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?

  • $\begingroup$ This reminds of a question that a friend and I tried to address some time ago. Namely, can one compute the polynomial $\prod(T-j(\alpha))$ where $\alpha$ varies over the fiber of such a modular parametrization. At the time, it seemed that computational resources weren't sufficient, but the situation may be different now. ...William? $\endgroup$
    – Ramsey
    Feb 13, 2011 at 17:44
  • $\begingroup$ I would say that this question would even be interesting in the rank 1 case. For example in the easiest rank one case $E=X_0^+(37)$ exactly $9$ of the points in $E(\mathbb Q)$ are heegner points and one point corresponds to the exceptional 37 isogeny. These 10 points are the only 10 with integer j-invariant. So for all the other $p \in E(\mathbb Q)$ the pullbacks to $X_0(37)$ will give quadratic numbers fields, but the curves over these points won't be CM. The points in $E(Q)$ whose lift is not CM might still be Heegner points though, since a Heegner point is a sum of CM points. $\endgroup$ Aug 24, 2014 at 21:12
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    $\begingroup$ See page 42 of math.harvard.edu/~elkies/modular.pdf $\endgroup$ Aug 24, 2014 at 21:13

2 Answers 2


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:).

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    $\begingroup$ Just a remark that Hao is my Ph.D. student, and one of the motivations for him doing the above is computation of things like the Mazur--Swinnerton-Dyer critical subgroup of E(Q)... $\endgroup$ Oct 12, 2014 at 0:08
  • $\begingroup$ If increasing the conductor is computationally hard, maybe you could enlarge the base field to get to rank 2 examples. Find a small conductor curve with large rank over a small quadratic field. $\endgroup$ Nov 20, 2014 at 9:16

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.

The question is similar to this .

  • $\begingroup$ Hey Chris. I agree that one interpretation of "modular symbols" doesn't help, because they'll build the cuspidal subspace of the Jacobian. I meant "modular symbols" more generally, in the sense that they can also be used to build the entire Jacobian. But you're a natural person to answer the question and if you say they don't help that makes me a bit more pessimistic. I agree that there's no reason to expect the degree to be small. You say that some generators for the 389 example were done already in your other answer---do you know a reference for this? [PS see you Wednesday, presumably] $\endgroup$ Feb 13, 2011 at 12:43
  • $\begingroup$ 389: maybe I remembered the wrong thing. Christophe Delaunay computed the critical points for this modular parametrisation at the end of archive.numdam.org/ARCHIVE/JTNB/JTNB_2005__17_1/…. I will ask him, if he knows more. [PS: Yes, see you on Wednesday, looking forward to your talk.] $\endgroup$ Feb 13, 2011 at 13:25
  • $\begingroup$ What do you mean by ""I meant "modular symbols" more generally, in the sense that they can also be used to build the entire Jacobian."" ? $\endgroup$ Feb 14, 2011 at 8:19
  • $\begingroup$ A little remark concerning the method outlined by Chris : it is known that if $N$ is an odd prime then $X_0(N)(\mathbf{R})$ is connected and consists of the paths $\{0,\infty\}$ and $\{\infty,\frac12\}$, with $\frac12$ being $\Gamma_0(N)$-equivalent to $0$. But I don't know if all preimages of $P \in E(\mathbf{R})$ are real. In other words, does the topological degree of the map $X_0(N)(\mathbf{R}) \to E(\mathbf{R})$ equal the degree of the covering $X_0(N) \to E$? $\endgroup$ Feb 14, 2011 at 13:09
  • $\begingroup$ I realize that in this case the topological degree of the map $X_0(N)(\mathbf{R}) \to E(\mathbf{R})$ is $0$, so real preimages exist only if $P$ is close to $0$ in $E(\mathbf{R})$, as mentioned in Chris' answer. $\endgroup$ Feb 15, 2011 at 0:49

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