Let $R$ be a domain with quotient field $K$, let $I\neq 0$ be an $R$-ideal, and write $I^{-i}=R\underset{K}{:} I^i$. Then $S=\bigcup\limits_{i\geq 0} I^{-i}$ is a subring of $K$ containing $R$. Show $$H^1_I(R) \cong S/R$$ as $R$-modules.
Proof: Let $$0\to R\to K \to K/R\to 0$$ be an exact sequence. Apply $-\underset{K}{:} I^i$ to the sequence $$0\to R\underset{K}{:} I^i\to K\underset{K}{:} I^i \to K/R\underset{K}{:} I^i\to 0$$ which remains exact. Then $$0\to I^{-i}\to I^i \to K/R\underset{K}{:} I^i\to 0$$ Apply $\bigcup\limits_{i\geq 0} -$ $$0\to \bigcup\limits_{i\geq 0} I^{-i}\to \bigcup\limits_{i\geq 0} I^i \to \bigcup\limits_{i\geq 0} K/R\underset{K}{:} I^i\to 0$$
So far am I correct? If so, what I need to get next is a long exact sequence. I am thinking of using the Tor Functor but I am not sure what the second entry would be?