Problem related to prime ideals of B and A where B is integral over A

Question

Let $ A $ be an entire ring, integrally closed. Let $ B $ be entire, integral over $A$. Let $ Q_1, Q_2$ be prime ideals of $B$ with $Q_1 \supseteq Q_2$ but $Q_1 \neq Q_2$. Let $P_i=Q_i \bigcap A$. Show that $ P_1\neq P_2$.

Please give some hints. Thanks in advance.

Answer 1

Assume that $P_1=P_2$. Then $Q_1$ and $Q_2$ are two different prime ideals lying over the same prime ideal $P$ in $A$. By the incomparability theorem, see for example Corollary $14.15$ here, neither $Q_1\subset Q_2$ nor $Q_2\subset Q_1$, a contradiction. Hence $P_1\neq P_2$.

Answer 2

Can you reduce to the case of $Q_2 = P_2 = 0$? Having done this, take a non-zero element $x$ of $Q_1$ and try to use an equation of integral dependence for $x$ over $A$ to get something in $P_1$. It doesn't seem to me that we need to start off assuming that $A$ and $B$ are integral domains.