It is a well-known result that the modular function $1728J(\tau) := \frac{1728E_4(\tau)^3}{E_4(\tau)^3-E_6(\tau)^2}$ has integral values if $\tau$ has class number 1 - for example at $\tau_{163}:=\frac{1+i\cdot\sqrt{163}}{2}$ you get $1728J(\tau_{163})=-640320^3$.

Now define the quasi-modular function $s_2(\tau):=\frac{E_4(\tau)}{E_6(\tau)}\cdot\left(E_2(\tau)-\frac{3}{\pi Im(\tau)}\right)$.

Then I have looked at all 13 class 1 discriminants and have verified numerically that $$M(\tau):=s_2(\tau)\cdot(1728J(\tau)-1728)=\frac{1728E_4(\tau)E_6(\tau)}{E_4(\tau)^3-E_6(\tau)^2}\left(E_2(\tau)-\frac{3}{\pi Im(\tau)}\right)$$ also has integral values for all these $\tau_N$.

For example $M(\tau_{163})=-2^{13}\cdot3^5\cdot5\cdot7\cdot11\cdot19\cdot23\cdot29\cdot127\cdot181$.

My Question: How can I prove that $M(\tau)\in\mathbb Z$ for all $\tau$ with class number 1? Or where can I find a proof for it?

Edit: After the answer of @Zavosh (thank you!!!) it remains to prove this question. Who can help?

  • $\begingroup$ With that Question, one could compute the coefficients arising in the Chudnovsky-Algorithm: $\frac{1-s_2(\tau_{163})}{6} =\frac{1-\frac{77265280}{90856689}}{6} = \frac{13591409}{545140134}$ $\endgroup$
    – L. Milla
    May 17, 2018 at 9:21
  • 1
    $\begingroup$ There may be some kind of Moonshine-like phenomenon behind that, as $ M(\tau_{163}) $ is divisible by the order of each Mathieu group. $\endgroup$ May 18, 2018 at 10:00

1 Answer 1


Algebraicity is proven in appendix one of [1]. The function considered there is

$$ \psi(\tau) = \frac{3E_4(\tau)}{2E_6(\tau)} (E_2(\tau) - \frac{3}{\pi \rm{Im} \tau}) = \frac{3}{2} s_2(\tau).$$

The proof is by establishing, for quadratic irrationals $\tau$, the identity $$ \psi(\tau) = 9j(\tau)\gamma + \frac{3(7j(\tau)-6912)}{2(j(\tau)-1728)}$$ where $\gamma$ is a rational function in the coefficients of the Taylor series of a certain polynomial $\Phi(X,Y)$ around the point $(j(\tau),j(\tau))$ where it vanishes. Then

$$M(\tau) = 1728 s_2(\tau) (j(\tau)-1) = 1728(j(\tau)-1) ( 6j(\tau)\gamma + \frac{(7j(\tau) - 4\cdot1728)}{j(\tau)-1728}.$$

There is also a different identity for calculations, for when $\tau$ is not equivalent to $i$ or $\rho$, and satisfies $$ A \tau^2 + B \tau + C = 0$$ with $A,B,C$ coprime positive integers:

$$\psi(\tau) = \frac{-g_2 S}{Cg_3 (2A+B\tau)}$$ where $S$ is the sum of $\wp(z)$ as $z$ ranges over $C\tau$-torsion points of $\mathbb{C}/\langle 1,\tau \rangle$.

Masser gives a table of $\psi(\tau)$, for a list of generators of rings of integers of imaginary quadratic fields with class number 1. They are all rational numbers, and one can indeed check that $$ M(\frac{-1+\sqrt{-163}}{2}) = 223263987730882560,$$ which agrees with the factorization in the question. Also

$$\begin{align}M(\frac{-1+\sqrt{-67}}{2})&=-112852776960=-2^{11}\cdot 3^5 \cdot 5 \cdot 7 \cdot 11 \cdot 19 \cdot 31\\ M(\frac{-1+\sqrt{-43}}{2})&=-627056640=-2^{13}\cdot 3^7 \cdot 5 \cdot 7\\ M(\frac{-1+\sqrt{-27}}{2})&=-7772161=-1181\cdot 6581\\ M(\frac{-1+\sqrt{-19}}{2})&=-497664=-2^{11} \cdot 3^5\\ M(\frac{-1+\sqrt{-11}}{2})&=-14336=-2^{11} \cdot 7\end{align},$$ etc.

Probably one can prove $M(\tau)$ is an algebraic integer by going through Masser's calculations and clearing denominators.

[1] Masser, David W. Elliptic Functions and Transcendence

  • 2
    $\begingroup$ Thanks a lot! Masser gives $\gamma$ as a quotient of some of the integral coefficients of $\Phi(X,Y)$, thus $\gamma\in\mathbb Q$. This implies that $M\in\mathbb Q$. Moreover, as @HenriCohen said here, $M\cdot\sqrt(D)$ is an algebraic integer. This implies that $M^2\cdot D$ is an algebraic integer and is in $\mathbb Q$, thus we have $M^2\cdot D\in\mathbb Z$. $\endgroup$
    – L. Milla
    May 22, 2018 at 11:28

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