Let $R$ be a commutative PID. Let $0\neq a \in R$.
Then for any $R$-module $N$, we should have $Ext_R^1(R/(a), N)\cong N/aN$
I guess this should be a standard exercise, but I have no idea on how to pick an injective resolution of $N$ to start...
EDIT:
Ok I think I got it. Please correct me if I am wrong. $0\rightarrow R_1\rightarrow_\phi R_0\rightarrow R/(a)\rightarrow 0$. $R_1, R_0$ both equal to $R$. $\phi: R_1\rightarrow R_0$ by $\phi(x)=ax$.
Applying $Hom(-, N)$ we have $0 \rightarrow Hom(R/(a), N) \rightarrow Hom(R_0,N)\rightarrow Hom(R_1,N)\rightarrow 0$.
Since $Ext^1(R/(a), N)=Hom(R_1,N)/Im(Hom(\phi, N))$, $Hom(R,N)\cong N$ and $Hom(\phi, N) = f \mapsto af$, we have $Ext^1(R/(a), N)=N/aN$.
Is the proof correct? I am not sure if I can have the rightmost arrow of the exact sequence, since $Hom(-, N)$ is left exact only.