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Given $I, J$ are ideals of $R$, what is a necessary condition (either on the ideals or the ring) to say the element-wise product $$\tilde {IJ} = \{ij : i\in I, j\in J\}$$ forms an ideal.

For example in $\mathbb{Z}$, $(2)(3) = (6)$; and this holds generally in PIDs since $$(a)(b) \ni ar_1 bs_1 + \cdots + ar_n bs_n = ab(r_1s_1+\cdots r_ns_n) \in (ab) $$

Xiao
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    Here may be a helpful fact: If you look at your example of $\mathbb{Z}$ you will see you just need one of the ideals to be principle. – Sheel Stueber Jun 19 '18 at 01:12
  • Do you really mean necessary condition or did you actually want sufficient conditions? – rschwieb Jun 19 '18 at 13:38
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    @rschwieb I meant the most general sufficient condition, since taking $R$ to be PID would be one sufficient condition. – Xiao Jun 19 '18 at 17:16
  • @SheelStueber hmm, what do you mean? Since $\mathbb{Z}$ is a PID. – Xiao Jun 19 '18 at 17:18
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    @Xiao Thanks for clarifying. I can't think of anything better than "at least one of the ideals is principal" like Sheel suggested. S/he just means $\mathbb Z$ is a special case. It'll work like that in any commutative ring (if one of the ideals is principal, then the element-wise product is an ideal.) – rschwieb Jun 19 '18 at 17:48

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