reduction $\widetilde{\mathcal E}_{5}$ of the curve modulo $5$.
\subpart Compute the number of points $\#\widetilde{\mathcal E}_{p}(\mathbb{F}_p)$ of the reduction $\widetilde{\mathcal E}_{p}$ of the curve modulo $p$.
\end{subparts}

\part[10] 
\begin{subparts}
	
\subpart Show that there is no point of ${\mathcal E}(\Q_p)$ reducing to the singular point \[(0,0) \in \widetilde{{\mathcal E}}_{p}(\mathbb{F}_p).\]
\subpart Deduce that $\#{\mathcal E}(\Q)_{{\rm tors}} = 1$.

\end{subparts}

\end{parts}

\noindent\textbf{Solution 2.}
\begin{itemize}
\item[(a)] [Bookwork, 7 marks] The discriminant of the cubic $x^3 + p$ is $27p^2$, so the primes $q \nmid 6p$ are the primes of good reduction. We recall from lectures that the reduction map defines a homomorphism 
\[\mathcal{E}(\Q_q) \to \widetilde{\mathcal{E}}_q(\mathbb{F}_q) \] with kernel the points $\mathcal{E}_1(\Q_q) = \{(x,y) : |x|_q > 1, |y|_q > 1\}$ (together with the point at infinity). We recall also that the group $\mathcal{E}_1(\Q_q)$ is isomorphic to the group $\mathcal{F}_{\mathcal{E}}(q\Z_q)$ associated to a formal group law over $\Z_q$. The crucial fact we use about formal groups is that if $z \in q\Z_q$ is a torsion point in $\mathcal{F}_{\mathcal{E}}(q\Z_q)$, then it has $q$-power order, and if its order is $q^n > 1$, then we have $|z|_q \ge q^{1/(q^n-q^{n-1})}$. Since this lower bound is $> q^{-1}$, the group $\mathcal{F}_{\mathcal{E}}(q\Z_q)$ is actually torsion-free. [It's also OK for them to state the result from lecturers that this group is torsion-free.]

Now we know that $\mathcal{E}_1(\Q_q)$ is torsion-free, so we deduce that the reduction homomorphism is injective on the torsion subgroup $\mathcal{E}(\Q_q)^{\mathrm{tors}}$. In other words, $\mathcal{E}(\Q_q)^{\mathrm{tors}}$ is isomorphic to a subgroup of $\widetilde{\mathcal{E}}_q(\mathbb{F}_q)$. $\mathcal{E}(\Q)_{\mathrm{tors}}$ is itself obviously a subgroup of $\mathcal{E}(\Q_q)^{\mathrm{tors}}$, so we get the desired conclusion that its order divides $\#\widetilde{\mathcal{E}}_q(\mathbb{F}_q)$.

\item[(b)(i)] [Easy seen similar, 3 marks] Modulo $5$, cubing is a bijection. So $x^3 + p$ takes on the $5$ different residue classes mod $5$ as $x$ varies. In particular it takes the square values $0, 1, -1$ once each. We deduce that there are $1 + 2\times 2 = 5$ solutions in $\mathbb{F}_5$ to $y^2 = x^3 + p$. Adding in the point at infinity gives $\#\widetilde{\mathcal E}_{5}(\mathbb{F}_5) = 6$.

\item[(b)(ii)] [Similar but less familiar, 5 marks.] Note that $p$ is a prime of bad reduction. We have to count solutions to $y^2 = x^3$ in $\mathbb{F}_p$. The point at infinity and $(0,0)$ contribute two points. Otherwise, writing $y = g^i$ and $x = g^j$ for a primitive root $g$, we need to solve $2i = 3j$ for $i,j \in \Z/(p-1)\Z$. The equation implies that $j$ is even, and for each even $j$ there are two possible values of $i$, given by $i = 3(j/2) \mod (p-1)/2$. That gives $p-1$ more solutions, for a total of $p+1$. 

[\textit{It was mentioned in lectures that the singular cubic $y^2=x^3$ with the singular point $(0,0)$ removed can be identified with the affine line, which explains the count of $p+1$.}]

\item[(c)(i)] [New but they have seen similar arguments with $p$-adic valuations, 5 marks.] The points reducing to $(0,0)$ are the $(x,y)$ with $|x|_p < 1$ and $|y|_p <1$. This should be explained somehow: for example, these points do reduce to $(0,0)$; if we have $|x|_p = 1$ or $|y|_p = 1$ then the other co-ordinate is necessarily a $p$-adic integer and the points then reduces to a non-singular point; if we have $|x|_p > 1$ or $|y|_p > 1$ then the point reduces to the point at infinity. 

