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\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}


