Lemma 33.25.2. Let k be a field. Let X be a scheme over k. Assume
X is locally of finite type over k,
\Omega _{X/k} is locally free,
X is reduced, and
k is perfect.
Then the structure morphism X \to \mathop{\mathrm{Spec}}(k) is smooth.
Lemma 33.25.2. Let k be a field. Let X be a scheme over k. Assume
X is locally of finite type over k,
\Omega _{X/k} is locally free,
X is reduced, and
k is perfect.
Then the structure morphism X \to \mathop{\mathrm{Spec}}(k) is smooth.
Proof. Let x \in X be a point. As X is locally Noetherian (see Morphisms, Lemma 29.15.6) there are finitely many irreducible components X_1, \ldots , X_ n passing through x (see Properties, Lemma 28.5.5 and Topology, Lemma 5.9.2). Let \eta _ i \in X_ i be the generic point. As X is reduced we have \mathcal{O}_{X, \eta _ i} = \kappa (\eta _ i), see Algebra, Lemma 10.25.1. Moreover, \kappa (\eta _ i) is a finitely generated field extension of the perfect field k hence separably generated over k (see Algebra, Section 10.42). It follows that \Omega _{X/k, \eta _ i} = \Omega _{\kappa (\eta _ i)/k} is free of rank the transcendence degree of \kappa (\eta _ i) over k. By Morphisms, Lemma 29.28.1 we conclude that \dim _{\eta _ i}(X_ i) = \text{rank}_{\eta _ i}(\Omega _{X/k}). Since x \in X_1 \cap \ldots \cap X_ n we see that
Therefore \dim _ x(X) = \text{rank}_ x(\Omega _{X/k}), see Algebra, Lemma 10.114.5. It follows that X \to \mathop{\mathrm{Spec}}(k) is smooth at x for example by Algebra, Lemma 10.140.3. \square
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