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?

**Answer**

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}

**Attribution***Source : Link , Question Author : palio , Answer Author : palio*