Generic smoothness

Remarks

The proof reduces to the case where $f$ is dominant and then compares tangent spaces. In characteristic $0$, the field extension $k(Y) \subseteq k(X)$ is separable, so the map on differentials is generically surjective. Then $\Omega_{X/Y}$ is locally free of the right rank over a dense open set, which gives smoothness there. For example, the map $\mathbb{A}^2_k \to \mathbb{A}^1_k$, $(x, y) \mapsto x^2 + y^2$, is smooth over $V = \mathbb{A}^1_k \setminus \set{0}$. Its fibre over $0$ is the pair of lines $x = \pm i y$, which is singular at the origin. The characteristic hypothesis is necessary. In characteristic $p > 0$, the Frobenius map $\mathbb{A}^1_k \to \mathbb{A}^1_k$, $t \mapsto t^p$, is a morphism of smooth curves. Its fibre over $a$ is $\operatorname{Spec} k[t]/(t^p - a) = \operatorname{Spec} k[t]/((t - b)^p)$ with $b^p = a$, a non-reduced point, so it is smooth over no nonempty open set.

Going deeper

For a student

Generic smoothness is the algebraic counterpart of Sard's theorem on critical values of smooth maps. It is the main tool behind Bertini's theorem: a general hyperplane section of a smooth projective variety in characteristic $0$ is smooth. It also shows that, in characteristic $0$, in a family with smooth total space over a curve, only finitely many fibres are singular. Its proof uses that every finitely generated field extension in characteristic $0$ is separably generated.