[Chapter IV Proposition 18.6.12, EGA4]

Lemma 15.106.3. Let $A$ be a local ring. Let $A^ h$ be the henselization of $A$. The following are equivalent

1. $A$ is unibranch, and

2. $A^ h$ has a unique minimal prime.

Proof. This follows from Lemma 15.106.2 but we will also give a direct proof. Denote $\mathfrak m$ the maximal ideal of the ring $A$. Recall that the residue field $\kappa = A/\mathfrak m$ is the same as the residue field of $A^ h$.

Assume (2). Let $\mathfrak p^ h$ be the unique minimal prime of $A^ h$. The flatness of $A \to A^ h$ implies that $\mathfrak p = A \cap \mathfrak p^ h$ is the unique minimal prime of $A$ (by going down, see Algebra, Lemma 10.39.19). Also, since $A^ h/\mathfrak pA^ h = (A/\mathfrak p)^ h$ (see Algebra, Lemma 10.156.2) is reduced by Lemma 15.45.4 we see that $\mathfrak p^ h = \mathfrak pA^ h$. Let $A'$ be the integral closure of $A/\mathfrak p$ in its fraction field. We have to show that $A'$ is local. 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 $\mathfrak m_1$, $\mathfrak m_2$. Choose elements $f_1, f_2 \in A'$ with $f_ i \in \mathfrak m_ i$ and $f_ i \not\in \mathfrak m_{3 - i}$. We find a finite subalgebra $B = A[f_1, f_2] \subset A'$ with distinct maximal ideals $B \cap \mathfrak m_ i$, $i = 1, 2$. Note that the inclusions

$A/\mathfrak p \subset B \subset \kappa (\mathfrak p)$

give, on tensoring with the flat ring map $A \to A^ h$ the inclusions

$A^ h/\mathfrak p^ h \subset B \otimes _ A A^ h \subset \kappa (\mathfrak p) \otimes _ A A^ h \subset \kappa (\mathfrak p^ h)$

the last inclusion because $\kappa (\mathfrak p) \otimes _ A A^ h = \kappa (\mathfrak p) \otimes _{A/\mathfrak p} A^ h/\mathfrak p^ h$ is a localization of the domain $A^ h/\mathfrak p^ h$. Note that $B \otimes _ A \kappa$ has at least two maximal ideals because $B/\mathfrak mB$ has two maximal ideals. Hence, as $A^ h$ is henselian we see that $B \otimes _ A A^ h$ is a product of $\geq 2$ local rings, see Algebra, Lemma 10.153.5. But we've just seen that $B \otimes _ A A^ h$ 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 inducing an isomorphism of residue fields which is a localization of an étale $A$-algebra. In particular $\mathfrak m_ B$ is the unique prime containing $\mathfrak m B$. Then $B' = A' \otimes _ A B$ is integral over $B$ and the assumption that $A \to A'$ is local 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 $A^ h$ is a filtered colimit of the local rings $B$ it follows that $A^ h$ has a unique minimal prime. Namely, if $fg = 0$ in $A^ h$ 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. $\square$

There are also:

• 2 comment(s) on Section 15.106: Local irreducibility

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).