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Say $R$ is an integral domain with field of fractions $F$. I need to show that, for any $R$-module $B$, $Tor_1^R(F/R, B)\cong t(B)$, where $t(B)$ is the torsion submodule of $B$.

So say $$\cdots\to P_2\to P_1\to P_0\to B$$ is a projective resolution for $B$. Then, to compute the Tor, I look at the tensored version of this: $$\cdots\to F/R\otimes P_2\to F/R\otimes P_1\to F/R\otimes P_0\to F/R\otimes B$$.

But I'm not sure how to proceed from here, since I'd need to compute the kernel mod the image at $P_1$, which is too far away from $B$ for me to be able to say anything...I can see that the torsion elements of $B$ become $0$ in $F/R\otimes B$, but I'm not sure how to work with the projective resolution.

Nishant
  • 9,155

1 Answers1

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Use the short exact sequence $0\to R\to F\to F/R\to 0$ and consider the long exact sequence you get when applying $-\otimes_R B$. Use the fact that $F$ is flat to get an exact sequence $0\to \mathrm{Tor}_1(F/R,B)\to B\to F\otimes_R B$.