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

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


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

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

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

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Is this a homework exercise? – S. Carnahan Apr 15 '13 at 12:28
up vote 3 down vote accepted

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 $I\otimes_A J \to IJ$.

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Please refrain from solving such basic exercises on mathoverflow. Otherwise the requests will remain. – Martin Brandenburg Apr 15 '13 at 15:42

I don't know about the second edition, but if you have, or can get, the first edition of Rotman, then the second one is there too, as Corollary 11.27(ii).

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In my edition there isn't 11.27 – user33122 Apr 15 '13 at 12:49

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