I am struggling with the following problem. Given a complex polynomial $f : \mathbb{C}^n \rightarrow \mathbb{C}$ such that $\nabla f$ doesn't vanish anywhere on the whole $\mathbb{C}^n$, $V = V(f)$ is irreducible.
I think we can solve the problem with a topological argument by saying that $f : \mathbb{C}^n \rightarrow \mathbb{C}$ is a fiber bundle for the Euclidean topology and since $\mathbb{C}$ is contractible, the bundle is trivial hence $\mathbb{C}^n \cong \mathbb{C} \times F$ where $F$ is the fiber. We deduce that $F$ is connected for the Euclidean topology thus it is irreducible for the Zariski topology (because it is smooth). However, it seems to be a convoluted argument and I would like the proof to be generalisable in any algebraically closed field.
I tried by contradiction, if $V$ is reducible, then $f = gh$ for some non constant $g,h$ and $\nabla f = g\nabla h + h\nabla g \neq 0$ implies that $V(g) \cap V(h) = \emptyset$ so there exists polynomials $u,v$ such that $gu + hv = 1$. We deduce that, $fu + h^2v = h$ so, on $D(uv)$, $$ f = -\frac{v}{u}\left(h - \frac{1}{2v}\right)^2 + \frac{1}{4uv}. $$ In particular, $\nabla f = \frac{1}{4}\nabla\!\left(\frac{1}{uv}\right) + \mathrm{O}\!\left(h - \frac{1}{2v}\right)$. The contradiction is obvious if $uv$ is constant but not in the general case and I am stuck here. Please tell me if you have any idea (or counterexample but I doubt it exists) !