Lemma 15.106.5. Let $A$ be a local ring. Let $A^{sh}$ be a strict henselization of $A$. The following are equivalent

$A$ is geometrically unibranch, and

$A^{sh}$ has a unique minimal prime.

[Lemma 2.2, Etale-coverings] and [Chapter IV Proposition 18.8.15, EGA4]

Lemma 15.106.5. Let $A$ be a local ring. Let $A^{sh}$ be a strict henselization of $A$. The following are equivalent

$A$ is geometrically unibranch, and

$A^{sh}$ has a unique minimal prime.

**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

\[ A/\mathfrak p \subset B \subset \kappa (\mathfrak p) \]

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

\[ A^{sh}/\mathfrak p^{sh} \subset B \otimes _ A A^{sh} \subset \kappa (\mathfrak p) \otimes _ A A^{sh} \subset \kappa (\mathfrak p^{sh}) \]

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

\[ B/\mathfrak pB \subset B \otimes _ A \kappa (\mathfrak p) = B' \otimes _{A'} (\text{fraction field of }A') \subset \text{fraction field of }B' \]

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. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: