For a field of order $2^n$ to contain a field of order $2^m$

Question

For a field of order $2^n$ to contain a field of order $2^m$, where $m<n$, must it be the case that $m$ divides $n$? I've been told this is indeed the case, however I can't figure out why. Without going into group theory, does a proof exist?

Answer 1

For any prime $p$, suppose we have a tower of fields $\mathbb{F}_{p} \subset \mathbb{F}_{p^m} \subset \mathbb{F}_{p^n}$. The multiplicativity formula for degrees tells us that $[\mathbb{F}_{p^n}: \mathbb{F}_p] = [\mathbb{F}_{p^n}: \mathbb{F}_{p^m}] \cdot [\mathbb{F}_{p^m}: \mathbb{F}_p]$.

What is $[\mathbb{F}_{p^m}: \mathbb{F}_p]$? What is $[\mathbb{F}_{p^n}: \mathbb{F}_p]$?

The approach in Ravi's answer [+1] is the idea one uses to prove this formula.

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As a side note, when you get to group theory, you'll find that this multiplicativity formula in the field theory setting is very similar to the multiplicativity of indices for groups: if we have $H \subset K \subset G$, then $[G:H] = [G:K] \cdot [K:H]$ (assuming these values are finite). This is not a mere coincidence; the fundamental theorem of Galois theory illuminates the connection.

Answer 2

In a sentence, if a $F\subseteq F'$ is a finite extension of fields, then $F'$ is a finite dimensional $F$-vector space. So $F'$ has a basis $v_1,\cdots ,v_k$ over $F$ so literally all the elements of $F'$ are uniquely expressible as $$\sum_{i=1}^k a_iv_i \text{ for } a_i\in F$$

If $F$ is finite of order $q$, we can count that the number of elements of $F'$ as $q\cdot q\cdots q$ for a total of $k$ times, for $q^k$. So of course $q$ divides $q^k$.