# $A\subseteq B$, $B$ integral over $A$, $\mathfrak{q}_{1}\subsetneq\mathfrak{q}_{2}$, then $A\cap\mathfrak{q}_{1}\subsetneq A\cap\mathfrak{q}_{2}$.

Let $A, B$ be commutative rings such that $A\subseteq B$ and $B$ is integral over $A$. I want to prove that if $\mathfrak{q}_{1},\mathfrak{q}_{2}$ are prime ideals of $B$ such that $\mathfrak{q}_{1}\subsetneq\mathfrak{q}_{2}$, then $A\cap\mathfrak{q}_{1}\subsetneq A\cap\mathfrak{q}_{2}$.

I get most of my examples of commutative algebra from the ring of integers of number fields, but now I am blocked, because the dimension of those rings is $1$

• Btw, this property of integral extensions is called incomparable. – user26857 Jun 3 '15 at 12:24

Have you tried to suppose $A\cap q_1=A\cap q_2$? Let's do it and set $p=A\cap q_1=A\cap q_2$. Since $A/p\subset B/q_1$ is integral we may reduce the question to the following:
$A\subset B$ is an integral extension of integral domains and $q$ is a prime ideal in $B$ with $q\cap A=0$. Show that $q=0$.
Let $b\in q$, and suppose $b\ne 0$. Then $b$ is integral over $A$, so $b^n+a_{n-1}b^{n-1}+\cdots+ a_1b+a_0=0$ with $a_i\in A$. We can assume $a_0\ne 0$ (why?). It follows $a_0\in q\cap A$, a contradiction.