As stated in the title I am trying to compute the transcendence degree over $\mathbb{C}$ of the field of fractions of $\mathcal{O}(\mathbb{C})$. (I know that it is infinity, but I need to prove it.) The teacher gave it to us an exercise. He advised us to use the set $E = \{ e^{z^n}, n \geq 1\}$. I still don't have found the proof of the algebraic independence of this set, but I thought the set $F = \{ z^n, n \geq 1\}$ should work as well. A finite sum of elements of $F$ with coefficient in $\mathbb{C}$ is a polynomial and to say it is identically zero means all the coefficient are zero. Does it work? If not, would someone give me some hint to prove the algebraic independence of $E$?
Thank you all!
Let me show that $e^{z^3}$and $e^{z^7}$ are algebraically independent.
Suppose some polynomial, say $f(x,y)=i\sqrt 2 xy^2 +\pi y^3-17x^4y$, kills that pair of entire functions, so that $$f(e^{z^3},e^{z^7})=i\sqrt 2 e^{z^3}(e^{z^7})^2 +\pi (e^{z^7})^3-17(e^{z^3})^4e^{z^7}=i\sqrt 2 e^{z^3+2z^7} +\pi e^{3z^7}-17e^{4z^3+z^7}=0$$.
Dividing out by $e^{3z^7}$ we get that for all $z\in \mathbb C$$$i\sqrt 2 e^{z^3-z^7} +\pi - 17e^{4z^3-2z^7}=0$$ Letting $z\in \mathbb R$ tend to $+\infty$, we get $\pi=0$, a well-known falsity.
In high school they learn how to replace explicit numbers like $7, \sqrt 2$ by letters $a,b,c,...$ and even by cascades of letters like $a_{i_1...i_k}x_{i_1}^{\nu_{i_1}}\cdots x_{i_k}^{\nu_{i_k}}$: ask those guys to replace numbers by letters in the above calculation to see how the result could be proved in the general case....