\documentclass[a4paper]{article} \usepackage{amsmath,amssymb,amsthm,enumerate,eucal,tikz} \input amssym.def \input amssym.tex %\def\Bbb{\bf} %\nopagenumbers %% \magnification=\magstep1 %\hoffset=1truecm %\voffset=2truecm %\baselineskip = 5.2 true mm %\def\notdiv{{\not\hskip-.5pt |\ }} %\font\frkkk=eufm10 %\font\twelverm=cmr12 %\font\tenrm=cmr10 %\font\ninerm=cmr9 %\font\ninebf=cmbx9 %\font\eightrm=cmr8 %\font\sixrm=cmr6 %\font\sevrm=cmr7 %\font\scrpp=eusm10 %\font\frkk=eufm10 %\font\deffont=cmssi10 %\font\chaptitle=cmbx10 at 14 pt %\tolerance=10000 \def\sqr{\ifmmode\square\else{$\square$}\fi} \def\square{\vcenter{ \hrule height.1mm \hbox{\vrule width.1mm height2.2mm\kern2.18mm\vrule width.1mm} \hrule height.1mm}} % This is a slimmer sqr. %\def\sqr{$\vcenter{\hrule height .3mm %\hbox {\vrule width .3mm height 2mm \kern 1.4mm %\vrule width .3mm} \hrule height .3mm}$} % \null % %\vsize=19.5 true cm %\hsize=11.5 true cm %\vskip 5 true cm \def\leqslant{\le} \def\geqslant{\ge} \def\c{{\cal C}} \def\d{{\cal D}} \def\C{{\cal C}} \def\D{{\cal D}} \def\pk{\phi _\kappa} \def\im{{\hbox{\sl im}}} \def\hs{H_{\varsigma}} \def\hpk{\hat \phi _\kappa} \font\sc=cmssqi8 \def\scc#1{\hbox{\sc #1}} \def\sf{{\scc F}} \def\pnbq{{\Bbb P}^n(\overline {\Bbb Q} )} \def\hk{{\hat \kappa}} \def\bq{{\overline {\Bbb Q}}} \def\hq{{\hat q}} \def\pv{\prod\limits_v } \def\pnk{{\Bbb P}^n(K)} \def\mnkvw{{\Bbb M}^n(K[{\bf v}^2,{\bf w}^2])} \def\pnkv{{\Bbb P}^n(K[{\bf v}^2])} \def\kj{\kappa (J)} \def \qmods {{\mathbb Q}^*/({\mathbb Q}^*)^2} \def \qmodss { {\Bbb Q}^*/({\Bbb Q}^*)^2 \times {\Bbb Q}^*/({\Bbb Q}^*)^2 } \def \qs{{\Bbb Q}^*} \def \qss{({\Bbb Q}^*)^2} \def\bbQ{\Bbb Q} \def\bbF{\Bbb F} \def\bbZ{\Bbb Z} \def\bbR{\Bbb R} \def\bbC{\Bbb C} \def\R{\Bbb R} \def\Q{\mathbb Q} \def\Z{\mathbb{Z}} \def\F{\Bbb F} \def\e{{\mathcal E}} \def\q{\mathbb Q} % \def\etq{{\mathcal E}_{\mathrm{tors}}({\mathbb Q})} \def\ctq{{\mathcal C}_{\mathrm{tors}}({\mathbb Q})} \def\cotq{{\mathcal C}_{\mathrm{oddtors}}({\mathbb Q})} \def\dotq{{\mathcal D}_{\mathrm{oddtors}}({\mathbb Q})} \def\g{{\cal G}} \def\h{{\cal H}} \begin{document} \noindent \centerline{{\bf Elliptic Curves: Solutions to Sheet 4}} \bigskip \begin{enumerate}[{\bf (1)}] \item Let $\phi : \c (\q ) \rightarrow \d (\q)$ be the usual isogeny, which is a group homomorphism with kernel: $\mathbf{o}, (0,0)$, and let $\hat\phi : \d (\q ) \rightarrow \c (\q)$ be the usual dual isogeny, which is also a group homomorphism with kernel: $\mathbf{o}, (0,0)$. Now, let $\cotq$ be the set of torsion elements of $\c (\q)$ which have odd order. First check that $\cotq$ is a subgroup: (1) The identity {\bf o} has order~1, which is odd, so ${\bf o}\in \cotq$, (2) If $P,Q$ have odd torsion orders $m,n$, respectively, then $mn(P+Q) = n(mP) + m(nQ) = {\bf o}$ and so the order of~$P+Q$ divides $mn$, giving that the order of $P+Q$ is odd, i.e.