The Galois field $GF(p^n)$ of order $p^n$ can be viewed as a vector space over the Galois field $GF(p)$ of order $p$. But I haven't understood why the degree of $GF(p^n)$ over $GF(p)$ is $n$ (that is the dimension of $GF(p^n)$ over $GF(p)$ is $n$).
The proof that I've found seems unsatisfactory for me. It says that the Galois field $GF(p^n)$ is group isomorphic to $\mathbb{Z}_p^n$ and since $\mathbb{Z}_p^n$ is a vector space over $\mathbb{Z}_p$, it's dimension over the field $\mathbb{Z}_p$ is $n$. I didn't have any problem understanding the proof till here.
The part of the proof which I had problem with is this: the proof says that because $GF(p^n)$ is group isomorphic to $\mathbb{Z}_p^n$ under addition, and it has dimension $n$ over $\mathbb{Z}_p$, which is field isomorphic to $GF(p)$, $GF(p^n)$ will have dimension $n$ over $GF(p)$. It's easy for me to understand why $\mathbb{Z}_p^n$ will be a vector space over $\mathbb{Z}_p$ with dimension, because the set $\left\{ e_1,e_2,\ldots,e_n\right\}$ will form a basis for $\mathbb{Z}_p^n$ over $\mathbb{Z}_p$, where $e_i \in \mathbb{Z}_p^n$, whose $i^{th}$ coordinate is $1$ while the rest of the coordinate is zero. The trouble is, I couldn't find an equivalent basis like this in the Galois field $GF(p^n)$ over $GF(p)$. I also couldn't create linear isomorphism from $GF(p^n)$ to $\mathbb{Z}_p^n$ since, from what I've learned from linear algebra, linear transformation is defined between two vector spaces over the same field. It need not be true that both $GF(p^n)$ and $\mathbb{Z}_p^n$ will share the same field, although they both have scalar field that is isomorphic to $\mathbb{Z}_p$. Because, if I can show that both $GF(p^n)$ and $\mathbb{Z}_p^n$ will have the same scalar field, then it won't be any problem for me to prove the final result. It'd help me if someone give me a hint to create a linear isomorphism from $GF(p^n)$ to $\mathbb{Z}_p^n$ under the same field.
I am not sure where the problem is. We have $GF(p)\cong \Bbb Z/p\Bbb Z\cong \Bbb F_p$. This is the ground field. Now $GF(p^n)$ has $p^n$ elements, so it is isomorphic to $\Bbb F_{p^n}$. This is not the ring $\Bbb Z/p^n\Bbb Z$, which has zero divisors. Any two (finite-dimensional) vector spaces over the same ground field are isomorphic if and only if they have the same dimension. So $$ \dim_{GF(p)}(GF(p^n))=\dim_{\Bbb F_p}(\Bbb F_{p^n})=n. $$