Let $I$ be an ideal in a Noetherian quasi-unmixed local ring $(R,\mathfrak m)$ of dimension $d.$ 


Let $q(I)=\overline{I}\cap I^{sat}$ where $I^{sat}=\cup\_{n\geq 0}I:\mathfrak m^n.$ Then $q(I)$ is called *relative integral closure* of $I.$


**Question** Is there any example of a principal ideal $I$ (i.e. $I=(a)$) such that $I\neq q(I)$ where $d\geq 2?$ 


I know that in normal domains, $I=q(I)$ for all principal ideals $I.$



Example of a principal ideal which is properly contained in its relative integral closure
There are examples in arbitrary dimension. Begin with the polynomial ring $S=k[N]=k[s\_1,\dots,s\_d]$, i.e., the semigroup $k$-algebra on the semigroup $N=(\mathbb{Z}\_{\geq 0})^d$ of exponent vectors for monomials in $S$. Now for any subsemigroup $M\subset N$, let $R=k[M]\subset k[N]$ be the corresponding $k$-subalgebra of $S$. These give plenty of examples. 


For instance, let $R$ be the $k$-subalgebra generated by monomials of total degree $\geq d+1$. Let $\mathfrak{m}$ be the maximal ideal generated by all monomials of total degree $\geq d+1$. Let $I$ be the principal ideal generated by $x=s\_1^2s\_2\cdots s\_d$. Then the element $y=s\_1^3s\_2\cdots s\_d$ in $R$ is in $q(I)$, but it is not in $I$. Indeed, $y^{d+1}$ equals $x^{d+1}\cdot s\_1^{d+1}$, so $y$ is in the integral closure of $I$. Also, for every monomial $m$ of degree $\geq d+1$, then $my$ equals $(s\_1m)x$, where $s\_1m$ is another monomial in $R$. Thus $y$ is also in the saturation of $I$ with respect to the maximal ideal $\mathfrak{m}$. Please note also: $R$ is regular in codimension $\leq d-1$, but it is definitely not $S2$. 


There are some exercises in Eisenbud's "Commutative Algebra" that discuss these types of rings and their integral closures. 


