Lemma 15.12.8. Let $I_1, I_2, \dots , I_ n \subset A$ be ideals in a ring $A$. Suppose $I_ i + I_ j = A$ for $i \not= j$, $1 \leq i, j \leq n$. Denote $A^ h$ the henselization of the pair $(A, I_1 \cap \ldots \cap I_ n)$ and $A_ i^ h$ the henselization of the pair $(A, I_ i)$. Then the natural morphism
is an isomorphism.
Proof. The map comes from the functoriality of the construction. If $n = 1$ the result is immediate. If $n > 1$, observe that $(I_1 \cap \ldots \cap I_{n - 1}) + I_ n = A$ by our condition on the ideals. Thus if we prove the lemma for $n = 2$, then the lemma follows for all $n$ by induction.
The case $n = 2$. Choose $f_ i \in I_ i$ such that $f_1 + f_2 = 1$. Recall that $A^ h$ is constructed as the colimit of $A$-algebras $B$ such that $A \to B$ is étale and $A/I_1 \cap I_2 \to B/(I_1 \cap I_2)B$ is an isomorphism. See proof of Lemma 15.12.1. Denote $Hens$ the category of such $A$-algebras $B$. Similarly, denote $Hens_ i$, $i = 1, 2$ the category of étale $A$-algebras $B$ such that $A/I_ i \to B/I_ iB$ is an isomorphism.
Consider the subcategory $Hens_1(f_2) \subset Hens_1$ consisting of those $B$ such that $f_2$ is invertible. This is a cofinal subcategory since if $B$ is in $Hens_1$, then $B_{f_2}$ is in $Hens_1$ as well (hint: $B_{f_2}/I_1B_{f_2} = (B/I_1B)_{f_2} = B/I_1B$ because $f_2$ is a unit modulo $I_1$). Similarly, the subcategory $Hens_2(f_1) \subset Hens_2$ is cofinal. Hence to compute $A_1^ h$ and $A_2^ h$ we may work with this smaller categories, see Categories, Lemma 4.17.2.
Consider the functor
This functor makes sense as
The second equality holds because $(I_1 \cap I_2)B_1 = I_1B_1 \cap I_2B_1 = I_1B_1$ as $f_2 \in I_2$ is invertible in $B_1$ (we have also used Algebra, Lemma 10.39.2). The last equality holds by Algebra, Lemma 10.15.4. To finish the proof, it suffices to prove that the displayed functor is cofinal (see lemma quoted above). The first condition of Categories, Definition 4.17.1 holds because an object $B$ of $Hens$ maps to $B_{f_2} \times B_{f_1}$ which is in the image of the functor. The second one always holds because if we have $B \to B_1 \times B_2$ and $B \to B'_1 \times B'_2$ with $B_1, B'_1 \in Hens_1(f_2)$ and $B_2, B'_2 \in Hens_2(f_1)$, then we can set $B''_1 = B_1 \otimes _ B B'_1$ which is in $Hens_1(f_2)$ and similarly $B''_2 = B_2 \otimes _ B B''_2$ in $Hens_2(f_1)$ to find a map $B \to B''_1 \times B''_2$ fitting into the commutative diagram
This finishes the proof. $\square$
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