The henselization of a pair only depends on the radical of the ideal

Lemma 15.12.6. Let $A$ be a ring with ideals $I$ and $J$. If $V(I) = V(J)$ then the functor of Lemma 15.12.1 produces the same ring for the pair $(A, I)$ as for the pair $(A, J)$.

Proof. Let $(A', IA')$ be the pair produced by Lemma 15.12.1 starting with the pair $(A, I)$, see Lemma 15.12.2. Let $(A'', JA'')$ be the pair produced by Lemma 15.12.1 starting with the pair $(A, J)$. By Lemma 15.11.7 we see that $(A', JA')$ is a henselian pair and $(A'', IA'')$ is a henselian pair. By the universal property of the construction we obtain unique $A$-algebra maps $A'' \to A'$ and $A' \to A''$. The uniqueness shows that these are mutually inverse. $\square$

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