# Rational points of order 2 on elliptic curves

I'm working on a question that's asking me to list all the $\mathbb Q$-rational points of order 2 and all the $\mathbb C$-rational points of order 2 for some elliptic curves.

I've made the following observation but can't be sure whether I'm right.

Is it true that I'm just looking for points on the curve where it's tangent is vertical?

Suppose point $P$ lies on elliptic curve $C$. Then $2P=o$, where $o$ is the point at infinity. However, how do we know that this tangent will not intersect another point $P'$ on $C$?

• Assuming you've embedded your elliptic curve in the usual way, if the tangent is vertical then that means it's order is at least two when mapping down to $\mathbb{P}^1$ via the natural projection. However this projection is a degree two map, so no other point maps to the same point. If another point on the elliptic curve intersected the vertical tangent, it would also map to that point. – user16544 Jan 22 '14 at 19:05
• Do you mean that the order is at most two? – Haikal Yeo Jan 22 '14 at 19:09
• Well, the order is at most two since the map is degree two. The order is at least two since if you pull back dx at a point with a vertical tangent, it will vanish since you're at a point where y=0 on the curve $y^2=x(x-1)(x-\lambda)$ (or $y^2=$ however you're representing your elliptic curve). – user16544 Jan 22 '14 at 19:34
• Sorry I realized I am overloading the term "order" here, and you are probably using it to mean the order of P in the group law. To use a different/better term, I mean we have the projection map $\pi:C\rightarrow \mathbb{P}^1$ and by order I meant the ramification index of $\pi$ at $P$ (p. 299 of Hartshorne). – user16544 Jan 22 '14 at 19:49

When looking for points of order at most $2$, that is $2P=\mathcal{O}$, it's simpler to look at the equivalent condition that $P=-P$. We assume that $P\neq \mathcal{O}$ because that point is obvious. With $\mathcal{O}$ being the additive group identity and being the point at infinity, the condition $P=-P$ means we are looking for the points such that $-(x,y)=(x,-y)$. Of course, these are the points where the $y$-coordinate is $0$.
So if our elliptic curve has $3$ real roots, these are the points $(r_1,0),(r_2,0)$, and $(r_3,0)$, where $r_1,r_2,r_3$ are the roots of the polynomial $f(x)=y^2$. Moreover, these form either the Klein-$4$ group or a cyclic group of order $2$, depending on if all the roots are real or not.
As for your last question on will the line intersect the curve at an addition point, think of the geometric representation of addition on an elliptic curve. A tangent line is a polynomial of degree one while the elliptic curve is formed using a polynomial of degree $3$. The fact that the tangent line intersects the elliptic curve vertically means it intersects the curve there with what multiplicity? It must intersect the curve at least once more at $\mathcal{O}$. This gives $3$ intersection points, could there be a $4$th? What would it mean if there were a $4$th?
This is in some sense why we add the point at infinity so things 'work the way we want them too.' Meaning, that if we have a curve of degree $n$ intersecting the elliptic curve with degree $3$, they will intersect at $3n$ points counting multiplicity (of course one or all of these points may be $\mathcal{O}$). This is a case of a more general theorem (I forget at the moment but if I recall Bezout's Theorem, or at least a generalized version of it) that two curves of degree $m$ and $n$ in the projective plane meet at $mn$ points.
• After all these years, looking back at this answer, I should have mentioned that I make use of the fact that (over $\mathbb{Q}$), we first write the elliptic curve as $y^2= x^3 + Ax + B$ for some $A,B \in \mathbb{Q}$. Otherwise, not everything I have said will be true for an elliptic curve written in its more general Weierstrass form. – mathematics2x2life May 24 at 3:12