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Let $R$ commutative ring and $I$ an ideal of $R$.

How do I prove that $\operatorname{Ext}^1_R(R/I,R/I)$ isomorphic to $\operatorname{Hom}_R(I/I^2,R/I)$ ?

This question is an exercise of the course but has a chance of being false.

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Have you tried playing with some basic examples first, $R=\mathbb Z$ and $I=p\mathbb Z$ for some prime $p$? I have no idea how to attack such problem, but in my experience trying to see how the basic problems work can give you quite some insight for later on. – Asaf Karagila Jun 16 '12 at 21:52
up vote 4 down vote accepted

There's really only one thing you can try when confronted with such a problem: find a suitable short exact sequence. Here's an obvious one: $$0 \to I \to R \to R/I \to 0$$ Applying $\textrm{Ext}_R^\bullet$, we get $$0 \to \textrm{Hom}_R(R/I, R/I) \to \textrm{Hom}_R(R, R/I) \to \textrm{Hom}_R(I, R/I) \to \textrm{Ext}_R^1(R/I, R/I) \to 0$$ since $R$ is a projective $R$-module. Now, observe that $\textrm{Hom}_R(R, R/I) \cong R/I$ by the universal property of $R$ as a free $R$-module, while $\textrm{Hom}_R(I, R/I) \cong \textrm{Hom}_R(I/I^2, R/I)$ by considering the action of any element in $I$. On the other hand, the map $\textrm{Hom}_R(R, R/I) \to \textrm{Hom}_R(I, R/I)$ must be the zero map. Hence, $$\textrm{Hom}(I/I^2, R/I) \cong \textrm{Ext}_R^1(R/I, R/I)$$ as required.

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