Exercise 7.13

Proof. Let $f(x)=x^{p^n}-x$. We know from Ex. 7.12 that there exists a finite extension $K$ of ${𝔽}_p$ such that $f$ splits in linear factors on $K$ :

The set $k=\{{\alpha }_1,\cdots ,{\alpha }_{p^n}\}\subset K$ of the roots of $x^{p^n}-x$ is a subfield of $K$. Indeed, if $\alpha ,\beta \in k$,

(a)

$f(1)=0$, so $1\in k$

(b)

${(\alpha -\beta )}^{p^n}={\alpha }^{p^n}-{\beta }^{p^n}=\alpha -\beta$, so $\alpha -\beta \in k$.

(c)

${(\mathit{\alpha \beta })}^{p^n}={\alpha }^{p^n}{\beta }^{p^n}=\mathit{\alpha \beta }$, so $\mathit{\alpha \beta }\in k$.

(d)

${({\alpha }^{-1})}^{p^n}={({\alpha }^{p^n})}^{-1}={\alpha }^{-1}$, so ${\alpha }^{-1}\in k$ if $\alpha \ne 0$.

As $f^{\prime }(x)=-1$, $f(x)\wedge f^{\prime }(x)=1$. Thus $f$ has no multiple root, and the cardinality of $k$ is $p^n$.

Let $g(x)\in {𝔽}_p[x]$ a factor of $f(x)$, irreducible in ${𝔽}_p[x]$, with $d=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)$. As $g\mid f$, $g$ splits in linear factors in $k[x]$. Let $\alpha$ be a root of $g(x)$ in $k$. As $g$ is irreducible on ${𝔽}_p$, $d=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)=[{𝔽}_p[\alpha ]:{𝔽}_p]$. Moreover $n=[k:{𝔽}_p]=[k:{𝔽}_p[\alpha ]][{𝔽}_p[\alpha ]:{𝔽}_p]$, so $d\mid n$.

Conversely, suppose that $g$ is any irreducible polynomial in ${𝔽}_p[x]$, with $d=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)\mid n$. Then $K_0={𝔽}_p[x]∕⟨g⟩$ contains a root $\alpha$ of $g$, and $[K_0:{𝔽}_p]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)=d$, so ${\alpha }^{p^d}=\alpha$.

As $d\mid n$ , then $p^d-1\mid p^n-1$ and $x^{p^d-1}-1\mid x^{p^n-1}-1$ (Lemma 2,3 in section 1), thus

$f(\alpha )={\alpha }^{p^n}-\alpha =0$ and $g$ is the minimal polynomial of $\alpha$, so $g\mid f$.

Since $f$ has no multiple roots in $K$, $g^2\nmid f$.

Conclusion :

where $F_d(x)$ is the product of the monic irreducible polynomial of degree $d$. □