I basically have to show that $d(R/(e))\cong R/(\frac{e}{g})$ where $R$ is a PID, and $\gcd(e,d)=(g)$.
I defined $\pi:d(R/(e))\rightarrow R/(\frac{e}{g})$ by $\pi:dr+dce\mapsto \frac{dr+dce}{g}$. Note that even though we are not in a field, we can divide this elements by $g$ since it divides $d$ and $e$. The function is clearly a homomorphism so I need to show that it is injective and surjective.
For injectivity I basically say that if $\frac{d}{g}r+dc\frac{e}{g}\in \left(\frac{e}{g}\right)$ happens if $\frac{d}{g}r\in \left(\frac{e}{g}\right)$ This is equivalent to saying that for some $k\in R$: $\frac{d}{g}r=k\frac{e}{g}$ Thus, $dr=ke\in (e)$, ergo, $dr+dce$ is zero in $d(R/(e))$ meaning that the kernel is trivial and thus the function is injective.
For surjectivity I let $r+c\frac{e}{g}\in R/(\frac{e}{g})$, then I want to find $r_1,c_1\in R$ such that $dr_1+dc_1e\mapsto r+c\frac{e}{g}$, but I get something like $r_1=r\frac{g}{d}$$c_1=\frac{c}{d}$The problem that I have is that dividing by $d$ in this case might not be in my $R$, so I am rather stuck here.
Any help?
Thanks