Considering the set $$X=\{(z,w)\in \mathbb{C}^2 \mid P(z,w)=0\},$$ $P:\mathbb{C}²\rightarrow \mathbb{C}$ holomorphic in each variable separately, including any polynomial in $z$ and $w$ and $P$ is non-singular. Consequently $X$ is a Riemann surface.
If I take $P(z,w)=w²-z$ it's clear that this function is holomorphic in each variable and $P$ is polynomial. But if I take $P(z,w)=\exp(w)-z$, is this also a polynomial in $z$ and $w$ (because $\exp(w)$ can be represented as a power series, but is this enough to say $\exp(w)$ is a polynomial?)
Furthermore is $P(z,w)=w²-z$ irreducible and why?
Thank you very much! :)
No, $\exp w$ is not a polynomial: if it were, its derivatives of sufficiently high order would be $0$ – and $\exp^{(n)}(w)=\exp w$ for all $n$.
As to $P(z,w)= w^2-z$ being irreducible; you can apply Eisenstein's criterion in the univariate polynomial ring $\mathbf C[z][w]$ over the P.I.D. $\mathbf C[z]$.