$E_4(z)= - \frac{B_4}{8}+ \sum_{n=1}^\infty \sigma_3(n) q^n$


$E_6(z)= - \frac{B_6}{12}+ \sum_{n=1}^\infty \sigma_5(n) q^n$

How does one show they are algebraically independant over $\mathbb{C}$ ?


First of all, $E_4$ and $E_6$ are modular forms of weights 4 and 6. Therefore, if we have $P(E_4,E_6)=0$ for some nonzero polynomial $P$, then there exists some polynomal $G$ such that $G(t^4x,t^6y)=t^kG(x,y)$ for all $x,y,t$. (That is true, because $P(E_4,E_6)$ is always equal to the sum of some modular forms of different weights, that are polynomials in $E_4$ and $E_6$, and modular forms of different weights are obviously linearly independent). So, we should have $G(E_4,E_6)=0$ for some polynomial $G$ of the form

$$G(x,y)=\sum_{4n+6m=k} a_{nm} x^ny^m,$$

where $k$ is some positive integer. Let us choose $G$ with minimal possible degree. Now, $E_4(e^{2\pi i/3})=0$ and $E_6(i)=0$. But $G(x,y)=Ax^u+By^v+xyF(x,y)$ for some complex $A$ and $B$ and $4u=6v=n$. If $A$ or $B$ is nonzero, then we would have $G(E_4,E_6)(e^{2\pi i/3})\neq 0$ or $G(E_4,E_6)(i)\neq 0$, which is impossible. Therefore, $G(x,y)=xyF(x,y)$, so $F(E_4,E_6)=0$, which contradicts the minimality.


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