Grade of an ideal in a Noetherian ring 
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 $\operatorname{grade}(I)≥2$. 

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. 
 A: I take this opportunity to show that your problem is a particular case of this one.

Let $R$ be an integral domain, $K$ its field of fractions, and $I$ a non-zero ideal of $R$. Then $I^{-1}=R$ if and only if the canonical homomorphism $R\to\operatorname{Hom}_R(I,R)$ given by $a\mapsto(i\mapsto ia)$ is an isomorphism.

"$\Rightarrow$" Since $R$ is an integral domain the canonical homomorphism $R\to\operatorname{Hom}_R(I,R)$ is injective. Let's prove that it is surjective. Chose $\psi\in\operatorname{Hom}_R(I,R)$. Since $\psi(ii')=i\psi(i')=i'\psi(i)$ for any $i,i'\in I$ we can deduce that the element of $\psi(i)/i\in K$ does not depend on $i$ when $i\in I$, $i\ne 0$. Set $a=\psi(i)/i$. We observe that $a\in I^{-1}$, and therefore $a\in R$. Consequently, $\psi$ is in the image of the canonical homomorphism $R\to\operatorname{Hom}_R(I,R)$.
"$\Leftarrow$" Let $x\in I^{-1}$. Then $xI\subseteq R$. Define $\psi:I\to R$ by $\psi(i)=ix$. Obviously $\psi\in\operatorname{Hom}_R(I,R)$, and therefore there is $a\in R$ such that $\psi(i)=ia$ for all $i\in I$. Thus we get $x=a\in R$.
A: 
If $R$ is a Noetherian integral domain and $I$ a nonzero ideal of $R$, then $I^{-1}=R$ if and only if $\operatorname{grade}(I)\ge 2$. 

"$\Rightarrow$" Let $a\in I$, $a\ne 0$. Suppose that all elements of $I$ are zero-divisors in $R/aR$. Then there is $b\notin aR$ such that $bI\subseteq aR\Leftrightarrow (b/a)I\subseteq R$. Since $b/a\in I^{-1}\setminus R$ we arrived to a contradiction. 
"$\Leftarrow$" Let $a,b$ be an $R$-sequence in $I$, and $x=u/v\in I^{-1}$, $u,v\in R$, $v\ne 0$. Since $xa,xb\in R$ we can write $ua=va'$ and $ub=vb'$. From $ua=va'$ we get $uab=va'b$, so $vb'a=va'b$. It follows $b'a=a'b\Rightarrow a'b\in aR\Rightarrow a'\in aR$ and therefore there is $a''\in R$ such that $a'=aa''$ $\Rightarrow$ $ua=vaa''$ $\Rightarrow$ $u=va''\Rightarrow x\in R$.
