How to prove that the additive group of a finite field of order $p^n$ is isomorphic to $Z_p^n$?

Question

Let $\mathbb{F}$ be a finite field of cardinality $p^n$ where $p$ is prime. How to prove that the additive group of $\mathbb{F}$ is isomorphic to $\mathbb Z_p^n$?

Answer 1

The point is that you don't need to use the explicit characterization of $\mathbb F$ as the splitting field of the polynomial $X^{p^n}$ over $\mathbb F_p$ - $\mathbb F$ is a vector space over $\mathbb F_p$ (since it's manifestly closed under addition, taking additive inverses and multiplying by elements of the field $\mathbb F_p$); now from linear algebra we know that any finite-dimensional vector space over a field $K$ is isomorphic to $K^n$ (and our field is finite, so it's certainly finite dimensional).

In this case, $K=\mathbb F_p$, and $\left|K^n\right|=p^n$. So just by considering cardinalities, we can say that $\mathbb F\cong\mathbb F_p^n$ as vector spaces. Since an isomorphism of vector spaces is a fortiori an isomorphism on additive groups, we have that $(\mathbb F,+)\cong (\mathbb F_p^n,+)\cong \mathbb Z_p^n$.

Note: $\mathbb F$ is not given by $\mathbb F_p[x]/(x^{p^n}-1)$ - the polynomial $(x^{p^n}-1)$ isn't irreducible in general, so this quotient needn't even be an integral domain, let alone a field. Instead, it's the field generated by all the roots of the polynomial $x^{p^n}-1$; alternatively, if $f\in\mathbb F_p[x]$ is irreducible of degree $n$, then it is a factor of $x^{p^n}-1$ and we have $\mathbb F\cong \mathbb F_p[x]/(f)$.

Answer 2

According to The fundamental theorem of finitely generated abelian groups I, there are additives groups $G_d1,G_d2,...,G_dt$ so the additive group $F$ is isomorphic to $G_d1 x G_d2 x ... x G_dt$ , where $d_1/d_2/d_3/.../d_t$ and $d_1d_2..d_t=p^n$. We know that the characteristic of the additive group F is p ($p*x=0$ for every $ x ε F $). So $d_1=d_2=..=d_t=p$ Furthermore, F is isomorphic to $ (Z_p)^n $