 Solution

The degree of a field extension $[K:M]$ is the dimension of a vector space of $K$ over $M$. If $({\alpha }_i)$ is an $M$-basis of $K$ and $({\beta }_j)$ a $K$-basis of $L$, then $({\alpha }_i{\beta }_j)$ spans $L$ over $M$. An $M$-linear relation among these products, regrouped by $j$, first vanishes coefficientwise by independence of the ${\beta }_j$, and then by independence of the ${\alpha }_i$. This proves the tower law $[L:M]=[L:K][K:M]$, including cardinal dimensions.

A finite field has prime characteristic $p$ and is a finite-dimensional vector space over its prime field ${\mathbb{F}}_p$. Thus its size is $p^n$. A subfield of size $p^m$ forces $m\mid n$ by the tower law. Conversely, if $m\mid n$, the polynomial $X^{p^m}-X$ divides $X^{p^n}-X$; all its $p^m$ distinct roots therefore lie in $K$. They are closed under addition, multiplication and inverses by the Finite-field Frobenius automorphism, forming the required subfield.

For an irreducible polynomial of degree $d$ over ${\mathbb{F}}_q$, a root generates ${\mathbb{F}}_{q^d}$. Its conjugates are $\alpha ,{\alpha }^q,\dots ,{\alpha }^{q^{d-1}}$, so this field is already the splitting field. The automorphism $x\mapsto x^q$ cycles these roots and has order $d$. Consequently