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 Answer 1


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).

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    $\begingroup$ Well, you reduced the question to «why would a point in the intersection not have its local field a domain?» :) $\endgroup$ Feb 5, 2011 at 13:44
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    $\begingroup$ 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. $\endgroup$ Oct 21, 2012 at 4:23
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    $\begingroup$ @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, $\endgroup$
    – Matt E
    Oct 21, 2012 at 12:43
  • $\begingroup$ Okay, so locally Noetherian is enough. Cool. Thanks! $\endgroup$ Oct 21, 2012 at 14:22

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