non-singular Riemann surface implies irreducible polynomial without connectedness?

Question

Let $$ F(w,z) = \sum_{i=0}^n a_i(z)w^{n-i}$$ be a polynomial in $z,w$. Define a Riemann surface as the set $$\Gamma:= \left\{ (z,w)\in \mathbb C^2 \mid F(z,w)=0 \right\} $$ and call it non-singular if $$\left(\frac{\partial F}{\partial z},\frac{\partial F}{\partial w}\right)(z_0,w_0) \neq 0$$ for every point $(z_0,w_0) \in \Gamma$.

Question: I don't see how the non-singularity condition implies irreducibility of $F$, i.e. does $\Gamma$ connected follow from any of the above assumptions?

Answer 1

Non-singularity does not imply irreducibility: Take $$F(z,w) := (z-z_0)(z-z_1), z_0 \neq z_1.$$ The zero set are two disjoint copies of $\mathbb C$.