Lemma 33.6.4. Let $k$ be a field of characteristic $p > 0$. Let $X$ be a scheme over $k$. The following are equivalent

1. $X$ is geometrically reduced,

2. $X_{k'}$ is reduced for every field extension $k'/k$,

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

4. $X_{k^{1/p}}$ is reduced,

5. $X_{k^{perf}}$ is reduced,

6. $X_{\bar k}$ is reduced,

7. for every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is geometrically reduced (see Algebra, Definition 10.43.1).

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 10.44.3 to $\mathcal{O}_ X(U)$ for $U \subset X$ affine open. $\square$

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