11
$\begingroup$

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).

1 Answers 1

17

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).

  • 7
    Well, you reduced the question to «why would a point in the intersection not have its local field a domain?» :)2011-02-05
  • 1
    Dear @Matt, Isn't some quasi-compactness hypothesis necessary here? I know that a connected, regular, Noetherian (i.e. quasi-compact) scheme $X$ is necessarily irreducible, using your argument, because one gets that $X$ is the disjoint union of its finitely many irreducible components, so by connectedness there can only be one. But what if there are infinitely many irreducible components? I definitely don't have a counter-example in mind. I just can't think of a proof.2012-10-21
  • 1
    @KeenanKidwell: Dear Keenan, Smoothness of a morphism includes a *locally of finite type* hypothesis, so a smooth scheme over a field is locally of finite type by definition, and so every point has a neighbourhood with only finitely many irreducible components. This should be enough. Regards,2012-10-21
  • 0
    Okay, so locally Noetherian is enough. Cool. Thanks!2012-10-21