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$I$ 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 $\text{Tor}$.

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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 $x=x(a+b)=ax+xb\in IJ$.

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  • $\begingroup$ Thanks. But Eisenbud ask to prove that homologicaly. $\endgroup$ – Macadam Jun 12 at 6:51
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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.

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  • $\begingroup$ 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? $\endgroup$ – Macadam Jun 12 at 19:23

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