The Stacks project

Proof. This follows from Lemma 15.106.4 but we will also give a direct proof; this direct proof is almost exactly the same as the direct proof of Lemma 15.106.3. Denote $\mathfrak m$ the maximal ideal of the ring $A$. Denote $\kappa $, $\kappa ^{sh}$ the residue field of $A$, $A^{sh}$.

Assume (2). Let $\mathfrak p^{sh}$ be the unique minimal prime of $A^{sh}$. The flatness of $A \to A^{sh}$ implies that $\mathfrak p = A \cap \mathfrak p^{sh}$ is the unique minimal prime of $A$ (by going down, see Algebra, Lemma 10.39.19). Also, since $A^{sh}/\mathfrak pA^{sh} = (A/\mathfrak p)^{sh}$ (see Algebra, Lemma 10.156.4) is reduced by Lemma 15.45.4 we see that $\mathfrak p^{sh} = \mathfrak pA^{sh}$. Let $A'$ be the integral closure of $A/\mathfrak p$ in its fraction field. We have to show that $A'$ is local and that its residue field is purely inseparable over $\kappa $. Since $A \to A'$ is integral, every maximal ideal of $A'$ lies over $\mathfrak m$ (by going up for integral ring maps, see Algebra, Lemma 10.36.22). If $A'$ is not local, then we can find distinct maximal ideals $\mathfrak m_1$, $\mathfrak m_2$. Choosing elements $f_1, f_2 \in A'$ with $f_ i \in \mathfrak m_ i, f_ i \not\in \mathfrak m_{3 - i}$ we find a finite subalgebra $B = A[f_1, f_2] \subset A'$ with distinct maximal ideals $B \cap \mathfrak m_ i$, $i = 1, 2$. If $A'$ is local with maximal ideal $\mathfrak m'$, but $A/\mathfrak m \subset A'/\mathfrak m'$ is not purely inseparable, then we can find $f \in A'$ whose image in $A'/\mathfrak m'$ generates a finite, not purely inseparable extension of $A/\mathfrak m$ and we find a finite local subalgebra $B = A[f] \subset A'$ whose residue field is not a purely inseparable extension of $A/\mathfrak m$. Note that the inclusions

give, on tensoring with the flat ring map $A \to A^{sh}$ the inclusions

the last inclusion because $\kappa (\mathfrak p) \otimes _ A A^{sh} = \kappa (\mathfrak p) \otimes _{A/\mathfrak p} A^{sh}/\mathfrak p^{sh}$ is a localization of the domain $A^{sh}/\mathfrak p^{sh}$. Note that $B \otimes _ A \kappa ^{sh}$ has at least two maximal ideals because $B/\mathfrak mB$ either has two maximal ideals or one whose residue field is not purely inseparable over $\kappa $, and because $\kappa ^{sh}$ is separably algebraically closed. Hence, as $A^{sh}$ is strictly henselian we see that $B \otimes _ A A^{sh}$ is a product of $\geq 2$ local rings, see Algebra, Lemma 10.153.6. But we've just seen that $B \otimes _ A A^{sh}$ is a subring of a domain and we get a contradiction.

Assume (1). Let $\mathfrak p \subset A$ be the unique minimal prime and let $A'$ be the integral closure of $A/\mathfrak p$ in its fraction field. Let $A \to B$ be a local map of local rings which is a localization of an étale $A$-algebra. In particular $\mathfrak m_ B$ is the unique prime containing $\mathfrak m_ AB$. Then $B' = A' \otimes _ A B$ is integral over $B$ and the assumption that $A \to A'$ is local with purely inseparable residue field extension implies that $B'$ is local (Algebra, Lemma 10.156.5). On the other hand, $A' \to B'$ is the localization of an étale ring map, hence $B'$ is normal, see Algebra, Lemma 10.163.9. Thus $B'$ is a (local) normal domain. Finally, we have

Hence $B/\mathfrak pB$ is a domain, which implies that $B$ has a unique minimal prime (since by flatness of $A \to B$ these all have to lie over $\mathfrak p$). Since $A^{sh}$ is a filtered colimit of the local rings $B$ it follows that $A^{sh}$ has a unique minimal prime. Namely, if $fg = 0$ in $A^{sh}$ for some non-nilpotent elements $f, g$, then we can find a $B$ as above containing both $f$ and $g$ which leads to a contradiction. $\square$