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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?

roo
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  • 1 is a multiplicative unit... it's hard to elaborate if you don't explain what particular issue you have about $y=y/1$. –  Mar 12 '12 at 04:29
  • Note that $\frac{1}{s}(\frac{ssy}{s})$ is certainly in $S^{-1}A$. – Arturo Magidin Mar 12 '12 at 04:32
  • I don't know how to elaborate because I find this issue confusing. I am always uncomfortable with the subtle identifications that are often happening. Sorry I can't be more clear.

    Do we identify $y/1$ with the product of the inverse of $1$ with $y$?

     – roo Mar 12 '12 at 04:35
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    When confusion reigns, you could always not identify -- e.g. the first sentence of your update, did you really mean $(s/1) y = a/1$ for some $a \in A$? Anyways, we're in a commutative ring and $1$ is invertible, so division by 1 means multiplication by the inverse of 1. Also, maybe it would help to think about the fraction field $F$? With care, you can choose $F$ so that $A$ literally is a subring. Then, $S^{-1} A$ can be chosen to be another subring of $F$. (you may wish to prove that last statement) –  Mar 12 '12 at 05:13
  • Looking over your worry and your response again, I think something you missed is that $y = a/t$ for some $a \in A$ and $t \in S$, simply because $y \in S^{-1} A$: you don't have to try to write $y$ as $y/1$, since it's already a fraction. –  Mar 12 '12 at 05:19
  • Thank you for your thorough response. I'd have to look it over again, but the core of the issue is that I'm trying to prove that $y\in S^{-1}A$. It's already been verified that $y/1\in S^{-1}A$. This is why I want to use the fact that $y = y/1$. – roo Mar 12 '12 at 05:40
  • @Hurkyl: With one more apology for being so unclear, just as I was drifting off last night I thought of how to pin down why I am confused.

    It seems to be obvious to everyone else that $y = \frac{y}{1}$. But my problem is that I can't use the definition of equality in $S^{-1}A$ ($a/b = c/d$ if and only if there exists $e\in S$ such that $e(ad - bc) = 0$ in $A$) because I am not able to write $y$ as a fraction unless I assume that which I am trying verify.

    I hope this makes more sense.

     – roo Mar 12 '12 at 18:09
  • My mistake on the type of $y$ -- it was assumed to be in the fraction field of $S^{-1} A$ rather than in $S^{-1} A$. If the notation is still confusing you, could you explain where $y/1$ came from? I would have done the calculation more like $\frac{1}{s} s y = (\frac{1}{s} s)y = 1 \cdot y$. –  Mar 12 '12 at 18:31
  • let us continue this discussion in chat – roo Mar 13 '12 at 00:22

2 Answers2

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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 $A$.

Dylan Moreland
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    ahh! i was multiplying by $s$ which wasn't doing the trick. Thanks very much I will try this! :) – roo Mar 12 '12 at 04:03
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    update: that was literally the step I needed. thank you! – roo Mar 12 '12 at 04:05
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    This is actually not completely true. You get equality with $0$ in the localization which implies that there exists $v\in S$ such that multiplying your polynomial with $v$ is $0$. So the same goes for multiplying your polynomial eith $v^n$ and then you shuffle stuff around and get $bsv$ is integral over $A$. – Shaked Nov 24 '19 at 17:50
  • @Shaked In this case $A$ is an integral domain and $0 \notin S$, so $A \rightarrow S^{-1}A$ is injective. This means that $p(sx) = 0$ in $S^{-1}A$ implies that $p(sx) = 0$ in $A$. – David Lui Jan 23 '23 at 13:45
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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.

Martin Sleziak
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Pete L. Clark
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