Galois group of $F_{p^2}$ over $\mathbb{Z}_p$ Q) For any prime $p$, what is the Galois group of $F_{p^2}$ over $\mathbb{Z}_p$, where $F_{p^2}$ is the field with $p^2$ elements. 
I know how to find Galois group of a particular polynomial over a field. For this question, should I think of $F_{p^2}$ as a splitting field extension of $\mathbb{Z}_p$ for some irreducible polynomial in $\mathbb{Z}_p$? If yes, the Galois group $G_{F_{p^2}/\mathbb{Z}_p}$ is the group of all automorphisms of $F_{p^2}$ which preserve $\mathbb{Z}_p$. I can permute $p^2 - p$ elements in $F_{p^2}$ in $(p^2 - p)!$ ways but have to preserve homomorphism structure too and am not sure how to constrain that? Thanks and appreciate a hint.
 A: Let's write $\mathbb{F}_p$ instead of $\mathbb{Z}_p$ for the field with $p$ elements.
Now $\mathbb{F}_{p^2}$ is a two-dimensional vector space over $\mathbb{F}_p$ (by counting), so any element $\alpha$ of $\mathbb{F}_{p^2} \setminus \mathbb{F}_p$ must satisfy an irreducible degree $2$ polynomial and therefore generate this field extension. Therefore, the Galois group must be $\mathbb{Z}/2\mathbb{Z}$.
(Note that this answer assumes one already knows a priori that a field with $p^2$ elements exists, which seems fair, given the way the question is worded. If you are asked to construct one, the problem becomes a little harder.)
A: Since this is a Galois extension we have $2=[\mathbb{F_{p^2}}:\mathbb{F_p}]=|Gal(\mathbb{F_{p^2}}/\mathbb{F_p})|$. Hence it is obvious that the Galois group must be isomorphic to $\mathbb{Z}/2\mathbb{Z}$. Now, if you want to find the automorphisms then it is easy as well. We have the trivial automorphism, and it is well known that the Frobenius endomorphism $x\to x^p$ is also an automorphism on finite fields of characteristic $p$. 
A more general result: the Galois group of $\mathbb{F_{p^n}}$ over $\mathbb{F_p}$ is the cyclic group generated by the Frobenius endomorphism. 
