Example 96.12.8. In this example we show that the local ring $\mathcal{O}_{S, s}$ has to be a G-ring in order for the result of Lemma 96.12.7 to be true. Namely, let $\Lambda$ be a Noetherian ring and let $\mathfrak m$ be a maximal ideal of $\Lambda$. Set $R = \Lambda _\mathfrak m^\wedge$. Let $\Lambda \to C \to R$ be a factorization with $C$ of finite type over $\Lambda$. Set $S = \mathop{\mathrm{Spec}}(\Lambda )$, $U = S \setminus \{ \mathfrak m\}$, and $S' = U \amalg \mathop{\mathrm{Spec}}(C)$. Consider the functor $F : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$ defined by the rule

$F(T) = \left\{ \begin{matrix} * & \text{if }T \to S\text{ factors through }S' \\ \emptyset & \text{else} \end{matrix} \right.$

Let $\mathcal{X} = \mathcal{S}_ F$ is the category fibred in sets associated to $F$, see Algebraic Stacks, Section 92.7. Then $\mathcal{X} \to (\mathit{Sch}/S)_{fppf}$ is limit preserving on objects and there exists an effective, versal formal object $\xi$ over $R$. Hence if the conclusion of Lemma 96.12.7 holds for $\mathcal{X}$, then there exists a finite type ring map $\Lambda \to A$ and a maximal ideal $\mathfrak m_ A$ lying over $\mathfrak m$ such that

1. $\kappa (\mathfrak m) = \kappa (\mathfrak m_ A)$,

2. $\Lambda \to A$ and $\mathfrak m_ A$ satisfy condition (4) of Algebra, Lemma 10.141.2, and

3. there exists a $\Lambda$-algebra map $C \to A$.

Thus $\Lambda \to A$ is smooth at $\mathfrak m_ A$ by the lemma cited. Slicing $A$ we may assume that $\Lambda \to A$ is étale at $\mathfrak m_ A$, see for example More on Morphisms, Lemma 37.34.5 or argue directly. Write $C = \Lambda [y_1, \ldots , y_ n]/(f_1, \ldots , f_ m)$. Then $C \to R$ corresponds to a solution in $R$ of the system of equations $f_1 = \ldots = f_ m = 0$, see Smoothing Ring Maps, Section 16.13. Thus if the conclusion of Lemma 96.12.7 holds for every $\mathcal{X}$ as above, then a system of equations which has a solution in $R$ has a solution in the henselization of $\Lambda _{\mathfrak m}$. In other words, the approximation property holds for $\Lambda _{\mathfrak m}^ h$. This implies that $\Lambda _{\mathfrak m}^ h$ is a G-ring (insert future reference here; see also discussion in Smoothing Ring Maps, Section 16.1) which in turn implies that $\Lambda _{\mathfrak m}$ is a G-ring.

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