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Assume $A \subseteq C \subseteq B$ are integral domains, with $B$ flat over $A$.

Generally, $B$ is not necessarily flat over $C$.

For example, see van den Essen's book "Polynomial Automorphisms and the Jacobian Conjecture", page 294, Exercise 4: $k[x^2] \subset k[x^2,x^3] \subset k[x]$. $k[x]$ is a free $k[x^2]$-module of rank $2$ (hence flat). But $k[x]$ is NOT a flat $k[x^2,x^3]$-module.

My question: What additional conditions are needed for flatness of $B$ over $C$?

Please notice that the above question is not the same as Projectivity of $B$ over $C$, given $A \subset C \subset B$, since here the extensions need not be finitely generated as modules, while there the extensions are assumed to be finitely generated as modules. Also here there is no assumption on $C$ over $A$, while there $C$ is assumed to be a free $A$-module.

On the other hand, here there is an additional condition that all rings are assumed to be integral domains, while there possibly not.

EDIT: Given $C \subseteq B$, there exist some really nice results showing when $B$ is flat over $C$ (for example, by Nagata, Chase, Ohm and Rush, Ohi). However, I wish to know how the additional information I have (namely, $A \subseteq C \subseteq B$ with $B$ flat over $A$) helps to decide when $B$ is flat over $C$.

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  • $\begingroup$ I cannot tell how well this would be received at MO. Looks like it could be ok, so if you feel like asking this there you can do so. The comparison to an older question in MSE can (should?) then be left out. Therefore it is IMO better that you post an appropriately edited version of this there yourself. Do remember to add links to each other to both posts so that future viewers of either version will not duplicate efforts to reproduce something already covered on the other site. $\endgroup$ – Jyrki Lahtonen Jun 27 '15 at 7:07
  • $\begingroup$ @JyrkiLahtonen Thank you very much! If I will post this question there (I guess I will), then I will take your advice (about adding links). $\endgroup$ – user237522 Jun 28 '15 at 0:46
  • $\begingroup$ This is not true. I am asking here, since I hope there are already known results which I can apply, and I am not aware of. (I mentioned the above papers only to save time for people here who are willing to help, to let them know that I already know those results and looking for something else). $\endgroup$ – user237522 Jun 30 '15 at 21:33

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