Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f: X \to Y$ be a morphism of schemes over a field $k$ such that $f$ induces (1) a bijection between their closed points, and (2) an isomorphism of their Zariski tangent spaces.

Under these conditions, is $f$ always an isomorphism? And if not, under what additional (mild, hopefully) assumptions could one conclude it is an isomorphism?

I'm looking, in particular, for references with a precise statements and proofs that (under some conditions) $f$ is an isomorphism.

For example, maybe $X$ and $Y$ need to be smooth, or satisfy some finiteness condition.. or maybe $f$ is always an isomorphism, I don't know.

share|improve this question
3  
Certainly not always iso. You can have ${\rm Spec}\, k[x]/<x^2>\to {\rm Spec}\, k[x]/<x^3>$. –  Lev Borisov Mar 12 at 12:40
3  
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. –  user76758 Mar 12 at 13:00
1  
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. –  Damian Rössler Mar 12 at 14:11
    
consider algebraic geometry a first course, p. 179. –  roy smith Mar 13 at 2:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.