TITLE: A complex algebraic variety which is connected in the usual topology
QUESTION [10 upvotes]: Hartshorne wrote in his book's Appendix B that it can be easily proved that a complex algebraic variety is connected in the usual topology if and only if it is connected in Zariski topology.
How can it be proved?

REPLY [10 votes]: Fredrik already explained why the connectedness for the complex topology implies the connectedness for the Zariski topology. 
The converse is harder. In the projective case, see Georges's wonderful answer. Let me give a reference for the general case. Let $X_1, \dots, X_n$ be the (Zariski) irreducible components of $X$. By Shafarevich, Basic Algebraic Geometry 2, Chapter VII, § 2.2, Theorem 1, $X_i(\mathbb C)$ is connected in complex topology for all $i\le n$. Let $Y$ be a connected component of $X(\mathbb C)$. Then $Y$ is the union of some $X_i(\mathbb C)$ and $X(\mathbb C)\setminus Y$ is the union of the other $X_i(\mathbb C)$'s. So $Y$ and $X\setminus Y$ are both Zariski closed and disjoint. This implies that $X$ is not Zariski connected if $X(\mathbb C)$ is not connected for the complex topology.