\ $P+Q \in \cotq$, (3) The order of $-P$ is the same as the order of $P$, so $P\in\cotq \Rightarrow -P \in \cotq$. Similarly define $\dotq$, a subgroup of $\d (\q)$. \par Now, consider any $P\in \cotq$. Then $P$ must have odd order~$m$, say. Let $R = \phi (P)$. Then $mR = m\phi(P) = \phi (mP) = \phi ({\bf o}) = {\bf o}$, which means that the order of~$R$ divides~$m$, and so the order of~$R$ is odd; that is: $R\in \dotq$. Hence, $\phi$ gives a map from $\cotq$ to $\dotq$, which certainly satisfies the homomorphism property $\phi (P+Q) = \phi(P) + \phi(Q)$ for all $P,Q\in \cotq$, since $\phi$ satisfies this property on the larger set $\c (\q )$. Furthermore, the kernel of $\phi : \cotq \rightarrow \dotq$ must only contain {\bf o} [since $(0,0) \not\in \cotq$], and so $\phi : \cotq \rightarrow \dotq$ is an injection. We have therefore established that there is an injective homomorphism from $\cotq$ to $\dotq$, which implies that $\cotq$ is isomorphic to a subgroup of $\dotq$. Applying the same argument to $\hat\phi : \dotq \rightarrow \cotq$ gives that $\dotq$ is isomorphic to a subgroup of $\cotq$. Since $\cotq$ and $\dotq$ are finite groups, we deduce that we have inequalities $|\cotq| \le |\dotq|$ and $|\dotq|\le |\cotq|$, hence $|\cotq| = |\dotq|$. Now $\phi : \cotq \rightarrow \dotq$ is an injective homomorphism between finite groups of the same size, and is hence an isomorphism. \item The preimages of~$(0,0)$ under $\hat\phi$ are given by the points of order~2 on $\d$ distinct from~$(0,0)$, namely $Q_1 = ((-a_1 + \sqrt{a_1^2 - 4b_1})/2,0) = (a + 2\sqrt{b},0)$ and $Q_2 = ((-a_1 - \sqrt{a_1^2 - 4b_1})/2,0) = (a - 2\sqrt{b},0)$. Recall the standard map from lectures $q : \d (\q) \rightarrow \qmods$ which takes ${\bf o} \rightarrow 1$, $(0,0) \rightarrow b_1$, and otherwise takes $(u,v) \rightarrow u$. We know from lectures that this map has kernel precisely $\phi ( \c (\Q ))$. Now, the multiplication by 2 map on $\c (\Q)$ is $\hat\phi \circ \phi$, and so $(0,0) \in 2\c (\Q)$ iff either $Q_1$ or $Q_2$ is a member of $\phi (\c (\Q) )$. In particular, the $Q_i$ need to be in $\d(\Q)$ which is equivalent to $b = m^2$ for some $m \in \Z$. The additional condition that $Q_i \in \phi(\c(\Q))$ is equivalent to $q(Q_i) = 1$, or $a \pm 2m \in (\Q^\ast)^2$. Finally, since $a \pm 2m \in \Z$, this condition is equivalent to the existence of $n \in \Z$ with $a \pm 2m = n^2$. Changing the sign of $m$ if necessary gives the desired result. \newpage \item {\bf (a).} The map is well-defined since $\hat{\phi} \circ \phi = [2]$ and is obviously surjective. Let $[P] \in \h /\phi(\g)$ map to $[o]$, i.e. $\hat{\phi}(P + \phi(\g)) = [o]$, that is $\hat{\phi}(P) + 2\g = 2\g$ so $\hat{\phi}(P) \in 2\g$. Then $\hat{\phi}(P) = 2R$ for some $R \in \g$. Hence $\hat{\phi}(P - \phi(R)) = o$ as $\hat{\phi} \circ \phi = [2]$. From lectures (Lemma 6.1) we know $P - \phi(R) = o$ or $P - \phi(R) = (0,0)$. Hence $[P]$ in $\h /\phi(\g)$ is $[o]$ or $[(0,0)]$. {\bf (b).} We thus get $ \hat{\phi}(\h) / 2 \g$ is $C_2^{b}$ or $C_2^{b-1}$ according to whether the class $[(0,0)]$ is trivial or not. The result now follows from the isomorphism of groups \[ \g / \hat{\phi}(\h) \cong \g / 2 \g \, / \, \hat{\phi}(\h) / 2 \g.\] {\bf (c).