\documentclass{oxmathexam}
\usepackage{standardmacros}





\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{10/12/2025}
%Please be sure to mark each draft submitted with the correct date.

\begin{document}

\makecoverpage

\input{C3.7_qns_2026_V2}

\begin{itemize}
\newpage
\item[Solution 1.]
(a)(i) [5 marks, Bookwork] As suggested in the hint, we use the normalized invariant differential $\omega_F = P(T)dT$ on $F$. Since $[p]$ is a homomorphism, $\omega_F\circ [p] = [p]'(T) P([p](T)) dT$ is a multiple of $\omega_F$. [\textit{2 marks}]

Comparing constant terms, since $[p](T) = pT \mod T^2$, we get $[p]'(T) P([p](T))  = p P(T)$. Since $P([p](T)) = 1 \mod T$ and is hence invertible in $\Z_p\lb T\rb$, we deduce that $[p]'(T) \in p\Z_p\lb T \rb$. In other words, every coefficient of $[p]'(T)$ is in $p\Z_p$, and we deduce that $[p](T)$ has the desired form. [\textit{3 marks}]

(a)(ii) [5 marks, Unseen but relatively easy. If they do (b)(ii) correctly they can apply it here for full credit.] We prove by induction on $n$ that $|[p^n](x)|_p \le p^{-(n+1)}$. The case $n=0$ is clear, because $|x|_p \le p^{-1}$. Suppose the inequality holds for $y = [p^n](x)$. Then $[p^{n+1}](x) = p f(y) + g(y^p)$, so $|[p^{n+1}](x)|_p \le \max(p^{-1}|f(y)|_p,|g(y^p)|_p)$. 

The power series $f(T), g(T)$ have no constant term, so $|f(y)|_p \le |y|_p$ and $|g(y^p)|_p \le |y|_p^p$. We deduce that $p^{-1}|f(y)|_p \le p^{-(n+2)}$ and $|g(y^p)|_p \le p^{-(n+1)p}\le p^{-(n+2)}$. Hence $|[p^{n+1}](x)|_p \le p^{-(n+2)}$ and we are done by induction. 

(b)(i) [4 marks, Unseen] There are two approaches here. The first, direct, approach is to consider the identity 
\[f(F(X,Y)) = F(f(X),f(Y)).\]
Suppose for a contradiction that the lowest degree non-zero term in $f(T)$ is $a T^k$, with degree $k \ge 2$. The degree $k$ terms on the LHS of the above identity are $a(X+Y)^k$, whilst on the RHS we get $aX^k + aY^k$. Since $k \ge 2$ (and $R = \Z_p$ has characteristic $0$), we get a contradiction. 

The second approach (which is not much different!) is to use the logarithm map for $F$ (defined over $\Q_p$). The composition $g = \log_F\circ f \circ \exp_F$ is an endomorphism of $\widehat{\mathbb{G}}_a$,  which again has $g(T) = 0 \mod T^2$. Now we have an identity $g(X+Y) = g(X) + g(Y)$ and we again get a contradiction if $g(T)$ is non-zero. 

(b)(ii) [4 marks, Unseen, refines (a)(ii)]
Part (a)(i) shows this is true for $n=1$. (Or we could use $n=0$ as base case.) We proceed by induction for general $n$. 

Suppose $h(T) = [p^n](T) \in (p,T)^{n+1}$. Then $[p^{n+1}](T) = p f(h(T)) + g(h(T)^p)$. Considering the two terms in turn, since $f(T)$ has no constant term, $f(h(T)) \in (p,T)^{n+1}$ and hence $p f(h(T)) \in (p,T)^{n+2}$. Similarly, since $h(T)^p \in (p,T)^{p(n+1)} \subset (p,T)^{n+2}$, we have $g(h(T)^p) \in (p,T)^{n+2}$, and we are done.  

