Let $R$ be a Dedekind domain. Given two non-zero ideals $\mathfrak a, \mathfrak b$ in $R$ there exists $c\in K^\times$ (the fraction field of $R$) such that $c\mathfrak a$ and $\mathfrak b$ are relatively prime.
We want to prove that there is $c\in K^\times$ such that $c\mathfrak a+\mathfrak b=R$. Let $0\ne a\in\mathfrak a$ and $\mathfrak c$ an ideal of $R$ such that $\mathfrak c\mathfrak a=aR$. (In a Dedekind domain every non-zero ideal is invertible, so one can chose $\mathfrak c=a\mathfrak a^{-1}$.) Moreover, $R/\mathfrak c\mathfrak b$ is a principal ideal ring (maybe here is where Lang's hint to use CRT applies), so $\mathfrak c/\mathfrak c\mathfrak b=(\bar b)$. Thus $\mathfrak c=\mathfrak c\mathfrak b+bR$, and multiplying this by $\mathfrak a$ and then dividing by $a$ we get $c\mathfrak a+\mathfrak b=R$ where $c=b/a$.
I don't understand few steps from this proof such as why crt apply here and part after that so $\mathfrak c/\mathfrak c\mathfrak b=(\bar b)$. Thus $\mathfrak c=\mathfrak c\mathfrak b+bR$, Please somebody help me out.