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Timeline for What are unramified morphisms like?

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Oct 1, 2015 at 2:18 comment added PrimeRibeyeDeal We have the normalization map $\mathbb{A}^1 \to C$ by $a\mapsto (a^2, a^3)$ of the cuspidal cubic as well as a map $C \to \mathbb{A}^1$ given by projection onto either factor. If I'm not mistaken, these give us a finite surjective map that is unramified (resp. smooth) but not smooth (resp. unramified). And the singularity is also geometrically unibranch.
Nov 17, 2010 at 16:32 comment added Laurent Moret-Bailly @BCnrd: a slightly more general hypothesis is "geometrically unibranch". See EGA IV-4, (18.10.1).
Nov 16, 2010 at 5:22 vote accept Yuhao Huang
Nov 16, 2010 at 4:46 comment added BCnrd Oh drat, I was being an idiot; somehow I fooled myself into thinking the normalization of the nodal cubic is not unramified. But it is, as Emerton properly explains. OK, so that's the end of this one: normality is the "right" hypothesis for flatness to be automatic.
Nov 16, 2010 at 4:33 history edited Emerton CC BY-SA 2.5
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Nov 16, 2010 at 4:17 comment added BCnrd I have been wondering about this for a bit of time (with reduced connected noetherian schemes), but am not yet satisfied. In the connected normal (implies irreducible) noetherian case flatness is easy to verify by passing to strict henselizations (which are domains, by normality) -- this argument already appears early in the Freitag-Kiehl book, for example -- and my feeling is that without normality there should be counterexamples, even with irreducibility. No success yet in finding one. Need some interesting non-normal singularities.
Nov 16, 2010 at 4:02 comment added Yuhao Huang See my edit: what if I add the condition on connectedness? I'm wondering if there is anything can imply the flat locus is both open and closed. Thank you.
Nov 16, 2010 at 3:00 history answered Emerton CC BY-SA 2.5