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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.

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    $\begingroup$ Certainly not always iso. You can have ${\rm Spec}\, k[x]/<x^2>\to {\rm Spec}\, k[x]/<x^3>$. $\endgroup$
    – Lev Borisov
    Commented Mar 12, 2014 at 12:40
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    $\begingroup$ 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. $\endgroup$
    – user76758
    Commented Mar 12, 2014 at 13:00
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    $\begingroup$ 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. $\endgroup$
    – Damian Rössler
    Commented Mar 12, 2014 at 14:11
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    $\begingroup$ consider algebraic geometry a first course, p. 179. $\endgroup$
    – roy smith
    Commented Mar 13, 2014 at 2:24
  • $\begingroup$ 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). $\endgroup$
    – Gro-Tsen
    Commented Jul 8, 2021 at 9:43

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