2
$\begingroup$

Let $R$ be any commutative ring, and $I,J\subseteq R$ two ideals. Then Exercise A3.17 of Eisenbud's Commutative Algebra claims that $$\mathrm{Tor}_1^R(R/I,R/J)=(I\cap J)/(IJ).$$ However, I've hit a snag when I tried to calculate a free (and therefore projective) resolution for $R/I$. This is what I have so far: $$\cdots\to(\ldots)\to\bigoplus_i R\to R\to R/I\to I\to 0$$ where the map $\bigoplus_i R\to R$ has image $I$. However, I'm pretty clueless about what's supposed to go in place of $(\ldots)$, and my approach doesn't seem to be giving enough information to actually calculate homology once we tensor with $R/J$.

Is there some other approach to solving this problem?

  • 0
    Have you tried to use the long exact homology sequence for Tor applied to the ses $0\to I\to R\to R/I\to 0$? – 2017-01-01
  • 0
    But how do we know that $I$ is a projective $R$-module? – 2017-01-01
  • 0
    $I$ isn't, but $R$ it is. – 2017-01-01

1 Answers 1

1

Consider the exact sequence $$0\rightarrow I \rightarrow R\rightarrow R/I\rightarrow 0$$ and tensor it by $R/J$ over $R$. We obtain the long exact sequence \begin{align}\DeclareMathOperator{\Tor}{Tor} \dots\rightarrow0=\Tor^R_1(R,R/J)\rightarrow\Tor^R_1(R/I,R/J)\rightarrow I\otimes_R R/J \rightarrow R/J\rightarrow R/I\otimes_R R/J \rightarrow 0 \end{align} This shows $\Tor^R_1(R/I,R/J)$ is the kernel of the morphism $ I\otimes_R R/J\simeq I/IJ \rightarrow R/J$. This morphism maps $i+IJ$ $\;(i\in I)$ to $\;i+J$. It maps to $J$ if and only if $i\in J$, hence in $I\cap J$. Thus, as claimed, the kernel is $$I\cap J/IJ.$$