# How to calculate Tor(R/I, R/J) ??

How can I prove that $\text{Tor}_1(R/I,R/J) = (I \cap J)/IJ$, where $R$ is a ring and $I, J$ ideals.

Moreover, if we suppose $R=I+J$, how do I prove that $\text{Tor}_1(R/I,R/J)=0$?

Ps: No, this is not a homework question.

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Your question is ambiguous: which Tor group do you mean? In any case, it really does sound like homework to me. Here's a hint: think about what happens if you apply Tor to an exact sequence like $0\to I\to R\to R/I\to 0$. – Charles Rezk Dec 8 '10 at 23:58
I mean Tor_1 over the ring R. That is exactly what I did, but I get Tor(R/I x R/J) = Ker(I x R/I ----> A x A/J), where x is the tensor product, but then I don't know how to prove the equality. – Kripton Dec 9 '10 at 0:06
Karl, thank you for the answer. It with the one below allowed me to have the solution. I really apreciate that. – Kripton Dec 9 '10 at 2:33
I answered this a few days ago here: answers.yahoo.com/question/… – Steve D Dec 9 '10 at 3:01
yes, I've already proved the first part, but now I can't see the second. Anyway thank you for your comment. – Kripton Dec 9 '10 at 3:10

Hints: 1) First prove that $I\otimes(R/J)=I/IJ$ . 2) If $I+J=R$, write $1=i+j$ and use the fact that $x=1x$.
$I+J = R \Rightarrow I \cap J = IJ$ is an easy exercise in the beginning of commutative algebra and does not involve any Tor-functors. – Martin Brandenburg Dec 9 '10 at 9:05
Although already pointed out by others that this is an easy exercise, the most obvious answer to the second question has been overlooked: since $I$ and $J$ are both in the annihilator of $\text{Tor}_1(R/I,R/J)$ (by functoriality) and $I+J=R$, the Tor module must be zero.