Lemma 33.12.3. Let $k$ be a field. Let $X$ be a locally Noetherian scheme over $k$. The following are equivalent

1. $X$ is geometrically regular,

2. $X_{k'}$ is a regular scheme for every finitely generated field extension $k'/k$,

3. $X_{k'}$ is a regular scheme for every finite purely inseparable field extension $k'/k$,

4. for every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is geometrically regular (see Algebra, Definition 10.166.2), and

5. there exists an affine open covering $X = \bigcup U_ i$ such that each $\mathcal{O}_ X(U_ i)$ is geometrically regular over $k$.

Proof. Assume (1). Then for every finitely generated field extension $k'/k$ and every point $x' \in X_{k'}$ the local ring of $X_{k'}$ at $x'$ is regular. By Properties, Lemma 28.9.2 this means that $X_{k'}$ is regular. Hence (2).

It is clear that (2) implies (3).

Assume (3) and let $U \subset X$ be an affine open subscheme. Then $U_{k'}$ is a regular scheme for any finite purely inseparable extension $k'/k$ (including $k = k'$). This means that $k' \otimes _ k \mathcal{O}(U)$ is a regular ring for all finite purely inseparable extensions $k'/k$. Hence $\mathcal{O}(U)$ is a geometrically regular $k$-algebra and we see that (4) holds.

It is clear that (4) implies (5). Let $X = \bigcup U_ i$ be an affine open covering as in (5). For any field extension $k'/k$ the base change $X_{k'}$ is gotten by gluing the spectra of the rings $\mathcal{O}_ X(U_ i) \otimes _ k k'$ (see Schemes, Section 26.17). Hence $X_{k'}$ is regular. So (1) holds. $\square$

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