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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 $I.$

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  • $\begingroup$ It looks to me like you already get an example for $R=(k[x,y]/\langle \ y^2-x^3 \rangle)_{\langle \overline{x},\overline{y} \rangle}$ and $I=\langle \overline{x} \rangle$. $\endgroup$
    – Jason Starr
    Commented Jan 20, 2017 at 8:48
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    $\begingroup$ Wait a minute: was that question supposed to be a riddle involving your username? $\endgroup$
    – Jason Starr
    Commented Jan 20, 2017 at 8:50
  • $\begingroup$ @JasonStarr actually I am trying to find an example in higher dimensional rings $\endgroup$
    – Cusp
    Commented Jan 20, 2017 at 9:14
  • $\begingroup$ When you say that you are trying to find an example in higher dimensional rings, are you trying to find an example that satisfies $R1$ but not $S2$? $\endgroup$
    – Jason Starr
    Commented Jan 20, 2017 at 9:19
  • $\begingroup$ @JasonStarr I want an example where the ring is such that the principal ideal is not integrally closed and I is different from its relative integral closure. $\endgroup$
    – Cusp
    Commented Jan 20, 2017 at 9:27

1 Answer 1

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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.

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  • $\begingroup$ Here by codimension did you mean embedding dimension (maximal homogeneous ideal ) minus dim R? $\endgroup$
    – Cusp
    Commented Jan 20, 2017 at 11:30
  • $\begingroup$ By codimension I really meant the height of the corresponding prime ideal. You are correct that I should localize the ring $R$ at the maximal ideal $\mathfrak{m}$. Since the finitely generated $R$-module $S/R$ is annihilated by $\mathfrak{m}$, for every prime ideal $\mathfrak{p}$ with $\mathfrak{p}\subsetneq \mathfrak{m}$, the localization $R_{\mathfrak{p}}$ equals $S_{\mathfrak{p}}$. Thus $R_{\mathfrak{p}}$ is regular for all prime ideals $\mathfrak{p}R_{\mathfrak{m}}$ of $R_\mathfrak{m}$ except the maximal ideal. $\endgroup$
    – Jason Starr
    Commented Jan 20, 2017 at 12:10

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