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Let $F$ be a field and let $\newcommand{\Fract}{\operatorname{Fract}}$ $\Fract(F)$ be the field of fractions of $F$; that is, $\Fract(F)= \{ {a \over b } \mid a \in F , b \in F \setminus \{ 0 \} \}$. I want to show that these two fields are isomorphic. I suggest this map

$ F \to \Fract(F) \ ; \ a \mapsto {a\over a^{-1}} ,$

for $a \neq 0$ and $0 \mapsto 0$, but this is not injective as $a$ and $-a$ map to the same image. I was thinking about the map $ \Fract(F) \rightarrow F ;\; a/b\mapsto ab^{-1}$ and this is clearly injective. It is also surjective as $a/1 \mapsto a$. Is this the desired isomorphism?

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    Note that the fraction field is not actually the set you describe. The fraction field of a domain $D$ is the set modulo the equivalence relation $\frac{a}{b}\sim \frac{c}{d}\Longleftrightarrow ad=bc$.2011-11-05

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Let $F$ be a field and $Fract(F)=\{\frac{a}{b} \;|\; a\in F, b\in F, b\not = 0 \} $ modulo the equivalence relation $\frac{a}{b}\sim \frac{c}{d}\Longleftrightarrow ad=bc$. We exhibit a map that is a field isomorphism between $F$ and $Fract(F)$. Every fraction field of an integral domain $D$ comes with a canonical ring homomorphism $\phi: D\rightarrow Fract(D);\; d\mapsto \frac{d}{1}$ This map is clearly injective.

In the case $D$ is a field $F$, this canonical map is an isomorphism with inverse $Fract(F)\rightarrow F;\; {a\over b} \mapsto ab^{-1}$

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    It does. Indeed, $x/1\sim y/1$ if and only if $x.1=y.1$ if and only if $x=y$2014-07-19