Let $R$ be a commutative Noetherian domain, $K$ its fraction field, and $J$ a fractional ideal (i.e. a finitely generated sub-$R$-module of $K$) such that $J^2=J$. Is it true that $J=0$ or $J=R$? If not in this generality, can someone provide a counterexample?

I can prove that the claim is true if we assume $R$ is locally factorial, e.g. regular or a UFD. Indeed, the claim is local so we can reduce to the case of a UFD. In this case choose $d\in R$ such that $I:=dJ\subset R$. We may and do assume that $d=\delta^n$ is a power of a squarefree element $\delta\in R$. The condition $J^2=J$ means that $I^2=dI$, and in particular for every $x\in I$ there is $y\in I$ such that $x^2=dy$. It follows that $\delta$ divides $x$. Let $e\geqslant 1$ be minimal with the property that $\delta^e$ divides all elements of $I$. If we choose an $x=\delta^ex'\in I$ whose $\delta$-valuation is equal to this minimum, and $y=\delta^fy'\in I$ such that $x^2=dy$, we find $2e=n+f$. Since $f\geqslant e$ we obtain $e\geqslant n$, or in other words $d$ divides all elements of $I$. Therefore $I\subset (d)$ as ideals of $R$, hence $J\subset R$. We conclude using the well-known case of an idempotent ideal in a Noetherian domain.

Remark: I put the Dedekind-domains tag because experts in Dedekind rings may have ideas, although the question is about more general domains.

[EDIT] In fact the conclusion that $J$ equals $0$ or $R$ is true under the weaker assumption that $R$ is normal. Indeed, from $J^2=J$ it follows that each $x\in J$ defines a multiplication map $x:J\to J$. The determinant trick proves that $x$ is integral over $R$. Hence $J\subset R$ if $R$ is normal, whence the conclusion again by the well-known case of an idempotent ideal in a Noetherian domain.

somewhat widely used tags. There is exactly one user that follows the tag dedekind domains, and it really does not add much regarding visibility. – user9072 Apr 19 '14 at 12:51