Let $R$ be a commutative ring with unit and $I$ and $J$ are nonzero ideals of $R$. Do we have a nice description for $\mathrm{Ext}^1_R(R/I,R/J)$?

What do I mean by a nice description? For example $$\mathrm{Tor}_1^R(R/I,R/J)=(I\cap J)/IJ,$$ so I would say that this is a nice description. So can we find something like this for $\mathrm{Ext}$, say for $J=0$? We do have a description for $\mathrm{Ext}$ when $J=I$, $$\mathrm{Ext}^1_R(R/I,R/I)=\mathrm{Hom}(\mathrm{Tor}^1_R(R/I,R/I), R/I)$$

  • 5
    $\begingroup$ Let $M$ be any $R$-module and $I$ an ideal. Then for each element $m\in M$, there is an $R$-linear map $g_m: I \rightarrow M$ given by $g_m(i) = im$. Let $M_I := \{g_m \mid m \in M\}$. Then $M_I$ is an $R$-submodule of $M$ and $$\operatorname{Ext}^1_R(R/I, M) \cong \operatorname{Hom}_R(I,M) / M_I.$$ Is that a nice description? $\endgroup$
    – neilme
    Apr 26, 2013 at 7:12
  • $\begingroup$ @neilme: Why don't you post this as an answer? $\endgroup$ Apr 26, 2013 at 8:27
  • $\begingroup$ @neilme, Since "nice description" is vague, i cannot say it is not "nice". I knew this description. But the problem in this description is that $Hom(I,R)$ is not that "easy" to describe. Do we have a good description of $Hom(I,R)$ for say a nice ring like Gorenstein ring? $\endgroup$
    – messi
    Apr 26, 2013 at 9:25
  • 4
    $\begingroup$ In order to get a better answer, you have to specify what you mean by "nice". $\endgroup$ Apr 26, 2013 at 22:54
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    $\begingroup$ @user251222 Reference? I dunno; it just comes from Homming the sequence $0 \rightarrow I \rightarrow R \rightarrow R/I \rightarrow 0$ into $M$. The map from $M$ to $\text{Hom}_R(I,M)$ sends $m$ to $g_m$, so your sequence of four splits into two short exact sequences, the second one of which is $0 \rightarrow M_I \rightarrow \text{Hom}_R(I,M) \rightarrow \text{Ext}^1_R(R/I,M) \rightarrow 0$. As for your second question, I don't know offhand. $\endgroup$
    – neilme
    Aug 9, 2016 at 23:33


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