The Stacks project

Proof. Assume (1). This in particular implies that $\mathcal{O}_{X, x}$ is reduced. Let $k'/k$ be a finite purely inseparable field extension. Consider the ring $\mathcal{O}_{X, x} \otimes _ k k'$. By Algebra, Lemma 10.46.7 its spectrum is the same as the spectrum of $\mathcal{O}_{X, x}$. Hence it is a local ring also (Algebra, Lemma 10.18.3). Therefore there is a unique point $x' \in X_{k'}$ lying over $x$ and $\mathcal{O}_{X_{k'}, x'} \cong \mathcal{O}_{X, x} \otimes _ k k'$. By assumption this is a reduced ring. Hence we deduce (2) by Algebra, Lemma 10.44.4.

Assume (2). Let $k'/k$ be a field extension. Since $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$ is surjective, also $X_{k'} \to X$ is surjective (Morphisms, Lemma 29.9.4). Let $x' \in X_{k'}$ be any point lying over $x$. The local ring $\mathcal{O}_{X_{k'}, x'}$ is a localization of the ring $\mathcal{O}_{X, x} \otimes _ k k'$. Hence it is reduced by assumption and (1) is proved. $\square$

Proof. The first statement follows from Lemma 33.6.2 and Algebra, Lemma 10.43.6 and the definition of a perfect field (Algebra, Definition 10.45.1). The second statement follows from the first. $\square$

Proof. Assume (1). Then for every field extension $k'/k$ and every point $x' \in X_{k'}$ the local ring of $X_{k'}$ at $x'$ is reduced. In other words $X_{k'}$ is reduced. Hence (2).

Assume (2). Let $U \subset X$ be an affine open. Then for every field extension $k'/k$ the scheme $X_{k'}$ is reduced, hence $U_{k'} = \mathop{\mathrm{Spec}}(\mathcal{O}(U)\otimes _ k k')$ is reduced, hence $\mathcal{O}(U)\otimes _ k k'$ is reduced (see Properties, Section 28.3). In other words $\mathcal{O}(U)$ is geometrically reduced, so (7) holds.

Assume (7). For any field extension $k'/k$ the base change $X_{k'}$ is gotten by gluing the spectra of the rings $\mathcal{O}_ X(U) \otimes _ k k'$ where $U$ is affine open in $X$ (see Schemes, Section 26.17). Hence $X_{k'}$ is reduced. So (1) holds.

This proves that (1), (2), and (7) are equivalent. These are equivalent to (3), (4), (5), and (6) because we can apply Algebra, Lemma 10.44.4 to $\mathcal{O}_ X(U)$ for $U \subset X$ affine open. $\square$

Proof. Note that if $k'/k$ is purely inseparable, then $X_{k'} \to X$ induces a homeomorphism on underlying topological spaces, see Algebra, Lemma 10.46.7. Whence the uniqueness of $x'$ lying over $x$ mentioned in the statement. Moreover, in this case $\mathcal{O}_{X_{k'}, x'} = \mathcal{O}_{X, x} \otimes _ k k'$. Hence the lemma follows from Lemma 33.6.2 above and Algebra, Lemma 10.44.4. $\square$

Proof. It is clear that (1) implies (2). Assume (2). Let $k''/k$ be a finite purely inseparable field extension and let $x'' \in X_{k''}$ be a point lying over $x$ (actually it is unique). We can find a common field extension $k'''/k$ (i.e. with both $k' \subset k'''$ and $k'' \subset k'''$) and a point $x''' \in X_{k'''}$ lying over both $x'$ and $x''$. Consider the map of local rings

This is a flat local ring homomorphism and hence faithfully flat. By (2) we see that the local ring on the right is reduced. Thus by Algebra, Lemma 10.164.2 we conclude that $\mathcal{O}_{X_{k''}, x''}$ is reduced. Thus by Lemma 33.6.5 we conclude that $X$ is geometrically reduced at $x$. $\square$

Proof. To prove (1) combine Lemma 33.6.2 with Algebra, Lemma 10.43.5. To prove (2) combine Lemma 33.6.4 with Algebra, Lemma 10.43.5. To prove (3) note that $(X \times _ k Y)_{\overline{k}} = X_{\overline{k}} \times _{\overline{k}} Y_{\overline{k}}$ and use (2) as well as Lemma 33.6.4. To prove (4) use (3) combined with Lemma 33.6.3. $\square$

Proof. Part (1) follows from Lemma 33.6.2 and the fact that if $A$ is a geometrically reduced $k$-algebra, then $S^{-1}A$ is a geometrically reduced $k$-algebra for any multiplicative subset $S$ of $A$, see Algebra, Lemma 10.43.3.

Let $A = \mathcal{O}_{X, x}$. The assumptions (a) and (b) of (2) imply that $A$ is reduced, and that $A_{\mathfrak q}$ is geometrically reduced over $k$ for every minimal prime $\mathfrak q$ of $A$. Hence $A$ is geometrically reduced over $k$, see Algebra, Lemma 10.43.7. Thus $X$ is geometrically reduced at $x$, see Lemma 33.6.2.

Part (3) follows trivially from part (2). $\square$

Proof. Assume $X$ locally Noetherian and geometrically reduced at $x$. By Properties, Lemma 28.29.8 we can find an affine open neighbourhood $U \subset X$ of $x$ such that $R = \mathcal{O}_ X(U) \to \mathcal{O}_{X, x}$ is injective. By Lemma 33.6.2 the assumption means that $\mathcal{O}_{X, x}$ is geometrically reduced over $k$. By Algebra, Lemma 10.43.2 this implies that $R$ is geometrically reduced over $k$, which in turn implies that $U$ is geometrically reduced. $\square$

Proof. To prove this lemma we may replace $X$ by its reduction $X_{red}$. Hence we may assume that $X$ is reduced and locally of finite type over $k$. Let $x_1, \ldots , x_ n \in X$ be the generic points of the irreducible components of $X$. Note that for every purely inseparable algebraic extension $k'/k$ the morphism $(X_{k'})_{red} \to X$ is a homeomorphism, see Algebra, Lemma 10.46.7. Hence the points $x'_1, \ldots , x'_ n$ lying over $x_1, \ldots , x_ n$ are the generic points of the irreducible components of $(X_{k'})_{red}$. As $X$ is reduced the local rings $K_ i = \mathcal{O}_{X, x_ i}$ are fields, see Algebra, Lemma 10.25.1. As $X$ is locally of finite type over $k$ the field extensions $K_ i/k$ are finitely generated field extensions. Finally, the local rings $\mathcal{O}_{(X_{k'})_{red}, x'_ i}$ are the fields $(K_ i \otimes _ k k')_{red}$. By Algebra, Lemma 10.45.3 we can find a finite purely inseparable extension $k'/k$ such that $(K_ i \otimes _ k k')_{red}$ are separable field extensions of $k'$. In particular each $(K_ i \otimes _ k k')_{red}$ is geometrically reduced over $k'$ by Algebra, Lemma 10.44.2. At this point Lemma 33.6.8 part (3) implies that $(X_{k'})_{red}$ is geometrically reduced. $\square$