Lemma 15.50.15. Let $A$ be a G-ring. Let $I \subset A$ be an ideal. Let $(A^ h, I^ h)$ be the henselization of the pair $(A, I)$, see Lemma 15.12.1. Then $A^ h$ is a G-ring.

** Being a G-ring is stable under Henselizations along ideals **

[Theorem 5.3 i), Greco]

**Proof.**
Let $\mathfrak m^ h \subset A^ h$ be a maximal ideal. We have to show that the map from $A^ h_{\mathfrak m^ h}$ to its completion has geometrically regular fibres, see Lemma 15.50.7. Let $\mathfrak m$ be the inverse image of $\mathfrak m^ h$ in $A$. Note that $I^ h \subset \mathfrak m^ h$ and hence $I \subset \mathfrak m$ as $(A^ h, I^ h)$ is a henselian pair. Recall that $A^ h$ is Noetherian, $I^ h = IA^ h$, and that $A \to A^ h$ induces an isomorphism on $I$-adic completions, see Lemma 15.12.4. Then the local homomorphism of Noetherian local rings

induces an isomorphism on completions at maximal ideals by Lemma 15.43.9 (details omitted). Let $\mathfrak q^ h$ be a prime of $A^ h_{\mathfrak m^ h}$ lying over $\mathfrak q \subset A_\mathfrak m$. Set $\mathfrak q_1 = \mathfrak q^ h$ and let $\mathfrak q_2, \ldots , \mathfrak q_ t$ be the other primes of $A^ h$ lying over $\mathfrak q$, so that $A^ h \otimes _ A \kappa (\mathfrak q) = \prod \nolimits _{i = 1, \ldots , t} \kappa (\mathfrak q_ i)$, see Lemma 15.45.12. Using that $(A^ h)_{\mathfrak m^ h}^\wedge = (A_\mathfrak m)^\wedge $ as discussed above we see

Hence, as one of the components, the ring

is geometrically regular over $\kappa (\mathfrak q)$ by assumption on $A$. Since $\kappa (\mathfrak q^ h)$ is separable algebraic over $\kappa (\mathfrak q)$ it follows from Algebra, Lemma 10.166.6 that

is geometrically regular over $\kappa (\mathfrak q^ h)$ as desired. $\square$

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