$S_m(F)=\{a\in F:a^{p^m}=a\}$. $S_m (F)$ is a subfield of the field $F$.

Question

Consider $S_m(F)=\{a\in F:a^{p^m}=a\}$

$F$ is a field of order $p^n$ and $m$ divides $n$.

how can you show that $S_m (F)$ is a subfield of the field $F$?

I can show that $S_m (F)-0$ is a multiplicative cyclic group. But how can I show that $S_m (F)$ is a additive group as well.

Can anyone please help me?

Answer 1

If $\alpha,\beta \in S_m(F)$, we have, by Freshman's dream, \begin{align*} (\alpha+\beta)^{p^k}-(\alpha+\beta) & = \alpha^{p^k}+\beta^{p^k} -(\alpha+\beta)\\ & =0, \\ (\alpha\beta)^{p^k} -\alpha\beta & =\alpha^{p^k}\beta^{p^k}-\alpha\beta \\ & = (\alpha^{p^k}-\alpha)\beta^{p^k}+\alpha(\beta^{p^k}-\beta)\\ & =0. \end{align*} Hence sum and product are again in $S_m(F)$. In the same way, $-\alpha\in S_m(F)$ for $p>2$. For $p=2$ we trivially have $-\alpha=\alpha$.