Suppose $A$ is a domain and $S\subseteq A$ is a multiplicative system. Show that $S\subseteq A^\times$ if and only if $S^{-1}A$ is integral over $A$.
I've started $\Leftarrow$ below...
Suppose that $S^{-1}A$ is integral over $A$. Then every element of $S^{-1}A$ is the root of a monic polynomial. For $s^{-1}a=\alpha\in S^{-1}A$, let's call such a polynomial $$p_{\alpha}(x)=\alpha_0+\alpha_1x+\cdots+\alpha_{n-1}x^{n-1}+x^n$$ We want to show that $\alpha\in A^\times$, i.e. that $s^{-1}a$ has a multiplicative inverse in $A$.
My first instinct was to write $$\frac{a}{s}\frac{b}{t}=\frac{ab}{st}=1\Rightarrow ab=st$$but I don't think I can justify writing the inverse in the form of $b/t$ here.
Where does the integral-ness come in? I could multiply $p_\alpha(\alpha)$ through by $s^n$ to show that $a$ is integral over $A$, but I don't think that gets me anywhere. Also, it doesn't use that $A$ is a domain.
Please try to use simple language in your answer as I am a beginner and am struggling to understand.
Try proving this proposition first (if you get stuck, you can look here):
Proposition: If $R \subseteq S$ is an integral extension, then $S^\times \cap R = R^\times$.
Now in your problem, since $A$ is a domain, the canonical map $A \to S^{-1}A$ is injective for any multiplicative set $S$, so one can view $A \subseteq S^{-1}A$. The proposition above will then give the implication $\Leftarrow$, and the implication $\Rightarrow$ is clear.