How to describe the quotient of an elliptic curve by a finite cyclic translation action?

Question

Let $E$ be an elliptic curve (possibly over $\mathbb C$), and let $p \in E$ be a point of order $n \in \mathbb N$. Then the group $\mathbb Z/n$ acts on $E$ by translation $$\sigma: \mathbb Z/n \times E \to E, (k,z) \mapsto z + k\cdot p.$$ Is there a general way to describe the quotient $F = E / (\mathbb Z / n)$? Since the action $\sigma$ is free, the quotient map $E \to F$ is étale, so by Hurwitz' theorem we can conclude that $F$ is again a curve of genus $1$. Is it true that $F \cong E$?

The specific example I'm interested in is $E = \mathbb C / (\mathbb Z + \zeta \mathbb Z)$, where $\zeta = e^{\frac{2 \pi i}{3}}$ and $p \equiv \frac 1 2 + \frac{\sqrt{3}}{6} i \mod \mathbb Z + \zeta \mathbb Z$.

Answer 1

You're simply taking the quotient of $E$ by the cyclic subgroup generated by $p$. The projection map itself is a cyclic $n$-isogeny and the quotient is the codomain of the isogeny. The moduli space of all such cyclic $n$-isogenies is $X_0(n)$. In your specific example, the quotient is isomorphic to the starting curve $E$ itself, although this property certainly does not always hold in general for arbitrary $E$ and $p$. We can see that they are isomorphic in your example purely from geometry. Your starting curve $E$ is a lattice $\Lambda_1$ with fundamental region as below:

and your codomain curve $F = \mathbb{C}/\Lambda_2$ (with the additional identification of the new points in red with the identity element) has fundamental region as below:

We see that these two fundamental regions are similar polygons in the sense of Euclidean plane geometry (in complex multiplication theory we would say that the lattices $\Lambda_1$ and $\Lambda_2$ are homothetic), so by complex multiplication the resulting elliptic curves $E = \mathbb{C}/\Lambda_1$ and $F = \mathbb{C}/\Lambda_2$ are isomorphic.

In algebraic terms, your starting curve $E = \mathbb{C}/\langle 1, \zeta_3\rangle$ has $j$-invariant $0$ (as $j(\zeta_3) = 0$). One possible model for this curve over $\mathbb{Q}$ is $E : y^2 = x^3 + 1$. Under this model, the $3$-torsion point $p$ has coordinates $p = (0,1)$, and the isogeny $\phi\colon E \to F$ has equations

$$ \phi(x,y) = \left(\frac{x^3 + 4}{x^2}, \frac{y(x^3 - 8)}{x^3}\right) $$

and codomain $F : y^2 = x^3 - 27$. We observe that $F$ also has $j$-invariant $0$, confirming that $E \cong F$. You can calculate these equations for isogenies using Vélu's formulas (Isogénies entre courbes elliptiques, C. R. Acad. Sci. Paris Sér. I Math. 273 (1971), pp. 238-241).

Answer 2

As other people answer i think that it could happens when you work with elliptic curve with CM. Another example is $E_i$ the elliptic curve with the invertible endomorphism of order four (it has the multiplication by $i$) let us consider $p=\dfrac{i+1}{2}$ then $E_i/t_p$ is isomorphic to $E_i$. And i think there is a similar phenomena in higher dimension ..

Moreover this phenomenon happens because we are choosing special point over special elliptic curve, in the sense that if you look at my example the point $p$ is fixed by $i$, similarly this happens in your example. In general take $E$ an elliptic curve with CM and $t$ a translation by a point $q$ in $E$ . Then $E/t$ is iso to $E$ iff they have the same CM so iff the extra automorphism pass trough the quotient. This happens iff the extra automorphism fits the point $q$. So also if you consider a group of translation acting on $E$ if it is "preserves" by the extra automorphism then the quotient $E/G$ is iso to $E$.