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Prove that $\mathbb F_{p^m} \subseteq \mathbb F_{p^n}$ iff $m$ divides $n$. Here, $\mathbb F_{p^m}$ is the finite field of $p^m$ elements.

My classmate suggested that I count the dimensions, considering both of them as vector spcaes over the base field $\mathbb F_p$. However, I am not sure if counting dimensions works. $\mathbb F_{p^m}$ has dimension $m$ over the base field, but for any vector spaces $V \subseteq W$, the dimension of $V$ does not have to divide the dimension of $W$. What am I missing here? I also know that $\mathbb F_{p^m}$ consists of all roots of polynomial $x^{p^m} - x$.

Squirrel-Power
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  • Your classmate is not quite right. You do want to consider that $\mathbb{F}p\subseteq \mathbb{F}{p^m}\subseteq \mathbb{F}{p^n}$, but the key fact here is not that both are vector spaces over $\mathbb{F}_p$, but rather that the larger is also a vector space over the smaller; i.e., that $\mathbb{F}{p^n}$ is a vector space over $\mathbb{F}_{p^m}$. – Arturo Magidin Mar 07 '24 at 19:31
  • The fundamental result is that $p^m-1\mid p^n-1$ iff $m\mid n.$ – Thomas Andrews Mar 07 '24 at 20:59

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