If $G/H$ is flat $A$-module, and $I$ is an ideal of $A$, $G$ is flat, then how to see $GI\cap H=HI$?

Question

If $G/H$ is a flat $A$-module, and $I$ is a finitely generated ideal of $A$, $G$ is a flat $A$-module, $H$ is an $A$-submodule of $G$, then how to see $GI\cap H=HI$?

Answer 1

It's a general fact that given an exact sequence $$ 0\rightarrow M\hookrightarrow F\rightarrow N\rightarrow 0 $$ of (right) $A$-modules with $F$ flat, then $N$ is flat iff $M\cap FI=MI$ for every finitely generated (left) ideal $I$. The proof is a bit involved, but it can be found as Proposition 3.60 of Rotman's Introduction to Homological Algebra, (page 139, in my copy at least.)

This applies to your situation, as you have an exact sequence $$ 0\rightarrow H\hookrightarrow G\rightarrow G/H\rightarrow 0, $$ where $G$ is a flat $A$-module. Since $G/H$ is given to be flat, one direction of the above fact gives you the desired conclusion.