The field of fractions of a field $F$ is isomorphic to $F$ 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?

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 
$$Fract(F)\rightarrow F;\; {a\over b} \mapsto ab^{-1}$$
