\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\c{{\cal C}} \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 {{\Bbb Q}^*/({\Bbb 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\le{\leqslant} \def\ge{\geqslant} \def\etq{{\cal E}_{\lower 1pt\hbox{\eightrm tors}}({\Bbb Q})} \def\etqp{{\cal E}_{\lower 1pt\hbox{\eightrm tors}}({\Bbb Q}_p)} \def\c{{\cal C}} \def\d{{\cal D}} \def\e{{\cal E}} \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 {{\Bbb Q}^*/({\Bbb 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\notdiv{{\not\hskip-.5pt |\ }} \def\Q{{\Bbb Q}} \def\F{{\Bbb F}} \def\Z{{\Bbb Z}} \def\R{{\Bbb R}} \def\C{{\Bbb C}} \begin{document} \noindent \centerline{{\bf Elliptic Curves MT25: Solutions to Sheet 1 sections A and C}} \bigskip \begin{enumerate}[{\bf (1)}] \item \noindent {\bf (a).} {We find rational points on each curve. This tells us that each curve is birational to $\mathbb{P}^1$, and hence birational to each other. For the first, we have $(1,1)$. For the second, we have $(1,0)$. So the points of the first conic are parameterised by $t = \frac{y-1}{x-1} \in \mathbb{P}^1(\Q)$. To find a corresponding point on the second curve, we intersect $Y = t(X-1)$ with the curve. We get the equation $X^2 + t^2(X-1)^2 - 6tX(X-1) = 1$. The coefficient of $X^2$ is $1+t^2 -6t$ and coefficient of $X$ is $-2t^2 + 6t$. So if the intersection point is $(x_1,y_1)$, we have $x_1 + 1 = \frac{2t^2 - 6t}{t^2-6t+1}$, and hence $x_1 = \frac{t^2 - 1}{t^2-6t+1}, y_1 = \frac{2t(3t-1)}{t^2-6t+1}$. Substituting $t = \frac{y-1}{x-1}$, we get the rather unpleasant birational transformation from the first curve to the second} {\footnotesize\[(x,y) \mapsto \left(\frac{(y-1)^2 - (x-1)^2}{(y-1)^2-6(y-1)(x-1)+(x-1)^2}, \frac{2(y-1)(3(y-1)-(x-1))}{(y-1)^2-6(y-1)(x-1)+(x-1)^2} \right).\] } \par\noindent {\bf (b).} {We can rewrite the first equation as $(\frac{Y}{(X+2)^3})^2 = X^3 + 1$. So we can take the birational transformation} \[(x,y) \mapsto (x,\frac{y}{(x+2)^3}).\] \par\noindent {\bf (c).} {Over $\C$, we have the birational transformation $(x,y) \mapsto (\sqrt{2}x,y)$. Over $\Q$, the second curve is reducible, with irreducible components $Y = \pm X$. The first curve is irreducible. So the two curves are not birational over $\Q$. Alternatively, the first curve's only rational point is $(0,0)$, whilst the second has infinitely many, so again they cannot be birational over $\Q$.} \item \noindent {\bf (a).} The only points on $Y^2 = X^3 + 2 X$ over $\bbF_5$ are: ${\bf o}, (0,0)$. The group table is the same as that of $C_2$ (``cyclic~2'' group; i.e. the integers modulo~2 under addition), with $0,1$ replaced by ${\bf o}, (0,0)$, respectively. \par \noindent {\bf (b).