The Stacks project

Proof. This follows from Algebra, Lemma 10.140.7. $\square$

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. $\square$

Proof. (See also Lemma 33.12.6.) By Algebra, Lemma 10.140.3 every local ring $\mathcal{O}_{X, x}$ is regular. And because $X$ is locally of finite type over $k$ it is locally Noetherian. Hence $X$ is regular by Properties, Lemma 28.9.2. $\square$

Proof. (See also Lemma 33.12.6.) Let $k'$ be a finite purely inseparable extension of $k$. It suffices to prove that $X_{k'}$ is regular, normal, reduced, see Lemmas 33.12.3, 33.10.3, and 33.6.5. By Morphisms, Lemma 29.34.5 the morphism $X_{k'} \to \mathop{\mathrm{Spec}}(k')$ is smooth too. Hence it suffices to show that a scheme $X$ smooth over a field is regular, normal, and reduced. We see that $X$ is regular by Lemma 33.25.3. Hence Properties, Lemma 28.9.4 guarantees that $X$ is normal. $\square$

Proof. After possible shrinking $W$ we may assume that $W = \mathbf{A}^ d_ k \setminus V(f)$ for some $f \in k[x_1, \ldots , x_ d]$. If the lemma is wrong then $f(a_1, \ldots , a_ d) = 0$ for all $(a_1, \ldots , a_ d) \in (k^{sep})^ d$. This is absurd as $k^{sep}$ is an infinite field. $\square$

Proof. It suffices to show that given a nonempty smooth $X$ over $k$ there exists at least one closed point whose residue field is finite separable over $k$. To see this, choose a diagram

with $\pi $ étale, see Morphisms, Lemma 29.36.20. The morphism $\pi : U \to \mathbf{A}^ d_ k$ is open, see Morphisms, Lemma 29.36.13. By Lemma 33.25.5 we may choose a closed point $w \in \pi (U)$ whose residue field is finite separable over $k$. Pick any $x \in U$ with $\pi (x) = w$. By Morphisms, Lemma 29.36.7 the field extension $\kappa (x)/\kappa (w)$ is finite separable. Hence $\kappa (x)/k$ is finite separable. The point $x$ is a closed point of $X$ by Morphisms, Lemma 29.20.2. $\square$

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 10.44.2. 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 10.140.5 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. $\square$

Proof. The equality of the two sets follows immediately from Algebra, Lemma 10.140.5 and the definitions (see Algebra, Definition 10.45.1 for the definition of a perfect field). The set is open because the set of points where a morphism of schemes is smooth is open, see Morphisms, Definition 29.34.1. Finally, we give two arguments to see that it is dense: (1) The generic points of $X$ are in the set as the local rings at generic points are fields (Algebra, Lemma 10.25.1) hence regular. (2) We use that $X$ is geometrically reduced by Lemma 33.6.3 and hence Lemma 33.25.7 applies. $\square$

Proof. It suffices to show that $Y$ is geometrically regular at $y$, see Lemma 33.12.6. This follows from Lemma 33.12.5 (and Lemma 33.12.6 applied to $(X, x)$). $\square$

Proof. Let $U \subset X$ be the smooth locus of $X$. By assumption $U$ is nonempty and hence dense and scheme theoretically dense. Then $U_{\overline{k}} \subset X_{\overline{k}}$ is dense and scheme theoretically dense as well (some details omitted). Thus it suffices to show that $U$ is geometrically integral. Because $U$ has a $k$-rational point it is geometrically connected by Lemma 33.7.14. On the other hand, $U_{\overline{k}}$ is reduced and normal (Lemma 33.25.4. Since a connected normal Noetherian scheme is integral (Properties, Lemma 28.7.6) the proof is complete. $\square$

Proof. We may use induction on $\dim (X)$. By Lemma 33.6.11 we can find a finite purely inseparable extension $k'/k$ such that $(X_{k'})_{red}$ is geometrically reduced over $k'$. By Lemma 33.25.7 there is a nowhere dense closed subscheme $X' \subset (X_{k'})_{red}$ such that $(X_{k'})_{red} \setminus X'$ is smooth over $k'$. Then $\dim (X') < \dim (X)$. By induction hypothesis there exists a finite purely inseparable extension $k''/k'$, an integer $t' \geq 0$, and closed subschemes

such that $Y_0 = (X'_{k''})_{red}$ and $Y_ i \setminus Y_{i + 1}$ is smooth over $k''$ for all $i$. Then we let $t = t' + 1$ and we consider

given by $Z_0 = (X_{k''})_{red}$ and $Z_ i = Y_{i - 1}$ for $i > 0$; this makes sense as $X'_{k''}$ is a closed subscheme of $X_{k''}$. We omit the verification that all the stated properties hold. $\square$