TAOCP 4.6.2 Exercise 19
Assume that $u(x)$ is reducible over the integers.
Section 4.6.2: Factorization of Polynomials
Exercise 19. [M20] (Eisenstein's criterion.) Perhaps the best-known class of irreducible polynomials over the integers was introduced by T. Schönemann in Crelle 32 (1846), 100, then popularized by G. Eisenstein in Crelle 39 (1850), 166–169: Let $p$ be prime and let $u(x) = u_n x^n + \cdots + u_0$ have the following properties: (i) $u_n$ is not divisible by $p$; (ii) $u_{n-1}, \ldots, u_0$ are divisible by $p$; (iii) $u_0$ is not divisible by $p^2$. Show that $u(x)$ is irreducible over the integers.
Verified: yes
Solve time: 1m08s
Solution
Assume that $u(x)$ is reducible over the integers. Since $u(x)$ is primitive, Gauss's lemma implies that there exist primitive polynomials
$$ a(x)=a_r x^r+\cdots+a_0,\qquad b(x)=b_s x^s+\cdots+b_0, $$
with positive degrees $r,s<n$, such that
$$ u(x)=a(x)b(x). $$
Because $u_n$ is not divisible by $p$, the leading coefficients satisfy
$$ a_r b_s=u_n, $$
hence neither $a_r$ nor $b_s$ is divisible by $p$.
Consider the coefficients of $u(x)$ modulo $p$. Since
$$ u_{n-1},u_{n-2},\ldots,u_0 $$
are all divisible by $p$, we have
$$ u(x)\equiv u_n x^n \pmod p. $$
Therefore
$$ a(x)b(x)\equiv u_n x^n \pmod p. $$
Since neither $a_r$ nor $b_s$ is divisible by $p$, the reductions of $a(x)$ and $b(x)$ modulo $p$ are nonzero polynomials whose product is a monomial. Hence every nonleading coefficient of $a(x)$ and $b(x)$ must be divisible by $p$:
$$ p\mid a_{r-1},\ldots,a_0,\qquad p\mid b_{s-1},\ldots,b_0. $$
In particular,
$$ p\mid a_0,\qquad p\mid b_0. $$
But
$$ u_0=a_0b_0, $$
therefore
$$ p^2\mid u_0. $$
This contradicts hypothesis (iii), which states that $u_0$ is not divisible by $p^2$.
The assumption that $u(x)$ is reducible is impossible. Therefore $u(x)$ is irreducible over the integers.
This completes the proof.
∎