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.

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.

Let $R$ be a commutative ring. 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.

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.