How to explicitly compute lifting of points from an elliptic curve to a modular curve?

Question

Say $E$ is an elliptic curve over the rationals, of conductor $N$. There's a covering of $E$ by the modular curve $X_0(N)$, and if you rig it right then you can define this map over $\mathbf{Q}$: there's a map $\pi:X_0(N)\to E$ of algebraic varieties over $\mathbf{Q}$.

Now say I have an explicit $\mathbf{Q}$-point $P\in E(\mathbf{Q})$. Its pre-image in $X_0(N)$ will be a finite set of points, all defined over number fields. Perhaps a bit more conceptually, the pullback of $\pi$ via the map $Spec(\mathbf{Q})\to E$ induced by $P$ is a scheme $Spec(A)$ where $A$ is a finite $\mathbf{Q}$-algebra.

How would one go about actually computing these number fields in an explicit example? (or computing $A$, if you like). One can do computations in Jacobians of modular curves so easily these days using modular symbols, so I would imagine this is easy for the experts.

As an explicit example let's take a non-torsion point $P$ on an elliptic curve of rank two (so one can't "cheat" and do the calculation using Heegner points or cusps)---for example let $P$ be some random non-zero small height element of Mordell-Weil mod torsion in the rank two curve of conductor 389. What number fields do the points in the modular curve that map to $P$ cut out?

Answer 1

I recently implemented an algorithm to determine these number fields by computing the $j$-polynomial: Let $\varphi: X_0(N) \to E$ be a fixed modular parametrization and $P \in E(\mathbb{Q})$. By j-polynomial I mean the polynomial $F_P(x) = \prod_{z : \varphi(z) = P}(x - j(z))$. There's a Laurent series $x(q)$ with integer coefficients, which is the modular function $\frac{1}{x \circ \varphi}$ on $X_0(N)$,

We compute $x(q)$ via the gp function ell.taniyama. Then set $u = \frac{1}{j(q)}$, which is also an element in $\mathbb{Z}[[q]]$. Then using Linear algebra, one can find an irreducible polynomial $F$ such that $F(x,u) = 0$.

Setting $x_0 = \frac{1}{x(P)}$ and the polynomial $j^{2 \deg \varphi}F(x_0,1/j)$ is a constant multiple of $F_P(j)F_{-P}(j)$. Then we could use complex analytic method to determine which factor corresponds to $P$ and which is $-P$.

As an example we take the elliptic curve 121b1 with rank 1 and trivial torsion. $E(\mathbb{Q})$ is generated by $P = (4:5:1)$. Then we compute some j-polynomials:

• $F_{-P}(x) = x^4 + 1421551441067913615636000 x^3 + 910640170936002098476853963114167004130307406250 x^2 - 55869041153225091624766256009488963324954953937500000 x + 1513370207928838475604980619812428055721351700525634765625$

• $F_{4P}(x) = x^{4} - \frac{1131444376477476487694208}{43} x^{3} + \frac{11389877706351841520907948036498862509059802748293120}{1849} x^{2} + \frac{831545351967828972021160038394755202358001953700040409088}{1849} x + \frac{16095967144279358005293903881120972455827496828529236714192896}{1849}$

• $F_{P}(x) = x^4 + 98823634118413525094400000 x^3 + 45688143672322270430861721600000000 x^2 - 496864268553728774541064273920000000000 x + 1577314437358442913340940353536000000000000$

Since atkin-lehner acts as +1 in this case the number fields in question are splitting fields of these polynomials, respectively.

I plan to compute with the rank 2 curve 389a mentioned in the original post. Doing so is hard with the current computing power I have, so I was thinking about replacing $u$ by an $\eta$-product with smaller valence.

Please let me know if this helps. I'll keep polishing this algorithm for the goal of including it in my coming up thesis:).

Answer 2

I do not believe that modular symbols will help. They are good to detect that the rank is positive, but since they only produce torsion points in $E$, they are unfortunately of no use to construct points of infinite order. I fear also that they can not give you the fields that you are after.

If you have the modular parametrisation from a model of $X_0(N)$ to $E$, you can do it easily. But this is not what your after, I know.

Here is another idea. Take a prime $p$ and suppose the point $P$ lies in the formal group. If not you could multiply it to lie in it, but you would change the field. Then $P$ is $p$-adically close to $O$. Now compute its image $z=\log_{\hat{E}}(P)$ under the formal logarithm map into the $p$-adic points of the Lie algebra. The formal version of the modular parametrisation from $X_0(N)$ to $E$ is just $q\mapsto\sum \frac{a_n}{n} q^n$, which has a local inverse around $O$. So you will find a value of $q\in p\mathbb{Z}_p$ that maps to $z$. There is a Tate elliptic curve $A$ defined over $\mathbb{Q}_p$ with multiplicative reduction whose $q$-parameter is this $q$. Computing $j(q)$ to sufficiently high precision, it should be possible to guess its minimal polynomial over $\mathbb{Z}$ and then, using this guess one can verify the claim. (But I have not attempted to do this, so I have no idea if it is feasible.)

(edit:) This can be done for $p=\infty$, too. If $P$ is sufficiently close to $O$ in $E(\mathbb{R})$ then one can recover the $\tau$ in the upper half plane on the imaginary axis with largest imaginary part as above. (This corresponds to integrating the modular form along the imaginary axis until we get $P$ for the first time.) Then one can recover the minimal polynomial of $j(\tau)$ by acting on it with the cosets representatives of $\Gamma_0(N)$.

Note also, once that $j(q)$ is known. We also have to check if the corresponding curve has a $N$-isogeny over its field of definition or otherwise one needs to enlarge it.

In any case, your field will have large degree. In fact, I do not see any reason to believe that the degree of this field should be smaller than the degree of the modular parametrisation.

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