Zero set of polynomials in two variables over complex field contain a non-trivial connected subset? Let  $f \in \mathbb C[x,y]$ be an irreducible polynomial , then is it true that the set $\{(a,b)\in \mathbb C^2 : f(a,b)=0\}$ contains a connected subset with more than one point ? If the statement is not in general true , then what happens if we also assume $f$ is homogenous ? ( I only know that the zero set must be uncountable )
 A: I have good news for you: given an irreducible polynomial $f\in \mathbb C[x,y]$, the whole of its zero set $$V(f)=\{(a,b)\in \mathbb C^2\vert f(a,b)=0\}\subset \mathbb C^2$$ is already connected  in the topology inherited from $\mathbb C^2$ (in its classical metric topology).
This is a special case of a vastly more general theorem: 
Theorem
Given an irreducible  algebraic variety $X$ over $\mathbb C$, its associated holomorphic variety $X(\mathbb C)$ is connected in its classical topology.  
(The classical topology of $X(\mathbb C)$ is the one it inherits from its local embeddings into $\mathbb C^n$)
This applies to your situation, since your irreducible polynomial $f$ is easily  seen (without invoking the Nullstellensatz!) to define an irreducible algebraic variety $X=V(f)$ .  
References
The Theorem is proved in Shafarevich's Basic Algebraic Geometry , volume 2, Chapter VII, Section 2, Theorem 1, page 127.
The irreducibility assertion is in Fulton's online notes, Chapter 1, Section 1.6, Corollary 3, page 9.
