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In a commutative ring $R$, when does the assumption $r_i\mid r$ for $1\le i\le n$ imply $\prod_{1\le i\le n} r_i\mid r$ (when $r_i$ are fixed)?

Does there exist any criterion for this implication that is related to regular sequences?

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    $\begingroup$ Taking $r_1=r_2=r$ you get $r^2\mid r$. This is equivalent to $R$ being absolutely flat, i.e. reduced and $0$-dimensional. $\endgroup$ Dec 3, 2022 at 16:17
  • $\begingroup$ Sorry; I want a criterion when the $r_i$ are fixed. This can be a generalization of the Chinese remainder theorem. $\endgroup$ Dec 3, 2022 at 16:35
  • $\begingroup$ I have added some explanations as an answer below. $\endgroup$
    – Mohan
    Dec 3, 2022 at 21:21

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This is just an expansion of my comment. Question is local, so we may assume that the ring is local. If $r_1,\ldots, r_n$ is a regular sequence, then they are so in any order and any subset forms a regular sequence. So, assume $r_i|r$ for all $i$ and assume we have shown $r=ar_1r_2\cdots r_i$ for some $i<n$, the case of $i=1$ being given.

Then $r_1,\ldots, r_i$ is a regular sequence modulo $r_{i+1}$. Then $r_1\cdots r_i$ is also a regular sequence modulo $r_{i+1}$ and then since $r_{i+1}|r$ implies $r_{i+1}|a$ and by induction we are done.

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You ask when the intersection of (principal) ideals is their product. It is well-known to be true if the ideals are pairwise coprime (and this condition is stronger than just demanding that the generators are coprime in case of principal ideals).

The condition for two ideals has been discussed here: When is the product of two ideals equal to their intersection?

In particular, $IJ$ equals $I \cap J$ iff $\mathrm{Tor}^1(A/I,A/J) = 0$.

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  • $\begingroup$ This is a nice observation; thank you! Yet I rather wonder whether something like the regularity of $(r_1,\dots,r_n)$ can help. $\endgroup$ Dec 3, 2022 at 17:31
  • $\begingroup$ @MikhailBondarko Yes, regularity (I presume that you meant a regular sequence) will imply what you want. $\endgroup$
    – Mohan
    Dec 3, 2022 at 17:42
  • $\begingroup$ Why does it (when there are more than two elements_?:) Dear Mohan, can you give a reference? $\endgroup$ Dec 3, 2022 at 17:58

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