Using Weierstrass' $\wp$ function it can be proved that the group of complex points on an elliptic curve $E /\mathbb{C}: y^2 = x^3 + ax + b$ satisfies $E(\mathbb{C}) \cong \mathbb{R}/\mathbb{Z} \oplus \mathbb{R}/\mathbb{Z}$.

Now, suppose that $E$ is defined over $\mathbb{R}$. Is there a way to prove directly from the isomorphism above that the group of real points on $E$ is

$$ E(\mathbb{R}) \cong \begin{cases} \mathbb{R}/\mathbb{Z} \quad \text{if $x^3 + ax +b$ has one real root,}\\ \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{R}/\mathbb{Z} \quad \text{if $x^3 + ax +b$ has three real roots} \end{cases} $$

I have seen some arguments that use results from the theory of Lie groups to prove this, but I was wondering if it can be proved from the "torus" description of $E(\mathbb{C})$.

Thanks a lot for any help.


1 Answer 1


There is a direct way of proving it (actually showing even more things) using uniformization theorem. I will just list the facts that have to be proven, if you don't already know them, none of them is really difficult.

Let $E\colon y^2=x^3+Ax+B$ and suppose that $\Lambda\subseteq \mathbb C$ is the lattice which parametrizes $E$, namely such that $A=g_2(\Lambda)$ and $B=g_3(\Lambda)$ where as usual $g_2(\Lambda)=60\sum_{0\neq\omega\in \Lambda}\omega^{-4}$ and $g_3(\Lambda)=140\sum_{0\neq\omega\in \Lambda}\omega^{-6}$. Since $E$ is defined over $\mathbb R$, then $g_2(\Lambda)=\overline{g_2(\Lambda)}$ and $g_3(\Lambda)=\overline{g_3(\Lambda)}$. But one checks that $\overline{g_i(\Lambda)}=g_i(\overline{\Lambda})$ for $i=2,3$.

Fact 1): if $\Lambda,\Lambda'$ are lattices in $\mathbb C$ with $g_i(\Lambda)=g_i(\Lambda')$ for $i=2,3$ then $\Lambda=\Lambda'$. This implies in our case that $\Lambda=\overline{\Lambda}$.

Fact 2): $\Lambda$ has a basis $\{\alpha,\beta\}$ such that $\alpha\in\mathbb R_{>0}$ and either $\beta=iy$ or $\beta=\alpha/2+iy$ for real $y>0$. The argument is the following: let $\{\omega_1,\omega_2\}$ be a basis of $\Lambda$ and set $\alpha=\min\{|\omega|\colon\omega\in\Lambda\cap (\mathbb R\setminus\{0\})\}$. Then $\alpha=a\omega_1+b\omega_2$ for some $a,b\in\mathbb Z$. It must be $(a,b)=1$ by the minimality of $\alpha$ and therefore there exist $c,d\in\mathbb Z$ s.t. $ac-bd=1$. Now set $\beta=c\omega_1+d\omega_2$ and we have a basis $\{\alpha,\beta\}$. Using the fact that $\overline{\beta}\in\Lambda$, with few computation we see that we can assume $\beta$ to be as in the claim.

Fact 3): consider the analytic isomorphism $\phi\colon \mathbb C/\Lambda\to E(\mathbb C)\subseteq \mathbb P^2(\mathbb C)$ given by $[0]\mapsto (0\colon 1 \colon 0)$ and $0\neq z\mapsto (\wp(z)\colon\wp'(z)\colon 1)$. Then $\phi(z)=\overline{\phi(z)}$ iff $z\equiv \overline{z}\bmod \Lambda$.

Fact 4): $E(\mathbb R)[2]$ is isomorphic to $\mathbb (Z/2\mathbb Z)^2$ when $\beta=iy$ while it is isomorphic to $\mathbb Z/2\mathbb Z$ otherwise. Just use fact 3) plus the fact that $E(\mathbb C)[2]$ is generated by $\phi(\alpha/2)$ and $\phi(\beta/2)$.

Fact 5): Let $S=[0,\alpha]\subseteq \mathbb R$ and $T=\{z\in \mathbb C\colon z=x+iy/2\colon 0\leq x\leq \alpha\}$. Then $\phi^{-1}(E(\mathbb R))=S$ if $\beta=\alpha/2+iy$, while $\phi^{-1}(E(\mathbb R))=S\cup T$ otherwise. This follows directly from Fact 3) as well.

Now Fact 4)+Fact 5) implies easily your claim.

If you want a different proof, check Silverman's "Advanced topics in the arithmetic of elliptic curves", corollary V.2.3.1


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