TITLE: $\text{Tor}$ and $IJ=I\cap J$: Eisenbud exercice A3.17
QUESTION [6 upvotes]: $I$ and $J$ are ideal of a ring $R$. From the short exact sequence
$$ 0\to I\to R\to R/I\to 0 $$
we have 
$$ 0\to \text{Tor}_1^R(R/I,R/J)\to I/(IJ)\to R/J\to R/(I+J) \to 0 $$
So
$$ \text{Tor}_1^R(R/I,R/J)=(I\cap J)/IJ $$
In exercice A3.17 Eisenbud ask to deduce from it that if $I+J=R$ then $I\cap J=IJ$. 
I see that we get
$$ 0\to \text{Tor}_1^R(R/I,R/J)\to I/(IJ)\to R/J\to 0 $$
But I don't see how it can help me.
I see that (it the same thing) that $R/I\otimes R/J=R/(I+J)=0$ so $\text{Tor}_0(R/I,R/J)=0$. Does it imply that $\text{Tor}_1(R/I,R/J)=0$?
There is a second question: prove (with the same trick $\text{Tor}_1^R(R/I,R/J)=(I\cap J)/IJ$) that if $I$ is generated by a sequence of elements that form a regular sequence mod $J$ (that is, I guess, there is in $I$ a regular $R/J$-sequence) then $IJ=I\cap J$. My problem here is that I don't see a single link between regular sequence and $\text{Tor}$.

REPLY [2 votes]: I think I have it (for the first)!
For the first: $\text{Tor}_i(M,N)$ is annihilated by $\text{Ann}(M)$ and $\text{Ann}(N)$ because element of $\text{Tor}$ are nothing else classe of elements of $P_i\otimes N$ or $P_i\otimes M$. Here $\text{Tor}_1(R/I,R/J)$ is annihilated by $I$ and $J$ so if $R=I+J$ it is annihilated by 1 and the conclusion come.