Q:

Finite Galois extensions of the form $\frac{\mathbb Z_p[x]}{\langle p(x)\rangle}:\mathbb Z_p$

All the extensions with this form $\frac{\mathbb Z_p[x]}{\langle p(x)\rangle}:\mathbb Z_p$ are Galois extensions?
where $p$ is prime, $p(x)$ is irreducible and the extension is finite. 

A:

To add on Tsemo's answer. 
$q = p^k$ where $p$ is prime.

If $f \in \mathbb{F}_q[x], deg(f) = n$ is irreducible then $K = \mathbb{F}_q[x]/(f(x))$ is a finite integral domain and hence a field, with $q^n$ elements. 
Since $p = 0$ in $K$, $(a+b)^p = a^p+b^p$ and $\phi(a) = a^p$ is an automorphism of $K$, as well as $\varphi(a) = a^q = \phi^k(a)$.
Since $\mathbb{F}_q^\times $ has $q-1$ elements, those satisfy $a^{q-1} = 1$ and hence $\varphi \in \text{Gal}(k/\mathbb{F}_q)$. 
So all the elements of $\mathbb{F}_q$ are roots of $x^q-x$. But this polynomial of degree $q$ has at most $q$ roots in the field $k$. Thus $\mathbb{F}_q$ is exactly the splitting field of $x^q-x$.
Let $G = \langle \varphi \rangle$ the cyclic group generated by $\varphi$. We obtained $$\mathbb{F}_q = K^G = \{ a \in K, \forall g \in G, g(a) = a\}= \{ a \in K, a^q-a=0\}$$ and 
hence $K/\mathbb{F}_q$ is a Galois extension with Galois group $G$.
If $\alpha \in K$ then its minimal polynomial over $\mathbb{F}_q$ is
$$g(X) = \prod_{\beta \in G (\alpha)} (X-\beta)=\prod_{m=0}^{l-1} (X-\varphi^m(\alpha))=\prod_{m=0}^{l-1} (X-\alpha^{q^m}) \qquad \in \mathbb{F}_q[X]$$
where $l$ is the least positive integer such that $\alpha^{q^l} = \alpha$

A:

A finite extension of $\mathbb{Z}_p$ is a finite vector space of dimension $n$ over $\mathbb{Z}_p$. It has $p^n$ elements and is the  splitting field of $X^{p^n}-X$, so it is a normal extension and separable extension since $X^{p^n}-X$ is separable so it is a Galois extension.

