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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.

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Have you tried rewriting the usual proof (that an idempotent ideal in any Noetherian ring is generated by an idempotent element), which uses the determinant trick? See for example people.fas.harvard.edu/~amathew/chvarious.pdf, Lemma 1.1 part (2). –  Graham Leuschke Apr 19 at 11:42
    
Yes I tried, but there are several triples $(A,I,M)$ (ring, ideal, f.g. module) that one can try and none gave the result. For instance taking $A=R+J$ and $I=M=J$ I could prove that $R\subset J$ but then?... –  Matthieu Romagny Apr 19 at 12:10
    
Tangentially: I added the commutative-rings While I think it is fine to have dedekind-domains too it is especially for the purpose you mention rather more relevant to use somewhat widely used tags. There is exactly one user that follows the tag dedekind domains, and it really does not add much regarding visibility. –  quid Apr 19 at 12:51

1 Answer 1

up vote 6 down vote accepted

This is not true. The ring $R = k[t^2, t^3]$ with $J = k[t]$ is a counter example.

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Great, thanks! I should have tried such an example... –  Matthieu Romagny Apr 19 at 12:43

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