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In the proof of Proposition 1.9 in Chapter VII of Algebra by Serge Lang, it seems to me that the following property is used.

Let $A$ be a commutative entire ring, $S$ a multiplicative subset of $A$, $0 \not \in S$. Let $\alpha$ be an element of the quotient field of $A$. If $s \alpha \in A$ for some $s \in S$, then $\alpha \in S^{-1}A$.

(in a proof that $A$ integrally closed implies $S^{-1}A$ integrally closed).

Is this bold statement true ?
Since $\alpha$ is expressed as $\alpha=a/s'$, where $a, s' \in A, s' \neq 0$, $s \alpha \in A$ implies that $s a = s' b$ for some $b \in A$.

From this, how can I prove that $s' \in S$ ?
Any help would be appreciated.

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Which edition of Algebra? – Jonas Meyer Jul 24 '12 at 13:28
It is the revised third edition. – Aki Jul 24 '12 at 13:32
You can't prove $s'\in S$ because it is false in general: see my answer. – Georges Elencwajg Jul 24 '12 at 13:40
up vote 3 down vote accepted

What is $\,S^{-1}A\,$ ? By definition, $$S^{-1}A:=\{a/s\;:\;a\in A\,,\,s\in S\}\,$$with the "usual" operations (the definition is there).

Thus, we can in fact say that for $\,\alpha:=\frac{a}{b}\in F\,\,,\,a,b\in A\,,\,b\neq 0$ , as $$\alpha\in S^{-1}A\Longleftrightarrow \frac{as}{b}=\alpha s\in A\,\,,\,\text{for some}\,\,s\in S$$

In fact, there's hardly anything to prove here...

Added: Let's see if the following clarifies a little:

$$\exists\,\,s\in S\,\,s.t.\,\,s\alpha=a\in A\Longrightarrow \alpha=\frac{a}{s}\in S^{-1}A\,\,,\,\text{per definition of fractions ring}$$ and we don't care what the "original" form of $\,\alpha\,$, as element of the fractions field of $\,A\,$ , is/was.

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I think I get it thanks to your additional comment "we don't care what the original form of $\alpha$ is/was". – Aki Jul 24 '12 at 14:15

You don't need to show that $s'\in S$, because you have already shown $\alpha \in S^{-1}A$. As you said, $sa=s'b$, therefore $\alpha=a/s'=b/s\in S^{-1}A$.

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I think you misread the question. $\alpha$ a priori is an element of the field of fractions of the integral domain $\alpha$ and not necessarily an element of the localisation. – user38268 Jul 24 '12 at 13:06
Yes, I did not see "entire" at all the first time around. I think the number of authors using "entire" instead of "domain" must be countable on one of Homer Simpson's hands. – rschwieb Jul 24 '12 at 13:19
I think Lang is the only one. I always found it funny that in a book so terse he spends a paragraph or so defending the terminology, although his argument was very convincing to me until I realized that no one would ever know what I was talking about unless I said “domain”. – Dylan Moreland Jul 24 '12 at 14:18
Lang's mother tongue was French. The French terminology for domain is anneau intègre. Since the adjective integral was already used for another concept in ring theory, I suppose Lang settled for entire as the closest substitute. – Georges Elencwajg Jul 24 '12 at 15:05

Suppose that for some $s \in S$ we have $\alpha s \in A$ . Then $\alpha s = a$ for some $a \in A$ and so viewing everything as happening inside the fraction field of $A$ we can do division so that

$$\alpha =\frac{\alpha}{1} = \frac{a}{s}.$$

But then by definition the guy on the right is in the localisation from which it follows that $\alpha \in S^{-1}A$.

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Lang is right (see the other answers) but not what you suggest, namely that if $\alpha=a/b, \;b\neq 0$ and $s\alpha \in A$ for some $s\in S$, then you can deduce $b\in S$.
Indeed you can always write $1=1\cdot b/b\in A$ for any $b\in A\setminus \lbrace 0\rbrace$ and if what you suggest were true, then (since $1\in S$) you would always have $b\in S$, i.e. $S=A\setminus \lbrace 0\rbrace$: an absurd statement.

NB I have changed your $s'$ into $b$. I think your notation is confusing because it tends to suggest $s'\in S$ and that this partly created your difficulty.

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Thank you, @GeorgesElencwajg. Now, I see that my assumption was wrong. – Aki Jul 24 '12 at 14:03
Great, Aki, bravo! – Georges Elencwajg Jul 24 '12 at 14:59

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