Proposition 88.6.3. Let $I$ be an ideal of a Noetherian G-ring $A$. Let $B$ be an object of (88.2.0.2). If $B$ is rig-smooth over $(A, I)$, then there exists a finite type $A$-algebra $C$ and an isomorphism $B \cong C^\wedge $ of $A$-algebras.
Proof. Choose a presentation $B = A[x_1, \ldots , x_ r]^\wedge /J$. Write $P = A[x_1, \ldots , x_ r]^\wedge $. 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
\[ P^{\oplus t} \xrightarrow {k_1, \ldots , k_ t} P^{\oplus m} \xrightarrow {g_1, \ldots , g_ m} P \to B \to 0 \]
is a resolution. Write $k_ i = (k_{i1}, \ldots , k_{im})$ so that we have
88.6.3.1
\begin{equation} \label{restricted-equation-relations-straight-up} \sum \nolimits _ j k_{ij}g_ j = 0 \end{equation}
for $i = 1, \ldots , t$. Denote $K = (k_{ij})$ the $m \times t$-matrix with entries $k_{ij}$.
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. We may and do think of $A[x_1, \ldots , x_ r]^ h$ as a subring of $P = A[x_1, \ldots , x_ r]^\wedge $, see More on Algebra, Lemma 15.12.4. Since $A$ is a Noetherian G-ring, so is $A[x_1, \ldots , x_ r]$, see More on Algebra, Proposition 15.50.10. Hence we have approximation for the map $A[x_1, \ldots , x_ r]^ h \to A[x_1, \ldots , x_ r]^\wedge = P$ with respect to the ideal generated by $I$, see Smoothing Ring Maps, Lemma 16.14.1. Choose a large enough integer $n$ as in Lemma 88.6.1. By the approximation property we may choose $g'_1, \ldots , g'_ m \in A[x_1, \ldots , x_ r]^ h$ and a matrix $K' = (k'_{ij}) \in \text{Mat}(m \times t, A[x_1, \ldots , x_ r]^ h)$ such that $\sum \nolimits _ j k'_{ij}g'_ j = 0$ in $A[x_1, \ldots , x_ r]^ h$ and such that $g_ i - g'_ i \in I^ nP$ and $K - K' \in I^ n\text{Mat}(m \times t, P)$. By our choice of $n$ we conclude that there is an isomorphism
\[ B \to P/(g'_1, \ldots , g'_ m) = \left(A[x_1, \ldots , x_ r]^ h/(g'_1, \ldots , g'_ m)\right)^\wedge \]
This finishes the proof by Lemma 88.6.2. $\square$
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