Geometric interpretation of the 9 order 3 points of Elliptic curves

elliptic-curves

Silverman in their book Rational Points on Elliptic
Curves have a theorem;

Theorem 2.1: Let $C$ be a non-singular cubic curve
$$C : y^2 = f (x) = x^3 + ax + bx + c.$$

  • (c) A point $P = (x, y) \neq \mathcal{O}$ on $C$ has order three if and only if $x$ is a root of the polynomial
    $$\psi_3 (x) = 3x^4 + 4ax^3 + 6bx^2 + 12c^x + 4ac − b^2.$$

  • (d) The curve $C$ has exactly nine points of order dividing three. These nine points form a group that is a product of two cyclic groups of order three.

We say that a point $P$ has order 3 if $[3]P = \mathcal{O}$, where $\mathcal{O}$ is the identity element, not necessarily the point at infinity. Then we have $[2]P = -P$ and looking at the $x$ coordinates $$x([2]P) = x(-P)=x(P)$$ then we can conclude that $x([2]P)=x(P)$ if $P \neq \mathcal{O}$

When we consider a curve like $y^2 +y = x^3$

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If the two intersection points of the curve with the $y$-axis are inflection points then we will have the 6 points.

Where are all nine points on the curve geometrically?

Best Answer

Based on the clarification in the comments I am going to prove the following result:

Lemma Let $E/\mathbb{R}$ be an elliptic curve. Further suppose that $E$ is given in simplified Weierstrass form $$E : y^2 = x^3 + ax + b$$ with $a,b \in \mathbb{R}$, and $4a^3 + 27b^2 \neq 0$. Then we have that $E(\mathbb{R})[3]$ is either trivial or $ \mathbb{Z}/3\mathbb{Z}$

Proof: The $x$-coordinates of the $3$-torsion points of $E$ are given by the roots of the $3$-division polynomial $$\psi_3(x) = 3x^4 + 6ax^2 + 12bx - a^2$$

Now consider the splitting field $K$ of $\psi_3$. Since we are working over $\mathbb{R}$, we have that either $K = \mathbb{R}$ or $\mathbb{C}$.

Now by general field theory $K$ contains $\mathbb{R}(\sqrt{\delta})$ where $\delta$ is the discriminant of the polynomial $\psi_3(x)$. But we can just compute that $$\delta = -2^8 3^3 (4a^3 + 27b^2)^2$$

But we may remove square factors, hence $K$ contains $\mathbb{R}(\sqrt{-3}) = \mathbb{C}$. In particular $\psi_3$ cannot split over $\mathbb{R}$ (indeed, an analysis of the subgroup lattice of $S_4$ shows that $\psi_3$ can have at most one root by Galois theory over the function field $\mathbb{R}(a,b)$).

But even without this remark we have proved that $E(\mathbb{R})[3]$ does not have order $9$ (since at least one of the $x$-coordinates cannot be contained in $\mathbb{R}$). It is easy to come up with examples of curves with both of the other cases.

This is the low key way of proving that $K(E[N])$ contains the $N^{th}$ roots of unity, which can in general be seen from the Weil pairing.

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