 Solution

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