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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 $p$.

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

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  • $\begingroup$ Thanks for the answer. $\endgroup$ – onurcanbektas Feb 18 at 16:29

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