Contractions and Extensions of Ideals and Faithful Flatness

Question

Let $A \subseteq B \subseteq C$ be rings with $B$ a faithfully flat $A$-module and $C$ a faithfully flat $B$-module which is an integral extension of $B$. Given a maximal ideal $I \subseteq C$, is it true that $$(I \cap A)B = I \cap B$$ I know that faithful flatness is transitive and also gives $(I \cap B)C = I$. I feel like this is true, but have been unable to prove it.

Answer 1

Faithful flatness implies that contraction is left inverse to extension of ideals, I think you had that backwards. In other words $IC \cap B = I$ for faithfully flat extensions, but not $(I\cap B)C = I$.

For a counterexample to your question, consider $k \subseteq k[x^2] \subseteq k[x]$ and the maximal ideal $xk[x]$. This also shows that even for the finite faithfully flat extension $k[x^2] \subseteq k[x]$, contraction of ideals is not left inverse to extension.