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Henselization of a ring inherits good properties of formal fibers

Lemma 15.51.7. Let $A$ be a $P$-ring where $P$ satisfies (B), (C), (D), and (E). 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 $P$-ring.

Proof. Let $\mathfrak m^ h \subset A^ h$ be a maximal ideal. We have to show that the fibres of $A^ h_{\mathfrak m^ h} \to (A^ h_{\mathfrak m^ h})^\wedge $ have $P$, see Lemma 15.51.4. 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

\[ A_\mathfrak m \to A^ h_{\mathfrak m^ h} \]

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

\[ \prod \nolimits _{i = 1, \ldots , t} (A^ h_{\mathfrak m^ h})^\wedge \otimes _{A^ h_{\mathfrak m^ h}} \kappa (\mathfrak q_ i) = (A^ h_{\mathfrak m^ h})^\wedge \otimes _{A^ h_{\mathfrak m^ h}} (A^ h_{\mathfrak m^ h} \otimes _{A_{\mathfrak m}} \kappa (\mathfrak q)) = (A_{\mathfrak m})^\wedge \otimes _{A_{\mathfrak m}} \kappa (\mathfrak q) \]

Hence, looking at local rings and using (B), we see that

\[ \kappa (\mathfrak q) \longrightarrow (A^ h_{\mathfrak m^ h})^\wedge \otimes _{A^ h_{\mathfrak m^ h}} \kappa (\mathfrak q^ h) \]

has $P$ as $\kappa (\mathfrak q) \to (A_\mathfrak m)^\wedge \otimes _{A_\mathfrak m} \kappa (\mathfrak q)$ does by assumption on $A$. Since $\kappa (\mathfrak q^ h)/\kappa (\mathfrak q)$ is separable algebraic, by (E) we find that $\kappa (\mathfrak q^ h) \to (A^ h_{\mathfrak m^ h})^\wedge \otimes _{A^ h_{\mathfrak m^ h}} \kappa (\mathfrak q^ h)$ has $P$ as desired. $\square$

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