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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?

kelalaka
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  • You say $3P = O$ with $O$ not nessicarily at infinity - then immediately proceed to make a claim about the $x$-coordinates which requires $O$ to be at infinity (and $E$ to have $a_1 = a_3 = 0$).... – Mummy the turkey Feb 19 '21 at 10:14
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    Moreover "if the two intersection points of the curve with the $y$-axis are inflection points, then we will have $6$ points - $6$ what points? You can't have $6$ $k$-rational $3$-torsion points, because $6$ does not divide $9$... – Mummy the turkey Feb 19 '21 at 10:17
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    In any case - the easiest geometric interpretation of the $9$ $3$-torsion points on an elliptic curve (where the given point $O$ is an inflection point e.g., when $E$ is in Weierstrass form) is the inflection points (equivalently the zero dimensional variety given as the intersection between $E$ and the vanishing of its hessian determinant). – Mummy the turkey Feb 19 '21 at 10:20
  • @Mummytheturkey when I talk about identity in the general case, the Edwards curve has an identity, not the point at infinity it is $(0,1)$. Yes, $6 |9$ that is why I was asking where the other 3, but I've figured out that those points exist if the curve is defined one the $\mathcal{C}$. – kelalaka Feb 19 '21 at 11:10
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    Yes, an elliptic curve can have any point taken to be the identity, but the remark you make (that the $x$-coordinates of $P$ and $-P$ are the same) is specific to the case when $E$ is in Weierstrass form (although i did make a mistake saying $a_1 = a_3 = 0$) - unless you are much more general in your definition of '$x$' (i.e., taking it to be the double cover of $\mathbb{P}^1$ induced by the involution) this is not true in general – Mummy the turkey Feb 19 '21 at 11:18
  • cont., no $6$ does *not* divide $9$ - hence it is impossible to have exactly $6$ $k$-rational $3$-torsion points - you may have $1$, $3$, or $9$. – Mummy the turkey Feb 19 '21 at 11:20
  • Yes, that is Lagrange, The question is not answerable in this form? – kelalaka Feb 19 '21 at 11:27
  • It's just the question is not very clear - you're asking for the "other 3" points, but we just established there are 6 more. Are you asking for a classification of $3$-torsion structures over $\mathbb{R}$ – Mummy the turkey Feb 20 '21 at 11:04
  • @Mummytheturkey The more I read, I've seen that there are 8 complex roots of $\psi_3 (x) $ which can have only zero, two, or four real roots and each can have the negative, too. It later says, that need proof for me, there must be exactly one real value of $x$ for which the two corresponding y’s are real. So we can have at most two real-valued points. Let say $P$, $-P$ and $\mathcal{O}$ make the subgroup of the 9 points of order 3. – kelalaka Feb 20 '21 at 11:28
  • @Mummytheturkey Yes, I was looking at the 3-Torsion structure, now it turns that it can have at most order 3, isn't it? – kelalaka Feb 20 '21 at 11:29

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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.

Mummy the turkey
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