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Given $A \subseteq C \subseteq B$ with $A \subseteq B$ flat, when $C \subseteq B$ is flat? This question appears here.

Now, I wonder if restricting the above question to to following special case would be easier to answer:

Let $R \subseteq S$ be two (commutative) integral domains, $I$ an ideal of $S$. Obviously, $R+I$ is a ring. We have, $R \subseteq R+I \subseteq S$. Assume that $R \subseteq S$ is flat.

Question: When $R+I \subseteq S$ is flat?

One plausible positive answer to this question is when all those three rings have the same field of fractions, see Richman Lemma 2(1), page 795.

Motivation: The following is a known result: Given $A \subseteq C \subseteq B$ with $A \subseteq B$ separable, implies that $C \subseteq B$ is separable.

Thank you very much!

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  • $\begingroup$ Why is $R+I$ flat over $R$? $\endgroup$ – Mohan yesterday
  • $\begingroup$ Thank you for the comment. I have seen this claim (that $R \subseteq R+I$ is flat) in one of MSE's questions, and truly, have not checked this myself, though I should have. Perhaps I should delete this. Please notice that in Richman's result for $A \subseteq B \subseteq C$, it is not assumed that $A \subseteq B$ is flat. $\endgroup$ – user237522 yesterday
  • $\begingroup$ Please, any ideas when flatness of $R \subseteq S$ implies flatness of $R+I \subseteq S$ are welcome. $\endgroup$ – user237522 yesterday
  • $\begingroup$ Perhaps flatness of $R \subseteq R+I$ (if this claim is true), follows from $\frac{R}{R \cap I}=\frac{R+I}{I}$. $\endgroup$ – user237522 yesterday
  • $\begingroup$ I think in Richman's result we can weaken the hypotheses substantially... Something like $A \subseteq B \subseteq C$ are arbitrary rings such that $A \subseteq C$ is flat and there exists a ring $D$ containing $C$ such that $A \subseteq D$ is an epimorphism of rings. What's the motivation for your question? Where does the ring $R+ I$ come up in practice? Can you link to the question claiming that $R + I$ is flat over $R$? $\endgroup$ – Badam Baplan yesterday

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