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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).

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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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    Okay, so locally Noetherian is enough. Cool. Thanks!2012-10-21