Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4 / 2F (Groups, Rings and Modules)

A polynomial in $\mathbb{Z}[x]$ is a primitive polynomial when the greatest common divisor of its coefficients is one. The Gauss lemma for polynomials states that a product of primitive integer polynomials is primitive, and that a primitive positive-degree integer polynomial is reducible over $\mathbb{Q}$ exactly when it factors into positive-degree integer polynomials. More generally, polynomial content is multiplicative up to sign.

The Eisenstein criterion states that a primitive polynomial $a_n{x}^n+\cdots +a_0\in \mathbb{Z}[x]$ is irreducible over $\mathbb{Q}$ if there is a prime $p$ such that $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$. These criteria permit integer coefficient tests for rational irreducibility; the following parts apply them with their exact divisibility hypotheses.

Use the Eisenstein criterion with prime $5$. The leading coefficient is one; every lower coefficient, including the zero coefficients, is divisible by $5$; and the constant coefficient $5$ is not divisible by $25$. The polynomial is monic and hence primitive. Therefore it is irreducible over $\mathbb{Q}$.

A cubic over a field is reducible exactly when it has a root in that field: a nontrivial factorization must contain a degree-one factor. By the rational root theorem, a rational root of this monic integer polynomial must be an integer dividing its constant coefficient, hence either $1$ or $-1$. Substitution gives $-2$ and $4$, respectively. Neither is zero. Thus the cubic is irreducible over $\mathbb{Q}$.

Let $P(x)$ denote the polynomial under consideration. The change $x\mapsto x+1$ is an automorphism of $\mathbb{Q}[x]$ and preserves degrees, so it preserves reducibility. Its transformed polynomial isIt is monic; every other coefficient is divisible by $p$, because $p$ is prime and $1\le j<p$; and its constant coefficient is $p$, not divisible by $p^2$. The Eisenstein criterion proves that $P(x+1)$ is irreducible, and therefore $P$ is irreducible. The argument also includes $p=2$, when the polynomial is linear. This is the prime-index case of an irreducible polynomial arising as a cyclotomic polynomial.