Elliptic curves in projective form question Let
$K$
be any field with Char
$K
\neq 2, 3$,
 and let
$\varepsilon
:
F
(
X_0
;X_1
;X_2
) =
X_1^2
X_2-
(
X_0^3
+AX_0
X_2^2
+
BX_2^3
)$
;
with
$A, B
\in
K$,
be an elliptic curve. Let
$P$
be a point on
$\varepsilon$.
(a).
Show that $3P = \underline{o}$, where $\underline{o}$ is the point at infinity ($(0,1,0)$) if and only if the tangent line to
$\varepsilon$
at
$P$
intersects
$\varepsilon$
only at
$P$
(b).
Show that if $3P
=
\underline{o}$
then the 3 x
3 matrix
$( \frac{\partial ^2 F}{\partial X_i \partial X_j}$)
has determinant $0$.
[This matrix is called the Hessian matrix].
(c).
Show that there are at most nine 3-torsion points over
$K$
I'm having trouble getting to grips with the projection notation - any help greatly appreciated!

(a) Note that $3P=O$ iff $2P=-P$. To compute $2P$ you have to intersect the tangent line $t$ in $P$ with $\varepsilon$. The line $t$ will meet $\varepsilon$ in two points, say $P$ and $Q$, because it already meet $\varepsilon$ at $P$ with multiplicity $2$. In any case, we know that $2P=-Q$. Therefore, if $2P=-P$ then it means that $Q=P$ and so $t$ meets $\varepsilon$ only at $P$, and conversely if $t$ meets $\varepsilon$ only at $P$ then $Q=P$ and $2P=-P$.
(c) The entries of the Hessian matrix $H$ are linear polynomials, because you're taking second derivatives of an homogeneous polynomial of degree $3$. By point (b), a necessary condition for $P$ to be a $3$-torsion point is that $(\det H)(P)=0$. Now, $\det(H)$ is an homogeneous polynomial of degree $3$, so a necessary condition for $P$ to be a torsion point is that it is a zero of two homogeneous polynomials of degree $3$: one is $\det (H)$ and the other is $F$. So you're looking at the intersection points of two cubics. By Bezout's theorem, there are at most $9$ such points provided that the two cubics don't have a component in common. But since $F$ is an irreducible curve, this can happen only if $\det (H)$ and $\varepsilon$ are the same curve. This cannot happen, as you can check that $(1\colon 0\colon 0)\notin \varepsilon$ while it belongs to the cubic defined by $\det (H)=0$.
