# How to get explicit unramified covers of an elliptic curve?

Say we begin with an explicit elliptic curve over $\mathbb{C}$, say: $y^2=x(x-1)(x-2)$. According to abstract reasoning this elliptic curve has an (several, in fact) unramified cover with group $C_n$. My question is, how can one find the equations that define this cover? To be very precise: how can one describe the function field extension of $Quot(\mathbb{C}[x,y]/(y^2-x(x-1)(x-2)))$ as the quotient field of a polynomial ring modulo an explicit ideal?

It would be nice to have a method that can generate all such unramified $C_n$ covers, but a method that generates some unramified $C_n$ cover would be helpful too.

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## 6 Answers

The covers you seek are exactly the elliptic curves which admit cyclic $n$-isogenies into $E$ (or out of $E$, it doesn't matter). Proof: if $X\to E$ is unramified, then $X$ is a genus one curve (by the Riemann-Hurwitz formula). Choose a base point on $X$ lying above the origin of $E$; then $X$ is an elliptic curve and $X\to E$ is an isogeny (standard theorem about morphisms between abelian varieties).

The curves $X$ you want are what you get when you quotient $E$ by each of its cyclic subgroups of order $n$. Finding these $X$ is exactly the purpose of the modular polynomial $\Phi_n(x,y)$. The way that this works is that if $j(E)$ is the $j$-invariant of $E$, then the roots of $\Phi_n(x,j(E))$ are exactly the quantities $j(E')$, where $E'$ runs over the elliptic curves related to $E$ by cyclic $n$-isogeny. The modular polynomial $\Phi_n(x,y)$ is difficult to calculate, and its coefficients grow very quickly with $n$; a Google search comes up with this reference on calculating them. Once you have the $j$-invariants, it's easy to find the corresponding elliptic curves (this is one nice thing about working over $\mathbf{C}$).

Incidentally, the degree of $\Phi_n(x,j)$, which is (generically) the number of $X$ that you want, is $$n\prod_{p\vert n}\left(1+\frac{1}{p}\right).$$

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This is the right way to look at these things. But I'm trying to work out a specific case, using Makhalan's example (it has complex multiplication by $i$, after all), which should come up with genuine formulas. It's taking fairly long, though. – Lubin Jan 14 '11 at 23:35
Yes, I should have addressed the CM issue in my answer. These other curves are going to have CM by $\mathbf{Z}[ni]$. In fact, the OP is essentially just asking for the minimal polynomial of $j(ni)$, which generates a ray class field of $\mathbf{Q}(i)$. I don't know a quick way to do this, but maybe it's addressed in Cox' book, "Primes of the form $x^2+ny^2$." – Jared Weinstein Jan 15 '11 at 0:12
By the way, modular equations in $j$ are a bit of a disaster. The coefficients are much smaller if one works with $\lambda$ instead, or related higher level hauptmoduli. This is discussed in the Borweins' book on $\pi$ and the AGM. Best wishes, Matt – Emerton Jan 15 '11 at 4:40

In my perhaps skewed experience, it's much easier, from a computational standpoint, to get isogenies out of an elliptic curve than into one. So I'd say that you're really looking into the more uncomfortable end of the telescope.

Again, computationally, once you identify a finite subgroup $S$ of your elliptic curve, it's not too hard to get formulas for the isogeny with $S$ as kernel: use translation (implemented by chord-and-tangent) by the elements of $S$ to define the elements of a Galois group, and find its fixed field inside $\mathbb{C}(x,y)$.

But the curve you gave is same as $y^2=x^3-x$, and clearly has complex multiplication by $i$: you have $\lbrack i\rbrack (\xi,\eta) = (-\xi, i\eta)$. This permits you to write down an example of a cyclic $n$-isogeny any time that $n$ is the sum of two squares in $\mathbb{Z}$, so let's do it for $n=2$. Use the endomorphism $1+i$, which has degree two, so that your curve covers itself in a cyclic $2$-cover. You add $(\xi,\eta)$ to $(-\xi, i\eta)$ on your curve and the result is $(-i/2)(\xi^2-1)/\xi$ for the $x$-coordinate, and $-((1+i)/4)\eta(1+1/\xi^2)$ for the $y$-coordinate. (This point really is on the curve.) The two preceding quantities generate a subfield over which $\mathbb{C}(x,y)$ is quadratic, the extension is unramified as desired.

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How do you computationally find the mentioned fixed field? Is this faster than computing Velu's formulas? – Dror Speiser Jan 15 '11 at 19:08
The fixed field is the one generated by the $x$ and the $y$-coordinates mentioned. V\'elu's formulas are probably faster. – Lubin Jan 24 '11 at 0:43

Let me expand Felipe's answer a bit.

Vélu's formulae given in the very readable short paper [1] are very easy to use. For instance if you are given a $n$-torsion point one can immediately write down the Weierstrass equation for the quotient by this point.

To get such a point, one can use the division polynomials which can be computed recursively very fast. The zeroes of this polynomial are exactly the $x$-coordinates of the $n$-torsion points. Oover the complex number it is not difficult to get them also using the parametrisation by the Weierstrass $\wp$-function.

There are quite a few paper improving Vélu's formula. Most importantly, there is Kohel's thesis [2]. Or for instance [3]. These were used in the implementation for isogenies in sage [sage] and in [magma].

[1] MR0294345, Jacques Vélu, Isogénies entre courbes elliptiques. C. R. Acad. Sci. Paris Sér. A-B 273, 1971.

[2] MR2695524, David Kohel, Endomorphism rings of elliptic curves over finite fields. Thesis (Ph.D.), University of California, Berkeley. 1996.

[3] MR2398793, Bostan, A.; Morain, F.; Salvy, B.; Schost, É. Fast algorithms for computing isogenies between elliptic curves. Math. Comp. 77 (2008).

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Velu's formulas. They are implemented, e.g., in sage.

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If you need a quick and dirty example, you can easily get the equation of one of the double covers. If $y^2=x(x-1)(x-\lambda)$ is the equation of your curve, then $$w^2 = (z^2-1)(z^2-\lambda)$$ is the equation of a double cover. In geometric terms, if $E\to\mathbb{P}^1$ is ramified at 0, 1, $\lambda$ and $\infty$, then we get an unramified double cover $E'\to E$ by pulling back over the map $\mathbb{P}^1 \to \mathbb{P}^1$ given by $x=z^2$. The curve $E'\to \mathbb{P}^1$ is ramified over the preimages of $1$ and $\lambda$.

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I suspect that all these covers are given by isogenies of elliptic curves. In order to find those, you should look for a (torsion) subgroup of $E(\mathbb{C})$ isomorphic to $C_n$ (this would give you a system of algebraic equations). Such a subgroup defines an action of $C_N$ on $\mathbb{C}(E)$; the fixed field of this action would give you the function field of the corresponding elliptic curve.

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Also, the system of equations for $C_n\subset E(\mathbb{C})$ has an abelian Galois group; so it shouldn't be too difficult to solve. – Mikhail Bondarko Jan 14 '11 at 23:07
Sorry for being slow, but I don't really see how this translates to a system of algebraic equations... Can you do a simple example so I can see what you mean? – Makhalan Duff Jan 14 '11 at 23:12
I am sorry, but I am too lazy to write down even the simplest example here.:) First you write the equation $[n]X=0$, where $[n]$ is the multipication by $n$ on $E$. Then you should consider the translates by elements of your chosen $C_n$ and do some explicit Galois theory. – Mikhail Bondarko Jan 15 '11 at 11:02