I'm having trouble with exercise 3.10 from Silverman's Arithmetic of Elliptic Curves. I include part (a) for context and part (d) because that's the part I'm stuck on. All fields in this section are algebraically closed.

Let $E/K$ be an elliptic curve with Weierstrass coordinate functions $x$ and $y$.

(a) Show that the map $$ \phi: E \to \mathbb{P}^{3} $$ defined by $$ \phi = [1 : x : y : x^{2}] $$ maps $E$ isomorphically onto the image of two quadrics in $\mathbb{P}^{3}$.

(d) Let $P \in E$. Prove that $[4]P = \mathcal{O}$ if and only if there exists a hyperplane $H \subset \mathbb{P}^{3}$ such that $\phi(E) \cap H = \{ P \}$. If char($K$) $\neq 2$, prove that there are exactly $16$ such hyperplanes, and hence that $\#E[4] = 16$.

I assume char$K$ $\neq 2, 3$ for notational convenience. Thus $E/K$ is isomorphic to a plane curve given by equation $$ y^{2} = x^{3} + ax + b $$ where $4a^{3} - 27b^{2} \neq 0$. The quadrics described in part (a) should then be given by $$ Q_{1}: z_{1} = z_{3}z_{0} $$ $$ Q_{2}: z_{2}^{2} = z_{3}z_{1} + az_{1}z_{0} + bz_{0}^{2} $$ where the $z_{i}$ are homogeneous coordinates on $\mathbb{P}^{3}$. The first part of (d) follows immediately from part (c). I'm having trouble with the hyperplanes, here is my latest attempt: We first note that there are $4$ immediate $4$-torsion points of $E$ given by $\mathcal{O}$ and the $3$ points of order $2$ corresponding to the roots of the equation $x^{3} + ax + b$. This gives us $4$ hyperplanes with the required intersection property. It can be checked that $[0 : 0 : 0 : 1] = \phi(\mathcal{O})$ and that this is the only point on both quadrics where $z_{0} = 0$, so we can assume $z_{0} \neq 0$. We can also assume $z_{2} \neq 0$ since these points are the images of the points of order $2$.

Now consider a general hyperplane: $$ H: c_{0}z_{0} + c_{1}z_{1} + c_{2}z_{2} + c_{3}z_{3}. $$ If $c_{3} = 0$ then $H$ passes through $\phi(\mathcal{O})$ and so if it intersects $\phi(E)$ once could only be one of the hyperplanes we have already counted, so we can assume $c_{3} \neq 0$ and since the hyperlanes are parameterised by $\mathbb{P}^{3}$, we can assume $c_{3} = 1$ . Thus setting $u = \frac{z_{1}}{z_{0}}$, $v = \frac{z_{2}}{z_{0}}$, $w = \frac{z_{3}}{z_{0}}$, we reduce to $$ Q_{1}: u^{2} = w $$ $$ Q_{2}: v^{2} = wu + au + b $$ $$ H: c_{0} + c_{1}u + c_{2}v + w = 0. $$ The first equation clearly makes $w$ redundant so we can reduce to two equations: $$ Q: v^{2} = u^{3} + au + b $$ $$ H: c_{0} + c_{1}u + c_{2}v + u^{2} = 0 $$ so finally I've reduced to the intersection of an elliptic curve and a quadratic. When we do the necessary substitutions we get a quartic in $u$, which we would need to have a repeated root. I've attempted various ways using the coefficients to obtain equations in the $c_{i}$ ( formally differentiating multiple times to obtain the root and then comparing coefficients with binomial expansion etc) but each has left me with a confusing mess, so I can't help thinking I've gone wrong somewhere above. Also I never used the $z_{2} \neq 2$ condition. Any help would be greatly appreciated!

Edit: The purpose of this question is to prove in the special case $m = 4$ that deg$[m] = m^{2}$, so ideally I'd like answers that don't quote this result.


For part (d) just note that any two hyperplane sections of the curve are linearly equivalent. In particualar, if $H$ and $H'$ are such that $\phi(E) \cap H = 4P$ and $\phi(E) \cap H' = 4P'$ then $4P \sim 4P'$. Since there is a hyperplane $H_0$ (given by $z_0 = 0$) such that $\phi(E) = 4P_0$, where $P_0$ is the origin of the curve, it follows that any for any $H$ as above $4P \sim 4P_0$.

Conversely, if $4p \sim 4P_0$ then there is a hyperplane $H$ such that $\phi(E) \cap H = 4P$ just because any divisor linearly equivalent to $4P_0$ is a hyperplane section of $E$ (since $\phi$ is given by a COMPLETE linear system).

  • $\begingroup$ Thanks for the response, how would you go about showing that there are precisely 16 hyperplanes that intersect at exactly one point? $\endgroup$ – Rob Rockwood May 12 '18 at 9:22
  • $\begingroup$ Just because each 4-torsion point creates such a hyperplane. $\endgroup$ – Sasha May 12 '18 at 10:00
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    $\begingroup$ and how do you know there are 16 4-torsion points? (I think the general thrust of this question is towards proving that deg[4] = 16 rather than quoting it, in which case the result does indeed follow trivially). $\endgroup$ – Rob Rockwood May 12 '18 at 11:33

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