Ring of fractions of $R$ is integral over $R$

Question

Let $R$ be a commutative ring and let $S\subseteq R$ be a multiplicative closed subset containing 1.

I need to show that if $S^{-1}R$ is integral over $R$, then $S^{-1}R=R$.

I don't see how to get started.

Answer 1

One direction is because $R\subset S^{-1}R$. Now if $ 1/s$ is integral over $R$, then it satisfies: $$ \frac{1}{s^n} +\frac{ a_{n-1}}{s^{n-1}} +\cdots+\frac{a_0}{1} = \frac01 $$ where $ a_i \in R$. Now thus we have $ 1 = -\sum_{i=0}^n s^{n-i}a_i \implies s(-\sum_{i=0}^n s^{n-1-i}a_i)= 1 \implies s$ is a unit in $R$. (If $s^ka_i = 0$ for all $i$, where $k < n$, then we get $1/s^{k-n} = 0$ which is not possible.) Thus $ 1/s \in R$. Thus $S^{-1}R \subset R$. Combining these two we get $S^{-1}R =R $.