Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $J=I$, $$\mathrm{Ext}^1_R(R/I,R/I)=\mathrm{Hom}(\mathrm{Tor}^1_R(R/I,R/I), R/I)$$

share|cite|improve this question
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 $$\Ext^1_R(R/I, M) \cong \Hom_R(I,M) / M_I.$$ Is that a nice description? – neilme Apr 26 '13 at 7:12
@neilme: Why don't you post this as an answer? – Martin Brandenburg Apr 26 '13 at 8:27
@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? – messi Apr 26 '13 at 9:25
@neilme, In my previous comment i meant $Hom(I,M)$. – messi Apr 26 '13 at 10:26
In order to get a better answer, you have to specify what you mean by "nice". – Martin Brandenburg Apr 26 '13 at 22:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.