In a lecture note that I have, it is written that
if $F$ is a field of $q$ elements of characteristic $p$, then $q =
p^m$ for some $m>0$.To show this, observe that $F$ is a vector space over the field $F_p
= \{n \cdot 1_F | n \in \mathbb{N}\} $ with $(n \cdot 1_F) * x = n \cdot x$ for $x\in F$. So the result directly follows.
I can't understand why the cardinality of the vector space over a finite field of characteristic $p$ has to be a power of $p$.
Best Answer
Suppose $F$ has basis $\{v_1, v_2, \ldots, v_m\}$ over $\mathbb{F}_p$. Then each element $x$ of $F$ is uniquely expressible as $$ x = c_1v_1 + c_2v_2 + \cdots +c_mv_m. $$ But there are $p=|\mathbb{F}_p|$ choices for $c_1$, $p=|\mathbb{F}_p|$ choices for $c_2$, ..., and $p=|\mathbb{F}_p|$ choices for $c_m$. Hence there are $p^m$ ways to build such linear combinations, so there are $p^m$ elements in $F$.