$\text{Tor}$ and $IJ=I\cap J$: Eisenbud exercice A3.17

$$I$$ 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 $$\text{Tor}$$.

If $$R$$ is commutative, the statement is completely elementary --- not homological at all.
Suppose $$I+J=R$$, so that $$a+b=1$$ for some $$a\in I$$, $$b\in J$$; we want to show $$I\cap J\subseteq IJ$$. Let $$x\in I\cap J$$. Then $$ax\in IJ$$ and $$xb\in IJ$$. So $$x=x(a+b)=ax+xb\in IJ$$.
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.
• For the second: some ideas: if one hase a $R/J$-regular sequence $(x_1,\ldots,x_n)$ in $I$ then the Koszul complex $R/J\otimes K(x_1,\ldots,x_n)$ is a free resolution of $R/J$. The Koszul complex $K(x_1,\ldots,x_n)$ end with $R/(x_1,\ldots,x_n)=R/I$ (hypothesis: the regular sequence generate $I$). So we can calculate the $\text{Tor}$ with this sequence... How to conclude? – Macadam Jun 12 at 19:23