Prove that if $B \subseteq A$ be an ideal of $D$ then exists unique an ideal $C$ of $D$ such that $B = CA$. 
Problem: Denote $D$ be an algebraic ring of algebraic number field. Let $A \ne 0$ be any ideal of $D$. Prove that if $B \subseteq A$ be an ideal of $D$ then exists unique an ideal $C$ of $D$ such that $B = CA$.

My attempt: Since $A$ be an ideal of $D$ and $A \ne 0$ then exists $k>0$ such that $A^k = (\alpha)$, $\alpha \in D, \alpha \ne 0$. Since $B \subseteq A \Rightarrow B A^{k-1} \subseteq A^k = (\alpha)$. We have $B A^{k-1} (\alpha^{-1}) \subseteq (\alpha)(\alpha^{-1}) = D$. Let $C = B A^{k-1} (\alpha^{-1})$. Then $C$ is an ideal of $D$ and $CA = B A^{k} (\alpha^{-1}) = AD = A$. 
How to prove the uniqueness of $C$? Thank all!

Suppose $AC'=AC$.
If $A$ is not a principal ideal, then we can cancel $A$ in the ideal class group to get $C'-C=(\beta)$ is a principal ideal. But then we get $\beta A=0$. If $\beta \neq 0$ then $A=\beta^{-1}\beta A=0 \Rightarrow \Leftarrow$.
If $A$ is a principal ideal, say $A=(\gamma)$ with $\gamma \neq 0$, then $\gamma C' = \gamma C$. Hence $C'=C$.
