TITLE: Is there a notion of Galois extension for Z / p^2?
QUESTION [8 upvotes]: The above title is in fact a special case of what I want to ask. 
Certainly we have a well defined notion of Galois extension for $ \mathbb{Q}_p $. The intersections of these extensions to the ring of integer of the absolute algebraic closure of $\mathbb{Q}_p$ give us a notion of Galois extensions for $\mathbb{Z}_p $. ( I know that there is a notion of Galois extension for commutative rings, and I believe that it should give us this. Am I correct?)
Let's go further. Let $A_K$ be the ring of integer in a finite Galois extension $K$ of $ \mathbb{Q}_p$. Let $e$ be the ramification degree of $K$ over $\mathbb{Q}_p$. The injection of $ \mathbb{Z}_p$ into $A_K$ will induce an injection of $ \mathbb{Z} / p^n $ into $ A_K / \mathfrak{p}^{en} $. In this picture, there seems to be some desire to say that $ A_K / \mathfrak{p}^{en} $ is the correct notion Galois  of extension of $ \mathbb{Z} / p^n $. But there are problems; taking this notion of Galois extension, if $K$ is has ramification degree $e >1$, the corresponding extension $ A_K /p^e $ is not a field (it is not even an integral domain).
Question 1: Is there any notion of Galois extensions corresponding to what I desire?
Question 2: Can a class field theory (i.e a nice description of absolute abelian Galois extension) of $ \mathbb{Z}/p^n$ be developed in this context? Is there any relationship between this and the local class field theory of $\mathbb{Q}_p$ ( which is the same as that of $\mathbb{Z}_p $)?

REPLY [4 votes]: Let $R$ be a commutative ring and $G$ a finite group. A Galois extension of $R$ with Galois group $G$ is a pair consisting of a morphism $f : R \to S$ and an $R$-linear action of $G$ on $S$ such that the natural map $R \to S^G$ is an isomorphism and the natural map
$$S \otimes_R S \ni s_1 \otimes s_2 \mapsto \prod_{g \in G} s_1 g s_2 \in \prod_{g \in G} S$$
is also an isomorphism. This is just a translation into algebra of the definition of a $G$-torsor over $\text{Spec } R$. With this definition, it is not true that Galois extensions of $\mathbb{Q}_p$ induce Galois extensions of $\mathbb{Z}_p$; I think only the unramified extensions do. 
Galois extensions of $R$ in this sense are classified by conjugacy classes of continuous homomorphisms $\pi_1(\text{Spec } R) \to G$ from the étale fundamental group of $\text{Spec } R$ to $G$. It's known that the étale fundamental group of $\text{Spec } \mathbb{Z}/(p^n)$ is $\hat{\mathbb{Z}}$, same as with $\mathbb{F}_p$. This says more or less that the underlying ring extension of every Galois extension is a finite product of the extensions described in Robin Chapman's answer.