11
$\begingroup$

Let $E_d$ be the group of rational points on the elliptic curve $$y^2=(x-d)x(x+d)$$

It is well-known – and easy to verify by elementary algebra – that $(x,y)\in E_d$ belongs to $2E_d$ precisely when all three of $x-d$, $x$, $x+d$ are squares.

This seems too neat to be a mere coincidence.

Is there a natural generalisation of this fact? Or a more conceptual way to understand why it ought to be true?

$\endgroup$
13
$\begingroup$

It looks as if you're working over $\mathbb Q$. And you are using an elliptic curve in which all of the 2-torsion is rational, so $E[2]$ is isomorphic to (say) $\boldsymbol\mu_2^2$ as a Galois module, where $\boldsymbol\mu_2=\{\pm1\}$ is the group of square roots of 1. Okay, now consider the injection (this comes from basic Kummer theory of elliptic curves) \begin{align*} E(\mathbb Q)/2E(\mathbb Q) &\hookrightarrow H^1(G_{\overline{\mathbb Q}/\mathbb Q},E[2]) \\ &\cong H^1(G_{\overline{\mathbb Q}/\mathbb Q},\boldsymbol\mu_2^2) \\ &\cong H^1(G_{\overline{\mathbb Q}/\mathbb Q},\boldsymbol\mu_2)^2 \\ &\cong \mathbb Q^*/(\mathbb Q^*)^2\times \mathbb Q^*/(\mathbb Q^*)^2. \end{align*} (The final isomorphism is standard Kummer theory.) So this at least tells you that whether $P$ is in $2E(\mathbb Q)$ should depend on whether certain values are squares. If you trace through all of the maps, which requires choosing a basis for $E[2]$, you'll find (for one of the choices) that it is given by $$ P \longmapsto \bigl(x(P),x(P)-d\bigr), $$ so $P$ is in $2E(\mathbb Q)$ if and only if both $x(P)$ and $x(P)-d$ are squares. Of course, from the equation of $E$, any two of $x(P)$, $x(P)-d$ and $x(P)+d$ being a square forces the third one to also be square. Finally, if $x(P)=0$ or $x(P)=d$, one finds during the analysis that one needs to use a different formula. For example, since $x$ and $(x-d)(x+d)$ differ multiplicatively by a square, one has $$ 0 \longmapsto (-d^2,-d) = (-1,-d) \in \mathbb Q^*/(\mathbb Q^*)^2\times \mathbb Q^*/(\mathbb Q^*)^2. $$

To answer your second question, if $E[m]\subset E(K)$, then one similarly gets an injection $$ E(K)/mE(K) \hookrightarrow K^*/(K^*)^m\times K^*/(K^*)^m, $$ so (more or less) there are rational functions $f$ and $g$ in $K(E)$ such that for $P\in E(K)$ we have $$ P \in mE(K) \;\Longleftrightarrow\; \text{$f(P)$ and $g(P)$ are $m$'th powers in $K^*$.} $$

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ Note in particular that for $m=2$ this works for all curves of the form $y^2 = x (x-a) (x-b)$, not just the "congruent number" curves $y^2 = x(x-d)(x+d)$. $\endgroup$ – Noam D. Elkies Aug 13 '16 at 5:14
14
$\begingroup$

A geometric view to complement Joe Silverman's cohomological answer:

Suppose more generally that $E$ is the "product-is-square" elliptic curve $$ y^2 = (x-e_1)(x-e_2)(x-e_3) $$ for some pairwise distinct $e_i$, and let $C$ be the "each-is-square" curve $$ \begin{cases} y_1^2 = x-e_1, \cr y_2^2 = x-e_2, \cr y_3^2 = x-e_3, \end{cases} $$ Then $E$ is a double cover of the $x$-line, and $C$ is a $\{\pm1\}^3$ cover with $E$ as an intermediate cover with Galois group isomorphic with $({\bf Z} / 2{\bf Z})^2$; explicitly the $4:1$ map $C \to E$ is $$(x,y_1,y_2,y_3) \mapsto (x, y_1 y_2 y_3).$$ By Riemann-Hurwitz, $E$, like $C$, has genus $1$, so this $({\bf Z} / 2{\bf Z})^2$ cover $C \to E$ is unramified. Hence if we choose for the origin of $C$ one of the preimages of the origin of $E$, the cover must be the doubling map.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Isn't it simpler just to show the multiplication by 2 map? $\endgroup$ – joro Aug 13 '16 at 5:37
  • 3
    $\begingroup$ Yes, but the OP already knows that calculation and asked for a more conceptual explanation. $\endgroup$ – Noam D. Elkies Aug 13 '16 at 5:40
  • $\begingroup$ I don't understand: you write that the cover "is" the doubling map, but for what identification of $C$ with $E$? $\endgroup$ – RP_ Aug 13 '16 at 15:54
  • 1
    $\begingroup$ The "magical" ingredient here is that there's only one unramified $({\bf Z} / 2{\bf Z})^2$ cover of $E$ (because $\pi_1 \cong {\bf Z}^2$ has a unique quotient isomorphic with $({\bf Z} / 2{\bf Z})^2$), so $C$ is automatically identified with $E$ once we choose an origin of the group law. $\endgroup$ – Noam D. Elkies Aug 13 '16 at 16:31
  • 2
    $\begingroup$ It is perhaps worth pointing out that, in order for your whole argument to work, it is very important that $C(\mathbb{Q}) \neq \emptyset$ (I mean, $E$ has many $(\mathbb{Z}/2\mathbb{Z})^2$-covers without rational points). $\endgroup$ – Daniel Loughran Aug 14 '16 at 13:35

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.