I have some questions regarding the proof of the going-up and down theorem (Theorem 0.2.6) from here (p. 4).
First,
Why are the prime ideals $\mathfrak q\in \operatorname{Spec}(B)$ for which $\mathfrak q\cap A= \mathfrak p$ exactly the elements in $\operatorname{Spec}(B\otimes_A(A/\mathfrak p)_{(0)})$?
Second, I do understand why $B\otimes_A (A/\mathfrak p)_{(0)}$ is a finite-dimensional vector space over $(A/\mathfrak p)_{(0)}$ (basically, by this argument), but I don't understand why it is a (finitely generated) $(A/\mathfrak p)_{(0)}$-algebra.
Update 1: Alright, I believe I managed to answer the second question. We have $B\otimes_A (A/\mathfrak p)_{(0)}=B_{(0)}\otimes_A (A/\mathfrak p)=B_{(0)}/\mathfrak pB_{(0)}$. Now $B_{(0)}/\mathfrak pB_{(0)}$ is a quotient ring, so multiplication is defined there. It is also an $(A/\mathfrak p)_{(0)}$-module (vector space) and hence an $(A/\mathfrak p)_{(0)}$-algebra. It is finitely generated as an algebra because it is finitely generated as a module (vector space).
Update 2: A misprint in the claim fixed (it is also present in the proof).