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> I want to prove that if $R$ is a Noetherian integral domain and $I$ is a nonzero ideal of $R$, then $I^{-1}=R$ if and only if $\operatorname{grade}(I)≥2$. 
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For the "only if" part, I say $II^{-1}=R$ yields $IR=R$, which gives $I=R$ since $I$ is an ideal of $R$. Now, $R$ is a domain, so each nonzero $r∈R$ is a nonzero divisor, whence $r$ is an $R$-sequence. But it can not be a maximal $R$-sequence in $I(=R)$ because being maximal means that $(I=)R⊆⋃P$, where the $P$ runs over $\operatorname{Ass}(R)$ which is not the case since by prime avoidance we would have $R=P$ for some $P$. Therefore, $\operatorname{grade}(I)≥2$.
I could not prove the converse and need some suggestion. Thanks for any help.