Past exam of the rings and modules course of the University of Oxford / 2021 / 3 / b

Every multiple of $dh$ is a multiple of both $d$ and $h$, so $dh\mathbb{Z}\subseteq d\mathbb{Z}\cap h\mathbb{Z}$.

For the reverse inclusion, let $z=du=hv$ lie in this intersection of ideals. Since $d,h$ are coprime integers, Bézout's identity supplies $r,s\in \mathbb{Z}$ with $rd+sh=1$. Multiplying by $u$ givesThus $h\mid u$, and $dh\mid du=z$. ConsequentlyHere Bézout's identity follows from the allowed principal ideal domain property of $\mathbb{Z}$: the ideal $d\mathbb{Z}+h\mathbb{Z}$ is generated by their greatest common divisor, which is $1$.

Define the group homomorphismBy part (b)(i), its kernel of a group homomorphism isIt is surjective. Indeed, choose $r,s\in \mathbb{Z}$ with $rd+sh=1$ by Bézout's identity. For any prescribed classes $u+d\mathbb{Z}$ and $v+h\mathbb{Z}$, the integersatisfies $t\equiv u(\mathrm{mod}d)$ and $t\equiv v(\mathrm{mod}h)$. The first isomorphism theorem for groups therefore givesThis proves the required isomorphism and is the Chinese remainder theorem for coprime integers. It also shows that $(1+d\mathbb{Z},1+h\mathbb{Z})$ generates the direct sum of modules when these additive groups are viewed as $\mathbb{Z}$-modules.