This problem is motivated by this question and by teaching modular polynomials for the classical modular invariant $j(\tau)$. The latter implies that if we consider the fields of modular functions $\mathcal K_n=\mathbb C[j(\tau),j(n\tau)]$, then $\mathcal K_n$ is an algebraic extension of $\mathcal K=\mathcal K_1=\mathbb C[j(\tau)]$. A standard application of this fact is in study of isogenies of elliptic curves but this has several links to transcendence theory (for example, Schneider's theorem on the transcendence of the values of $j(\tau)$ at non-CM points and a more recent result about transcendence of the values of $J(q)=j(e^{2\pi i\tau})$).

The field $\mathcal K$ (of transcendence degree 1 over $\mathbb C$) is a subfield of a larger field $\mathcal F=\mathbb C(E_2(\tau),E_4(\tau),E_6(\tau))$ (which has transcendence degree 3 and is in fact the differential closure of $\mathcal K$), where $$ E_2(\tau)=1-24\sum_{n=1}^\infty\sigma_1(n)q^n, \quad E_4(\tau)=1+240\sum_{n=1}^\infty\sigma_3(n)q^n, \qquad E_6(\tau)=1-504\sum_{n=1}^\infty\sigma_5(n)q^n $$ are the Eisenstein series of weight 2, 4 and 6, respectively, $\sigma_k(n)=\sum_{d\mid n}d^k$ and $q=e^{2\pi i\tau}$. The field $\mathcal F$ is differentially closed due to Ramanujan's system of algebraic differential equations $$ DE_2=\frac1{12}(E_2^2-E_4), \quad DE_4=\frac13(E_2E_4-E_6), \quad DE_6=\frac12(E_2E_6-E_4^2), \qquad D=\frac1{2\pi i}\frac{d}{d\tau}=q\frac d{dq}. $$ Because of the modular origin and as the field $\mathcal F$ can be defined as $\mathbb C(j(\tau),Dj(\tau),D^2j(\tau))$, it is more or less clear that $E_2(n\tau)$, $E_4(n\tau)$ and $E_6(n\tau)$ are algebraic over $\mathcal F$ for every positive $n$. (In fact, $E_4(n\tau)$ and $E_6(n\tau)$ are algebraic over the smaller field $\mathcal F'=\mathbb C(E_4(\tau),E_6(\tau))$.) In what follows, I denote $\mathcal R=\mathbb Q[E_2(\tau),E_4(\tau),E_6(\tau)]$ and $\mathcal R'=\mathbb Q[E_4(\tau),E_6(\tau)]$ the corresponding rational rings.

Problem. For each $n>1$, construct polynomials $A_n\in\mathcal R[x]$ and $B_n,C_n\in\mathcal R'[x]$ such that $$ A_n\bigl(E_2(n\tau)\bigr)=0, \quad B_n\bigl(E_4(n\tau)\bigr)=0, \quad C_n\bigl(E_6(n\tau)\bigr)=0, $$ and derive the arithmetic bounds for them (for degree and size of their coefficients).

Is this problem ever studied?

Some years ago I published relations expressing $E_2(\tau/2)$, $E_4(\tau/2)$ and $E_6(\tau/2)$ through the logarithmic derivatives of the thetanulls (Ramanujan Journal 7:4 (2003) 435--447, Section 4). The expressions for $E_2(\tau)$, $E_4(\tau)$ and $E_6(\tau)$ through the same set of (three) functions are classical (Sbornik: Mathematics 191:12 (2000) 1827--1871, Eqs. (0.11) and (0.12)), so I am aware of solution in the case $n=2$. But the approach is hardly generalizable.

