Exercise 7.14

Proof. Let $F={𝔽}_q$ a field with $q=p^m$ elements, and $n$ a positive integer.

By Theorem 2 Corollary 3, there exists an irreducible polynomial $f(x)\in {𝔽}_p[x]$ of degree $\mathit{nm}$. Let $g$ be an irreducible factor of $f$ in ${𝔽}_q[x]$, and let $\alpha$ be a root of $g$ in an extension of ${𝔽}_q$.

We show that ${𝔽}_q\subset {𝔽}_p[\alpha ]$.

${𝔽}_q$ and ${𝔽}_p[\alpha ]$ are two subfields of the same finite field ${𝔽}_q[\alpha ]$. Moreover, $|{𝔽}_q|=p^m$, and $|{𝔽}_p[\alpha ]|=p^{\mathit{nm}}$. As $m\mid \mathit{nm}$, ${𝔽}_q\subset {𝔽}_p[\alpha ]$ .

Indeed, for all $\gamma \in {𝔽}_q[\alpha ]$,

So ${𝔽}_q\subset {𝔽}_p[\alpha ]$.

We show that ${𝔽}_q[\alpha ]={𝔽}_p[\alpha ]$.

As ${𝔽}_p\subset {𝔽}_q$, ${𝔽}_p[\alpha ]\subset {𝔽}_q[\alpha ]$.

Let $\beta \in {𝔽}_q[\alpha ]$. Then $\beta =\underset{i=1}{\overset{k}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{a}_i{\alpha }^i$, where $a_i\in {𝔽}_q\subset {𝔽}_p[\alpha ]$, thus $a_i=p_i(\alpha ),p_i\in {𝔽}_p[x]$. Therefore

We have proved ${𝔽}_q[\alpha ]={𝔽}_p[\alpha ]$.

Since $|{𝔽}_p[\alpha ]|=p^{\mathit{nm}}$,

Thus $[{𝔽}_q[\alpha ]:{𝔽}_q]=n$. Moreover $g$ is the minimal polynomial of $\alpha$ on ${𝔽}_q$, thus $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)=n$.

Conclusion: if $F$ is a field with $q=p^m$ elements, there exist irreducible polynomials in $F[x]$ of degree $n$ for all positive integers $n$. □