Finitely generated extension of integral domains I have come across the following problem that I have not been able to solve. Especially, the implication (i) to (ii) remains elusive to me. Hints and/or solutions would be greatly appreciated!

Let $A \subset B$ be integral domains, with $B$ ﬁnitely generated as an $A$-algebra.
Show that the following are equivalent:
(i) The set of primes of $B$ lying over the prime $(0)$ of $A$ is ﬁnite.
(ii) $B$ is ﬁnite over $A[1/f]$ for some nonzero $f \in A$.

 A: The basic fact (not too difficult) you need is the following. If $K$ is a field, $R$ an integral domain which is a finitely generated $K$-algebra and if $R$ has only finitely many prime ideals, then $R$ is a field and $[R:K]<\infty$.
In your case, localize everything at $S=A-\{0\}$. Then, $S^{-1}A$ is a field, $S^{-1}B$ is an integral domain which is finitely generated as an algebra over $S^{-1}A$ and it has only finitely many primes by assumption i). So, we see that $S^{-1}B$ is a field and of finite dimension over $S^{-1}A$. We may pick a finite set of generators and clear denominators and assume that they are in $B$ and let them be the finite set $E\subset B$. First consider the $A$-algebra generated by $E$, say $C\subset B$. If $b\in B$, we can find an $s\in S$ such that $sb\in C$. Doing this for finitely many generators of $B$, we see that we may invert an $s\in S$ and then $C_s=B_s$. Since the elements of $E$ are finite and they generate $S^{-1}C$, one sees that each of the element in $E$ is algebraic over $S^{-1}A$ and thus by inverting an element $t\in S$, they are integral. Thus, the extension $A_{st}\to B_{st}$ is integral. But, $B_{st}$ is finitely generated as an algebra and thus it is a fintely generated $A_{st}$ module, which is what you wanted.