It remains to check that there are no such points. We would have $y^2 = x^3 + p$ with $|x|_p < 1$. It follows that $|x^3|_p < p^{-1}$, so the valuation of the right hand side is $p^{-1}$. This is not the valuation of a square in $\Q_p$, so there are no such points.

\item[(c)(ii)] [Somewhat new, since they are asked to consider reduction at a prime of bad reduction, 5 marks.] 

There is a short exact sequence 
\[0 \to \mathcal{E}_1(\Q_p) \to \mathrm{red}^{-1}(\widetilde{\mathcal E}_{p}(\mathbb{F}_p)^{ns}) \to \widetilde{\mathcal E}_{p}(\mathbb{F}_p)^{ns} \to 0\] with $\widetilde{\mathcal E}_{p}(\mathbb{F}_p)^{ns} =  \widetilde{\mathcal E}_{p}(\mathbb{F}_p) - (0,0)$ is the group of nonsingular points. Part (c)(i) shows that $\mathcal{E}(\Q_p)=\mathrm{red}^{-1}(\widetilde{\mathcal E}_{p}(\mathbb{F}_p)^{ns}).$ Using the same result on formal groups as in part (a), we know that $\mathcal{E}_1(\Q_p)$ is torsionfree. So $\mathcal{E}(\Q)_{\mathrm{tors}}$ injects into $\widetilde{\mathcal E}_{p}(\mathbb{F}_p)^{ns}$, and therefore has order dividing $p$. We know from part (b)(i) that its order also divides 6. Since $p$ and $6$ are coprime, we deduce that $\mathcal{E}(\Q)_{\mathrm{tors}}$ is trivial.

[\textit{Note that they can't just quote part (a) here, since $p$ is a prime of bad reduction.}]
\end{itemize}

\newpage
\question  
\begin{parts}
	\part[10] Consider an elliptic curve over $\Q$ defined by the equation \[\mathcal{C}: y^2 = x(x^2 + B),\] with $B$ a non-zero integer, together with the 2-isogeny $\hat{\phi}: \mathcal{D}\to \mathcal{C}$ from the curve
	\[\mathcal{D}: v^2 = u(u^2 - 4B)\] given by \[\hat{\phi}(u,v) = \left(\frac{1}{4}\left(\frac{v}{u}\right)^2,\frac{1}{8}\left(v + \frac{4Bv}{u^2}\right)\right).\] 
	
	Show that if there is a point $(u,v) \in \mathcal{D}(\Q)$ with $u = rt^2$, for $r$ a squarefree integer and $t \in \Q^\times$, then $r$ divides $4B$ and there are integers $l, m, n$, not all $0$, with $\gcd(l,m) = 1$, satisfying the equation
	\[rl^4 - \left(\frac{4B}{r}\right)m^4 = n^2.\]
	\part[10] Suppose $p$ is a prime with $p \equiv 3 \mod 8$. Show that the elliptic curve over $\Q$ \[{\mathcal E} : y^2 = x(x^2-p^2)\] has rank zero.
	
	[\textit{You may use any results from lectures provided they are clearly stated. It may also help to	recall that for an odd prime $p$ we have that $2$ is a quadratic residue modulo $p$ if and only if $p \equiv \pm1 \mod{8}$ and $-1$ is a quadratic residue modulo $p$ if and only if $p \equiv 1 \mod{4}$.}]
	\part[5] Find a point of infinite order on the curve with equation $y^2 = x(x^2 - 7^2)$, justifying why it has infinite order.
\end{parts}
\end{questions}
\textbf{Solution 3.}
\begin{itemize}
	\item[(a)] [Bookwork, 10 marks]
	Since $u(u^2 -4B) = v^2$, $u$ and $u^2 -4B$ are the same modulo squares,
	which means we can write:
	$$ u^2 -4B = rs^2,\ \ \ u = rt^2,\ \ \ \hbox{ for some } s \in \Q.
	$$
	Hence: $ (rt^2)^2  -4B = rs^2$. Let $t = \ell/m$,
	where $\ell, m \in \Z$ and gcd$(\ell,m) = 1$.
	Then: $ r^2 \ell^4 / m^4 -4B = rs^2$,
	and so: $ r^2 \ell^4 -4Bm^4 = r(m^2s)^2$.
	Now the LHS of this last equation
	is in~$\Z$, and so the RHS is also in~$\Z$; that is: $r(m^2s)^2\in \Z$.
	Since $r$ is square free, we must therefore have $m^2 s \in \Z$.
	Define: $n = m^2 s \in \Z$. Then our equation becomes:
	$$
	r^2 \ell^4 - 4Bm^4 = r n^2,\ \ 
	\hbox{ for some } \ell, m, n \in \Z, \hbox{ gcd}(\ell,m) = 1,
	$$
which gives the desired equation. It remains to prove that $r$ divides $4B$. It is sufficient to show, for any prime~$p$, that $p | r \Rightarrow p | 4B$. Suppose $p | r$. Then $p | 4Bm^4$. If $p$ does not divide $4B$, then $p|m$ and we see that $p^2$ divides $r^2 \ell^4 - 4Bm^4$ and hence $p|n$. So $p^3|rs^2$. Finally, we deduce that $p|\ell$ which contradicts coprimality of $\ell$ and $m$. So $p$ did divide $4B$ after all and we are done. 

