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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 $p$.

Semiclassical
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Our
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    Take a basis and start counting linear combinations. – Randall Feb 18 '19 at 16:15
  • @Randall Well,.. actually I couldn't find a basis. – Our Feb 18 '19 at 16:16
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    You don't need to find a basis. You just need to know one exists. – Wojowu Feb 18 '19 at 16:17
  • @Randall But then, I should also need to know the size of the basis. – Our Feb 18 '19 at 16:22
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    No you don't.. Call it $m$.... ("for some $m$....") – Randall Feb 18 '19 at 16:22
  • Sorely tempted to close this as a duplicate of this. One of the most common questions about finite fields on our site :-( – Jyrki Lahtonen Feb 18 '19 at 16:48
  • @JyrkiLahtonen I actually search for the question, but apparently google failed me (I know from expericen that in site search is worst than google. ) – Our Feb 18 '19 at 16:59
  • Google cannot grok TeX. Neither can the local search. But a local search within the tag [tag:finite-fields] gave me a lot of hits in response to a few appropriate keywords (cardinality IIRC). Occasionally you absolutely need to search for a TeX-string. Approach0 often helps there, but may need a bit of fine-tuning. – Jyrki Lahtonen Feb 20 '19 at 07:52
  • @JyrkiLahtonen Actually, I tried Approach0 many times before, but wasn't successful, so I stop trying, but I will try next time; thanks. – Our Feb 20 '19 at 07:59
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    Best of luck. Yes, we could use a better search engine. – Jyrki Lahtonen Feb 20 '19 at 08:27

1 Answers1

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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 $F$.

Randall
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