Wikipedia defines a transcendental entire function as an entire function that is not polynomials. Is the following true?
Given $f(z)$ an entire function, if there exists a polynomial $P \in \mathbb{C}[X,Y]$ with the property that $P(z,f(z)) = 0$ but $P$ is not identically zero, then $f(z)$ is a polynomial.
My initial thought is yes, since any entire function has an infinite Taylor series that converges everywhere in the complex to the function, and so any polynomial $P\in \mathbb{C}[X,Y]$ of finite degree in $X$ will not have the property that $P(z,f(z)) = 0$. Is there any way to take this idea and make it into a more concrete proof? If not, is there a reference I could find with this result?
Here is a proof (there is probably a more elementary argument, but I don't see one at the moment). If $f$ is not a polynomial, then it has an essential singularity at $\infty$, so by Big Picard $f$ takes all but at most one value infinitely many times. So, for any fixed $a\in\mathbb{C}$ (with at most one exception), $P(z,a)$ has infinitely many roots, namely all the values of $z$ such that $f(z)=a$. But $P(z,a)$ is just a polynomial in $z$, so this means that $P(z,a)=0$ Thinking of $P(z,w)$ as a polynomial in $z$ with coefficients which are polynomials in $w$, this means that each coefficient vanishes for all values of $w$ with at most one exception, and so each coefficient must be the $0$ polynomial. Thus $P$ must be the $0$ polynomial.