**Proof of Lemma 86.7.3 in case $A$ is a G-ring.**
This proof is easier in that it does not depend on the somewhat delicate deformation theory argument given in the proof of Lemma 86.7.2, but of course it requires a very strong assumption on the Noetherian ring $A$.

Choose a presentation $B = A[x_1, \ldots , x_ r]^\wedge /J$. Choose generators $g_1, \ldots , g_ m \in J$. Choose generators $k_1, \ldots , k_ t$ of the module of relations between $g_1, \ldots , g_ m$, i.e., such that

is exact in the middle. Write $k_ i = (k_{i1}, \ldots , k_{im})$ so that we have

for $i = 1, \ldots , t$. Let $I^ c = (a_1, \ldots , a_ s)$. For each $l \in \{ 1, \ldots , s\} $ we know that multiplication by $a_ l$ on $\mathop{N\! L}\nolimits ^\wedge _{B/A}$ is zero in $D(B)$. By Lemma 86.3.4 we can find a map $\alpha _ l : \bigoplus B\text{d}x_ i \to J/J^2$ such that $\text{d} \circ \alpha _ l$ and $\alpha _ l \circ \text{d}$ are both multiplication by $a_ l$. Pick an element $f_{l, i} \in J$ whose class modulo $J^2$ is equal to $\alpha _ l(\text{d}x_ i)$. Then we have for all $l = 1, \ldots , s$ and $i = 1, \ldots , r$ that

for some $h_{l, i}^{j', i'} \in A[x_1, \ldots , x_ r]^\wedge $. We also have for $j = 1, \ldots , m$ and $l = 1, \ldots , s$ that

for some $h_{l, j}^ i$ and $h_{l, j}^{j', j''}$ in $A[x_1, \ldots , x_ r]^\wedge $. Of course, since $f_{l, i} \in J$ we can write for $l = 1, \ldots , s$ and $i = 1, \ldots , r$

for some $h_{l, i}^ j$ in $A[x_1, \ldots , x_ r]^\wedge $.

Let $A[x_1, \ldots , x_ r]^ h$ be the henselization of the pair $(A[x_1, \ldots , x_ r], IA[x_1, \ldots , x_ r])$, see More on Algebra, Lemma 15.12.1. Since $A$ is a Noetherian G-ring, so is $A[x_1, \ldots , x_ r]$, see More on Algebra, Proposition 15.49.10. Hence we have approximation for the map $A[x_1, \ldots , x_ r]^ h \to A[x_1, \ldots , x_ r]^\wedge $ with respect to the ideal generated by $I$, see Smoothing Ring Maps, Lemma 16.14.1. Choose a large integer $M$. Choose

such that analogues of equations (86.8.0.1), (86.8.0.3), and (86.8.0.4) hold for these elements in $A[x_1, \ldots , x_ r]^ h$, i.e.,

and such that we have

where we take liberty of thinking of $A[x_1, \ldots , x_ r]^ h$ as a subring of $A[x_1, \ldots , x_ r]^\wedge $. Note that we cannot guarantee that the analogue of (86.8.0.2) holds in $A[x_1, \ldots , x_ r]^ h$, because it is not a polynomial equation. But since taking partial derivatives is $A$-linear, we do get the analogue modulo $I^ M$. More precisely, we see that

for $l = 1, \ldots , s$ and $i = 1, \ldots , r$.

With these choices, consider the ring

and denote $C^\wedge $ its $I$-adic completion, namely

In the following paragraphs we establish the fact that $C^\wedge $ is isomorphic to $B$. Then in the final paragraph we deal with show that $C^ h$ comes from a finite type algebra over $A$ as in the statement of the lemma.

First consider the cokernel

This $C^\wedge $ module is generated by the images of the elements $\text{d}x_ i$. Since $F_{l, i} \in J'$ by the analogue of (86.8.0.4) we see from (86.8.0.5) we see that $a_ l \text{d}x_ i \in I^ M\Omega $. As $I^ c = (a_ l)$ we see that $I^ c \Omega \subset I^ M \Omega $. Since $M > c$ we conclude that $I^ c \Omega = 0$ by Algebra, Lemma 10.19.1.

Next, consider the kernel

By the analogue of (86.8.0.3) we see that $a_ l J' \subset (F_{l, i}) + (J')^2$. On the other hand, the determinant $\Delta _ l$ of the matrix $(\partial F_{l, i}/ \partial x_{i'})$ satisfies $\Delta _ l = a_ l^ r \bmod I^ M C^\wedge $ by (86.8.0.5). It follows that $a_ l^{r + 1} H_1 \subset I^ M H_1$ (some details omitted; use Algebra, Lemma 10.14.5). Now $(a_1^{r + 1}, \ldots , a_ s^{r + 1}) \supset I^{(sr + 1)c}$. Hence $I^{(sr + 1)c}H_1 \subset I^ M H_1$ and since $M > (sr + 1)c$ we conclude that $I^{(sr + 1)c}H_1 = 0$.

By Lemma 86.3.5 we conclude that multiplication by an element of $I^{2(sr + 1)c}$ on $\mathop{N\! L}\nolimits ^\wedge _{C^\wedge /A}$ is zero (note that the bound does not depend on $M$ or the choice of the approximation, as long as $M$ is large enough). Since $G_ j - g_ j$ is in the ideal generated by $I^ M$ we see that there is an isomorphism

As $M$ is large enough we can use Lemma 86.7.1 with $d = d(I \subset A \to B)$, with $C^\wedge $ playing the role of $B$, with $2(rs + 1)c$ instead of $c$, to find a morphism

which agrees with $\psi _ M$ modulo $I^{q - 2(rs + 1)c}$ where $q$ is the quotient of $M$ by the number of generators of $I$. We claim $\psi $ is an isomorphism. Since $C^\wedge $ and $B$ are $I$-adically complete the map $\psi $ is surjective because it is surjective modulo $I$ (see Algebra, Lemma 10.95.1). On the other hand, as $M$ is large enough we see that

as graded $\text{Gr}_ I(A[x_1, \ldots , x_ r]^\wedge )$-modules by More on Algebra, Lemma 15.4.2. Since $\psi $ is compatible with this isomorphism as it agrees with $\psi _ M$ modulo $I$, this means that $\text{Gr}_ I(\psi )$ is an isomorphism. As $C^\wedge $ and $B$ are $I$-adically complete, it follows that $\psi $ is an isomorphism.

This paragraph serves to deal with the issue that $C^ h$ is not of finite type over $A$. Namely, the ring $A[x_1, \ldots , x_ r]^ h$ is a filtered colimit of étale $A[x_1, \ldots , x_ r]$ algebras $A'$ such that $A/I[x_1, \ldots , x_ r] \to A'/IA'$ is an isomorphism (see proof of More on Algebra, Lemma 15.12.1). Pick an $A'$ such that $G_1, \ldots , G_ m$ are the images of $G'_1, \ldots , G'_ m \in A'$. Setting $C = A'/(G'_1, \ldots , G'_ m)$ we get the finite type algebra we were looking for. $\square$

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