\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}} \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 (a).} Case 1. $m$ is odd. The birational transformation $(X,Y) \mapsto (X/f_m , Y/f_m^{(m+1)/2})$ [with inverse transformation: $(X,Y) \mapsto (Xf_m , Yf_m^{(m+1)/2})$] maps the given curve to: $f_m^{(m+1)} Y^2 = f_m^{(m+1)} X^m + \ldots + f_0$, which is the same as: $Y^2 = X^m + c_{m-1} X^{m-1} + \ldots + c_0$, where $c_i = f_i/f_m^{m+1-i} \in \bbQ$ [N.B. our birational transformation is legitimate, since $(m+1)/2$ is an integer]. So, our new right hand side is monic and is still defined over~$\bbQ$. Now, write each $c_i = n_i/d_i$, where each $n_i,d_i\in \bbZ$, with each $n_i,d_i$ coprime. Let $d$ be the least common multiple of $d_0,\ldots d_{(m-1)}$; that is to say, $d$ is the smallest integer divisible by all of the denominators of the coefficients~$c_i$. Now, the birational transformation $(X,Y) \mapsto (Xd^2 , Yd^m)$ [with inverse transformation: $(X,Y) \mapsto (X/d^2 , Y/d^m)$] maps to the curve: $Y^2/d^{2m} = X^m/d^{2m} + c_{m-1} X^{m-1}/d^{2m-2} \ldots + c_0$, which is the same as: $Y^2 = X^m + g_{m-1} X^{m-1} + \ldots + g_0$, where $g_i = c_i d^{2(m-i)}$. But now, each $d_i | d$ and so each $c_i d \in \bbZ$, giving that each $g_i \in \bbZ$, as required. \par Case 2. $f_m$ is a square. Let $w$ be such that $f_m = w^2$. The birational transformation $(X,Y) \mapsto (X , Y/w)$ [with inverse transformation: $(X,Y) \mapsto (X, Yw)$] maps the given curve to: $f_m Y^2 = f_m X^m + \ldots + f_0$, which is the same as: $Y^2 = X^m + c_{m-1} X^{m-1} + \ldots + c_0$, where $c_i = f_i/f_m \in \bbQ$. We are again in the situation where our right hand side is monic and is still defined over~$\bbQ$. So now apply the same second birational transformation as in Case~1. \par\noindent {\bf (b).} The birational transformation $(X,Y) \mapsto (5X , 5^2Y )$ [with inverse $(X,Y) \mapsto ((1/5)X , (1/5^2)Y )$] maps the given curve to the curve: $Y^2 = X^3 + 75 X^2 + 625$. The birational transformation $(X,Y) \mapsto (X + 25 , Y )$ [with inverse $(X,Y) \mapsto (X - 25 , Y )$] maps this to the curve: $Y^2 = X^3 - 1875 X + 31875$, as required. \medskip \item \par\noindent To compute the group inverse, we need to find the tangent line through $(1,-1,0)$. Using Lemma 1.40 in lecture notes as suggested, we compute that this tangent is $3X+3Y=0$,i.e.~$Y = -X$. Substituting this into the equation of the curve gives $aZ^3 = 0$. The only point with $Z=0$ and $Y=-X$ is $(1,-1,0)$, so the tangent intersects the curve at $(1,-1,0)$ with multiplicity three. Moreover, the inverse of a point $(x,y,z)$ is given by the third point of intersection on the line joining $(x,y,z)$ to $(1,-1,0)$. This line has equation $z(X+Y) - (x+y)Z = 0$. We can check (or deduce from the symmetry in $X$ and $Y$) that the third point of intersection with the curve is $(y,x,z)$. To compute the doubling map, we first note that the tangent line through $(x,y,z)$ has equation \[x^2 X + y^2 Y + az^2 Z =0.