Lemma 37.38.1. Let $S$ be a locally Noetherian scheme. Let $X$, $Y$ be schemes locally of finite type over $S$. Let $x \in X$ and $y \in Y$ be points lying over the same point $s \in S$. Assume $\mathcal{O}_{S, s}$ is a G-ring. Assume further we are given a local $\mathcal{O}_{S, s}$-algebra map

$\varphi : \mathcal{O}_{Y, y} \longrightarrow \mathcal{O}_{X, x}^\wedge$

For every $N \geq 1$ there exists an elementary étale neighbourhood $(U, u) \to (X, x)$ and an $S$-morphism $f : U \to Y$ mapping $u$ to $y$ such that the diagram

$\xymatrix{ \mathcal{O}_{X, x}^\wedge \ar[r] & \mathcal{O}_{U, u}^\wedge \\ \mathcal{O}_{Y, y} \ar[r]^{f^\sharp _ u} \ar[u]^\varphi & \mathcal{O}_{U, u} \ar[u] }$

commutes modulo $\mathfrak m_ u^ N$.

Proof. The question is local on $X$ hence we may assume $X$, $Y$, $S$ are affine. Say $S = \mathop{\mathrm{Spec}}(R)$, $X = \mathop{\mathrm{Spec}}(A)$, $Y = \mathop{\mathrm{Spec}}(B)$. Write $B = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. Let $\mathfrak p \subset A$ be the prime ideal corresponding to $x$. The local ring $\mathcal{O}_{X, x} = A_\mathfrak p$ is a G-ring by More on Algebra, Proposition 15.50.10. The map $\varphi$ is a map

$B_\mathfrak q^\wedge \longrightarrow A_\mathfrak p^\wedge$

where $\mathfrak q \subset B$ is the prime corresponding to $y$. Let $a_1, \ldots , a_ n \in A_\mathfrak p^\wedge$ be the images of $x_1, \ldots , x_ n$ via $R[x_1, \ldots , x_ n] \to B \to B_\mathfrak q^\wedge \to A_\mathfrak p^\wedge$. Then we can apply Smoothing Ring Maps, Lemma 16.13.4 to get an étale ring map $A \to A'$ and a prime ideal $\mathfrak p' \subset A'$ and $b_1, \ldots , b_ n \in A'$ such that $\kappa (\mathfrak p) = \kappa (\mathfrak p')$, $a_ i - b_ i \in (\mathfrak p')^ N(A'_{\mathfrak p'})^\wedge$, and $f_ j(b_1, \ldots , b_ n) = 0$ for $j = 1, \ldots , n$. This determines an $R$-algebra map $B \to A'$ by sending the class of $x_ i$ to $b_ i \in A'$. This finishes the proof by taking $U = \mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(B)$ as the morphism $f$ and $u = \mathfrak p'$. $\square$

Comment #5397 by Will Chen on

I don't think you want to say "between the complete local rings" right?

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