\item[(b)][Seen many similar examples of 2-descent, 10 marks. The same question appears in the 2013 exam paper.]

We rename $\mathcal{E}$ as $\mathcal{C}$ and use the setup of part (a). First we consider the group $\mathcal{D}(\Q)/{\phi}(\mathcal{C}(\Q))$. We recall from lectures that this admits an injective homomorphism \[q:\mathcal{D}(\Q)/{\phi}(\mathcal{C}(\Q)) \to \Q^\times/(\Q^\times)^2\] with image described in part (a). In particular, the image is a subgroup of $\{\pm1,\pm2,\pm p,\pm2p\}$, and we have to consider the solvability of equations $rl^4 + (4p^2/r)m^4 = n^2$. We see that for $r < 0$ there are no non-trivial solutions (since the left hand size is then $\le 0$). 

We need to analyse the remaining possibilities for $r$:

($r=2$) Consider $2l^4 + 2p^2 m^4 = n^2$. We deduce that $2l^4 \equiv n^2 \mod{p}$. Since $p \equiv 3 \mod{8}$ this implies that $p|l$ and $p|n$. We can then re-arrange the equation to $2p^2(l/p)^4 + 2 m^4 = (n/p)^2$. The same reasoning shows that $p|m$, contradicting coprimality of $l$ and $m$. So $2$ is not in the image of $q$. 

($r=p$) Consider $pl^4 + 4p m^4 = n^2$, equivalently $l^4 + 4 m^4 = p(n/p)^2$. This shows that $l^4 = -4m^4 \mod{p}$. Since $l$ and $m$ are coprime, we deduce that $-4$ (or, equivalently, $-1$) is a square mod $p$. But this contradicts $p \equiv 3 \mod{8}$. So $p$ is not in the image of $q$. 

($r=2p$) Consider $2pl^4 + 2p m^4 = n^2$, equivalently $l^4 + m^4 = 2p(n/2p)^2$. Again this contradicts the fact that $-1$ is not a square mod $p$. 

So we conclude that $\mathcal{D}(\Q)/{\phi}(\mathcal{C}(\Q))$ is trivial. 

On the other hand, the size of $\mathcal{C}(\Q)/\hat{\phi}(\mathcal{D}(\Q))$ is bounded by $4$, with image under the homomorphism $\hat{q}$ a subgroup of $\{\pm1,\pm p\} \subset \Q^\times/(\Q^\times)^2$. All $4$ elements appear as the images of the four $2$-torsion points $\infty,(0,0),(p,0),(-p,0)$. We conclude that $\mathcal{C}(\Q)/2\mathcal{C}(\Q)$ has size $4$, and is generated by the $2$-torsion points. It follows from the Mordell--Weil theorem that the rank of $\mathcal{C}(\Q)$ is zero. 

\item[(c)] [New, 5 marks] One way to find such a point is via $2$-descent. Following the working of part (b), we find that we can solve the equation $2l^4 + 2\cdot 7^2 m^4 = n^2$, with $l = m = 1$, $n = 10$. This gives us the point $(2,20)$ on the curve $v^2 = u(u^2 + 4\cdot 7^2)$, which is of infinite order since we can compute that the torsion points on this curve are just $\{O,(0,0)\}$. Indeed, reducing mod $3$ we see that the torsion points are isomorphic to a subgroup of $C_4$ (hence cyclic) whilst reducing mod  $5$ we see that the torsion points are isomorphic to a subgroup of $C_2 \times C_2$. 

We then get a point of infinite order $\hat{\phi}(2,20) = (25,-120)$ on our original curve. 

[\textit{If they somehow spot the rational point $(25,\pm120)$ I'll give 2 marks, but they need to prove that it has infinite order to get full marks. They are given the formula for $\hat{\phi}$ earlier to help with this part of the question.}] 
\end{itemize}

\end{document}