\] We need to compute the third point of intersection of this line with the curve $X^3+Y^3+aZ^3=0$. Substituting $Z = \frac{-x^2X - y^2 Y}{az^2}$ gives \[X^3(1-\frac{x^6}{a^2z^6}) - X^2Y\frac{3x^4y^2}{a^2z^6} + ... = 0.\] This polynomial factorises as \[(1-\frac{x^6}{a^2z^6})(X-(x/y)Y)^2(X-(x'/y')Y)\] and we need to compute $x'/y'$. We get $2(x/y) + (x'/y') = \frac{3x^4y^2}{a^2z^6-x^6} = \frac{3x^4y^2}{(az^3+x^3)(az^3-x^3)} = \frac{3x^4}{y(x^3-az^3)}.$ Rearranging gives \[(x'/y') = \frac{x(x^3+2az^3)}{y(x^3-az^3)} = \frac{x(az^3-y^3)}{y(x^3-az^3)}.\] We deduce from this that the third point of intersection is $(x',y',z')$ with \[(z'/y') = \frac{-x^2x'/y' - y^2}{az^2} = \frac{z(y^3-x^3)}{y(x^3-az^3)}.\] Rewriting the projective point by rescaling gives \[(x',y',z') = (x(az^3-y^3),y(x^3-az^3),z(y^3-x^3)).\] We have $[2](x,y,z) = -(x',y',z') = (y',x',z')$ which gives \[[2](x,y,z) = (y(x^3-az^3),x(az^3-y^3),z(y^3-x^3))\] as required. \emph{It is probably easier to establish the formula for points $(x,y,1)$ and then homogenize to get the general formula.} \item \par\noindent {\bf (a).} Since $p \equiv 2$~(mod~$3$), we have $\hbox{gcd}(3,p-1) = 1$, and so there exist $\lambda, \mu \in \bbZ$ such that $3\lambda + (p-1)\mu = 1$. For any $x,y\in\bbF_p^*$, we have $x^{p-1} = y^{p-1} = 1$ [by Fermat's Little Theorem], and so: $$ x^3 = y^3 \Rightarrow \bigl( x^3 \bigr)^\lambda \cdot 1^\mu = \bigl( y^3 \bigr)^\lambda \cdot 1^\mu \Rightarrow \bigl( x^3 \bigr)^\lambda \cdot \bigl(x^{p-1}\bigr)^\mu = \bigl( y^3 \bigr)^\lambda \cdot \bigl(y^{p-1}\bigr)^\mu $$ $$ \ \ \ \ \ \Rightarrow x^{ 3\lambda + (p-1)\mu } = y^{ 3\lambda + (p-1)\mu } \Rightarrow x = y. $$ Thus the map $x \mapsto x^3$ is injective on $\bbF_p^*$ and so is a bijection. Since $0^3 = 0$, the map is also a bijection on $\bbF_p$. Now, rearrange the equation of the curve as: $X^3 = Y^2 - A$. We see that, for each $Y = 0,\ldots, p-1$ in $\bbF_p$, there is precisely one $X \in \bbF_p$ such that $X^3 = Y^2 - A$, giving rise to a total of precisely~$p$ affine points, which becomes $p+1$ points when we include the point at infinity. \smallskip \par\noindent {\bf (b).} Since $p\equiv 3$~(mod~$4$), we have that $\bigl( {-1 \over p} \bigr) = -1$ [i.e.\ $-1$ is not congruent to a square mod~$p$]. First note that $X=0$ gives rise to precisely one point~$(0,0)$. Now, pair each $X\in\bbF_p^*$ with its negative $-X$ mod~$p$. If $X(X^2 + B) = 0$ then the pair $X,-X$ gives rise to precisely two points: $(X,0),(-X,0)$. If $X(X^2 + B) \not= 0$ then one of $X(X^2 + B)$ and $-X(X^2 + B)$ will be a nonzero quadratic residue mod~$p$ and the other will be a quadratic nonresidue mod~$p$ [using the fact that $-1$ is a quadratic nonresidue mod~$p$ and the standard property of Legendre symbols that: $\bigl( {mn \over p} \bigr) \bigl( {m \over p} \bigr) \bigl( {n \over p} \bigr)$]. Whichever one is a nonresidue will contribute no points; which one is a residue will contribute two points: $(X,Y),(X,-Y)$. In all cases, each pair $X,-X$ will contribute two points. This gives rise to a total of precisely~$p$ affine points, which becomes $p+1$ points when we include the point at infinity. \medskip \item \par\noindent {\bf (a).