# Why does the cardinality of the vector space over a finite field of characteristic $p$ have to be a power of $p$?

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 $$p$$.

• Take a basis and start counting linear combinations. – Randall Feb 18 at 16:15
• @Randall Well,.. actually I couldn't find a basis. – onurcanbektas Feb 18 at 16:16
• You don't need to find a basis. You just need to know one exists. – Wojowu Feb 18 at 16:17
• No you don't.. Call it $m$.... ("for some $m$....") – Randall Feb 18 at 16:22
• Best of luck. Yes, we could use a better search engine. – Jyrki Lahtonen Feb 20 at 8:27

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 $$F$$.