Galois Theory for Finite Local Commutative Rings

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.

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.