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 quasifinite, 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.)

1$\begingroup$ When you say "normal affine domain over an algebraically closed field", do you mean that the ring is finitely generated over the field? For general rings containing a field, I am not sure this is true. However, in the case of finitely generated $k$algebras, this is true. The simplest argument involves blowing up the point and passing to the stalk at each generic point of the exceptional divisor: this reduces your problem to a problem about extensions of DVRs. $\endgroup$ – Jason Starr Mar 21 '15 at 12:12

4$\begingroup$ Other method: by Zariski's Main Theorem we may assume $B$ finite over $A$. Let $A'$ be the completion of $A$ and $B'=A'\otimes_A B$. Then $B'$ is a product of $r$ complete noetherian local rings $B_i$ with $\dim B_i=\dim B=\dim A$. Clearly $r$ is the number of points in the closed fiber, and for dimension reasons each $\mathrm{Spec}\,(B_i)$ must dominate $\mathrm{Spec}\,(A)$, hence $r\leq d$. $\endgroup$ – Laurent MoretBailly Mar 21 '15 at 15:01

$\begingroup$ @LaurentMoretBailly: How do you use the normality of $A$? (This, or something like it, is obviously necessary.) $\endgroup$ – naf Mar 22 '15 at 9:34

2$\begingroup$ @ulrich: Normality guarantees that $\mathrm{Spec}\,(A')$ is irreducible, which is necessary to conclude that it is dominated by each $\mathrm{Spec}\,(B_i)$. So, I guess "$A$ unibranch" would suffice. $\endgroup$ – Laurent MoretBailly Mar 22 '15 at 16:53
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 $P$⊂$A$ are permuted by $G:=Gal(L/Q(A))$ (AtiyahMacdonald 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 ≤ the number of cosets $H$\ $G$ = degree($Q(B)/Q(A))$. For the purely inseparable case: there is only one prime of $B$ over $P$ (AtiyahMacdonald 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.

1$\begingroup$ A comment on your edit. If $B'$ is the integral closure of $A$ in $Q(B)$, $A\subset B'\subset B$. The quasifiniteness implies by Zariski's main theorem that $\mathrm{Spec}\, B$ is an open subset of $\mathrm{Spec}\, B'$ (see e.g. Mumford's red book). $\endgroup$ – Mohan Jun 2 '16 at 16:57