} $(2,0) + (-(2,0)) = {\bf o}$, where $-(2,0)$ is the inverse of $(2,0)$; but we know that $-(2,0) = (2,-0) = (2,0)$ and so $(2,0) + (2,0) = {\bf o}$; that is, $(2,0)$ is of order~$2$, as required. \par\noindent {\bf (b).} In general, a point $(x,y)$ on $Y^2 = X^3 + f_2X^2 + f_1X + f_0$ has inverse: $(x,-y)$ and so $(x,y)$ is of order~2 exactly when $(x,y) + (x,y) = {\bf o} \iff (x,y) = (x,-y)$. But $y=-y$ precisely when $y=0$. So, we see that all points of order~2 are of the form~$(x,0)$ where~$x$ is a root of the cubic $X^3 + f_2X^2 + f_1X + f_0$. There are at most~3 such points over any field, since a cubic has at most 3~roots in any field. In the complex numbers, of course, there will always be exactly~3 such points. Applying this to the curve $Y^2 = X(X^2-3)$, we see that $X(X^2 - 3)$ has one $\bbQ$-rational root $X=0$, giving $(0,0)$ as the only $\bbQ$-rational point of order~2; the other two roots give the points $(\sqrt{3}, 0)$ and $(-\sqrt{3},0)$, which are the remaining two $\bbC$-rational points of order~2. The curve $Y^2 = X^3 - 7$ has no $\bbQ$-rational point of order~2 (since $X^3 - 7$ has no $\bbQ$-rational roots); over~$\bbC$, there are the 3~points of order~2 given by: $(\omega_1,0),(\omega_2,0),(\omega_3,0)$, where $\omega_k = 7^{1/3}e^{2\pi i k /3}$, for $k=0,1,2$. Finally, the curve $Y^2 = X(X-1)(X-7)$ has the points of order~2: $(0,0),(1,0),(7,0)$ all of which are both $\bbC$-rational and $\bbQ$-rational. Group structures of the $2$-torsion groups over~$\bbQ$ for the three curves are clearly: $C_2, C_1$ and $C_2 \times C_2$, respectively (since all members except~{\bf o} have order~2). \medskip \item Using $2YY' = 3X^2$, we see that the slope of the curve at~$(0,2)$ is $3\cdot 0^2 / (2\cdot 2) = 0$, and so the tangent to the curve at $(0,2)$ is the line: $Y=2$. The $x$-coordinate of the third point of intersection is then: $0^2 - 0 - 0 = 0$, with corresponding $y$-coordinate~$2$. In summary: The line $Y=2$ intersects the curve at $(0,2)$ with multiplicity~3. Hence: $(0,2) + (0,2) + (0,2) = {\bf o}$, and so: $3(0,2) = {\bf o}$. But, $1(0,2) = (0,2)$ and $2(0,2) = (0,-2)$, and so~$3$ is the smallest $k \ge 1$ such that $k(0,2) = {\bf o}$; that is, $(0,2)$ has order~$3$. \medskip %\item %\par\noindent {\bf (a).