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Aug 24, 2017 at 21:50 vote accept dorebell
Aug 24, 2017 at 15:08 review Close votes
Aug 25, 2017 at 9:10
Aug 24, 2017 at 13:29 answer added Jason Starr timeline score: 5
Aug 24, 2017 at 4:02 comment added dorebell Thanks, Jason! If you'll submit this as an answer, I'll accept it.
Aug 22, 2017 at 10:44 comment added Jason Starr For the field $\mathbb{R}$, the affine $\mathbb{R}$-scheme $\text{Spec} \ \mathbb{R}[x,y]/\langle x^2+y^2\rangle$ is integral and geometrically connected, but it is not geometrically irreducible. If $X_k$ is an integral, locally finite type $k$-scheme that is geometrically connected and normal, then $X_{\overline{k}}$ is irreducible. For geometric irreduciblity it is irrelevant whether $k$ is perfect: the field extension $\overline{k}/k^{\text{sep}}$ is a universal homeomorphism. However, perfectness is relevant for reducedness.
Aug 22, 2017 at 10:34 history asked dorebell CC BY-SA 3.0