TITLE: Why is a smooth connected scheme irreducible?
QUESTION [14 upvotes]: Why is a smooth connected scheme (say over a field) necessarily irreducible? 
Intuitively it makes sense because we might very well expect points in the intersection of two irreducible components to be singular points.
But what is a proof? Feel free to add any extra hypotheses if needed (e.g., separated if that is required).

REPLY [23 votes]: The local rings of a smooth scheme over a field are regular, and a regular local ring is a domain.  Thus a smooth scheme over a field has all local rings being domains.  Thus the intersection of any two components must be empty (a point lying on the intersection would not have its local ring being a domain).