Lemma 88.6.2. Let $I$ be an ideal of a Noetherian ring $A$. Let $C^ h$ be the henselization of a finite type $A$-algebra $C$ with respect to the ideal $IC$. Let $J \subset C^ h$ be an ideal. Then there exists a finite type $A$-algebra $B$ such that $B^\wedge \cong (C^ h/J)^\wedge $.
Proof. By More on Algebra, Lemma 15.12.4 the ring $C^ h$ is Noetherian. Say $J = (g_1, \ldots , g_ m)$. The ring $C^ h$ is a filtered colimit of étale $C$ algebras $C'$ such that $C/IC \to C'/IC'$ is an isomorphism (see proof of More on Algebra, Lemma 15.12.1). Pick an $C'$ such that $g_1, \ldots , g_ m$ are the images of $g'_1, \ldots , g'_ m \in C'$. Setting $B = C'/(g'_1, \ldots , g'_ m)$ we get a finite type $A$-algebra. Of course $(C, IC)$ and $C', IC')$ have the same henselizations and the same completions. It follows easily from this that $B^\wedge = (C^ h/J)^\wedge $. $\square$
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