Let $R$ be a Noetherian domain and $I\neq 0$ an ideal. Show that $H_I^1(R)=0$ if and only if $I$ is not contained in an associated prime of any principal ideal.
Proof: $\Rightarrow$ Suppose that $H^1_I(R)=0$. Previously, I have shown that $H_I^1(R)\simeq S/R$, where $S=\bigcup\limits_{i\geq 0} I^{-i}$ is a subring of $K$ containing $R$ and $I^{-i}=R\underset{K}{:} I^i$. This implies \begin{equation*} \begin{aligned} S = R & \Rightarrow R = \bigcup\limits_{i\geq 0} I^{-i} \Rightarrow R = \bigcup\limits_{i\geq 0} R \underset{K}{:} I^i \Rightarrow R = {\{x\in K | Ix\subset R\}} \end{aligned} \end{equation*}
At this point I suppose there is something that will imply that $I$ is not contained in an associated prime of any principal ideal. But what? I think what it giving me a hard time is that I do not understand why "$I$ is not contained in an associated prime of any principal ideal" is related to the cohomology?