$I\cdot J$ principal implies $I$ and $J$ principal? 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 ?
 A: 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.
A: In ${\bf Z}[\sqrt{-5}]$,
$$(2)=(2,1+\sqrt{-5})(2,1-\sqrt{-5})$$
A: 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.
