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 $p$.

  • 4
    $\begingroup$ 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
  • 4
    $\begingroup$ You don't need to find a basis. You just need to know one exists. $\endgroup$ – Wojowu Feb 18 at 16:17
  • 2
    $\begingroup$ No you don't.. Call it $m$.... ("for some $m$....") $\endgroup$ – Randall Feb 18 at 16:22
  • 1
    $\begingroup$ Best of luck. Yes, we could use a better search engine. $\endgroup$ – 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 $F$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.