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 $n \geq 1$.

## 10.146 Local homomorphisms

Some lemmas which don't have a natural section to go into. The first lemma says, loosely speaking, that an étale map of local rings is an isomorphism modulo all powers of a nonunit principal ideal.

**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

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. $\square$

Lemma 10.146.2. 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 $R$. Then there exists a finite, finitely presented, faithfully flat ring map $R \to S'$ such that for every maximal ideal $\mathfrak m'$ of $S'$ there is a factorization

of the ring map $R \to S'_{\mathfrak m'}$.

**Proof.**
Write $S = T_{\mathfrak q}$ for some étale $R$-algebra $T$. By Proposition 10.144.4 we may assume $T$ is standard étale. Apply Lemma 10.144.5 to the ring map $R \to T$ to get $R \to S'$. Then in particular for every maximal ideal $\mathfrak m'$ of $S'$ we get a factorization $\varphi : T \to S'_{g'}$ for some $g' \not\in \mathfrak m'$ such that $\mathfrak q = \varphi ^{-1}(\mathfrak m'S'_{g'})$. Thus $\varphi $ induces the desired local ring map $S \to S'_{\mathfrak m'}$.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)