Subfields of a finite field

Statement

Let $p$ be a prime number and let $n \ge 1$. For each divisor $m$ of $n$, the field $\mathbb{F}_{p^n}$ has exactly one subfield of order $p^m$, namely $\{ a : a^{p^m} = a \}$. It has no other subfields.

Remarks

If $K$ is a subfield of $\mathbb{F}_{p^n}$ of order $p^m$, then $n = m \cdot [\mathbb{F}_{p^n} : K]$ by the tower law, so $m$ divides $n$. For example, the subfields of $\mathbb{F}_{64} = \mathbb{F}_{2^6}$ are $\mathbb{F}_2$, $\mathbb{F}_4$, $\mathbb{F}_8$ and $\mathbb{F}_{64}$. The subfield of order $p^m$ is the fixed field of $\varphi^m$, where $\varphi$ is the Frobenius automorphism.

Explanations

L1 human

The subfields of a finite field match the divisors of the exponent. The field with $2^6$ elements contains the fields with $2^1$, $2^2$ and $2^3$ elements, but not one with $2^4$ elements.