Example 1

For $A = \mathbf{Z}$, the $\mathbf{Z}$-modules of finite length are precisely the finite abelian groups; then $K = \mathbf{Q}$, so $K/A = \mathbf{Q}/\mathbf{Z}$. Then to define $D(M)$, we sometimes take, instead of $\mathbf{Q}/\mathbf{Z}$, a $\mathbf{Z}$-module isomorphic to it, such as (V, p. 79, Prop. 2) the group $R$ of roots of unity (under multiplication) in an algebraically closed field of characteristic 0 ; we then put $D(M) = \operatorname{Hom}_{\mathbf{Z}}(M, R)$. We leave the reader to rewrite the preceeding results for this special case in the corresponding notation.