Example 97.12.8. 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 97.12.7 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

Let $\mathcal{X} = \mathcal{S}_ F$ is the category fibred in sets associated to $F$, see Algebraic Stacks, Section 93.7. 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 97.12.7 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

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

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

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 37.38.5 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 97.12.7 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.

## Comments (0)