Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If we take our elliptic curve over $K$ to be the zero set of $$ F(X_1, X_2, X_3) = X_2^2 X_3 - (X_1^3 + AX_1X_3^2 + BX_3^2), $$ which is in projective form with $X = X_1, Y = X_2, Z=X_3$, then I have been able to show that for any point $P$ on the curve, if $3P = \mathbf{o}$ then the Hessian matrix $$ \bigg(\frac{\partial F}{\partial X_i \partial X_j}\bigg) $$ has determinant $0$ at $P$.

I am then asked on this exercise to show that there are at most nine 3-torsion points over $K$. Is this an obvious deduction? I am afraid I cannot see how to do it.

share|cite|improve this question
The determinant of the Hessian is a cubic polynomial. $F$ is also a cubic polynomial. So...? – Qiaochu Yuan Jun 10 '11 at 20:27
I must be missing something. Surely this means they will have 3 solutions each, if $K$ is complete...? Even then I'm not sure how to verify these solutions will simultaneously solve both. – Fahad Sperinck Jun 10 '11 at 21:54
They're homogeneous equations in $3$ variables, not in $2$ variables. – Qiaochu Yuan Jun 10 '11 at 22:49
what Qiaochu is getting at is known as Bezout's theorem. Ever heard of it? – Jyrki Lahtonen Jun 15 '11 at 18:29

To prove this I'm going to appeal to some results in algebraic geometry. We have that $P$ is a simple point on $F$ if and only if $\mathcal{O}_P(F)$, the local ring of $F$ at $P$, is a discrete valuation ring (see Fulton theorem 1 section 3.2. A simple point $P$ on $F$ is called an ordinary flex if $ord_p^F(L) = 3$ where $L$ is the tangent line to $F$ at $P$.

By a theorem in section 5.3 (again see Fulton), $I(P, H \cap F) = 1$ if and only if $P$ is an ordinary flex. Since an elliptic curve is nonsingular, all of its points are simple. Since we are looking for the number of $3$-torsion points, these are exactly the points that are ordinary flexes. By Bezout's theorem we have that, since the Hessian is a cubic polynomial and elliptic curves have degree $3$, $$ \sum_P I(P, H \cap F) = 9 $$ so $F$ has at most nine ordinary flexes.

share|cite|improve this answer
Would the downvoter care to comment? – robjohn Jun 13 '12 at 15:03
Downvoter, any feedback? – Pedro Tamaroff Jun 15 '12 at 22:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.