Let \(m,k\ge1\) and set \[ R=\ZZ/(mk)\ZZ, \qquad M=\ZZ/m\ZZ. \]
Show that \[ [r]_{mk}\cdot [x]_m=[rx]_m \] defines an \(R\)-module structure on \(M\).
Compute \(\operatorname{Hom}_R(M,R)\) and show that it is isomorphic to \(M\) as an \(R\)-module.
Solution
1. The displayed scalar action is well-defined.
If \([r]_{mk}=[r']_{mk}\), then \(mk\mid r-r'\), hence in particular \(m\mid r-r'\). Therefore \[ [rx]_m=[r'x]_m. \] If \([x]_m=[x']_m\), then \(m\mid x-x'\), so \[ [rx]_m=[rx']_m. \] Thus the value depends only on the residue classes.
2. This action makes \(M\) an \(R\)-module.
For \(r,s\in R\) and \(x,y\in M\), all module axioms follow from multiplication and addition in the integers modulo \(m\): \[ r(x+y)=rx+ry, \qquad (r+s)x=rx+sx, \qquad (rs)x=r(sx), \qquad 1x=x. \]
3. An \(R\)-linear map \(f:M\to R\) is determined by \(f([1]_m)\), and this element must lie in\[ \operatorname{Ann}_R(m)=\{a\in R:ma=0\}=kR. \]
Since \(M\) is generated by \([1]_m\) as an \(R\)-module, \(f\) is determined by \(a=f([1]_m)\). The relation \[ m[1]_m=0 \] forces \[ ma=f(m[1]_m)=0. \] Conversely, if \(a\in R\) satisfies \(ma=0\), then \[ f_a([x]_m)=xa \] is well-defined and \(R\)-linear. In \(R=\ZZ/(mk)\), the congruence \(ma\equiv0\pmod{mk}\) is equivalent to \(k\mid a\), so the annihilator is exactly \(kR\).
4. Hence\[ \operatorname{Hom}_R(M,R)\cong kR\cong \ZZ/m\ZZ=M. \]
3. The map\[ \ZZ/m\ZZ\longrightarrow kR, \qquad [\ell]_m\longmapsto[k\ell]_{mk} \]is well-defined, bijective, and \(R\)-linear. Composing gives the claimed isomorphism.