Is an irreducible regular algebraic curve, connected? Motivated by this MO post we ask:
Assume that $H(x,y)$ is  a real polynomial which is irreducible as a complex polynomial. Is it true to say that $H^{-1}(0)$  ($H:\mathbb{C}^{2} \to \mathbb{C}$) is  a  connected subset of $\mathbb{C}^{2}$?
Moreover what is an example of a connected regular curve in $\mathbb{C}^{2}$ which intersection with $\mathbb{R}^{2}$ contains at least 2 distinct closed curve?
By "closed curve" I mean a simple Jordan curve.
 A: 1) The question has nothing to do with algebraic geometry but is pure topology: every irreducible topological space is connected.
Indeed, by definition  an irreducible space $X$ is not  the union of two closed strict subsets of $X$. 
Hence a fortiori it is not a union of two disjoint closed strict subsets of $X$, which by definition of "connected" proves that $X$ is connected. 
2) The curve $C$ given by $y^2 =x(x-1)(x-2)(x-3)(x-4)$ in $\mathbb C^2$ is smooth and irreducible hence connected.
Its real trace $C\cap \mathbb R^2$ however has two compact connected components, each homeomorphic to a circle.
A: $H^{-1}(0)$ is clearly the curve defined by $H(x,y)=0$ and any irreducible variety over $\mathbb{C}$ is connected. For the second question, elliptic curves over $\mathbb{C}$ are connected, but can have two connected components over $\mathbb{R}$. For example, consider the curve $y^2=x^3-x$. Over $\mathbb{C}$ this curve is connected since the equation is irreducible. But over reals, since the map to the $x$-axis is proper, suffices to show that the image is not connected. But the image consists of $x$ such that $x^3-x\geq 0$, since it must have a square root. Easy to check that these have two pieces, $x\geq 1$ and $-1\leq x\leq 0$.
