The order of a finite field

Question

I'm reading a theorem about the order of a finite field:

Here is the proof:

At the end, the author said

It follows that $\mathbb{F}$ is a vector space over $\mathbb{F}_{p}$, implying that its size $q$ is equal to $p^{m}$ for some $m>0$.

I do not understand how $\mathbb{F}$ is a vector space over $\mathbb{F}_{p}$ implies that its size $q$ is equal to $p^{m}$. Could you please elaborate on it?

Answer 1

Given any field $k$ and a vector space $V$ of finite dimension $n$ over $k$, the cardinality of $V$ is exactly $\lvert k\rvert^n$.

To see this, just fix any basis of $V$ and use the fact that any element of $V$ can be uniquely expressed as a $k$-linear combination of basis vectors, and there are clearly $k^n$ possible linear combinations of the $n$ basis vectors.

(More generally, if $k$ is any field and $V$ is a $k$-vector space of dimension $\lambda$, then by the same argument, the cardinality of $V$ is $\sum_n \binom{\lambda}{n}(\lvert k\rvert-1)^n$, which for infinite $\lambda$ is equal to $\lambda \cdot \lvert k \rvert$.)

Answer 2

The author has proved $\mathbf F$ satisfies the axioms for being a vector space over $\mathbf F_p$. As it is a finite field, it is ipso facto finite dimensional, hence isomorphic to $\mathbf F_p^m$ for some $m$ and the latter has cardinality $p^m$.

Answer 3

Any field $F$ of characteristics $p$ contains the subfield $\Bbb{F}_p$ (and also any field of ch. $0$ contains $\Bbb{Q}$).

Now just check, if $K$ is a subfield of $F$ then $F$ is a vector space over $K$ by the action $K\times F\rightarrow F$ defined by, $(k,v)\mapsto kv$.