I'm trying to prove the following statement:
Let $R$ be an integral domain and let $S\subset R$ be a multiplicative closed subset. If $R\subset R'$ is an extension ring of $R$ with $\overline{R}$ the integral closure of $R$ over $R'$, then $S^{-1}\overline{R}$ is the integral closure of $S^{-1}R$ in $S^{-1}R'$.
My attempt
I want to prove that $S^{-1}\overline{R}=\overline{S^{-1}R}$.
$\subseteq)$ Let $\frac{b}{s}\in S^{-1}\overline{R}$. Since $b\in \overline{R}$, $\exists$ a monic relation $b^n+c_{n-1}b^{n-1}+\cdots+c_0=0$ with $c_0,\cdots,c_{n-1}\in R$ and thus
$$\left(\frac{b}{s}\right)^n+\frac{c_{n-1}}{s}\left(\frac{b}{s}\right)^{n-1}+\cdots+\frac{c_1}{s^{n-1}}\left(\frac{b}{s}\right)+\frac{c_0}{s^n}=0$$
where $\frac{c_{n-i}}{s^i}\in S^{-1}R$, $i=1,\cdots,n$. Then $\frac{b}{s}\in\overline{S^{-1}R}$.
I tried to prove $\supseteq)$ starting in this way: Let $\frac{b}{s}\in\overline{S^{-1}R}$. Then $\exists$ a monic relation
$$\left(\frac{b}{s}\right)^n+z_{n-1}\left(\frac{b}{s}\right)^{n-1}+\cdots+z_1\left(\frac{b}{s}\right)+z_0=0$$ with $z_0,\cdots,z_{n-1}\in S^{-1}R$.
At this point I don't know how to continue the proof. I am grateful for any ideas you can give me to continue it.
Let $b \in \overline{S^{-1} R}$. Then there is some monic polynomial $p \in S^{-1} R[X]$, say of degree $n$, with $p(b) = 0$. Now let $s \in S$ such that $s \cdot p(X) \in R[X]$. Then $s^n \cdot p(X) = q(s X)$ for an appropriate monic polynomial $q \in R[X]$ (write it down). Now $q(sb) = 0$, hence $sb \in \overline R$ and $b = \frac{sb}{s} \in S^{-1} \overline{R}$.