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I've managed [with help from the wonderful people lurking on this site :)] to prove that for an integral domain $A$, if $A$ is integrally closed, then $S^{-1}A$ is integrally closed for all multiplicatively closed subsets $S$ of $A$.

The problem that I want to apply this to is:

$A$ is integrally closed if and only if $A_{P}$ is integrally closed for all maximal ideals $P$ of $A$.

So far:

$(\Rightarrow)$ If $P$ is a maximal ideal of $A$, then $A\setminus P$ is multiplicatively closed, because maximal ideals are prime. So by the result mentioned above, $A_{P} = (A \setminus P)^{-1}A$ is integrally closed.

Update: Attempting to prove the contrapositive.

Assume there is some $\frac{r}{s}$ in $F$ (the field of fractions of $A$) such that $\frac{r}{s}$ is integral over $A$ but not in $A$. I feel like I haven't done hardly anything at all with your hint, but I suppose that I would want to somehow get a maximal ideal of $A$ from this $\frac{r}{s}$ such that $A_{P}$ is not maximal, thus proving the converse.

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    Prove the contrapositive. (This is a general strategy which often helps because it flips around quantifiers and sometimes gives you a more obvious first step of the proof.)2012-03-13
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    So assume $A$ is not integrally closed and then show that there is a maximal ideal $P$ of $A$ which is not integrally closed. I will give this another try. I actually started out this way but didn't get very far.2012-03-13
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    Hum. So the proof sketch I had in mind doesn't seem to actually work. My apologies.2012-03-13
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    No problem at all Qiaochu. You had something I certainly didn't: an idea. :)2012-03-13
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    The answer by Rankeya just solves your problem!!2013-01-05

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There is also another nice proof that if each $A_m$ is integrally closed, then so is $A$. It goes as follows:

Suppose $\exists t \in Frac(A)-A$ such that $t$ is integral over $A$. Define the set $I = \{a \in A: at \in A \}$. Then $I$ is clearly an ideal of $A$. Since $t \notin A$, it follows that $1 \notin A$. So, $I$ is a proper ideal of $A$, hence contained in some maximal ideal $m$. Then one can easily see that $t \notin A_m$, but $t$ is integral over $A_m$ (remember $A$ can be identified with a subring of $A_m$). This contradicts the fact that $A_m$ is integrally closed.

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    At first glance this proof seems to be independent of Pete L. Clark's or Merlin's proof. But, may be it is just a different way of stating their result. Either way, choose your poison.2012-03-26
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    The ideal I in your answer is nothing but the "set of denominators of $t$", hence it says that $t$ is not in $A_m$, but is integral over $A_m$. More explicitly, the key in the above proof is that every $t$ not in $A$ is not in some $A_m$; an equivalent formulation is thus $\bigcap_{\mathfrak{m} \in \operatorname{MaxSpec} A} A_{\mathfrak{m}} = A$. So the two proofs actually coincide.2013-01-05
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You will need the fact that for any domain $A$, $\bigcap_{\mathfrak{m} \in \operatorname{MaxSpec} A} A_{\mathfrak{m}} = A$. If you are not familiar with this, see e.g. $\S 7.7$ of my commmutative algebra notes.

Armed with this fact, the proof is virtually immediate: let $K$ be the fraction field of $A$, and suppose that an element $x \in K$ is integral over $A$. It must then be integral over the larger ring $A_{\mathfrak{m}}$ for each $\mathfrak{m} \in \operatorname{MaxSpec} A$, so...?

(Note that this approach does not involve proving the contrapositive, although presumably one can make that work as well...)

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    Yeah once the facts were pointed out I saw the argument immediately. Thank you very much for your material on the first fact. As for the second fact (that localization commutes with integral closure), I'm looking for it in my notes but still haven't found it (that doesn't mean it isn't there!). Thanks again I'm going to keep browsing.2012-03-13
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    Unfortunately, we don't have any such result in our development (that localization commutes with integral closure). But I don't think it's actually needed, unless I'm missing something?2012-03-13
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    @borninthe80s: no, that result is not needed for this direction of the argument. (Or maybe "Yes, that result is not needed..." :-).)2012-03-13
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Hint: (I am assuming from your previous question that you mean $A$ is a domain). In this case, $A=\cap_{P\text{ maximal}} A_P$. Use this and the fact that integral closure commutes with localization.

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    You are correct. I am sorry for not including this hypothesis. (Actually I didn't notice it!)2012-03-13
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    Everything seems to fall into place given your hints. I am going to look and see if I can find those facts in my notes now. If I can, then I think I've got it. Thanks very much.2012-03-13
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    Pete L. Clark gives a reference below.2012-03-13