Let 


$E\_4(z)= - \frac{B\_4}{8}+ \sum\_{n=1}^\infty \sigma\_3(n) q^n$ 


and 


$E\_6(z)= - \frac{B\_6}{12}+ \sum\_{n=1}^\infty \sigma\_5(n) q^n$ 


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



Eisenstein series $E_4(z)$ and $E_6(z)$ algebraically independent 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.


