Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 2 / 2G (Groups, Rings and Modules)

A complex number is an algebraic number over $\mathbb{Q}$ when it is a root of a nonzero rational polynomial. Its minimal polynomial of an algebraic element is the unique monic polynomial in $\mathbb{Q}[X]$ of smallest degree annihilating it; polynomial division shows that it divides every rational polynomial annihilating the element, and it is irreducible.

Consider the surjective evaluation ring homomorphism ${\mathrm{ev}}_{\alpha }:\mathbb{Z}[X]\to \mathbb{Z}[\alpha ]$. Certainly $(f)\subseteq \ker {\mathrm{ev}}_{\alpha }$. Because the nonconstant $f$ is irreducible over $\mathbb{Z}$, its content is a unit: otherwise its content and its nonconstant primitive part give a factorization into nonunits. By the Gauss lemma for polynomials, $f$ is irreducible in $\mathbb{Q}[X]$ and is a scalar multiple of the minimal polynomial.

If $g\in \mathbb{Z}[X]$ vanishes at $\alpha$, then $f$ divides $g$ in $\mathbb{Q}[X]$. Primitive divisibility in Gauss lemma for polynomials implies division in $\mathbb{Z}[X]$ as well. Explicitly, write the rational quotient as $(a/b)h$ with $h$ primitive integral and $\gcd (a,b)=1$; the primitive product $fh$ shows that integrality of $(a/b)fh$ forces $b=1$. Thus the kernel is precisely $(f)$, and the first isomorphism theorem for rings yieldsNo monicity of $f$ is needed; this is an isomorphism of rings, not an assertion that either ring is a field.