# Why is a smooth connected scheme irreducible?

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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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. – Keenan Kidwell Oct 21 '12 at 4:23