Past exam of the algebra 2 course of the University of Oxford / 2014 / 1 / c / Solution

Take a reduced fraction $a/b\in R$, with $b\ne 0$. Since $\gcd (a,b)=1$, Bézout's identity supplies integers $u,v$ with $ua+vb=1$. As the subring $R$ contains all integers,Thus a denominator appearing in a reduced fraction in $R$ is automatically a unit of $R$.

Now let $I\cap \mathbb{Z}=d\mathbb{Z}$. Because $d\in I$, the principal ideal $dR$ is contained in $I$. For the reverse inclusion, take $x\in I$ and write $x=a/b$ in lowest terms. Multiplying by $b\in R$ gives $a=bx\in I\cap \mathbb{Z}=d\mathbb{Z}$, so $a=dc$ for some integer $c$. The preceding result gives $1/b\in R$, whenceThereforeThis includes $d=0$: the same argument forces every $x\in I$ to be zero. This is the ideals of subrings of the rational numbers property; every such ring is a principal ideal domain.