$\DeclareMathOperator{\coker}{coker}$ In Weibel's intro to Homological Algebra one has the following exercise :
Show that $\operatorname{Tor}_1^R(R/I,R/J)\simeq \frac{I\cap J}{IJ}$ for every right ideal $I$ and every left ideal $J$ of $R$. In particular, $\operatorname{Tor}_1^R(R/I,R/I)\simeq \frac{I}{I^2}$ for every two-sided ideal $I$. Hint: Apply the snake lemma to $\require{AMScd}$
\begin{CD}
0@>>>IJ@>>>I@>>>I\otimes R/J @>>>0 \\
@. @VaVV @VbVV @VcVV \\
0 @>>>J @>>> R @>>> R\otimes R/J @>>> 0
\end{CD}
The question of computing $\operatorname{Tor}_1^R(R/I,R/J)$ has already been treated on this website, for example here. In this answer, the question is treated with what Weibel calls 'dimension shifting', that is, given a right exact functor $F$ and an exact sequence $0\to M \to P \to A \to 0$ with $P$ projective, one has $L_iF(A)\simeq L_iF(M)$ for $i\geq 2$ and $L_1F(A)$ is the kernel of $F(M)\to F(P)$. The reason this is true is what's pointed out in the post I linked, i.e. given the short exact sequence $0\to M \to P \to A\to 0$, we get a long exact sequence \begin{align*}
\cdots&&\to &&L_{2}F(M)\to &&L_{2}F(P)\to &&L_{2}F(A)&& \\
&&\to &&L_{1}F(M)\to &&L_{1}F(P)\to &&L_{1}F(A)&& \\
&&\to &&F(M)\to &&F(P)\to &&F(A)&&\to0.
\end{align*}
where the terms $L_iF(P)$ are all $0$ because $P$ is projective.
With all this in mind, applying the following paragraph to $0\to I\to R\to R/I\to 0$ and the functor $-\otimes R/J$ we have $\operatorname{Tor}_1^R(R/I,R/J)\simeq \ker(I\otimes R/J\to R\otimes R/J)$. I think this is where the hint comes in, because then $\operatorname{Tor}_1^R(R/I,R/J)$ appears as the right-most kernel in the diagram. Indeed using the snake lemma we have $$ 0\to \ker a \to \ker b \to \operatorname{Tor}_1^R(R/I,R/J) \to \coker a \to \coker b \to \coker c\to 0$$
From now on, everything I write is a guess : I think $\ker b = 0$, $\coker b = R/I $ the rest is unclear. To be honest, I'm not entirely sure what the map $a$ and the map $IJ\to I$ are. I believe $a$ might be a quotient map $IJ \to IJ/(IJ\cap I)$, this would mean $\ker a = IJ\cap I $. Since $\ker b$ is (I believe) $0$, then $\ker a$ is also $0$ by exactness. This would give the sequence $$ 0 \to 0 \to 0 \to \operatorname{Tor}_1^R(R/I,R/J) \to \coker a \to R/I \to \coker c\to 0 $$
I'm not sure what the remaining cokernels are, I feel like $\coker c$ might be something like $R/I$ but this is very much a gut feeling and not something I have a proof of.
How does one conclude, once we know $\operatorname{Tor}_1^R(R/I,R/J)\simeq \ker(I\otimes R/J\to R\otimes R/J)$ ?
Best Answer
The snake lemma does indeed give us an exact sequence $0\to \operatorname{Tor}_1^R(R/I,R/J)\to\operatorname{coker} a \to R/I$, hence $\operatorname{Tor}_1^R(R/I,R/J)$ is also the kernel of $\operatorname{coker} a \to R/I$, but notice that $\operatorname{coker} a$ is actually just $J/IJ$. If $\pi(x) \in J/IJ$ ($\pi$ denotes the map $J\to \operatorname{coker} a=J/IJ$) is mapped to $0$, then this means $x\in I$, hence it turns out that indeed the kernel of this map is $\left\lbrace \pi(x)\in J/IJ, x\in I\right\rbrace=(I\cap J)/IJ$