Let $K$ be finite field and $L$ be an extension of $K$ of degree $n$. Fix
a monic irreducible polynomial $f(x)\in K[X]$ of degree d dividing n. Show
that there is element $\alpha \in L $ which has minimal polynomial
$f$ over $K$.
I know that $K$ is isomorphic to field $\mathbb{F}_{p^m}$ for some $m$. If $m=1$ then $K=\mathbb{F}_{p}$ and we get result from the fact that $L$ is given by roots of polynomial $X^{p^n}-X$, which is product of all irreducible polynomials over $\mathbb{F}_{p}$ of degree $d$ dividing $n$, and hence has to contain roots of any irreducible polynomial of such degree.
I have trouble with general case when $m\neq1$.
If I take a root $\alpha$ of polynomial $f$ then I get extension $K(\alpha)$ of degree $d$ over $K$, which is isomorphic to $\mathbb{F}_{p^{md}}$. Field $\mathbb{F}_{p^{md}}$ is given by roots of polynomial $X^{p^{md}}-X$ which is product of all irreducible polynomials over $\mathbb{F}_{p}$ of degree dividing $md$. Hence minimal polynomial of $\alpha$ over $\mathbb{F}_{p}$ has to be of degree dividing $md$, and hence also dividing $mn$. Because of that, similarly as in case $m=1$, $L$ has to contain $\alpha$. Is my reasoning correct?
Is there is another, quicker approach?
Best Answer
It is most convenient to consider an algebraic closure $F$ of $L$; then $F$ will automatically be an algebraic closure of $K$ containing $L$ as a subextension. The structure of the algebraic extension of a finite field is remarkably simple: for any $n \in \mathbb{N}^*$ there exists a unique subextension $E_n$ of degree $n$ over $K$, given explicitly as the set of all roots of the polynomial (separable over $K$) $X^{q^n}-X$, where $q=|K|$; furthermore, one has
$$F=\bigcup_{n \in \mathbb{N}^*} E_n$$
and
$$E_m \subseteq E_n \Leftrightarrow m|n$$
Considering an arbitrary root $x \in F$ of your given polynomial $f$, it is clearly the case that $[K(x):K]=d$ whence $K(x)=E_d \subseteq E_n=L$; therefore, all the roots of $f$ lie in the subextension $L$.