Lemma 33.25.2. Let $k$ be a field. Let $X$ be a scheme over $k$. Assume

1. $X$ is locally of finite type over $k$,

2. $\Omega _{X/k}$ is locally free,

3. $X$ is reduced, and

4. $k$ 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 29.14.6) 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 10.24.1. Moreover, $\kappa (\eta _ i)$ is a finitely generated field extension of the perfect field $k$ hence separably generated over $k$ (see Algebra, Section 10.41). 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 29.27.1 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

$\text{rank}_ x(\Omega _{X/k}) = \text{rank}_{\eta _ i}(\Omega _{X/k}) = \dim (X_ i).$

Therefore $\dim _ x(X) = \text{rank}_ x(\Omega _{X/k})$, see Algebra, Lemma 10.113.5. It follows that $X \to \mathop{\mathrm{Spec}}(k)$ is smooth at $x$ for example by Algebra, Lemma 10.139.3. $\square$

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