If $R$ is integral over $S$, then $\operatorname{Frac}(R)/\operatorname{Frac}(S)$ is finite extension of fields

Question

How to show that:

If $R\supset S$ are integral domains, $R$ is integral over $S$, $K$ and $L$ the fraction fields of $R$ and $S$ respectively, then $K/L$ is a finite extension of fields.

Answer 1

First, you need to assume that $R$ and $S$ are integral domains. Second, I will prove a more general claim:

Let $A$ and $B$ be rings and $S \subset A$ multilicatively closed subset of $A$. Assume $B$ is integral over $A$. Then $S^{-1}B$ is integral over $S^{-1}A$.

Proof. Let $x/s \in S^{-1}B$ with $x \in B$ and $s \in S$. Let $$ x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 = 0 $$ be the polynomial equation that $x$ satisfies (since $B$ is integral over $A$). Then $x/s$ satisfies $$ x^n/s^n + a_{n-1} x^{n-1}/s^n + ... + a_1 x/s^n + a_0/s^n = 0 $$ which is $$ (x/s)^n + (a_{n-1}/s) (x/s)^{n-1} + ... + (a_1/s^{n-1}) (x/s) + (a_0/s^n) = 0 $$ with $a_{n-1}/s , ... , a_0/s^n \in S^{-1}A$ so $x/s$ is integral over $S^{-1}A$. $\quad \square$