} Assume $\phi(P) = (0,0)_{\h}$ for some $P \in \g$. Applying $\hat{\phi}$ to both sides we see $2 P = \hat{\phi}((0,0)_\h) = o_\g$. But $P$ cannot be $(0,0)_\g$ for then we'd have $\phi(P) = o_\h$. So we have an additional $2$-torsion point in $\g$ and the $2$-torsion is $C_2^2$. Conversely, assume that the $2$-torsion in $\g$ is $C_2^2$ and let $P \in \g$ with $P \ne (0,0)_\g$ and $2P = o_\g$. Then we have $\hat{\phi} (\phi(P)) = o_\g$ and by Lemma 6.1 again we see $\phi(P) = o_\h$ or $(0,0)_\h$. But since $P \ne (0,0)_\g$ we cannot have the former. Recall that $\g/2\g \cong C_2^{r+t}$ where $r$ is the rank and $\g[2] \cong C_2^t$. We have either $t = 1$ or $2$. In the first case, $(0,0) \not \in \phi(\g)$, so the rank is $(a+b-1)-t = a + b -2$. In the second case, $(0,0) \in \phi(\g)$, so the rank is $(a + b) - t = a+b-2$. \item \par\noindent{\bf (a).} Consider $\c : Y^2 = X(X^2 + aX + b) = X(X^2 + 2X + 3)$, where $a=2, b=3$, and the isogenous curve $\d : Y^2 = X(X^2 + a_1X + b_1) = X(X^2 - 4X - 8)$, where $a_1 = -4, b_1=-8$, with the usual isogeny $\phi : \c (\Q ) \rightarrow \d (\Q) : (x,y) \mapsto (y^2/x^2 , y - 3y/x^2)$, and dual isogeny $\hat\phi : \d (\Q ) \rightarrow \c (\Q) : (u,v) \mapsto ( {1\over 4} v^2/u^2 , {1\over 8}( v + 8v/u^2) )$. %{\bf [3~marks]} \par The map $q : \d (\Q) / \phi (\c (\Q)) \rightarrow \qmods : (u,v) \mapsto u$, for $(u,v) \not= (0,0)$, with $q : (0,0) \mapsto b_1$ and $q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}(q)$ contained in $\{ d : d\hbox{ is square free and } d | b_1\} = \{ \pm 1 , \pm 2 \}$. Also, $(0,0) \mapsto -8 = -2$, so that $\{ 1,-2 \} \subset \hbox{im} q \subset \{ \pm 1 , \pm 2\}$. %{\bf [3 marks]} \par There is only one coset to check, represented by $-1$, say. We know that $-1 \in \hbox{im} q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $(-1)\cdot \ell^4 + a_1 \ell^2m^2 + (b_1/(-1))\cdot m^4 = n^2$ that is: $-\ell^4 - 4 \ell^2m^2 + 8 m^4 = n^2$. Rewrite as: $-(\ell^2 + 2m^2)^2 + 12 m^4 = n^2$. Reducing modulo~3 gives $ - (\ell^2 + 2m^2 )^2 \equiv n^2$ (modulo~3). If $n$ were coprime to~$3$, then this would give: $(( \ell^2 + 2 m^2 )/n)^2 = -1$ in $\F_3$, contradicting the fact that~$-1$ is not a quadratic residue modulo~3. So, $3 | n$ and so $3 | (\ell^2 + 2m^2 )$ also. This means that $9 | (\ell^2 + 2m^2 )^2$ and $9 | n^2$, which can be combined with $-(\ell^2 + 2m^2)^2 + 12 m^4 = n^2$ to give: $9 | 12 m^4$ and so $3 | m$. Combining $3 | m$ with $3 | (\ell^2 + 2m^2 )$ gives that $3 | \ell$. This contradicts the fact that $\hbox{gcd}(\ell,m) = 1$. Hence our equation is impossible in~$\Q_3$, and so impossible in~$\Q$. Hence $-1 \not\in \hbox{im} q$. %{\bf [6 marks]} \par We conclude that $\hbox{im} q = \{ 1,-2 \}$, and so $\d (\Q) / \phi (\c (\Q))$ is generated by~$(0,0)$. %{\bf [1 mark]} \par The map $\hat q : \c (\Q ) / \hat\phi (\d (\Q)) \rightarrow \qmods : (x,y) \mapsto x$, for $(x,y) \not= (0,0)$, with $\hat q : (0,0) \mapsto b$ and $\hat q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}\hat q$ contained in $\{ d : d\hbox{ is square free and } d | b\} = \{ \pm 1 , \pm 3 \}$. Also, ${\bf o} \mapsto 1$ and $(0,0) \mapsto 3$, so that $\{ 1,3 \} \subset \hbox{im} \hat q \subset \{ \pm 1 , \pm 3\}$. %{\bf [3 marks]} \par There is only one coset to check, represented by $-1$, say. We know that $-1 \in \hbox{im} \hat q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $(-1)\cdot \ell^4 + a \ell^2m^2 + (b/(-1))\cdot m^4 = n^2$; that is: $-\ell^4 + 2\ell^2m^2 - 3 m^4 = n^2$. Rewrite as: $ - ( \ell^2 - m^2 )^2 - 2 m^4 = n^2$. This is impossible in~$\R$ (the left hand side is $\leqslant 0$ and the right hand side is $\ge 0$, and equality only occurs when $\ell^2 - m^2 = m^4 = n^2 = 0$, implying $\ell = m = n = 0$, which is not allowed). Hence $-1 \not\in \hbox{im} \hat q$. %{\bf [4 marks]} \par We conclude that $\hbox{im} \hat q = \{ 1,3 \}$, and so $\c (\Q) / \hat\phi (\d (\Q))$ is generated by~$(0,0)$. %{\bf [1 mark]} \par Finally, since multiplication by 2 in $\c (\Q)$ is $\hat\phi \circ \phi$, we have that $\c (\Q) / 2\c (\Q)$ is generated by: generators for $\c (\Q) / \hat\phi (\d (\Q))$ [namely: $(0,0)$] together with the images under $\hat \phi$ of generators for $\d (\Q) / \phi (\c (\Q))$ [namely, $\hat\phi \bigl( (0,0) \bigr) = {\bf o}$]. Conclusion: $\c (\Q ) / 2\c(\Q )$ is generated by $(0,0)$, and so is isomorphic to $C_2$. We also know that $\c (\Q ) / 2\c(\Q )$ is isomorphic to $\ctq / 2\ctq \times C_2^r$, which is isomorphic to $\c (\Q )[2] \times C_2^r$, where $\c (\Q )[2]$ is the $2$-torsion group and $r$ is the rank. We know that $(0,0) \in \c (\Q )[2]$, so we deduce that $r = 0$. \par\noindent {\bf (b).} Let $\c : Y^2 = X(X^2 + aX + b) = X(X^2 + 14X + 1)$, where $a=14, b=1$, and isogenous curve $\d : Y^2 = X(X^2 + a_1X + b_1 ) = X(X^2 - 28X +192)$, where $a_1 = -28, b_1=192$, with the usual isogeny $\phi : \c (\Q ) \rightarrow \d (\Q) : (x,y) \mapsto (y^2/x^2 , y - y/x^2)$, and dual isogeny $\hat\phi : \d (\Q ) \rightarrow \c (\Q) : (u,v) \mapsto ( {1\over 4} v^2/u^2 , {1\over 8}( v - 192 v/u^2) )$. \par The map $q : \d (\Q) / \phi (\c (\Q)) \mapsto \qmods : (u,v) \mapsto u$, for $(u,v) \not= (0,0)$, with $q : (0,0) \mapsto b_1$ and $q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}q$ contained in $\{ d : d\hbox{ is square free and } d | b_1\} = \{ \pm 1 , \pm 2, \pm 3, \pm 6 \}$. Also, ${\bf o} \mapsto 1$, $(0,0) \mapsto 192 = 3$ and the point $(8,16) \mapsto 2$ [N.B. when searching for a point in~$\d (\Q)$ which might map to~$2$ under~$q$, one need only try points with $x$-coordinate equal to~$2$ modulo squares, e.g.~2,8,1/2,18,etc, so one should find the point $(8,16)$ quickly; also, it's a good idea at the outset just to look for ``obvious'' members of $\d (\q )$ with $x$ coordinates being integers in the range from $-10$ to $10$; doing this at the outset would also reveal the point $(8,16)$.] This means that $1,3,2 \in \hbox{im} q$; but $\hbox{im} q$ is a group, so $6 \in \hbox{im} q$ also (indeed $(0,0) + (8,16) = (24,-48)$, which maps to $6$ under $q$. Hence, $\{ 1,2,3,6 \} \subset \hbox{im} q \subset \{ \pm 1 , \pm 2,\pm 3, \pm 6\}$. \par There is only one coset to check, represented by $-1$, say. We know that $-1 \in \hbox{im} q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $(-1)\cdot \ell^4 + a_1 \ell^2m^2 + (b_1/(-1))\cdot m^4 = n^2$ that is: $-\ell^4 - 28 \ell^2m^2 - 192 m^4 = n^2$. We can see that the LHS is $\leqslant 0$ and the RHS is $\geqslant 0$, and there is equality iff $\ell = m = n = 0$, a contradiction. Hence $-1 \not\in \hbox{im} q$. We conclude that $\hbox{im} q = \{ 1,2,3,6 \}$ and that $\d (\Q) / \phi (\c (\Q)) = \{ {\bf o}, (0,0), (8,16), (24,-48) \}$, and so $\d (\Q) / \phi (\c (\Q))$ is generated by $(0,0)$ and $(8,16)$. \par The map $\hat q : \c (\Q ) / \hat\phi (\d (\Q)) \mapsto \qmods : (u,v) \mapsto u$, for $(u,v) \not= (0,0)$, with $\hat q : (0,0) \mapsto a_1^2 - 4b_1 = b$ and $\hat q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}\hat q$ contained in $\{ d : d\hbox{ is square free and } d | b\} = \{ \pm 1 \}$. Also, ${\bf o} \mapsto 1$ and $(0,0) \mapsto 1$, so that $\{ 1 \} \subset \hbox{im} \hat q \subset \{ \pm 1 \}$. \par There is only one coset to check, represented by $-1$, say. We know that $-1 \in \hbox{im} \hat q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $(-1)\cdot \ell^4 + a \ell^2m^2 + (b/(-1))\cdot m^4 = n^2$; that is: $-\ell^4 + 14\ell^2m^2 - m^4 = n^2$. Rewrite as: $ - ( \ell^2 - 7 m^2 )^2 + 48 m^4 = n^2$. Reducing modulo~3 gives: $ - ( \ell^2 - 7 m^2 )^2 \equiv n^2$ (modulo~3). If $n$ were coprime to~$3$, then this would give: $(( \ell^2 - 7 m^2 )/n)^2 = -1$ in $\F_3$, contradicting the fact that~$-1$ is not a quadratic residue modulo~3. So, $3 | n$ and so $3 | ( \ell^2 - 7 m^2 )$, also. Hence $9$ divides $( \ell^2 - 7 m^2 )^2$ and $n^2$ and so must divide $48 m^4$; but $48$ is only divisible by $3$ (not by~$9$) so that $3$ must divide $m^4$; hence $3$ divides $m$. Combining this with the fact (already found) that $3 | ( \ell^2 - 7 m^2 )$ gives that $3 | \ell$ also. We've shown that $3$ divides all of $\ell,m,n$, a contradiction. Hence $-1 \not\in \hbox{im} \hat q$. \par We conclude that $\hbox{im} \hat q = \{ 1 \}$, and so $\c (\Q) / \hat\phi (\d (\Q))$ contains only {\bf o}. [N.B. $(0,0)$ should not also be included as a separate member of $\c (\Q) / \hat\phi (\d (\Q))$ even though it is a rational point; $(0,0)$ maps to~1 under~$\hat q$, and so $(0,0) \in \hat\phi (\d (\Q))$; that is $(0,0) = {\bf o}$ in $\c (\Q) / \hat\phi (\d (\Q))$; the same comment applies to the obvious point $(1,4)$; in general, for each distinct member $r$ of $\hbox{im} \hat q$, you should only include exactly one point in $\c (\Q)$ which maps to~$r$]. \par Finally, since multiplication by 2 in $\c (\Q)$ is $\hat\phi \circ \phi$, we have that $\c (\Q) / 2\c (\Q)$ is generated by: generators for $\c (\Q) / \hat\phi (\d (\Q))$ [namely: {\bf o}] together with the images under $\hat \phi$ of generators for $\d (\Q) / \phi (\c (\Q))$ [namely, $\hat\phi \bigl( (0,0) \bigr) = {\bf o}$ and $\hat\phi \bigl( (8,16) \bigr) = (1,-4)$]. Conclusion: $\c (\Q ) / 2\c(\Q )$ is generated by $(1,-4)$, and so is the group $C_2$. Now $\c (\q)/2\c (\q)$ is isomorphic to $\ctq/2\ctq \times C_2^{rank}$, and $\ctq /2\ctq$ is isomorphic to the $2$-torsion group of $\ctq$ which is $C_2$ (consisting only of {\bf o} and $(0,0)$), so that $\ctq/2\ctq$ is isomorphic to~$C_2$. Conclusion: rank~$=0$. [If this seems surprising, then note that $(1,4),(1,-4)$ are points of order~$4$ in $\c (\q)$]. %\par\noindent {\bf (d).