Exercise 7.23

Proof. Let $f(x)=x^p-x-\alpha \in F[x]$. There exists an extension $K\supset F$ with finite degree on $F$ which contains a root $\gamma$ of $f$.

As ${\gamma }^p-\gamma -\alpha =0$, then for all $i\in {𝔽}_p$,

So $f$ has $n$ distinct roots in $K$ : $\gamma ,\gamma +1,\dots ,\gamma +p-1$, and so

$F[\gamma ]$ contains all roots of $f$.

$\text{∙}$ If $\gamma \in F$, $f(x)$ splits in linear factors in $F$. $f(x)$ is not irreducible, since $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=p>1$.

$\text{∙}$ If $\gamma \notin F$, we will show that $f$ is irreducible in $F[x]$.

If not, then $f(x)=g(x)h(x)$ is the product of two monic polynomials $g,h\in F[x]$ such that $1\le \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)\le p-1$.

The unicity of the decomposition in irreducible factors in $F[\gamma ][x]$ shows that

where $A$ is a subset of ${𝔽}_p$, with $A\ne \varnothing ,A\ne {𝔽}_p$. As $g(x)\in F[x]$, $\underset{i\in A}{\sum <!--\; FUNCTION\; APPLICATION\; -->}(\gamma +i)=\mathit{k\gamma }+l\in {𝔽}_p$, where $1\le k=|A|\le p-1$ and $l=\underset{i\in A}{\sum <!--\; FUNCTION\; APPLICATION\; -->}i\in {𝔽}_p$.

So $\mathit{k\gamma }\in {𝔽}_p$. Since $\gamma \notin {𝔽}_p$, $k$ is not invertible in ${𝔽}_p$, in contradiction with $1\le k\le p-1$. Consequently, $f(x)$ is irreducible.

We conclude that $x^p-x-\alpha \in F[x]$ is irreducible iff $\gamma \notin F$.

Let $F$ be the Frobenius automorphism of $K$ (cf. Ex. 7.21).

The sum of these equalities gives

As the cardinality of $F$ is $q=p^n$,

Conclusion : $x^p-x-\alpha$ is irreducible iff $\mathrm{tr}(\alpha )\ne 0$. If $\mathrm{tr}(\alpha )=0$, $x^p-x-\alpha$ splits in linear factors in $F[x]$. □