Let $A \subset B$ be a finite (finite type + integral) extension of integral domains and let $\mathfrak{p} \subset A, \mathfrak{q} \subset B$ be prime ideals such that $\mathfrak{q} \bigcap A =\mathfrak{p}$. Let $A/\mathfrak{p} \subset B/\mathfrak{q}$ be the induced inclusion of domains. Suppose $A$ is integrally closed.
Here is my question:
Is $A/\mathfrak{p} \subset B/\mathfrak{q}$ always a finite extension? If so, do we have $[K(B):K(A)] \ge [K(B/\mathfrak{q}):K(A/\mathfrak{p})]$?
PS: I only know this result is trivial in algebraic geometry (for varieties over a field $k$, [Fulton]'s intersection theory has a very nice formula on comparing the degrees with the ramification indices. I believe this is also easy when $K(B/\mathfrak{q})/K(A/\mathfrak{p})$ is a simple extension. But I am curious how general can it be. )
PS2: I still don't know how to prove the general case. If $K(B/\mathfrak{q})/K(A/\mathfrak{p})$ is separable, then this is a result of [Atiyah,MacDonald, Proposition 5.15].
