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.