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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?

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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/2756/discussion-between-borninthe80s-and-hurkyl)2012-03-13

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Let $x$ be an element of the fraction field of $A$ which is integral over $S^{-1}A$. Thus $x$ satisfies some equation \[ x^n + \frac{a_{n - 1}}{s_{n - 1}}x^{n - 1} + \cdots + \frac{a_0}{s_0} = 0 \] with $a_i \in A$ and $s_i \in S$. Now, $s = s_{n - 1} \cdots s_0$ is an element of $S$. If you multiply the integral equation by $s^n$ and shuffle some things around, you should find that $sx$ is integral over $A$, hence is an element of $A$.

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    update: that was literally the step I needed. thank you!2012-03-12
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The Theorem in $\S 14.2$ of my commutative algebra notes gives a slightly more general result:

Let $S/R$ be an extension of integral domains, and let $T \subset R$ be a multiplicatively closed subset. Then the integral closure of $T^{-1} R$ in $T^{-1} S$ is $T^{-1}$ (the integral closure of $R$ in $S$).

Applying this with $S$ equal to the fraction field of $R$ and $T$ equal to the multiplicative subset $S$ recovers the result you are asking about.