Let $P,Q,R$ be the Fourier series of the Eisenstein series $E_2,E_4,E_6$, that is,

$$ P(q)=1-24\sum_{n=1}^{\infty}\sigma_1(n)q^n, $$

$$ Q(q)=1+240\sum_{n=1}^{\infty}\sigma_3(n)q^n, $$

$$ R(q)=1-504\sum_{n=1}^{\infty}\sigma_5(n)q^n. $$

I heard that $P,Q,R$ are algebraically independent over $\mathbb C(q)$, due to Mahler, but I cannot find a reference. Could you help me if you know the reference?

Meanwhile, I found 'On Algebraic Differential Equations Satisfied by Automorphic Functions', by Mahler, 1969. I do not think that this paper addresses this topic, but if it does, why does it imply the algebraic independence of $P,Q,R$?

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    $\begingroup$ The problem for just $Q$ and $R$ is settled at mathoverflow.net/questions/282554/… $\endgroup$ – Gerry Myerson Oct 30 '18 at 11:55
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    $\begingroup$ What about $\mathbb C (z) $? $\endgroup$ – LWW Oct 30 '18 at 12:05
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    $\begingroup$ @GerryMyerson OP's question is about independence over $\mathbb{C}(z)$, not $\mathbb{C}$, so a harder question. It may be that using the functional equation for the Eisenstein series will reduce to the easier problem but it's not obvious. A harder question, that probably gives an answer to OP's question, is the Mahler-Manin conjecture, proved by K. Barré-Sirieix et al., (Invent. Math. 124 (1996), 1-9). That paper might contain a proof or a reference to OP's question. $\endgroup$ – Felipe Voloch Oct 30 '18 at 16:08
  • $\begingroup$ Voloch // why Mahler-Manin conjecture implies the question?? I didn't understand. $\endgroup$ – LWW Oct 31 '18 at 2:25
  • $\begingroup$ If you find $z$ such that $P(z),Q(z),R(z)$ are all algebraic, then by Mahler-Manin, $z$ is transcendental. This wouldn't happen if the functions were algebraically dependent. Have a look at the comment after Theorem 4 of Waldschmidt's Bourbaki talk n° 824. I think it completely answers your question. $\endgroup$ – Felipe Voloch Oct 31 '18 at 4:53

I think the follwing is the answer of this question. It is not depend on the above comments because I didn't understand them. This is just my approach. (But not fully my idea because it depends on a strong proposition which is already known, and the remaining part is just a corollary.)

Proposition. Let $f$ be a modular form of weight $k$, defined over $\mathbb Q^{\mathrm{alg}}$. If $f$ is non constant, then the functions $f, Df$ and $D^2 f$ are algebraically independent over the function field $\mathbb C(q)$.

The proof is in the book, 'Introduction to Algebraic Independence Theory', Y.V.Nesterenko and P.Philippon, (Eds), Ch.1 prop 1.1. Here $q=e^{2\pi i \tau}$ and $D=\frac{1}{2\pi i}{\tau}=q\frac{d}{dq}$.

Let $\Delta = E_4^3-E_6^2$, cusp form of weight 12. Due to Ramanujan, we have the following formulas :

$$ DP=\frac{1}{12}(P^2-Q), \quad DQ=\frac{1}{3}(PQ-R), \quad DR=\frac{1}{2}(PR-Q^2). $$

Thus we have $D\Delta = P\Delta$. Take $D$ in both side and do some elementary algebra, we have

$$ Q=13\frac{(D\Delta)^2}{\Delta^2}-12\frac{D^2 \Delta}{\Delta}. $$

By the proposition, $\{\Delta, D\Delta, D^2\Delta\}$ is algebraically independent over $\mathbb C(q)$, so $\{Q^3-R^2,P,Q\}$ is algebraically independent over $\mathbb C(q)$, so $\{P,Q,R^2\}$ is algebraically independent over $\mathbb C(q)$. This is equivalent to say that $R^2$ is transcendental over $\mathbb C(q)[P,Q]$, and this implies that $R$ is transcendental over $\mathbb C(q)[P,Q]$, i.e. $\{P,Q,R\}$ is algebraically independent over $\mathbb C(q)$.

  • $\begingroup$ It's worth mentioning Zagier's generalization of Jacobi and Ramanujan's D-algebraic dependence, see people.mpim-bonn.mpg.de/zagier/files/doi/10.1007/… $\endgroup$ – Igor Pak Nov 4 '18 at 20:06
  • $\begingroup$ Pak // Could you let me know where can I find those contents? I tried to find it in the textbook you refered but fail... $\endgroup$ – LWW Nov 4 '18 at 22:46
  • $\begingroup$ Don't remember the exact place. Start with p. 49 for D-alg eq. but also see short section on p. 85 for alg independence. $\endgroup$ – Igor Pak Nov 4 '18 at 22:52

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