Say $R$ is an integral domain with field of fractions $F$. Further, say there is an $R'$ (integral domain) such that $R\subset R'\subset F$. We want to prove that the field of fractions $F'$ of $R'$ is isomorphic to $F$.

I was thinking something along the lines, let $x,y\in R'$ with $y\neq 0$. Then $x,y\in F$, so $\frac{x}{y}\in F$ since the localization of $F$ is $F$ itself. Hence, $F'\subset F$. Also, since $R\subset R'$ we ought to have $F\subset F'$ and hence, $F=F'$.

Why is the question asking for isomorphic. I think that equality ought to hold. Am I doing something wrong?


  • $\begingroup$ I think that the question that you quote was carelessly framed. Like you, I would have said that $F$ is the fraction field of $R'$. $\endgroup$ – Lubin Nov 8 '12 at 6:40
  • 1
    $\begingroup$ It is a matter of definition. If the field of fraction is the set of pairs modulo relation, then the field of fraction of $R$, as a set, is different from the field of fraction of $R'$. However, a field of fraction $F$ has the (universal) property that any embedding of $R$ into a field $E$ extends in a unique way to embedding of $F$ to $E$. This property defines $F$ up to a unique isomorphism, and then, if one adopts the definition by the property, it is correct to say that the fields of fraction of $R$ and $R'$ equal. $\endgroup$ – Lior B-S Nov 8 '12 at 6:47
  • $\begingroup$ Thank you. I was thinking the same thing. I think its just a matter of technicalities. $\endgroup$ – Daniel Montealegre Nov 8 '12 at 7:11

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