unramified morphisms?

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Mar 17, 2021 at 10:13	comment	added	Peter Scholze		In characteristic $p$, the map $\mathbb F_p\to \mathbb F_p[T^{1/p^\infty}]$ is formally etale. In fact, any map between perfect rings of char $p$ is formally etale. I don't think they should count as "etale" or "smooth".
S Oct 7, 2020 at 7:40	history	suggested	KReiser	CC BY-SA 4.0	Added link to MSE post
Oct 7, 2020 at 6:16	review	Suggested edits
S Oct 7, 2020 at 7:40
Oct 6, 2020 at 18:00	comment	added	D.-C. Cisinski		@LanceWu More precisely, the property that the cotangent complex is perfect is equivalent to the property that, locally, the scheme is lci, which precisely means that we are defined by equations. If you add formal smoothness to that, you get classical smoothness.
Oct 6, 2020 at 17:29	comment	added	D.-C. Cisinski		@Lance Wu I insist on the finiteness, not the projectiveness! This is essential to prove finiteness results (e.g. that suitable cohomology are of finite type, or that some sheaves are constructible).
Oct 6, 2020 at 16:45	comment	added	Z Wu		@Denis-CharlesCisinski From Tag06B5, we know the cotanget sheaf of formally smooth morphism is locally projective, hence locally free. I believe this is nice enough to be called smooth-like. But the coordinate idea is good.
Oct 5, 2020 at 23:21	comment	added	D.-C. Cisinski		It is often useful in geometry to be allowed to work with coordinates, at least locally. If you think of smooth morphisms as being (Zariski) locally defined by a system of equations satisfying the jacobian criterion, then you end up with the notion of smooth morphism with the classical finitess properties. Put in other words, having a perfect cotangent complex is a very useful property, and if you are smooth-like, such a cotangent complex will be concentrated in degree zero, hence a locally free module of finite type. Such properties are useful in many ways in practice.
Oct 5, 2020 at 17:42	answer	added	Harry Gindi		timeline score: 6
Oct 5, 2020 at 16:12	comment	added	Harry Gindi		I guess I'll add it as an answer since people agreed with it.
Oct 5, 2020 at 16:06	comment	added	Z Wu		@HarryGindi Thanks, I didn't get it the first time, then I googled that submersions and covering spaces are open maps, now I get it.
Oct 3, 2020 at 21:50	comment	added	Harry Gindi		Chevalley's theorem guarantees that smooth and étale morphisms will be open maps (as they are flat, locally of fin pres), which isn't true for formally smooth/étale.
Oct 3, 2020 at 21:23	history	asked	Z Wu	CC BY-SA 4.0