Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 4 / 20G / Solution

A rational root of the monic cubic would be one of $\pm 1,\pm 3$, and substitution excludes all four. A reducible cubic over $\mathbb{Q}$ would have a rational root, so the polynomial is irreducible andIts polynomial discriminant is $-4(-1{)}^3-27(3{)}^2=-239$. The integer 239 is prime: it has no prime divisor among $2,3,5,7,11,13$, the primes not exceeding its square root.

The discriminant-index formula for an integral lattice for an integral primitive element isBoth discriminants are integers, and the left side is $-239$. Its squarefreeness forces the positive index to be one. Since $\alpha$ is an algebraic integer by its monic equation,