For background, I'm just starting to learn a little bit about classical algebraic geometry, so I'm trying to look at some simple problems that are as concrete as possible to develop an intuition.
This question is similar to this one, but I'm interested in when I get multiple disconnected segments in the solution set as opposed to whether the solution set is discrete or not.
Let $p(x, y)$ be a polynomial in at most $\{x, y\}$ with a zero constant term. $0$ is a valid value for $p(x, y)$, as is $x^2$, $y^3$ and $xy+4x^2y^7$. $4$ is not a valid value for $p(x, y)$. I'm looking at sets of the form $\{ (x, y) : p(x, y) = c \}$.
Suppose $X$ is a subset of $\mathbb{C}^2$. I'll say $X$ is path-connected if and only if, for every pair points $(p, q)$, there exists a continuous function $f : [0, 1] \to \mathbb{C}^2$ such that $f(0) = p$ and $f(1) = q$ and $f([0, 1]) \subset X$. Using a result here, a subset of $\mathbb{C}^n$ is connected if and only if it is path connected, so I'll just say that $X$ is connected for simplicity.
The solution set of $x^2 = 1$ is the union of $\{ (1, y) : y \in \mathbb{C}\}$ and $\{(-1, y) : y \in \mathbb{C}\}$. This set is not connected.
It seems pretty clear that when $p(x, y)$ is of the form $x^n$ or $y^n$ for $n \ge 2$ and $c \neq 0$, we get a disconnected set. So these conditions are sufficient.
However, I'm wondering what conditions are necessary and sufficient.