3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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