Is it true that if $\alpha \in \operatorname{Frac}(A)$ and $s\alpha \in A$, then $\alpha \in S^{-1}A$? 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.
 A: 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.
A: 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.
A: 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 $\alpha=a/s'=b/s\in S^{-1}A$.
A: 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 $\alpha \in S^{-1}A$.
