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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3$\begingroup$ 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 $0\to I\to R\to R/I\to 0$. $\endgroup$– Charles RezkCommented Dec 8, 2010 at 23:58
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$\begingroup$ 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. $\endgroup$– KriptonCommented Dec 9, 2010 at 0:06
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$\begingroup$ Karl, thank you for the answer. It with the one below allowed me to have the solution. I really apreciate that. $\endgroup$– KriptonCommented Dec 9, 2010 at 2:33
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$\begingroup$ I answered this a few days ago here: answers.yahoo.com/question/… $\endgroup$– Steve DCommented Dec 9, 2010 at 3:01
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$\begingroup$ yes, I've already proved the first part, but now I can't see the second. Anyway thank you for your comment. $\endgroup$– KriptonCommented Dec 9, 2010 at 3:10
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2 Answers
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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 $x=1x$.
answered Dec 9, 2010 at 0:32
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$\begingroup$ Thank you very much for your help. It was very useful. Concerning to 2), I think you don't understand what I asked. I need to prove that (I∩J) = IJ, using that Tor1(R/I,R/J)=(I∩J)/IJ and the fact that R=I+J. So I think if we prove that Tor vanishes in this case, we have the problem solved. Once again thank you. $\endgroup$– KriptonCommented Dec 9, 2010 at 2:25
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$\begingroup$ yes, that I know since the basic. lolol What I don't know is to prove that indirectly, I mean, using the fact that Tor(R/I,R/J)=(I∩J)/IJ. I must prove that tor vanishes to conclude that I∩J = IJ, see? $\endgroup$– KriptonCommented Dec 9, 2010 at 3:16
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8$\begingroup$ "I think you don't understand what I asked" is not the most polite way to put it. $\endgroup$– AngeloCommented Dec 9, 2010 at 6:57
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5$\begingroup$ $I+J = R \Rightarrow I \cap J = IJ$ is an easy exercise in the beginning of commutative algebra and does not involve any Tor-functors. $\endgroup$ Commented Dec 9, 2010 at 9:05
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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.
