I am trying to convince myself that for any ring $R$ (commutative, so I don't have to bother with left-or-right modules) and ideals $I$, $J$ we have $\operatorname{Tor}_1^R(R/I,R/J)=I\cap J/IJ$.

I have found several solutions on the internet using the projective resolution $0\to I\to R\to R/I\to 0$. I am having a hard time understanding why $I$ is projective?

After that I'm fine with the argument: Tensoring $0\to I\to R\to 0$ by $R/J$ and applying some canonical isomorphisms yields $0\to I/IJ\to R/J\to 0$ and the kernel of the middle map is what I want it to be.

Thank you very much in advance.


$\newcommand{\Tor}{\operatorname{Tor}}$The ideal $I$ is not projective in general, and therefore $0 \to I \to R \to R/I \to 0$ is not a projective resolution of $R/I$ in general. However, this sequence is a short exact sequence. Tensoring it with $R/J$ gives rise to the long exact sequence $$ \ldots \to \Tor_1(R,R/J) \to \Tor_1(R/I,R/J) \to I \otimes R/J \to R \otimes R/J \to R/I \otimes R/J \to 0. $$ Here $\Tor_1(R,R/J)=0$ because $R$ is projective. Now you should be able to continue your argument with this sequence.

  • 1
    $\begingroup$ Well, if 0->I->R->R/I->0 was a projective resolution, then I could compute Tor1 by tensoring 0->I->R->0 by R/J and taking 1st homolgy. This is the only way I had in mind to compute Tor, since I forgot about the long exact sequence. But anyway, thank you for your answer! $\endgroup$
    – user114885
    Sep 16 '15 at 11:38
  • $\begingroup$ Ah, now I see what you had in mind! I removed my comment in this direction from the post. $\endgroup$
    – moonlight
    Sep 16 '15 at 11:45

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