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Consider the ring $R=\mathbb C[X,Y]$. For every two ideals $I,J$ of $R$, define $I*J:=\{ij : i\in I, j\in J\}$.

Now definitely, $I*J=J*I$ always holds. If $I$ is principal, then actually $I*J$ is an ideal of $R$.

My question is:

If $I$ is a proper, non-zero ideal of $R=\mathbb C[X,Y]$ such that for every ideal $J$ of $R$, $I*J$ is also an ideal of $R$ (i.e. $I*J=IJ$), then does it imply that $I$ is principal ? Or at least $I$ is contained in a principal prime ideal ? If neither of these are true, then can we characterize all ideals $I$ with the said property in some other way ?

Some thoughts towards possibly showing $I$ is principal : To show $I$ is principal, enough to show $I$ is free, then by Quillen-Suslin, enough to show $I$ is projective, and since we are in Noetherian, finitely generated case, enough to show $I$ is flat over $R$. So it is enough to show $I \otimes_R J \cong IJ =I*J$ for every ideal $J$. No idea how to show that though ...

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  • $\begingroup$ If $I$ is contained in a principal prime ideal, then it would be same as saying it has height $1$, which in the context of your ring, would be same as saying $I$ is contained in a proper principal ideal ... $\endgroup$
    – user521337
    Commented Jan 30, 2019 at 18:14
  • $\begingroup$ (contd. from last comment ) ... probably that doesn't help in any way though ... $\endgroup$
    – user521337
    Commented Jan 30, 2019 at 22:19
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    $\begingroup$ since $I*J=IJ$. so $IJ$ contains no irreducible element for every proper ideal $J$, so taking $J=IK$, we get $I^2K$ does not contain any irreducible, hence does not contain any prime ideal for every ideal $K$ ... $\endgroup$
    – user521337
    Commented Jan 31, 2019 at 22:11
  • $\begingroup$ so in particular $I^2$ does not contain any non-zero prime ideal ... $\endgroup$
    – user521337
    Commented Jan 31, 2019 at 22:32

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