Please help me solve this exercise,
Suppose $R$ is an integral domain and $S$ is a multiplicatively closed subset. Prove that if every element of $R_{S}$ is a root of a monic polynomial of $R[x]$ then $R= R_{S}$.
It's trivial that $R\subset R_{S}$, but inverse is true when $S=\{1\}$. But I can't prove this!
I know, if $\frac{r}{s} \in R_{S}$ then by hypothesis there exists a polynomial $$f(x)=x^{n}+a_{n-1}x^{n-1}+\cdots + a_{o}$$ in $R[x]$ such that $f(\frac{r}{s})=0$.
So $$r^{n}+a_{n-1}r^{n-1}s+\cdots + a_{0}s^{n}=0.$$ But this doesn't help me.
Let $\frac{1}{s}\in R_{S}$. By hypothesis and the above mentioned proof, $$1=st$$ for some $t\in R$ (note that $R$ is domain).
This means every elements of $S$ belong to $U(R)$.
But by definition, $R_S$ is smallest ring such that every elements of $S$ is a unit. So we showed $R=R_S$.