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Let $R$ be a Noetherian domain, and let $I$ and $J$ be two ideals of $R$ such that their product $I\cdot J$ is a non-zero principal ideal. Is it true that $I$ and $J$ are principal ideals ? This seems an easy question to settle, but I can't find an answer.

Any idea is welcome, thanks !


Thanks a lot for your enlightening answers. I admit I'm more interested in a geometric setting (i.e. when $R$ is an algebra, finitely generated, or a localization of that). I fail to adapt examples coming from number theory to this setting. What do you think ?

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Dear Lierre: You may want to assume that your ideals are nonzero. –  Pierre-Yves Gaillard Aug 1 '12 at 9:33
    
Thanks, I wrote proper meaning non-zero. –  Lierre Aug 1 '12 at 9:41
    
$(x,y)z=0$ in $\mathbb Z[X,Y,Z]/(XZ,YZ)$ (obvious notation). –  Pierre-Yves Gaillard Aug 1 '12 at 9:59
    
Thanks, but the product ideal is zero. –  Lierre Aug 1 '12 at 10:06
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In an abelian semi-group a product of elements is invertible if and only if all factors are invertible - this is elementary. Apply this to the semi-group of fractional ideals of $R$, in which the principal ideals are invertible. You don't need the assumption that $R$ is noetherian. –  Hagen Aug 1 '12 at 10:30
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3 Answers 3

up vote 7 down vote accepted

At Lierre's request , here is a geometric example.

Consider an elliptic curve $\bar E$ (say over $\mathbb C$), a point $P\in \bar E$ of order $2$ in the group $\bar E(\mathbb C)$ (there are 3 such) and the complement $E=\bar E \setminus \lbrace O\rbrace $ of the origin in $\bar E$.
Like all non complete integral curves $E$ is affine, with ring $R=\Gamma (E, \mathcal O_E)$.
The ideal $I=\mathfrak m_P\subset R$ of functions vanishing at $P$ is not principal because $\mathcal O_E(-P)$ is a non-trivial line bundle (use Abel-Jacobi's theorem).
However $I^2=\mathfrak m_P^2$ is principal because the line bundle $\mathcal O_E(-P)$ has as its square $\mathcal O_E(-2P)=\mathcal O_E(0)=\mathcal O_E= $ the trivial line bundle, so that $I$ is an example of non-principal ideal with $I\cdot I$ principal.

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Nice example ! Am I right if I say that $I$ is locally principal ? –  Lierre Aug 1 '12 at 10:05
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Dear Lierre: yes, you are absolutely right. If $z=z_P\in Frac(R)$ is a uniformizing parameter for $\mathcal O_{E,P}$, then $I=(z)$ on any open neighbourhood of $P$ containing none of the finitely many poles of $z$ and none of the finitely many zeros of $z$ distinct from $P$. –  Georges Elencwajg Aug 1 '12 at 10:40
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In ${\bf Z}[\sqrt{-5}]$,

$$(2)=(2,1+\sqrt{-5})(2,1-\sqrt{-5})$$

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It is false.

The counterexample of Gerry Myerson in the ring $\Bbb Z[\sqrt{-5}]$ is actually a typical situation. Consider a field $K$ which is a finite algebraic extension of $\Bbb Q$ and let $\cal O_K$ its ring of integers, i.e. the integral closure of $\Bbb Z$ in $K$. Then we know that the non-zero ideals in $\cal O_K$ generate an abelian group $\cal I_K$ under ideal multiplication. The principal ideals generate a subgroup $\cal P_K<\cal I_K$.

A famous basic result in algebraic number theory says that the quotient group $\cal C_K=\cal I_K/\cal P_K$ (the class group of $K$) is actually finite.

Since any element in $\cal C_K$ can be represented by an ideal, this has the following two immediate consequences:

  • given any ideal $I$ in $\cal O_K$ there's always an ideal $J$ in $\cal O_K$ such that $IJ$ is principal;

  • there's a natural number $h$ depending only on $K$ such that for any ideal $I$ in $\cal O_K$, the ideal $I^h$ is principal.

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