If we take our elliptic curve over $K$ to be the zero set of $$ F(X_1, X_2, X_3) = X_2^2 X_3 - (X_1^3 + AX_1X_3^2 + BX_3^3), $$ 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.

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    $\begingroup$ The determinant of the Hessian is a cubic polynomial. $F$ is also a cubic polynomial. So...? $\endgroup$ – Qiaochu Yuan Jun 10 '11 at 20:27
  • $\begingroup$ 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. $\endgroup$ – Sputnik Jun 10 '11 at 21:54
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    $\begingroup$ They're homogeneous equations in $3$ variables, not in $2$ variables. $\endgroup$ – Qiaochu Yuan Jun 10 '11 at 22:49
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    $\begingroup$ what Qiaochu is getting at is known as Bezout's theorem. Ever heard of it? $\endgroup$ – 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.

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    $\begingroup$ Would the downvoter care to comment? $\endgroup$ – robjohn Jun 13 '12 at 15:03
  • $\begingroup$ Downvoter, any feedback? $\endgroup$ – Pedro Tamaroff Jun 15 '12 at 22:17

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