Hello,
I asked this question already on StackExchange with no answer, maybe it will be better suited here. It is well known that $\operatorname{Ext}^1\_\mathbb{Z}(\mathbb{Z}/p\mathbb{Z},A) \simeq A/pA$. Is there a generalization of this formula outside PIDs? I mean $\operatorname{Ext}^1\_R(R/I,M) \simeq M/IM$ for maximal or prime ideals? Does it hold for Dedekind domains? The classic proof I know for abelian groups seem to depend on principality of the ideal and I was unable to either think of different way make it work or to find an counterexample. Thanks for any input.



Ext of cyclic module
In general $Ext^1\_R(R/I,M) \not\cong M/IM$. 


As an example take a finite abelian group $G$ and set $R := \mathbb{Z}G$ and let $I:= I\_G = \ker(\mathbb{Z}G \to \mathbb{Z},\; g \mapsto 1)$ be the augmentation ideal. $I\_G$ is a prime ideal since $\mathbb{Z}G/I\_G \cong \mathbb{Z}$. Then, with trivial coefficients 
$$Ext\_R^1(R/I,\mathbb{Z})=Ext\_{ZG}^1(\mathbb{Z},\mathbb{Z})=H^1(G;\mathbb{Z})=Hom(G,\mathbb{Z})=0$$
while $\mathbb{Z}/I\_G\mathbb{Z}=\mathbb{Z}$. The latter holds because $I\_G$ is generated by $g-1$ $(g \in G)$ and $(g-1) \cdot 1 = 0$. 




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In general, the following holds:
$$Ext^1\_R(R/I,M)=\dfrac{Hom\_R(I,M)}{i(M)}$$
where $i: M \to Hom\_R(I,M),\; m \mapsto (x \mapsto xm)$. 


If $I$ is finitely generated by $a\_1,...,a\_n$ we can choose a presentation $R^n \to I \to 0$ that induces an embedding $0 \to Hom\_R(I,M) \to Hom\_R(R^n,M) \cong M^n$. Thus 
$$Ext\_R^1(R/I,M) \le \dfrac{M^n}{\lbrace (a\_1m,...,a\_nm) \mid m \in M \rbrace}.$$


