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.