Past exam of the options I course of the University of Oxford / 2010 / A (Introduction to Fields)

In characteristic $p$, the binomial theorem gives $(\alpha +1{)}^p={\alpha }^p+1$. HenceIterating shows that all $\alpha +j$, with $j\in {\mathbb{F}}_p$, are roots. They are distinct, since the prime-field elements $j$ are distinct. There are $p$ of them, equal to the polynomial's degree. Thus if $\alpha \in F$,splits into distinct linear factors in $F[X]$. Equivalently, its derivative is $-1$, so no repeated roots are possible.

Interpret the displayed factorization as a proper factorization, with $1\le r\le p-1$; that restriction is necessary for the conclusion. In a splitting field, the monic degree-$r$ factor has $r$ of the distinct roots of the original polynomial. Comparing its coefficient of $X^{r-1}$ givesfor an $r$-element subset $J\subseteq {\mathbb{F}}_p$. Both $a_1$ and the final sum lie in $F$. Since $0<r<p$, the element $r$ is invertible in $F$, soPart (1) then gives complete splitting into distinct linear factors. Thus an Artin–Schreier polynomial of this form either is irreducible or splits completely; a trivial factorization with $r=0$ or $r=p$ would not imply a root in $F$.

Reduce the monic integer polynomial modulo the prime $p$. For every $x\in {\mathbb{F}}_p$, $x^p=x$, soIf the reduction were reducible, a proper monic factorization would, by part (2), force a root in ${\mathbb{F}}_p$, a contradiction. Its reduction is therefore irreducible of degree $p$. The monic reduction criterion, obtained from Gauss lemma for polynomials by reducing a hypothetical integer factorization, now proves