Lemma 15.106.4. Let $A$ be a Noetherian ring and $I$ an ideal. Let $B$ be a finite type $A$-algebra. Let $B^\wedge \to C$ be a surjective ring map with kernel $J$ where $B^\wedge$ is the $I$-adic completion. If $J/J^2$ is annihilated by $I^ c$ for some $c \geq 0$, then $C$ is isomorphic to the completion of a finite type $A$-algebra.

Proof. Let $f \in I$. Since $B^\wedge$ is Noetherian (Algebra, Lemma 10.97.6), we see that $J$ is a finitely generated ideal. Hence we conclude from Algebra, Lemma 10.21.5 that

$C_ f = ((B^\wedge )_ f)_ e$

for some idempotent $e \in (B^\wedge )_ f$. By Lemma 15.106.3 we can find a surjection $B^\wedge \to C'$ such that $B^\wedge \to C \times C'$ becomes an isomorphism after inverting any $f \in I$. Observe that $C \times C'$ is a finite $B^\wedge$-algebra.

Choose generators $f_1, \ldots , f_ r \in I$. Denote $\alpha _ i : (C \times C')_{f_ i} \to B_{f_ i} \otimes _ B B^\wedge$ the inverse of the isomorphism of $(B^\wedge )_{f_ i}$-algebras we obtained above. Denote $\alpha _{ij} : (B_{f_ i})_{f_ j} \to (B_{f_ j})_{f_ i}$ the obvious $B$-algebra isomorphism. Consider the object

$(C \times C', B_{f_ i}, \alpha _ i, \alpha _{ij})$

of the category $\text{Glue}(B \to B^\wedge , f_1, \ldots , f_ r)$ introduced in Remark 15.88.10. We omit the verification of conditions (1)(a) and (1)(b). Since $B \to B^\wedge$ is a flat map (Algebra, Lemma 10.97.2) inducing an isomorphism $B/IB \to B^\wedge /IB^\wedge$ we may apply Proposition 15.88.15 and Remark 15.88.19. We conclude that $C \times C'$ is isomorphic to $D \otimes _ B B^\wedge$ for some finite $B$-algebra $D$. Then $D/ID \cong C/IC \times C'/IC'$. Let $\overline{e} \in D/ID$ be the idempotent corresponding to the factor $C/IC$. By Lemma 15.9.10 there exists an étale ring map $B \to B'$ which induces an isomorphism $B/IB \to B'/IB'$ such that $D' = D \otimes _ B B'$ contains an idempotent $e$ lifting $\overline{e}$. Since $C \times C'$ is $I$-adically complete the pair $(C \times C', IC \times IC')$ is henselian (Lemma 15.11.4). Thus we can factor the map $B \to C \times C'$ through $B'$. Doing so we may replace $B$ by $B'$ and $D$ by $D'$. Then we find that $D = D_ e \times D_{1 - e} = D/(1 - e) \times D/(e)$ is a product of finite type $A$-algebras and the completion of the first part is $C$ and the completion of the second part is $C'$. $\square$

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