} The only points on $Y^2 = X^3 + 1$ over $\bbF_5$ are: ${\bf o}, (0,1), (0,4), (4,0), (2,2), (2,3)$. The point~$(2,2)$ has order~6, and the group table is the same as that of $C_6$ (``cyclic~6'' group; i.e. the integers modulo~6 under addition), with $0,1,2,3,4,5$ replaced by ${\bf o}, (2,2), (0,4), (4,0), (0,1), (2,3)$, respectively. Note that it's also correct to say that the group is $C_2 \times C_3$, since $C_2\times C_3$ is isomorphic to $C_6$. \medskip \item We have: $2YY' = 3X^2 + 4$, and so the slope at the point~$(2,4)$ is: $(3\cdot 2^2 + 4)/(2\cdot 4) = 2$. The line tangent to the curve at~$(2,4)$ is: $Y=2X$. The $x$-coordinate of the third point of intersection is therefore: $2^2 - 2 - 2 = 0$, with $y$-coordinate $y=2\cdot 0 = 0$. Hence, $(2,4) + (2,4) + (0,0) = {\bf o}$, and so $(2,4) + (2,4) = (0,-0) = (0,0)$. Now, note that we always have $-(x,y) = (x,-y)$ and so $-(0,0) = (0,0)$, giving: $2(0,0) = {\bf o}$. Hence, $4(2,4) = 2(0,0) = {\bf o}$. Also, check that $3(2,4) = 4(2,4) - (2,4) = -(2,4) = (2,-4)$. Summarising: the first~4 multiples of~$(2,4)$ are: $1(2,4) = (2,4),\, 2(2,4) = (0,0),\, 3(2,4) = (2,-4),\, 4(2,4) = {\bf o}$, and so~4 is the smallest positive multiple of $(2,4)$ equal to~{\bf o}. \newpage \item[{\bf (9)}] {\bf (a).} Let $L$ be the tangent line to the curve at~$P$, and let~$R$ be the third point of intersection; that is, the line and the curve meet at $P,P,R$. Then $P+P+R = {\bf o}$. Then, $3P = {\bf o} \iff R=P \iff$ $L$ intersects $\cal E$ only at~$P$ (3 times). \par\noindent {\bf (b).} The Hessian matrix is: \[ \left( \begin{array}{ccc} -6X_0 & 0 & -2AX_2 \\ 0 & 2X_2 & 2X_1 \\ -2AX_2 & 2X_1 & -2AX_0 - 6BX_2 \end{array} \right) \] \noindent The determinant is: \par $-6X_0\bigl( 2X_2 ( -2AX_0 - 6BX_2 ) - (2X_1)(2X_1) \bigr) -2AX_2 \bigl( 0 - (2X_2)(-2AX_2) \bigr)$ \par $= 8(3AX_0^2X_2 + 9BX_0X_2^2 + 3X_0X_1^2 - A^2 X_2^3)$. \par\noindent The only projective point $(X_0,X_1,X_2)$ on the curve with $X_2=0$ is $(0,1,0) = {\bf o}$, for which the statement is true, since $3{\bf o} = {\bf o}$ and $(X_0,X_1,X_2) = (0,1,0)$ makes the Hessian determinant~$0$. When $X_2 \not= 0$, we can write everything in affine form, with $x = X_0/X_2$ and $y = X_1/X_2$, when we see that the Hessian determinant (after dividing through by~$8 X_2^3$) is~0 exactly when: \medskip \hskip 5 cm $3Ax^2 + 9Bx + 3xy^2 - A^2 = 0$.\hfill (1) \medskip \noindent Also, the point $P = (x,y)\not= {\bf o}$, written in affine form, has order~3 exactly when $2(x,y) = (x,-y)$ which happens if and only if the $x$-coordinate of $2(x,y)$ is~$x$. But, as usual, the $x$-coordinate of $2(x,y)$ is $m^2 - 2x$, where $m = (3x^2 + A)/(2y)$ (note that $y\not= 0$ here, since $y=0$ would make $P$ be of order~$2$). %So, $P$ is of order~$3$ exactly when So, $P$ being of order~$3$ implies $\bigl( (3x^2 + A)/(2y) \bigr)^2 - 2x = x$, that is: \medskip \hskip 5 cm $-9 x^4 - 6 A x^2 + 12xy^2 - A^2 = 0$. \hfill (2) \medskip \noindent Finally, we use the fact the $(x,y)$ is a point on the curve, so that $y^2 = x^3 + Ax + B$, and so we can replace $y^2$ by $x^3 + Ax + B$ in equations~(1),(2). This makes both equations become the same equation: $3x^4 + 6Ax^2 + 12 B x - A^2 = 0$, as required. \smallskip \par{\bf (c).} By part~(b), the $x$-coordinate of any point of order~3 must be a root of the quartic $3x^4 + 6Ax^2 + 12 B x - A^2$, which has at most~4 roots $x_1,\ldots ,x_4$. Each~$x_i$ gives rise to at most two points $(x_i,y_i), (x_i,-y_i)$ on the curve, giving at most $8$ points of order~$3$. Together with {\bf o}, this gives at most $9$ points that are $3$-torsion. \medskip \end{enumerate} \end{document}