# Flatness of $R+I \subseteq S$, where $R \subseteq S$ is flat

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!

• Why is $R+I$ flat over $R$? – Mohan yesterday
• 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. – user237522 yesterday
• Please, any ideas when flatness of $R \subseteq S$ implies flatness of $R+I \subseteq S$ are welcome. – user237522 yesterday
• Perhaps flatness of $R \subseteq R+I$ (if this claim is true), follows from $\frac{R}{R \cap I}=\frac{R+I}{I}$. – user237522 yesterday
• 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 $R$? – Badam Baplan yesterday