Past exam of the algebra 2 course of the University of Oxford / 2015 / 2 / b

Let $x,y\in \mathbb{Q}$. Choose a positive integer $D$ clearing their denominators and write $x=m/D$, $y=n/D$ with $m,n\in \mathbb{Z}$. If $x$ and $y$ are not both zero, thenis a relation with integer coefficients not both zero. If both vanish, use $1\cdot x+0\cdot y=0$. Thus every pair is linearly dependent over $\mathbb{Z}$.

A submodule $M\subseteq \mathbb{Q}$ is a torsion-free module: if $kz=0$ for a nonzero integer $k$ and a rational number $z$, then $z=0$. If $M$ is also a finitely generated module, part (a), applied to the Euclidean domain $\mathbb{Z}$, makes it a free module. Its rank cannot exceed one, since two different members of a basis of a module would be linearly independent, contradicting the relation just proved. Since $M\ne 0$, its rank cannot be zero. ThereforeThe denominator-and-greatest common divisor construction in part (c) gives an explicit version of this result for a finitely generated subgroup of the rational additive group.