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

A Euclidean domain is a principal ideal domain. The structure theorem for finitely generated modules over a principal ideal domain therefore gives, for every finitely generated module $M$ over $R$,where the $d_i$ are nonzero nonunits and $d_1\mid d_2\mid \cdots \mid d_s$. The factors $d_i$ are unique up to multiplication by units; $r$ is uniquely determined.

The rank of a module is this integer $r$. Equivalently, with $K$ the field of fractions of $R$,The tensor product of modules kills each displayed torsion summand after extending scalars to $K$, while sending $R^r$ to $K^r$, explaining the equality.

If $M$ is a torsion-free module, no summand $R/(d_i)$ can occur: its nonzero element $1+(d_i)$ is killed by the nonzero scalar $d_i$. Hence $s=0$ andThis also covers $M=0$, with $r=0$.

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.

Using the common denominator $70$, the three given generators becomeThus their submodule is contained in $(1/70)\mathbb{Z}$. For the reverse inclusion, the following explicit integer linear combination supplies the missing generator:soTherefore $1/70$ belongs to the generated submodule, and all its integer multiples belong as well. We have established equality in both directions:Finally, $k/70=0$ for an integer $k$ forces $k=0$, so the single generator is linearly independent. Hence the required basis of a module is

Suppose a finite set of rational numbers $q_1,\dots ,q_m$ generated $\mathbb{Q}$ as a module over $\mathbb{Z}$. A positive common denominator $D$ makes every $D{q}_i$ an integer. For every integer linear combination, therefore,The entire generated submodule is contained in $(1/D)\mathbb{Z}$. But $1/(2D)$ is a rational number outside that set: $1/(2D)=k/D$ would require the noninteger $k=1/2$. This contradicts generation of all of $\mathbb{Q}$. ConsequentlyEach finite set of generators imposes a common denominator; the full rational number system has no such bound.