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Let $R\subseteq S$ be two finite commutative local rings with unique maximal ideals $m$ and $M$, respectively. We say that $S$ is a separable extension of $R$ if $mS=M$. We also say that $S$ is a Galois extension of $R$ with Galois group $G$ if $S$ is a separable extension of $R$ and $G$ is a group of $R$-automorphisms of $S$ such that $S^G=R$, where

$S^G:=\{s\in S~|~\sigma(s)=s, \forall s\in G\}$.

I am reading the book "Finite Commutative Rings and Their Applications". In Chapter 5 of the book, the authors prove the following theorem:

Let $S$ be a separable extension of $R$. Then $S$ is a Galois extension of $R$ with Galois group $G_R(S)$ isomorphic to the Galois group $G_K(\mathbb{K})$, where $K=R/m$ and $\mathbb{K}=S/M$.

However, I cannot understand a part of their proof.

Let $G=G_R(S)$. It suffices to show that $S^G\subseteq R$, since $R\subseteq S^G$ is obvious and we can find a bijection between $G_R(S)$ and $G_K(\mathbb{K})$ by "lifting". In order to prove $S^G\subseteq R$, they say that if $s\in S\backslash R$ and $s$ is a unit of $S$, then

$\bar{\sigma}(\mu(s))\neq\mu(s),$ for some $\bar{\sigma}\in G_K(\mathbb{K})$,

where $\mu:S\rightarrow\mathbb{K}=S/M$ is a natural projection.

If $\mu(s)\notin K$, then such $\bar\sigma$ certainly exists. But I cannot show that $\mu(s)\notin K$. Could anyone tell me why $\mu(s)\notin K$ holds? Or please show me some alternative proof of the theorem if it exists.

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This is an old question, but I recently came across this result and used it repeatedly before realizing that it is actually false. This is inconvenient, so in the interest of preventing this from happening to other people, I will post an answer.

The correct result is that $S$ is Galois over $R$ if and only if $S$ is separable and free (as an $R$-module), so I'm glad to hear that you're having trouble proving it. For a general class of counterexamples, let $S$ be a finite local ring (not a field) with maximal ideal $M$ such that the residue field $S/M$ has degree $n>1$ over the prime subfield $\mathbb{F_p}$. Define $R$ to be the preimage under the canonical projection $S \to S/M$ of $\mathbb{F_p}$. Then $R$ is local with unique maximal ideal $M$, and thus the extension $R \subseteq S$ is unramified (separable) since the maximal ideals coincide. This extension is, however, not Galois. Any $R$-automorphism $\phi$ of $S$ necessarily fixes $M$ pointwise, but it is fairly easy to show that an automorphism of a finite local ring (that is not a field) which fixes the unique maximal ideal pointwise is the identity. Thus, if $H$ is the group of $R$-automorphisms of $S$, then $H$ is trivial, so $S^H= S$. But $S \neq R$, as they have different residue fields.

This claim appears in this paper, which was how I initially found it. You can view another paper, published some time later here, disproving it in a spirit that is similar to the argument above.

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