Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 3 / 11G (Groups, Rings and Modules)

Gauss lemma for polynomials says that the product of two primitive polynomials in $\mathbb{Z}[X]$ is primitive. Equivalently, a primitive polynomial in $\mathbb{Z}[X]$ is reducible over $\mathbb{Q}$ exactly when it is reducible over $\mathbb{Z}$.

Eisenstein criterion states that ifhas a prime number $p$ such that $p\nmid a_n$, $p\mid a_i$ for every $i<n$, and $p^2\nmid a_0$, then $f$ is irreducible over $\mathbb{Q}$.

After dividing out the polynomial content, suppose for contradiction that the resulting primitive polynomial factors as $f=gh$ with positive-degree $g,h\in \mathbb{Z}[X]$, using Gauss lemma for polynomials. Reduction modulo $p$ gives $\bar{f}=a_n{X}^n$, so in the polynomial ring ${\mathbb{F}}_p[X]$ both $\bar{g}$ and $\bar{h}$ are monomials. Because both factors have positive degree, their constant coefficients are divisible by $p$. Their product $a_0$ is then divisible by $p^2$, a contradiction.

Translate the variable by one:This satisfies Eisenstein criterion with $p=5$: every nonleading coefficient is divisible by $5$, while $25\nmid 30$. Translation $X\mapsto X+1$ is an automorphism of a polynomial ring, so it preserves irreducibility. Hence $f$ is irreducible over $\mathbb{Q}$.