Wikipedia says that if an integral domain $A$ is integrally closed, then $S^{-1}A$ is integrally closed if $S$ is a multiplicatively closed subset of $A$. They state it as a reason for another argument but I can't figure out how to verify it as a standalone statement.
Using the hypotheses, it is straightforward (but notationally cumbersome so please forgive me for not posting it here) to show that if $y$ is integral over $S^{-1}R$, then $y$ is algebraic over $R$. But this doesn't seem to help get me what I want.
I'm pretty sure that I need to use this fact for an equivalence of statements (for an integral domain) proof in a homework problem:
The homework problem: $A$ is integrally closed if and only if $A_{P}$ is integrally closed for every maximal prime ideal $P$ of $A$. (Note I am not looking for help with this part quite yet as I think I can get it if I can verify the claim above.)
UPDATE: Based on the argument below, I can conclude that if $y$ is integral over $S^{-1}A$ then there exists an $s\in S$ such that $sy$ is an element of $A$. From here I want to conclude that $y = \frac{1}{s}sy \in S^{-1}A$. But I'm a bit uncomfortable with the claim that $y = \frac{1}{s}sy$. Unless I can write $y = y/1$, I cannot conclude this. But I don't know anything about $y$ except that it is in the field of fractions in $S^{-1}A$. Am I missing something trivial?