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I have seen that (for $p$ prime) $\mathbb{F}_{p^m} $ is a subfield of $\mathbb{F}_{p^n}$ if and only if $m$ divides $n$. But my question is in what sense? Surely as long as $m\leq n $ then it is a subfield since $\mathbb{F}_{p^m} \subset \mathbb{F}_{p^n}$ and $\mathbb{F}_{p^m}$ is a field in it's own right - so it is a subfield?

Why is this wrong?

Anonmath101
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    No. For example $F_4$ is not a subset of $F_8$. $F_4$ has elements of multiplicative order $3$, $F_8$ only has elements of multiplicative order diving $7$. – ancient mathematician May 30 '21 at 12:40
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    If $m\le n$ but does not divide $n$, there is no subfield of $F_{p^n}$ of size $p^m$. There are only vector subspaces over $F_p$ of that size, but it's not the same thing. –  May 30 '21 at 12:43
  • This has been handled many times on the site. My choice of duplicate target is one of the oldests (but not necessarily the best fit for you). If you can make a more specific question, a better explanation can be given. Do continue! What made you think that $\Bbb{F}{p^2}$ would be a subset of $\Bbb{F}{p^3}$? It is roughly equivalent to suspecting that $\Bbb{Q}(\sqrt2)$ would be a subset of $\Bbb{Q}(\root3\of2)$. – Jyrki Lahtonen May 30 '21 at 17:45

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