The Stacks project

Lemma 33.25.7. Let $X$ be a scheme over a field $k$. If $X$ is locally of finite type and geometrically reduced over $k$ then $X$ contains a dense open which is smooth over $k$.

Proof. The problem is local on $X$, hence we may assume $X$ is quasi-compact. Let $X = X_1 \cup \ldots \cup X_ n$ be the irreducible components of $X$. Then $Z = \bigcup _{i \not= j} X_ i \cap X_ j$ is nowhere dense in $X$. Hence we may replace $X$ by $X \setminus Z$. As $X \setminus Z$ is a disjoint union of irreducible schemes, this reduces us to the case where $X$ is irreducible. As $X$ is irreducible and reduced, it is integral, see Properties, Lemma 28.3.4. Let $\eta \in X$ be its generic point. Then the function field $K = k(X) = \kappa (\eta )$ is geometrically reduced over $k$, hence separable over $k$, see Algebra, Lemma 10.43.1. Let $U = \mathop{\mathrm{Spec}}(A) \subset X$ be any nonempty affine open so that $K = A_{(0)}$ is the fraction field of $A$. Apply Algebra, Lemma 10.139.5 to conclude that $A$ is smooth at $(0)$ over $k$. By definition this means that some principal localization of $A$ is smooth over $k$ and we win. $\square$

Comments (1)

Comment #1931 by Keenan Kidwell on

It seems like reducing to the quasi-compact, irreducible case isn't strictly necessarily and doesn't dramatically affect the argument. Maybe the following variant is simpler (clearer?). The smooth locus is open by the definition of a smooth morphism (since is locally of finite type), and as is geometrically reduced, if is the generic point of an irreducible component of , then is a finitely generated separable extension field of (see 10.43.1). Thus is regular at , and therefore smooth at by 10.139.5. So the smooth locus contains all generic points, and hence is dense.

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