(b)(iii) [7 marks, Unseen]
Uniqueness follows from (b)(i): suppose there are two homomorphism $f, g: F \to F$ with $f(T), g(T) = aT \mod T^2$. Then the difference $F(f(T),i g(T))$ between the two homomorphisms is a homomorphism with linear term $aT - aT = 0$. So (b)(i) implies that $F(f(T),i g(T)) = 0$ and hence $g(T) = f(T)$ (by the uniqueness of inverses). [\textit{3 marks}]

Existence follows from (b)(ii), which shows that if we have a sequence $a_n$ of integers with $a_n = a \mod p^n$, then we get a well-defined power series \[[a](T) = \lim_{n \to \infty}[a_n](T).\] Moreover, $[a](T) = [a_n](T) \mod (p,T)^{n+1}$. Clearly $[a](T)$ has linear term $aT$. To show that it's a homomorphism we note that, since $[a_n]$ is a homomorphism, we have $[a](F(X,Y)) - F([a](X),[a](Y)) = 0 \mod(p,T)^{n+1}$ for every $n$, and hence $[a](F(X,Y)) - F([a](X),[a](Y))=0$. [\textit{4 marks}]
	
	\newpage
	\item[Solution 2.]
(a) [7 marks, Bookwork] We use the fact that the groups $(\mathcal{F}(p\Z_p),+_\mathcal{F})$ associated to formal groups over $\Z_p$ are torsion-free for odd primes, whilst for $p=2$ every torsion point is $2$-torsion. [\textit{2 marks for recalling this fact or similar}]

Suppose $O \neq P = (x_0,y_0) \in E(\Q)_{\mathrm{tors}}$. We deduce that for odd $p$, $P$ is not in the kernel of reduction $E_1(\mathbb{Q}_p)$, so $|x_0|_p\le 1$ and $|y_0|_p \le 1$. For $p = 2$, if $P$ has order $2$ then its coordinates are in $\Z$ (e.g.~Gauss's lemma tells us that the rational roots of $x^3+Ax+B$ are in $\Z$). If it does not have order $2$, then we again deduce that $P$ is not in the kernel of reduction, so $|x_0|_2 \le 1$ and $|y_0|_2 \le 1$. Putting everything together, $|x_0|_p \le 1$ and $|y_0|_p \le 1$ for all primes $p$, so $x_0,y_0 \in \Z$. [\textit{4 marks for argument at odd primes, 1 mark for considering $p=2$ separately}]

(b)(i) [5 marks, Seen Similar] The tangent line through $P$ has equation 
\[ y-y_0 = \frac{3 x_0^2}{2 y_0}(x-x_0).\] [\textit{2 marks for computing tangent}]

Intersecting with the curve gives a cubic equation in $x$: \[(y_0 + \frac{3 x_0^2}{2 y_0}(x-x_0))^2 = x^3 +B\] whose roots, with multiplicity, are $x_0, x_0$ and $x(2P)$. Considering the coefficient of $x^2$, we deduce that \[2x_0 + x(2P) = \frac{9x_0^4}{4y_0^2},\] so $x(2P) = \frac{9x_0^4 - 8x_0y_0^2}{4y_0^2}$. We rewrite using $y_0^2 = x_0^3 + B$: \[x(2P) = \frac{x_0^4 - 8Bx_0}{4(x_0^3+B)}.\] [\textit{3 marks for rest of computation}]

(b)(ii) [4 marks, Seen Similar] Multiplying out the suggested polynomial, we get $3x^2 a(x) - (3x^3-27B) b(x) = 3x^6 - 24Bx^3 -3x^6 - 3Bx^3 + 27Bx^3 + 27 B^2 = 27B^2$. 

We deduce that $y_0^2(12x_0^2 x(2P) - (3x_0^3 - 27B)) = b(x_0)(3 x_0^2 \frac{a(x_0)}{b(x_0)} - (3x_0^3 - 27B)) = 27B^2$, as desired. 

(b)(iii) [1 mark, Seen Similar] We use part (a): this implies that both factors on the left hand side of the equation in (b)(ii) are integers, so $y_0^2 | 27B^2$.

(c)(i) [4 marks, Unseen] First we note that there are no points of order two in $E(\Q)$, since $x^3 + p$ is irreducible over $\Q$. So the order of $E(\Q)_{\mathrm{tors}}$ is odd. Now suppose $O \neq P \in E(\Q)$ is a point of finite order. Part (b)(iii) gives us $8$ possibilities for the value of $y(P)$: it is an integer with $y(P)^2 | 27p^2$, so it is equal to $\pm 1, \pm 3, \pm p, \pm 3p$. For each value of $y(P)$, we have at most 3 possibilities for $x(P)$ (the three roots of a cubic), which gives at most $24$ points. Adding in $O$, we get at most $25$ elements in $E(\Q)_{\mathrm{tors}}$. 
 
