# Prove that $X$ is a Riemann surface using basic definitions

$f$ is a polynomial function; $a$, $b$ are complex variables.

$$X =\{(a,b)|f(a,b)=0\}\subset \mathbb C\times\mathbb C$$

When $f=0$, the complex gradient of $f$ is non zero:

$$\forall (a,b), s.t. (\frac{\partial f}{\partial a}\frac{\partial f}{\partial b})(a,b)\neq 0 \ \ \text{when} f(a,b)=0.$$

Prove that $X$ is a Riemann surface.

I tried to use the fundamental definition of Riemann surface. I think $f$ must be a holomorphic function, so we could find a collection of open sets s.t. $\cup _\alpha U_\alpha = X$, and we could holomorphically map the $U_\alpha$ onto open sets of $\mathbb C$

By Implicit function THM we could show that there exists a function $g$, such that $X$ coincides with the set of all points $(a,g(a))$. But how to prove that $(a,g(a))$ is locally homeomorphic to $\mathbb C$ everywhere?

Probably,

$$h:X\to\mathbb C$$ $$(a,g(a))\mapsto a$$ is a diffeomorphism from $X$ to $\mathbb C$.

But why do we need the slope of tangent lines to be non-zero?

Since we are only using one chart here, we do not have to prove that the transition map is holomorphic?

• Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. Jan 31, 2018 at 0:34
• @JoséCarlosSantos Edited...
– dodo
Feb 1, 2018 at 6:44

First, is false that "exists a function $g$, such that $X$ coincides with the set of all points $(a,g(a))$". The condition only is true locally (in a "chunk" of $X$). And possibly other chunk of $X$ is the image set of a function $b\mapsto(g(b),b)$. Think in the simpler (and real) example of the unit circle $x^2 + y^2 = 1$.
Second, a Riemann surface is more than a set. You need a complex structure: a family of charts with holomorphic transition maps. The functions $(a,g(a))\mapsto a$ and $(g(b),b)\mapsto b$ are the obvious choice.
• @dodo, yes. But Hausdorff is trivially true (subspace of $\Bbb C\times\Bbb C$) and connected must be easy. Feb 1, 2018 at 9:11