Lemma 15.12.2. Let $(A, I)$ be a pair. Let $(A^ h, I^ h)$ be as in Lemma 15.12.1. Then $A \to A^ h$ is flat, $I^ h = IA^ h$ and $A/I^ n \to A^ h/I^ nA^ h$ is an isomorphism for all $n$.

Proof. In the proof of Lemma 15.12.1 we have seen that $A^ h$ is a filtered colimit of étale $A$-algebras $B$ such that $A/I \to B/IB$ is an isomorphism and we have seen that $I^ h = IA^ h$. As an étale ring map is flat (Algebra, Lemma 10.143.3) we conclude that $A \to A^ h$ is flat by Algebra, Lemma 10.39.3. Since each $A \to B$ is flat we find that the maps $A/I^ n \to B/I^ nB$ are isomorphisms as well (for example by Algebra, Lemma 10.101.3). Taking the colimit we find that $A/I^ n = A^ h/I^ nA^ h$ as desired. $\square$

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Suggested slogan: "Henselization is a flat map that is an isomorphism mod I^n"

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