Martin Rubey computed explicitly the polynomials for $n\le7$. Here are $n=5$ instances: $$\begin{align*} A_5(x)&= 5^{11}x^6 - 2\cdot 3\cdot 5^{10}E_2x^5 + (3\cdot 5^{10}E_2^2-2^4\cdot 3\cdot 5^8E_4)x^4 \cr &\; + (-2^2\cdot 5^9E_2^3+2^6\cdot 3\cdot 5^7E_4E_2-2^9\cdot 5^6E_6)x^3 \cr &\; + (3\cdot 5^8E_2^4-2^5\cdot 3^2\cdot 5^6E_4E_2^2+2^9\cdot 3\cdot 5^5E_6E_2-2^8\cdot 3^2\cdot 5^4E_4^2)x^2 \cr &\; + (-2\cdot 3\cdot 5^6E_2^5+2^6\cdot 3\cdot 5^5E_4E_2^3-2^9\cdot 3\cdot 5^4E_6E_2^2+2^9\cdot 3^2\cdot 5^3E_4^2E_2-2^{13}\cdot 3\cdot 5E_6E_4)x \cr &\; + (5^5E_2^6-2^4\cdot 3\cdot 5^4E_4E_2^4+2^9\cdot 5^3E_6E_2^3-2^8\cdot 3^2\cdot 5^2E_4^2E_2^2+2^{13}\cdot 3E_6E_4E_2-2^{12}E_6^2), \cr B_5(x)&= 5^{20}x^6 - 2\cdot 3^2\cdot 5^{17}\cdot 7E_4x^5 + 3\cdot 5^{13}\cdot 11\cdot 19E_4^2x^4 \cr &\; + (2^2\cdot 5^9\cdot 7\cdot 210241E_4^3-2^{11}\cdot 5^{12}\cdot 23E_6^2)x^3 \cr &\; + (3^3\cdot 5^5\cdot 18858713E_4^4-2^11\cdot 3^2\cdot 5^8\cdot 13\cdot 17E_6^2E_4)x^2 \cr &\; + (2\cdot 3\cdot 7\cdot 11\cdot 59\cdot 71\cdot 24943E_4^5-2^{11}\cdot 3\cdot 5^4\cdot 13\cdot 967E_6^2E_4^2)x \cr &\; + (11^2\cdot 59^2\cdot 71^2E_4^6-2^{11}\cdot 5\cdot 389\cdot 971E_6^2E_4^3+2^{18}\cdot 5\cdot 11^3E_6^4), \cr C_5(x)&= 5^{30}x^6 - 2\cdot 3\cdot 5^{25}\cdot 521E_6x^5 + (-2^9\cdot 3^3\cdot 5^{19}\cdot 7^2\cdot 23E_4^3+3\cdot 5^{21}\cdot 269\cdot 773E_6^2)x^4 \cr &\; + (2^9\cdot 3^3\cdot 5^{13}\cdot 7^2\cdot 31123E_6E_4^3-2^2\cdot 5^{16}\cdot 521\cdot 80929E_6^3)x^3 \cr &\; + (-2^8\cdot 3^6\cdot 5^7\cdot 7^4\cdot 11^2\cdot 19^2E_4^6+2^8\cdot 3^5\cdot 5^8\cdot 7^2\cdot 11\cdot 17\cdot 13171E_6^2E_4^3 \cr &\;\quad -3\cdot 5^{11}\cdot 11\cdot 59\cdot 71\cdot 269\cdot 773E_6^4)x^2 \cr &\; + (-2^9\cdot 3^6\cdot 7^4\cdot 11^2\cdot 17\cdot 19^2\cdot 31E_6E_4^6+2^{10}\cdot 3^3\cdot 5^3\cdot 7^2\cdot 157\cdot 191\cdot 8147E_6^3E_4^3 \cr &\;\quad -2\cdot 3\cdot 5^5\cdot 11^2\cdot 59^2\cdot 71^2\cdot 521E_6^5)x \cr &\; + (-2^8\cdot 3^6\cdot 5\cdot 7^4\cdot 11^2\cdot 19^2E_6^2E_4^6+2^8\cdot 3^3\cdot 5\cdot 7^2\cdot 5237\cdot 22067E_6^4E_4^3-11^3\cdot 59^3\cdot 71^3E_6^6). \end{align*} $$ The conjecture about degree is $\psi(n)=n\prod_{p\mid n}(1+1/p)$, the same as for the modular polynomials. In fact, if we assign weight 2, 4 and 6 to the variable $x$, then $A_n$, $B_n$ and $C_n$ happen to be homogeneous polynomials of degree $2\psi(n)$, $4\psi(n)$ and $6\psi(n)$, respectively.

