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Note that we want $\mathbb{F}_p[a] \cong \mathbb{F}_p[b]$ as subfields of $K$.

I was trying to solve the problem by first noting that $|K|=p^n$ for some $n\in \mathbb{N}$. Then I wanted to count the elements in both $\mathbb{F}_p[a]$ and $\mathbb{F}_p[b]$ and show that they are the same (at that point it must follow they are isomorphic as subfields of $K$). However, I can't seem to use the fact that they are elements of the same order in the group of units. I know that if $a,b\in K^\times$ then they are roots of the polynomial $x^{p^n - 1}$ and so their order must divide $p^n-1$. Is there a better way to approach this problem?

user13121312
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  • I don't understand your notation. It's not necessarily the case that there is a natural map $a \mapsto \Bbb Z$ or $p[a] \mapsto \Bbb Z$, and without some kind of map (natural or otherwise), I don't undersand what you mean by $\Bbb Z/p[a]$. – Robert Shore Nov 08 '23 at 01:59
  • $\mathbb{Z}/p [a]$ smallest subring (subfield) of $K$ containing $\mathbb{Z}/p$ and $a$. This is similar notation to, for example, $\mathbb{Z}[\sqrt{2}]$. – user13121312 Nov 08 '23 at 03:37
  • Now I understand. I'd probably have used the notation $\Bbb F_p[a]$. – Robert Shore Nov 08 '23 at 07:47
  • Each an every element of order $m$ multiplicatively generates the (known to by cyclic) subgroup $\mu_m\le K^*$ of solutions of $x^m=1$. In other words, $a$ and $b$ are each others powers. – Jyrki Lahtonen Nov 08 '23 at 08:48
  • Thank you for your comment. Yes, I understand that. But how does that help me show the respective subfields are isomorphic? – user13121312 Nov 08 '23 at 15:54
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    @user13121312 do you really want to show that they're isomorphic, or that they define the same subfield of $K$? Jyrki's comment shows you that $a = b^{n}$ for some $n$, which means that $a \in \mathbb{F}{p}[b]$, and so $\mathbb{F}{p}[a] \subseteq \mathbb{F}{p}[b]$ by your own definitions above. Likewise, $b = a^{k}$ for some $k$, so $b \in \mathbb{F}{p}[a]$, and we get the symmetric inclusion $\mathbb{F}{p}[b] \subseteq \mathbb{F}{p}[a]$. – Alex Wertheim Nov 09 '23 at 01:22
  • Alex, I guess you are right that I want to show they are the same subfield. I could have interpreted the question from my original problem wrong, since it does say to show $\mathbb{F}_3[a]=\mathbb{F}_3[b]$, not $\cong$. Thank you for your comment – user13121312 Nov 09 '23 at 01:38

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