} As usual, $-(\alpha , 0) = (\alpha , 0)$, %and so $(\alpha , 0) + (\alpha , 0) = {\bf o}$; that is, $(\alpha , 0)$ %is a point of order~2, as required. %We now compute $(x',y') %= (x,y) + (\alpha , 0)$. First, the line through $(x,y)$ and %$(\alpha , 0)$ is: $Y = (y/(x-\alpha )) (X-\alpha)$. Substituting %this into the curve gives: $X^3 - (m^2 + \alpha - a) X^2 + \ldots = 0$, %where $m = y/(x-\alpha )$. The sum of the roots of this cubic %is: $m^2 + \alpha - a$, and so the $x$-coordinate of the %third point of intersection is given by: %$x' = m^2 + \alpha - a - x - \alpha = m^2 - x - a %= y^2/(x-\alpha)^2 - x - a = (\alpha x + a\alpha + b)/(x-\alpha)$. %So, we have: %$$ T : x \mapsto {\alpha x + a\alpha + b\over x-\alpha}.$$ %\noindent We now check $T^2(x) = T(T(x))$, which is: %$$ T : x \mapsto %{\alpha (\alpha x + a\alpha + b)/(x-\alpha) + a\alpha + b %\over (\alpha x + a\alpha + b)/(x-\alpha)-\alpha}.$$ %\noindent On multiplying numerator and denominator by %$x-\alpha$ and simplifying, we get: %$(\alpha^2 + a\alpha + b)x/(\alpha^2 + a\alpha + b)x$, %which is~$x$, as required [Note that $\alpha^2 + a\alpha + b \not= 0$, %since the cubic $(X-\alpha)(X^2 + aX + b)$ has distinct roots]. %\par\noindent {\bf (b).} For~$Y^2 = (X-\alpha_1)(X-\alpha_2) %(X-\alpha_3)$, the formula for~$T_1$ is the same as for~$T$ %in part~(a), but with $\alpha,a,b$ replaced by $\alpha_1, %-\alpha_2 - \alpha_3, \alpha_2\alpha_3$, respectively. That is: %$$ T_1 : x \mapsto {\alpha_1 x - \alpha_1\alpha_2 - \alpha_1\alpha_3 %+ \alpha_2\alpha_3 \over x - \alpha_1}.$$ %\noindent The formula for $T_2$ is the same as for~$T_1$, but %with $\alpha_1,\alpha_2,\alpha_3$ replaced by $\alpha_2,\alpha_3,\alpha_1$, %respectively. The formula for $T_3$ is the same as for~$T_2$, but (again) %with $\alpha_1,\alpha_2,\alpha_3$ replaced by $\alpha_2,\alpha_3,\alpha_1$, %respectively. These are then: %$$ T_2 : x \mapsto {\alpha_2 x - \alpha_2\alpha_3 - \alpha_2\alpha_1 %+ \alpha_3\alpha_1 \over x - \alpha_2}.$$ %$$ T_3 : x \mapsto {\alpha_3 x - \alpha_3\alpha_1 - \alpha_3\alpha_2 %+ \alpha_1\alpha_2 \over x - \alpha_3}.$$ %On computing $T_1 T_2 (x) = T_1 (T_2 (x))$, and %simplifying (by removing a common factor of $\alpha_1 - \alpha_2$ %from the numerator and denominator), we get the formula for~$T_3$; %that is: $T_1 T_2 = T_3$. The formula for $T_3$ is invariant %under $\alpha_1 \leftrightarrow \alpha_2$ and so $T_2 T_1 = T_3$, also. %Hence $T_1 T_2 = T_3 = T_2 T_1$, and so $T_1$ and $T_2$ commute. %By symmetry, $T_1$ and $T_3$ commute, as do $T_2$ and $T_3$. %Furthermore, $T_1 T_2 T_3 = T_3^2$ (since $T_1T_2 = T_3$), and we know %from part~(a) that $T_3^2 : x\mapsto x$; hence $T_1 T_2 T_3 : x\mapsto x$, %as required. %\par Finally, the fixed points of $T_1$ are the solutions %to $T_1(x) = x$, that is: %$$ %\alpha_1 x - \alpha_1\alpha_2 - \alpha_1\alpha_3 %+ \alpha_2\alpha_3 = (x - \alpha_1) x,$$ %which has solutions: %$$ x_1 = \alpha_1 + \sqrt{(\alpha_1 - \alpha_2)(\alpha_1 - \alpha_3)}, %\ \ %x_2 = \alpha_1 - \sqrt{(\alpha_1 - \alpha_2)(\alpha_1 - \alpha_3)}. %$$ %Computing $T_2(x_1)$ and $T_2(x_2)$ gives $x_2,x_1$, respectively. %\medskip \item {\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}