Proof of the reduction modulo p test for irreducibility

Proof

1. Let $f \in \Z[x]$ be monic, let $p$ be a prime, and assume that $\bar{f}$ is irreducible in $\F_p[x]$. Here $\F_p = \Z/p\Z$ is a field.
2. Let $\pi \colon \Z[x] \to \F_p[x]$ reduce each coefficient modulo $p$. By the universal property of the polynomial ring, applied to $\Z \to \F_p$ and $x \mapsto x$, the map $\pi$ is a ring homomorphism. Also $\pi(f) = \bar{f}$.
3. If $g \in \Z[x]$ is monic, then $\pi(g)$ is monic with $\deg \pi(g) = \deg g$, because the leading coefficient $1$ reduces to $1 \ne 0$.
4. Over a field $F$, the units of $F[x]$ are the nonzero constants. Indeed, $u v = 1$ forces $\deg u = 0$ by the degree of a product, and a nonzero constant has an inverse in $F$.
5. By the same degree argument, the units of $\Z[x]$ are the units of $\Z$, that is, $\pm 1$.
6. Since $\bar{f}$ is irreducible, it is not a unit, so $\deg \bar{f} \ge 1$ by Step 4. By Step 3, $\deg f = \deg \bar{f} \ge 1$. So $f$ is nonzero and not a unit of $\Z[x]$.
7. Suppose that $f = g h$ with $g, h \in \Z[x]$. Their leading coefficients multiply to $1$, so both are $1$ or both are $-1$.
8. In the second case, replace $g, h$ by $-g, -h$. This does not change the product, and $g$ is a unit if and only if $-g$ is. So we may assume that $g$ and $h$ are monic.
9. By Step 2, $\bar{f} = \pi(g) \, \pi(h)$. Since $\bar{f}$ is irreducible, $\pi(g)$ or $\pi(h)$ is a unit, say $\pi(g)$. By Step 4, $\deg \pi(g) = 0$.
10. By Step 3, $\deg g = 0$. A monic polynomial of degree $0$ is the constant $1$, so $g$ is a unit.
11. By Steps 6 to 10, $f$ is irreducible in $\Z[x]$.
12. The monic polynomial $f$ is primitive: an integer that divides every coefficient divides the leading coefficient $1$, so it is $\pm 1$.
13. The ring $\Z$ is a Euclidean domain, hence a principal ideal domain, hence a unique factorisation domain, with field of fractions $\Q$.
14. By Gauss's lemma, with Steps 11 to 13, $f$ is irreducible in $\Q[x]$.