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

In field characteristic $p$, the intermediate binomial coefficients in the $p$th power vanish. HenceRepeated translation gives all $p$ roots $\alpha +j$, $j\in {\mathbb{F}}_p$. They are distinct because the prime field embeds in $F$. They exhaust the roots by degree; also the formal derivative of $X^p-X-a$ is $-1$, so this is a separable polynomial. If one root belongs to $F$, they all do, andThis is the root-translation property of an Artin–Schreier polynomial.

The intended hypothesis must be a proper monic factor, with $1\le r\le p-1$. Without that restriction the printed deduction is false: every polynomial has the displayed trivial factorization with $r=p$, even when it has no root in $F$.

By part 1, the roots in a splitting field are distinct and have the form $\alpha +j$. The degree-$r$ monic factor is the product of $r$ corresponding linear factors. Comparing the coefficient of $X^{r-1}$ using Vieta's formulas givesBoth $a_1$ and the sum of prime-field elements lie in $F$. Since $1\le r<p$, the scalar $r$ is nonzero and invertible in $F$. Therefore $\alpha =(-a_1-{\sum }_{j\in J}j)/r\in F$, and part 1 proves the polynomial splits into distinct linear factors over $F$. In particular an Artin–Schreier polynomial of prime degree is either irreducible or splits completely.

Reduce the monic integer polynomial modulo $p$. Every $u\in {\mathbb{F}}_p$ satisfies $u^p=u$, so $u^p-u-1=-1\ne 0$. Thus the reduction has no root in ${\mathbb{F}}_p$. If it had a proper factor, part 2 would force such a root, a contradiction. Hence it is irreducible over ${\mathbb{F}}_p$.

The criterion of irreducibility by degree-preserving reduction modulo a prime, which follows from Gauss's lemma, now applies because the polynomial is monic and its degree is preserved on reduction. Consequently $X^p-X-1$ is irreducible in $\mathbb{Q}[X]$.