Let A⊂B be rings,with B integral over A and B is finitely generated over A. Let P be a prime ideal of B and p be a prime ideal of A such that p＝P∩A , then B/P is an extension of A/p of finite degree.
I can't think of how to proof this claim. Why is ［B/P:A/p］finite?
integral + finitely generated (as an algebra) = finite (as a module), i.e. $B$ is a finite $A$-module. Let $b_1, \dotsc, b_n$ be generators. Then the residue classes $\overline b_1, \dotsc, \overline b_n$ are generators of the $A/p$-vector space $B/P$.
Note the following: Even is $B$ is $A$-free and $b_1, \dotsc, b_s$ a basis, the residue classes might be not linear independent.
For example take $\mathbb Z \subset \mathbb Z[i]$ with basis $1,i$ and take $P=(1-i) \subset \mathbb Z[i]$. The residue classes $\overline 1$ and $\overline i$ are equal in $\mathbb Z[i]/(1-i)$. The field extension $\mathbb F_2 = \mathbb Z/(P\cap A) \subset \mathbb Z[i]/(1-i) = \mathbb F_2$ is of degree $1$, while the free module was of rank $2$.