$A \subset B$ be integral and flat extension then it is faithfully flat.

Question

I want to prove the following:

$A \subset B$ be integral and flat extension of rings then it is faithfully flat.

Clearly enough to show that for every ideal $I$ of $A$, $I^{ec}=I$. Since the extension is integral for every prime ideal $p$ of $A$ $p^{ec}=p$. I can’t show that it also holds for arbitrary ideal. I need some help. Thanks.

Answer 1

Hint:

It is enough to prove that for any prime ideal $\mathfrak p\in\operatorname{Spec} A$, there exists a prime ideal $\mathfrak q\in\operatorname{Spec} B$ such that $\;\mathfrak q\cap A=\mathfrak p$.