  • $\begingroup$ Hi Will. The relations for $\tau/2$ are in the Ramanujan Journal 7:4 (2003), 435--447, Section 4, and the classical ones in Sbornik: Mathematics 191:12 (2000), 1827--1871, Eqs. (0.11), (0.12). (Both papers are available from my webpage.) $\endgroup$ May 5, 2010 at 3:50
  • $\begingroup$ Where can I find the proof of the fact that $E_2 (2\tau)$ is algebraic over $F $? Could you let me know the reference? $\endgroup$
    – LWW
    Nov 12, 2018 at 2:17

2 Answers 2


For E22 = E2(q^2), E23 = E2(q^3), E24 = E2(q^4) I obtain the equations below using the (slightly modified) guessing package in FriCAS. Unfortunately, I currently have only adapted the slow algorithm, guessing the last of the three took already an unreasonable amount of time (roughly 14 hours). Using the modular algorithm it should not take more than ten minutes, so I probably should pull myself together and just do it...

I think it's fair to conjecture that there are equations homogeneous with respect to the usual weight, where $E2(q^k)$ and E2 have weight 2, E4 has weight 4 and E6 has weight 6. (Taking this into account should also cut down the time for guessing dramatically, yet another thing to implement...) Thus, the total degree is 6,8,12,...

Another observation: the leading coefficients are $2^5$, $3^7$, $4^{11}$, which gives an obvious correspondence to the total degree.

 5   3    4    2      3       2                2  3
2 E22  - 2 3E22 E2 + 2 3E22 E2  - 2 3E22 E4 - 2 E2  + 3E2 E4 - E6 = 0
 7   4    2 6   3        6   2  2    3 4   2      2 4      3
3 E23  - 2 3 E23 E2 + 2 3 E23 E2  - 2 3 E23 E4 - 2 3 E23 E2


 4 3             6           3  4    3 2  2      6         4  2
2 3 E23 E2 E4 - 2 3E23 E6 + 3 E2  - 2 3 E2 E4 + 2 E2 E6 - 2 E4 = 0
 22   6    21    5      18      4  2    15 2    4      18    3  3
2  E24  - 2  3E24 E2 + 2  3 5E24 E2  - 2  3 5E24 E4 - 2  5E24 E2


 15 2    3         12 3    3      14      2  4    12 3    2  2
2  3 5E24 E2 E4 - 2  3 5E24 E6 + 2  3 5E24 E2  - 2  3 5E24 E2 E4


 10 4    2         6 5    2  2    13       5    11 2       3
2  3 5E24 E2 E6 - 2 3 5E24 E4  - 2  3E24 E2  + 2  3 5E24 E2 E4


   8 4       2      5 5          2    4 5             10  6    7 2   4
- 2 3 5E24 E2 E6 + 2 3 5E24 E2 E4  - 2 3 E24 E4 E6 + 2  E2  - 2 3 5E2 E4


 6 3   3      2 5   2  2    2 5            3  3    3  2
2 3 5E2 E6 - 2 3 5E2 E4  + 2 3 E2 E4 E6 + 3 E4  - 3 E6 = 0
  • $\begingroup$ I TeXify: in the notation $E_{22} = E_2(q^2)$, $E_{23} = E_2(q^3)$ one has $-E6 + (- 6E_{22} + 3E_2)E_4 + 32E_{22}^3 - 48E_2 E_{22}^2 + 24E_2^2 E_{22} - 4E_2^3 = 0$ and $(- 192E_{23} + 64E_2)E_6 - 16E_4^2 + (- 648E_{23}^2 + 432E_2 E_{23} - 72E_2^2 )E_4 + 2187E_{23}^4 - 2916E_2 E_{23}^3 + 1458E_2^2 E_{23}^2 - 324E_2^3 E_{23} + 27E_2^4 = 0$. Yes, that what I mean to get for $E_{kn}$ ($k=2,4,6$ and $n\ge2$). The coefficients look reasonably small (compared to the ones for level 2 modular polynomial). I guess this is an experimental discovery. How to do for the general $n$? $\endgroup$ May 6, 2010 at 3:50
  • $\begingroup$ @Martin: Thanks! I just checked your relations (up to $O(q^{101})$, the first one can be compared with the one I know): perfect agreement. But I still wonder how to do the things non-experimentally and in general. I can't believe there were no work in this direction... $\endgroup$ May 6, 2010 at 4:12
  • $\begingroup$ Yes, I did this using my (slightly modified) guessing package. (Unfortunately, I did not yet have the time to modify it properly, and very likely I'll have to wait for Waldek to get full speed) However, with what I have I was able to check E24, and I did not find an equation of degree at most 5 and constant coefficients. I'll see what I can do... $\endgroup$ May 6, 2010 at 12:10
  • $\begingroup$ @Martin: Thanks for the $E_{24}$ news. It really seems that aricthmetically these objects are better than $j(\tau)$ as the coefficients involve small primes only. I would be happier if the things are written with "*" for multiplication and "^" for exponentiation. As for the homogeneous degree, I would expect (for $E_2(q^n)$) to be $2\psi(n)=2n\prod_{p\mid n}(1+1/p)$. So, again six for $n=5$. Have you tried $E_4(q^n)$ and $E_6(q^n)$ as well? $\endgroup$ May 7, 2010 at 9:21

