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Let $A \subset B$ be commutative integral domains with $\operatorname{Quot}(A) = \operatorname{Quot}(B).$

Now consider the multiplicatively closed subset $S = \{x\in A\setminus\{0\}: x^{-1}\in B\}$.

I want to show that $S^{-1}A = B$. I would appreciate any help.

Bernard
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mathcourse
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    Isn't $A = K[x,y]$, $B=K[x,y, y/x]$ a counter-example? As far as I can tell, it seems that $S = A^*$ in this case, so that $S^{-1} A = A \neq B$ in this case. – Marc Jul 31 '18 at 20:35

1 Answers1

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It suffices to prove $B\subseteq S^{-1}A$, since we already have $A\subseteq B\subseteq \mathrm{Frac}(A)$. For any $x\in B$, $x=\frac{a}{b}$ for some $a,b\in A$. By definition, correct.

Guanyu Li
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