(ii) [4 marks, Unseen] There are many choices here. For example we could take $p = 3$ and apply (b)(iii), checking none of the possible values for $y(P)^2$,  $(1, 9, 81)$, give rational values for $x(P)$. Alternatively, for $p = 7$, or indeed any prime $= 2 \mod 5$: check by reduction mod $5$ that $E(\Q)_{\mathrm{tors}}$ has order dividing $4$, which implies it has order $1$, since we have already observed it has odd order. The slightly tricky part here is that the first prime $p = 2$ has an integral point $(-1,1)$ satisfying the Nagell-Lutz bound. 

	\newpage
\item[Solution 3.] (a) [5 marks, Seen Similar] First we compute $2P$. The tangent line at $P$ has gradient $(13/6)$. So the $x$-coordinate of $2P$ satisfies $(13/6)^2 = 2 + x(2P)$. We deduce that $x(2P) = 123 \mod{4331}$. To compute $2P + P$, we need to invert $(x(2P)-x(P)) = 122$ to get the gradient of the line joining $P$ and $2P$. We apply the Euclidean algorithm:
\[ 4331 = 36*122 + 61\]
\[ 122 = 2*61\] 
So $61 | 4331$ and we have a factorization $4331 = 61 \times 71$. [\textit{2 marks for each point addition, 1 mark for correct conclusion about factorization}]

(b) [5 marks, Seen Similar] Since $3$ is a prime of good reduction, $C(\Q)_{\mathrm{tors}}$ is isomorphic to a subgroup of $C(\mathbb{F}_3)$, which has size $4$ (each value of $x$ gives a $2$-torsion point). On the other hand, $C(\Q)$ contains full $2$-torsion: $\{O,(0,0), (\pm 5, 0)\}$. So we deduce that these four points give the rational torsion subgroup. [\textit{3 marks for mod 3 computation, 2 more marks for conclusion. Full credit for alternative solutions using Lutz--Nagell}]

(c) [15 marks, Seen Similar, although not so many examples with positive rank.] We run the standard $2$-descent process, so we have maps $\phi, \hat{\phi}$ between $C$ and $D$ and we compute $G/\hat{\phi}(H)$, $H/{\phi}(G)$, where $G = C(\Q)$ and $H = D(\Q)$. 

The map $q: H/{\phi}(G) \to \Q^\times/(\Q^\times)^2$ has image contained in (the cosets of) squarefree integers dividing 100. It is therefore contained in the group generated by $(-1,2,5)$. A squarefree integer is contained in the image of $q$ if and only if the equation
\[W_r: r l^4 + (100/r)m^4 = n^2 \] has integer solutions with $(l,m)$ coprime. For $r=-1$, comparing signs of the left and right hand signs tells us there are no such solutions. [\textit{5 marks for correct set-up and translation in terms of $W_r$}]

For $r = 2$ the equation says $2l^4 + 50 m^4 = n^2$. Considering the equation mod $5$, using the fact that $2$ is not a square mod $5$, tells us that $5$ must divide $l$ and $n$. Rewriting the equation in terms of $l' = l/5$ and $n' = n/5$ gives $50 (l')^4 + 2m^4 = (n')^2$. Repeating the same argument shows that $5|m$, which contradicts coprimality of $l$ and $m$. So we deduce that $2$ is not in the image of $q$. [\textit{3 marks}]

For $r = 5$ the equation says $5l^4 +20 m^4 = n^2$. This has the solution $(l,m,n) = (1,1,5)$. We get a point $P = (5,25)$ in $D(\Q)$ with image $5$ under $q$. Its image in $C(\Q)$ is $(25/4 , -75/8)$ which necessarily has infinite order because it is not integral. [\textit{2 marks}]

At this point we have shown that $H/{\phi}(G)$ has size $2$, generated by the point $P$ of infinite order. 

Now we consider $G/\hat{\phi}(H)$. The image of the map $\hat{q}$ must be contained in squarefree integers dividing $25$. But the 2-torsion points already give the full group of order 4 generated by $-1, 5$. [\textit{2 marks}]

We conclude that $G/2G$ has order at most $8$, with the torsion subgroup contributing $4$. It follows that the rank is at most one, and the existence of the infinite order point $\hat{\phi}(P)$ shows that the rank is exactly one. [\textit{3 marks}]
\end{itemize}
\end{document}




