[Lemma on page 321, Lindel], [Lemma 4.1.5, KC]

Lemma 10.146.1. Let $(R, \mathfrak m_ R) \to (S, \mathfrak m_ S)$ be a local homomorphism of local rings. Assume $S$ is the localization of an étale ring extension of $R$ and that $\kappa (\mathfrak m_ R) \to \kappa (\mathfrak m_ S)$ is an isomorphism. Then there exists an $t \in \mathfrak m_ R$ such that $R/t^ nR \to S/t^ nS$ is an isomorphsm for all $n \geq 1$.

Proof. Write $S = T_{\mathfrak q}$ for some étale $R$-algebra $T$ and prime ideal $\mathfrak q \subset T$ lying over $\mathfrak m_ R$. By Proposition 10.144.4 we may assume $R \to T$ is standard étale. Write $T = R[x]_ g/(f)$ as in Definition 10.144.1. By our assumption on residue fields, we may choose $a \in R$ such that $x$ and $a$ have the same image in $\kappa (\mathfrak q) = \kappa (\mathfrak m_ S) = \kappa (\mathfrak m_ R)$. Then after replacing $x$ by $x - a$ we may assume that $\mathfrak q$ is generated by $x$ and $\mathfrak m_ R$ in $T$. In particular $t = f(0) \in \mathfrak m_ R$. We will show that $t = f(0)$ works.

Write $f = x^ d + \sum _{i = 1, \ldots , d - 1} a_ i x^ i + t$. Since $R \to T$ is standard étale we find that $a_1$ is a unit in $R$: the derivative of $f$ is invertible in $T$ in particular is not contained in $\mathfrak q$. Let $h = a_1 + a_2 x + \ldots + a_{d - 1} x^{d - 2} + x^{d - 1} \in R[x]$ so that $f = t + xh$ in $R[x]$. We see that $h \not\in \mathfrak q$ and hence we may replace $T$ by $R[x]_{hg}/(f)$. After this replacement we see that

$T/tT = (R/tR)[x]_{hg}/(f) = (R/tR)[x]_{hg}/(xh) = (R/tR)[x]_{hg}/(x)$

is a quotient of $R/tR$. By Lemma 10.126.9 we conclude that $R/t^ nR \to T/t^ nT$ is surjective for all $n \geq 1$. On the other hand, we know that the flat local ring map $R/t^ nR \to S/t^ nS$ factors through $R/t^ nR \to T/t^ nT$ for all $n$, hence these maps are also injective (a flat local homomorphism of local rings is faithfully flat and hence injective, see Lemmas 10.39.17 and 10.82.11). As $S$ is the localization of $T$ we see that $S/t^ nS$ is the localization of $T/t^ nT = R/t^ nR$ at a prime lying over the maximal ideal, but this ring is already local and the proof is complete. $\square$

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