I just stumbled across this old question. I don’t know if you are still interested in it, but here's some thoughts.

If $F(z)$ is any modular form for $\Gamma_0(N)$ of weight $k$, then $F(z)$ has a modular polynomial given by $$\Phi_F(X)= \prod_{\gamma \in \Gamma_0(N)\backslash \textrm{SL}_2(\mathbb {Z})} (X- F(z)|_k \gamma ),$$ where $|_k$ is the usual weight $k$ Petersson slash operator. The action of $\textrm{SL}_2(\mathbb {Z})$ permutes the roots of this polynomial transitively, and so its coefficients are all modular forms for $\textrm{SL}_2(\mathbb {Z})$, and it is either irreducible or a perfect power of an irreducible polynomial over the ring of modular forms for $\textrm{SL}_2(\mathbb {Z})$. The degree the of the polynomial is the index $\Gamma_0(N)$ in $\textrm{SL}_2(\mathbb {Z})$, which is $N\prod_{p|N} (1+\frac1p). $ If you assign $X$ the weight $k$, the polynomial is clearly homogeneous, since it is a product of homogeneous terms. If $F(z)$ the expansion of $F$ at every cusp has integer coefficients, then the coefficient forms of the polynomial will have integer coefficients.

In your case, we’re starting with $F$ a modular form for $\textrm{SL}_2(\mathbb {Z})$, and considering the polynomial for $F(Nz)$. In this case, the polynomial must be irreducible rather than a perfect power, since the individual roots can be clearly seen to be distinct. If $N$ is square free, the second coefficient will be a power of $N$ times $-F(z)|T_N$, where $T_N $ is the $N$-th hecke operator (more generally, linear combinations of $T_{N/d^2}$ where $d|N$). Using Newton’s identities for symmetric functions, you can write every coefficient in terms of the Hecke operators acting on powers of $F(z)$, so bounds for the coefficients come down to bounds on Hecke eigenvalues. The expansion of $F(Nz)$ at other cusps will generally have rational coefficients, not integer, so you may need to multiply by powers of $N$ to clear denominators in the polynomial.

We can do the same construction thing for $E_2$, by considering instead $E_2^*(z):= E_2(z)-\frac{3}{\pi y}$, which is modular for $\textrm{SL}_2(\mathbb {Z})$, not just quasi-modular. Here $y$ is the imaginary part of $z$. The polynomial for $E_2^*(Nz)$, is in $\mathbb{Q}[X,E_2, E_4,E_6, \frac{1}{\pi y}]$. If we simply ignore every component with a positive power of $\frac{1}{\pi y},$ we obtain the quasi-modular polynomial for $E_2(Nz).$

  • 1
    $\begingroup$ Hi Michael, I am still very interesting in this development (and therefore verifying your "drop positive positive powers of $1/(\pi y)$ claim") when I get a moment for that. Thanks for pointing out this simple-looking option to me. $\endgroup$ Mar 17, 2019 at 21:04

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