# Flatness of a certain ring extension

Let $$R \subseteq S$$ be two commutative rings. Let $$I$$ be an ideal of $$S$$. Clearly, $$R+I \subseteq S$$ is a subring of $$S$$. Assume that $$S$$ is flat over $$R$$. Then (by a known fact), $$R+I$$ is also flat over $$R$$.

In this situation, what can be said about $$I$$ or $$R$$ or $$S$$?

Moreover, if we further assume that $$I=Sw$$, for some $$w \in S$$ (namely, $$I$$ is principal), is there a more precise answer to my question?

Remark: We have $$\frac{R}{I \cap R} \cong \frac{R+I}{I} \subseteq \frac{S}{I}$$, but I do not see how this helps in answering my question.

Thank you very much.

Not much. For instance if $$R$$ is a field, $$S$$ can be any commutative $$R$$-algebra and $$I$$ any ideal. Conversely, for arbitrary $$R$$, we can take a prime ideal $$Q$$ of $$R$$ and then $$S=R_Q$$, $$I=0$$ satisfy your conditions.