**Proof.**
Note that any finite purely inseparable extension $k \subset k'$ embeds in $k^{perf}$. Moreover, $k^{1/p}$ embeds into $k^{perf}$ which embeds into $\overline{k}$. Thus it is clear that (5) $\Rightarrow $ (4) $\Rightarrow $ (3) $\Rightarrow $ (2) and that (3) $\Rightarrow $ (1).

We prove that (1) $\Rightarrow $ (5). Assume $k' \otimes _ k S$ is reduced for every finite purely inseparable extension $k'$ of $k$. Let $k \subset K$ be an extension of fields. We have to show that $K \otimes _ k S$ is reduced. By Lemma 10.42.4 we reduce to the case where $k \subset K$ is a finitely generated field extension. Choose a diagram

\[ \xymatrix{ K \ar[r] & K' \\ k \ar[u] \ar[r] & k' \ar[u] } \]

as in Lemma 10.41.4. By assumption $k' \otimes _ k S$ is reduced. By Lemma 10.42.6 it follows that $K' \otimes _ k S$ is reduced. Hence we conclude that $K \otimes _ k S$ is reduced as desired.

Finally we prove that (2) $\Rightarrow $ (5). Assume $k^{1/p} \otimes _ k S$ is reduced. Then $S$ is reduced. Moreover, for each localization $S_{\mathfrak p}$ at a minimal prime $\mathfrak p$, the ring $k^{1/p}\otimes _ k S_{\mathfrak p}$ is a localization of $k^{1/p} \otimes _ k S$ hence is reduced. But $S_{\mathfrak p}$ is a field by Lemma 10.24.1, hence $S_{\mathfrak p}$ is geometrically reduced by Lemma 10.43.1. It follows from Lemma 10.42.7 that $S$ is geometrically reduced.
$\square$

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