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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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The field of fractions comes equipped with a canonical map $F\rightarrow \operatorname{Frac}(F)$. You should check if your map in the opposite direction is the inverse. –  Joe Johnson 126 Nov 5 '11 at 12:08
    
@ Joe Johnson 126 : i think you mean the canonical map $a\mapsto a/1$. The composite with the map i gave is the identity since $a/b\mapsto ab^{-1}$ which by the canonical map goes to $ab^{-1}/1$ and this is in the same equivalence class of $ab^{-1}.b/1.b=a/b$ so yes they are mutually inverse to each other. –  palio Nov 5 '11 at 12:24
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If your map is injective and surjective, you need only check that it's a homomorphism and thus an isomorphism. –  Joe Johnson 126 Nov 5 '11 at 12:29
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Alternatively, you could also check that the canonical map is an isomorphism. Check that every $x\in\operatorname{Frac}(F)$ is equivalent to one of the form $\frac{a}{1}$ for some $a\in F$. –  Joe Johnson 126 Nov 5 '11 at 12:34
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palio: If you understood the idea, please consider writing it as an answer. Then others can not only check your work, but also upvote the answer. You may even accept your own answer. –  Srivatsan Nov 5 '11 at 16:16

1 Answer 1

up vote 2 down vote accepted

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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How did you show that an element $a/b$ in $Fract ~F$ is equal to $c/1$ for some $a,b,c \in F$? Thanks –  Panther Jul 18 at 5:08
    
you can take $c=ab^{-1}$ then $\phi(ab^{-1})=ab^{-1}/1\sim ab^{-1}b/b=a/b$. –  palio Jul 18 at 15:34
    
Thanks. I need a little explanation though. Why is $ab^{-1}/1 \sim ab^{-1}b/b$ ? –  Panther Jul 18 at 18:22
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This is by definition of the equivalence relation $x/y\sim z/t$ if and only if $xt=yz$. –  palio Jul 18 at 19:57
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It does. Indeed, $x/1\sim y/1$ if and only if $x.1=y.1$ if and only if $x=y$ –  palio Jul 19 at 1:31

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