In Properties, Definition 28.7.1 we have defined the notion of a normal scheme. This notion is defined even for non-Noetherian schemes. Hence, contrary to our discussion of “geometrically regular” schemes we consider all field extensions of the ground field.
Definition 33.10.1. Let $X$ be a scheme over the field $k$.
Let $x \in X$. We say $X$ is geometrically normal at $x$ if for every field extension $k'/k$ and every $x' \in X_{k'}$ lying over $x$ the local ring $\mathcal{O}_{X_{k'}, x'}$ is normal.
We say $X$ is geometrically normal over $k$ if $X$ is geometrically normal at every $x \in X$.
Lemma 33.10.2. Let $k$ be a field. Let $X$ be a scheme over $k$. Let $x \in X$. The following are equivalent
$X$ is geometrically normal at $x$,
for every finite purely inseparable field extension $k'$ of $k$ and $x' \in X_{k'}$ lying over $x$ the local ring $\mathcal{O}_{X_{k'}, x'}$ is normal, and
the ring $\mathcal{O}_{X, x}$ is geometrically normal over $k$ (see Algebra, Definition 10.165.2).
Proof. It is clear that (1) implies (2). Assume (2). Let $k'/k$ be a finite purely inseparable field extension (for example $k = k'$). 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 10.18.2). 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 normal ring. Hence we deduce (3) by Algebra, Lemma 10.165.1.
Assume (3). 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 normal by assumption and (1) is proved. $\square$
Lemma 33.10.3. Let $k$ be a field. Let $X$ be a scheme over $k$. The following are equivalent
$X$ is geometrically normal,
$X_{k'}$ is a normal scheme for every field extension $k'/k$,
$X_{k'}$ is a normal scheme for every finitely generated field extension $k'/k$,
$X_{k'}$ is a normal scheme for every finite purely inseparable field extension $k'/k$,
for every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is geometrically normal (see Algebra, Definition 10.165.2), and
$X_{k^{perf}}$ is a normal scheme.
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 normal. By definition this means that $X_{k'}$ is normal. Hence (2).
It is clear that (2) implies (3) implies (4).
Assume (4) and let $U \subset X$ be an affine open subscheme. Then $U_{k'}$ is a normal scheme for any finite purely inseparable extension $k'/k$ (including $k = k'$). This means that $k' \otimes _ k \mathcal{O}(U)$ is a normal ring for all finite purely inseparable extensions $k'/k$. Hence $\mathcal{O}(U)$ is a geometrically normal $k$-algebra by definition. Hence (4) implies (5).
Assume (5). 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 normal. So (1) holds.
The equivalence of (5) and (6) follows from the definition of geometrically normal algebras and the equivalence (just proved) of (3) and (4). $\square$
Lemma 33.10.4. Let $k$ be a field. Let $X$ be a scheme over $k$. Let $k'/k$ be a field extension. Let $x \in X$ be a point, and let $x' \in X_{k'}$ be a point lying over $x$. The following are equivalent
$X$ is geometrically normal at $x$,
$X_{k'}$ is geometrically normal at $x'$.
In particular, $X$ is geometrically normal over $k$ if and only if $X_{k'}$ is geometrically normal over $k'$.
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 normal. Thus by Algebra, Lemma 10.164.3 we conclude that $\mathcal{O}_{X_{k''}, x''}$ is normal. By Lemma 33.10.2 we see that $X$ is geometrically normal at $x$. $\square$
Lemma 33.10.5. Let $k$ be a field. Let $X$ be a geometrically normal scheme over $k$ and let $Y$ be a normal scheme over $k$. Then $X \times _ k Y$ is a normal scheme.
Proof. This reduces to Algebra, Lemma 10.165.5 by Lemma 33.10.3. $\square$
Lemma 33.10.6. Let $k$ be a field. Let $X$ be a normal scheme over $k$. Let $K/k$ be a separable field extension. Then $X_ K$ is a normal scheme.
Proof. Follows from Lemma 33.10.5 and Algebra, Lemma 10.165.4. $\square$
Lemma 33.10.7. Let $k$ be a field. Let $X$ be a proper geometrically normal scheme over $k$. The following are equivalent
$H^0(X, \mathcal{O}_ X) = k$,
$X$ is geometrically connected,
$X$ is geometrically irreducible, and
$X$ is geometrically integral.
Proof. By Lemma 33.9.5 we have the equivalence of (1) and (2). A locally Noetherian normal scheme (such as $X_{\overline{k}}$) is a disjoint union of its irreducible components (Properties, Lemma 28.7.6). Thus we see that (2) and (3) are equivalent. Since $X_{\overline{k}}$ is assumed reduced, we see that (3) and (4) are equivalent too. $\square$
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