} $Y^2 = X(X^2 + 2X + 9)$. %\par\noindent {\bf (e).} $Y^2 = X(X^2 + 9X - 1)$. %\par\noindent {\bf (f).} $Y^2 = X(X-12)(X-36)$. \medskip \item Let $\C : Y^2 = X(X^2 + aX + b) = X(X^2 + p^2)$, where $a=0, b=p^2$, and isogenous curve $\D : Y^2 = X(X^2 + a_1X + b_1 ) = X(X^2 - 4p^2)$, where $a_1 = -2a = 0, b_1= a^2 - 4b = -4p^2$, with the usual isogeny $\phi : \C (\Q ) \rightarrow \D (\Q) : (x,y) \mapsto (y^2/x^2 , y - p^2y/x^2)$, and dual isogeny $\hat\phi : \D (\Q ) \rightarrow \C (\Q) : (u,v) \mapsto ( \frac{1}{4} v^2/u^2 , \frac{1}{8}(v + 4p^2v/u^2))$. \par The map $q : \D (\Q) / \phi (\C (\Q)) \mapsto \qmods : (u,v) \mapsto u$, for $(u,v) \not= (0,0)$, with $q : (0,0) \mapsto b_1$ and $q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}q$ contained in $\{ d : d\hbox{ is square free and } d | b_1\} = \{ \pm 1 , \pm 2, \pm p, \pm 2p \}$. Also, ${\bf o} \mapsto 1$, $(0,0) \mapsto -1$, $(2p,0)\mapsto 2p$, $(-2p,0)\mapsto -2p$, so that $\{ \pm 1,\pm 2p \} \subset \hbox{im} q \subset \{ \pm 1 , \pm 2, \pm p, \pm 2p \}$. There is only one coset to check, represented by~$-2$, say. We know that $-2 \in \hbox{im} q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $(-2)\cdot \ell^4 + a_1 \ell^2m^2 + (b_1/(-2))\cdot m^4 = n^2$; that is: $-2\ell^4 + 2 p^2 m^4 = n^2$. Since $p\equiv 5$~(mod~$8$), $\bigl( \frac{-2}{p} \bigr) = -1$, and so $p | \ell$, $p | n$. Let $\ell = p \ell'$, $n = p n'$. Then $-2 p^2 (\ell')^4 + 2 m^4 = (n')^2$. Since also $\bigl( \frac{2}{p} \bigr) = -1$, this implies $p | m$, $p | n'$. This gives a contradiction, since $p| \ell, p | m$ and $\hbox{gcd}(\ell,m) = 1$. Hence~$-2\not\in \hbox{im} q$, giving that $\hbox{im} q = \{ \pm 1 , \pm 2p \}$, and $\D (\Q) / \phi (\C (\Q)) = \{{\bf o}, (0,0), (2p,0), (-2p,0)\}$.\\ \par The map $\hat q : \C (\Q ) / \hat\phi (\D (\Q)) \mapsto \qmods : (u,v) \mapsto u$, for $(u,v) \not= (0,0)$, with $\hat q : (0,0) \mapsto a_1^2 - 4b_1 = b$ and $\hat q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}\hat q$ contained in $\{ d : d\hbox{ is square free and } d | b\}$ $= \{ \pm 1 , \pm p\}$. Also, ${\bf o} \mapsto 1$ and $(0,0) \mapsto p^2 = 1$, so that $\{ 1 \} \subset \hbox{im} \hat q \subset \{ \pm 1 , \pm p\}$. \par We know that $-1 \in \hbox{im} \hat q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $(-1)\cdot \ell^4 + a \ell^2m^2 + (b/(-1))\cdot m^4 = n^2$; that is: $-\ell^4 - p^2 m^4 = n^2$, clearly impossible in $\R$. Hence~$-1\not\in \hbox{im} \hat q$. Similarly, $-p\not\in\hbox{im} \hat q$. We know that $p \in \hbox{im} \hat q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $p\cdot \ell^4 + a \ell^2m^2 + (b/p)\cdot m^4 = n^2$; that is: $p\ell^4 + p m^4 = n^2$, and so $p | n$. Letting $n = p n'$, the equation becomes $\ell^4 + m^4 = p^2 (n')^2$, which in turn implies $p | \ell, p | m$, since $-1 \not\equiv a^4$ for any~$a$ [as can be seen by taking both sides of $-1 \equiv a^4$ to the power of $(p-1)/4$]. This again contradicts $\hbox{gcd}(\ell,m) = 1$, and so $p\not\in \hat q$. We conclude that $\hbox{im} \hat q = \{ 1 \}$, and so $\C (\Q) / \hat\phi (\D (\Q)) = \{ \bf o\}$.\\ \par Finally, since multiplication by 2 in $\C (\Q)$ is $\hat\phi \circ \phi$, we have that $\C (\Q) / 2\C (\Q)$ is generated by: generators for $\C (\Q) / \hat\phi (\D (\Q))$ [namely: $\bf o$] together with the images under $\hat \phi$ of generators for $\D (\Q) / \phi (\C (\Q))$ [namely, $\hat\phi \bigl( (0,0) \bigr) = {\bf o}$, and $\hat\phi \bigl( (2p,0) \bigr) = (0,0)$]. Conclusion: $\C (\Q ) / 2\C(\Q )$ is generated by $(0,0)$, and so is the group $C_2$. Now $\C (\Q)/2\C (\Q)$ is isomorphic to $\C(\Q)_{tors}/2\C(\Q)_{tors} \times C_2^{rank}$, and $\C(\Q)_{tors} /2\C(\Q)_{tors}$ is isomorphic to the $2$-torsion group of $\C(\Q)_{tors}$ which is $C_2$ (consisting only of {\bf o} and $(0,0)$), so that $\C(\Q)_{tors}/2\C(\Q)_{tors}$ is isomorphic to $C_2$. Conclusion: rank $=0$. \item We are given that $A,+$ is an Abelian group, and that $h : A \rightarrow \R_{\ge 0}$ satisfies: \par\noindent \ \ \ \ (I) There exists a constant~$C$, independent of~$P,Q$, such that \par \ \ \ \ $ | h(P+Q) + h(P-Q) - 2 h(P) - 2 h(Q) | \le C $, for all $P,Q \in A$, \par\noindent \ \ \ \ (II) For any~$B\in\R$, the set $\{P\in A:h(P)\le B\}$ is finite. \par From~(I) we obtain: $h(P+Q) + h(P-Q) - 2 h(P) - 2 h(Q) \le C $ and so (since $h(P-Q) \ge 0$): $h(P+Q) \le h(P+Q) + h(P-Q) \le 2 h(P) + 2 h(Q) + C \le 2 h(P) + C_1(Q)$, where $C_1(Q) = 2 h(Q) + C$. Hence Property~(1) in the definition of height function is satisfied. \par Letting $Q=P$ in~(I), we obtain: \par \ \ \ \ $ | h(2P) + h(e) - 4 h(P) | \le C, \ \ \ \ (*) $ \par\noindent where~$e$ denotes the identity element of the group~$A$. This gives: $h(2P) + h(e) - 4 h(P) \ge -C$, and so: $h(2P) \ge 4 h(P) - C_2$, where $C_2 = C + h(e)$. Hence Property~(2) in the definition of height function is satisfied. Furthermore, (II) is the same as Property~(3) in the definition of height function. Hence all~3 required properties are satisfied, giving that~$h$ is a height function, as required. \par Replacing $P,Q$ in~(I) with $2P,P$, respectively, gives: \par \ \ \ \ $ | h(3P) - 2 h(2P) - h(P) | \le C $. \par\noindent Multiplying $(*)$ by~$2$ gives: \par \ \ \ \ $ | 2 h(2P) + 2 h(e) - 8 h(P) | \le 2C$. \par\noindent These last two equations then give: \par $ | h(3P) - 9 h(P) | = | h(3P) - 2 h(2P) - h(P) + 2 h(2P) + 2 h(e) - 8 h(P) - 2 h(e) |$ \par\ \ \ \ \ $\le | h(3P) - 2 h(2P) - h(P) | + | 2 h(2P) + 2 h(e) - 8 h(P) | + 2 | h(e) | \le C_3,$ \par\noindent where $C_3 = 3C + 2 | h(e) |$ (which is independent of~$P$). \medskip \item In all of the following, each step multiplies numbers $\leqslant N$ (followed by a possible reduction modulo~$N$), and so we are guaranteed that everything can be done on an $9$-digit calculator, since $N^2$ has only 9~digits. \par\noindent {\bf (a).