Quotient of ring is flat gives an identity of ideals

Question

I have problem to understand and solve the exercise 1.2.14 on Qing Liu's book "Algebraic Geometry and Arithmetic Curves". It goes as follows:

Let $A\to B$ be a ring homomorphism, and let $J$ be an ideal of $B$ such that $B/J$ is flat over $A$. Show that for any ideal $I$ of $A$, we have that $(IB)\cap J=IJ$ (tensor the injection $I\to A$ by $B/J$).

Where do we need the homomorphism $A\to B$ as Liu does not use it explicitly in the problem? Could someone elaborate why tensoring the injection helps to prove the theorem?

Answer 1

I don't get Liu's hint, but the following it's useful and not hard to prove:

Let $M$ be a flat $A$-module. Then if tensor a short exact sequence of $A$-modules $0\to M_1\to M_2\to M\to 0$ by any $A$-module it remains exact.

Now consider $0\to J\to B\to B/J\to 0$ a short exact sequence of $A$-modules. By tensoring this with $A/I$ we get $0\to J\otimes_AA/I\to B\otimes_AA/I\to B/J\otimes_AA/I\to 0$. We know that $J\otimes_AA/I\simeq J/IJ$, $B\otimes_AA/I\simeq B/IB$, and $B/J\otimes_AA/I\simeq B/(IB+J)$. (I've used here that $M\otimes_AA/I\simeq M/IM$.) This shows that $(IB+J)/IB\simeq J/(IB\cap J)=J/IJ$, and we are done.