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The Stacks project

Lemma 33.10.3. Let k be a field. Let X be a scheme over k. The following are equivalent

  1. X is geometrically normal,

  2. X_{k'} is a normal scheme for every field extension k'/k,

  3. X_{k'} is a normal scheme for every finitely generated field extension k'/k,

  4. X_{k'} is a normal scheme for every finite purely inseparable field extension k'/k,

  5. for every affine open U \subset X the ring \mathcal{O}_ X(U) is geometrically normal (see Algebra, Definition 10.165.2), and

  6. 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


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