Timeline for Given a morphism of schemes, when does bijective + isomorphic tangent spaces = isomorphism?
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7 events
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| Jul 8, 2021 at 9:43 | comment | added | Gro-Tsen | For completeness of MathOverflow links: this question is pretty much the same, and has meaningful answers (a positive answer under smoothness conditions, and a negative answer merely with integrality). | |
| Mar 13, 2014 at 2:24 | comment | added | roy smith | consider algebraic geometry a first course, p. 179. | |
| Mar 12, 2014 at 14:11 | comment | added | Damian Rössler | See Lemma 2.4, p. 172 in Cornell-Silverman, 'Arithmetic Geometry' (article by Milne) for a statement in the direction of what you are looking for. | |
| Mar 12, 2014 at 13:00 | comment | added | user76758 | Presumably you mean for $X$ and $Y$ to be of finite type over $k$, but even when $k$-smooth it isn't true: consider ${\rm{Spec}}(k') \rightarrow {\rm{Spec}}(k)$ for a nontrivial finite extension $k'/k$ (separable to ensure smoothness, say), as the tangent spaces are then 0. But if also isomorphism between residue fields at closed points then it suffices that $Y$ is smooth and $X$ is either smooth or geometrically connected over $k$. This is an exercise with etale morphisms. | |
| Mar 12, 2014 at 12:43 | review | First posts | |||
| Mar 12, 2014 at 12:58 | |||||
| Mar 12, 2014 at 12:40 | comment | added | Lev Borisov | Certainly not always iso. You can have ${\rm Spec}\, k[x]/<x^2>\to {\rm Spec}\, k[x]/<x^3>$. | |
| Mar 12, 2014 at 12:27 | history | asked | user48134 | CC BY-SA 3.0 |