} First compute (modulo $N=10481$): $2^1 \equiv 2$, $2^2 \equiv 4$, $2^4 \equiv 16$, $2^8 \equiv 256$, $2^{16} \equiv 2650$, $2^{32} \equiv 230$ (where each of these was obtained be squaring the previous one, and reducing modulo~$N$). Now, we write $46$ in base~2: $46 = 2 + 4 + 8 + 32$ and so $2^{46} \equiv 2^2 2^4 2^8 2^{32} \equiv 4\cdot 16 \cdot 256 \cdot 230 \equiv 64\cdot 6475 \equiv 5641$ modulo~$N$, so that $2^{46} - 1 \equiv 5640$ modulo~$N$. \par Now, compute $\hbox{gcd}(5640, N)$ by Euclid's Algorithm: $10481 = 1 \cdot 5640 + 4841$; $5640 = 1\cdot 4841 + 799$; $4841 = 6\cdot 799 + 47$, $799 = 17\cdot 47 + 0$. So, $47$ is a factor of $N$. Compute $10481/47 = 223$, giving the factorisation $N = 10481 = 47 \cdot 223$. %{\bf [7~marks]} \par\noindent {\bf (b).} The line tangent to~$\e$ at $P=(5,11)$ has slope $y'$ given by $2yy' = 3x^2 - 1$, with $x=5,y=11$; that is, the slope is $74/22 = 37/11$. This tangent line also goes through $(5,11)$ and so has equation: $Y = (37/11)X - 64/11$. The $x$-coordinate of $2P$ is therefore $(37/11)^2 - (5+5) = 159/121$. [It will turn out not to be necessary to evaluate this mod~$N$, although if this is done using EA, then it is~7364], and the $y$-coordinate is: $-((37/11)\cdot (159/121) - (64/11)) = 1861/1331$ [again, although unnecessary in this example, this can be computed by EA to be: 6679 mod~N], so that $Q = 2P = (159/121 , 1861/1331)$. We now wish to compute $3P = P + Q$, and so again the first step is to find the line joining $P$ and $Q$. This has slope given by $(1861/1331 - 11)/(159/121 - 5) = 6930/2453$, and so we need to compute $6930/2453$ (modulo~$N=10481$), for which the first step is to find the inverse of $2453$ (modulo~$N=10481$). Using Euclid's Algorithm: $10481 = 4\cdot 2453 + 669$; $2453 = 3 \cdot 669 + 446$; $669 = 1\cdot 446 + 223$; $446 = 2\cdot 223 + 0$. So, we cannot find the inverse of $2453$ (modulo~$N=10481$), and this step has given us a factor~$223$ of~$N$. As before, compute $10481/223 = 47$, giving the factorisation $N = 10481 = 47 \cdot 223$. %{\bf [10~marks]} \par\noindent {\bf (c).} Since $N = 47 \cdot 223$, we have $\phi (N) = 46 \cdot 222 = 10212$. Compute the gcd of $d=4085$ and $\phi(N)$, we see: $10212 = 2\cdot 4085 + 2042$; $4085 = 2\cdot 2042 + 1$, so that $\hbox{gcd}(10212,4085) = 1$. Reversing the steps: $1 = 4085 - 2\cdot 2042 = 4085 - 2\cdot (10212 - 2\cdot 4085) = 5\cdot 4085 - 2\cdot 10212$. Hence, $5$ is the inverse of $4085$ modulo~$10212$. The decoding operation is therefore $X \mapsto X^{5} \hbox{ mod }N$. Computing $6012^{5} = (6012^2)^2 \cdot 6012 \equiv 5656^2\cdot 6012 \equiv 2324\cdot 6012 \equiv 715$. (modulo~$N = 10481$). Also: $3236^{5} = (3236^2)^2 \cdot 3236 \equiv 1177^2 \cdot 3236 \equiv 1837 \cdot 3236 \equiv 1805$ (modulo~$N = 10481$). The decoded message is therefore: $0715,\, 1805$; that is: GORE. \medskip \end{enumerate} \end{document}