# 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!

• I guess that you wanted to write $F(X_0,X_1,X_2)=X_1^2X_2-(X_0^3+AX_0X_2^2+BX_2^2)$, no? Ohterwise you have a reducible curve... Jan 27, 2014 at 12:17
• yes, thank you! Jan 27, 2014 at 12:52

(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 $\det (H)=0$.