Number of elements in a fiber Let $A\subseteq B$ be normal affine domains over an algebraically closed field of characteristic 0. If it is given that the corresponding morphism of schemes Spec $B\rightarrow$ Spec $A$ is quasi-finite, and the degree of the field extension [$\mathbb{Q}(B):\mathbb{Q}(A)]$ is $d$, how can one show that over each maximal ideal of $A$, there exist at most $d$ many maximal ideals of $B$? 
(Any elementary proof and/or a proper reference will be appreciated.)
 A: For the finite separable case: embed $Q(B)$ in a finite Galois extension $L/Q(A)$.  The primes of the integral closure of $A$ in $L$ over a prime $P$&subset;$A$ are permuted by $G:=Gal(L/Q(A))$ (Atiyah-Macdonald exercise 13 p. 68), and the number of primes of $B$ over $P$ is the number of double cosets $G$=∪$HgD(Q/P)$ where $H:=Gal(L/Q(B))$ and $D(Q/P)$:=decomposition group of a prime $Q$ of $L$ over $P$.  This number of double cosets is &leq; the number of cosets $H$\ $G$ = degree($Q(B)/Q(A))$.  For the purely inseparable case:  there is only one prime of $B$ over $P$ (Atiyah-Macdonald exercise 15 p. 69). Combining these cases gives the result.
Aside: the residue field extension can be infinite inseparable, see: Divisibility of the degree of an extension by the degree of its residual field
Edit:  I think I misunderstood the original question and I assumed that B is the integral closure of A in Q(B).  Additional hypotheses would be needed to get that B is an open immersion in this integral closure of A.  I'll remove this answer if requested to do it.
