# When the element-wise product of two ideals produces an ideal

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 ...

• 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 ... Commented Jan 30, 2019 at 18:14
• (contd. from last comment ) ... probably that doesn't help in any way though ... Commented Jan 30, 2019 at 22:19
• 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 $K$ ... Commented Jan 31, 2019 at 22:11
• so in particular $I^2$ does not contain any non-zero prime ideal ... Commented Jan 31, 2019 at 22:32