Let $I$ and $J$ be two ideals in $A$. Show that

$\operatorname{Tor}_{1} (A/I, A/J) =\frac {I \cap J} { IJ}$

and

$Tor_{2} (A/I, A/J) =\ker(I \otimes_ {A}J \to IJ )$.

The first Tor is not a problem, this is also in Rotman, An introduction to homological algebra, but the second ?

-

closed as off topic by Steven Landsburg, Martin Brandenburg, Graham Leuschke, Chris Gerig, S. Carnahan♦Apr 16 '13 at 1:03

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

Is this a homework exercise? –  S. Carnahan Apr 15 '13 at 12:28
Also posted here: math.stackexchange.com/questions/362206/an-exercise-about-tor –  user26857 Apr 15 '13 at 15:51

Consider the short exact sequence $0 \to I \to A \to A/I \to 0$. Tensoring with the $A$-module $A/J$ gives the long exact sequence $\cdots \to 0 \to \mathrm{Tor}_2^A(A/I,A/J) \to \mathrm{Tor}_1^A(I,A/J) \to 0 \to \mathrm{Tor}_1^A(A/I,A/J) \to I/IJ \to A/J \to (A/I)\otimes_A (A/J)\to 0$. Here I use the fact that $A$ is a free $A$ module, so all its Tor's are 0. The right part of the sequence gives the first equality you mention. The left part identifies $\mathrm{Tor}_2^A(A/I,A/J)$ with $\mathrm{Tor}_1^A(I,A/J)$.
Now tensor $0 \to J \to A \to A/J \to 0$ with the $A$-module $I$. This gives $\cdots \to 0 \to \mathrm{Tor}_1^A(A/J,I) \to J\otimes_A I \to I \to I/IJ \to 0$, so $0 \to \mathrm{Tor}_1^A(A/J,I) \to J\otimes_A I \to IJ \to 0$, and since Tor and $\otimes$ are symmetric, $\mathrm{Tor}_2^A(A/I,A/J)$ is the kernel of $I\otimes_A J \to IJ$.