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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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    The determinant of the Hessian is a cubic polynomial. $F$ is also a cubic polynomial. So...?2011-06-10
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    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.2011-06-10
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    They're homogeneous equations in $3$ variables, not in $2$ variables.2011-06-10
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    what Qiaochu is getting at is known as Bezout's theorem. Ever heard of it?2011-06-15

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