How to prove a number system is a fraction field of another?

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For example, how can I show that $\mathbb{Q}$ is the fraction field of $\mathbb{Z}$? Or that $\mathbb{C}$ is the fraction field of $\mathbb{R}$?

I understand that $\mathbb{Z}$ is a subring of $\mathbb{Q}$ & each r in $\mathbb{Q}$ can be written as a fraction r = a/b with a,b in $\mathbb{Z}$ and no proper subfield of $\mathbb{Q}$ has that property. But is there some general way to show this for the other number systems?

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The universal mapping property of localizations (e.g. fraction fields) yields an easy test for (unique) isomorphism, cf. $3.2$ below, from Atiyah & MacDonald, Commutative Algebra, p. $39$.

In OP: $\:\!A\,$ is a domain and $\,S\,$ is the set of nonzero elements in $A,\,$ so $\,S\,$ contains no zero-divisors, so condition $(ii)$ in $3.2$ simplifies to $\,g:A\to B\,$ is an injection, so $3.2$ specializes to

Corollary $\,B\,$ is isomorphic to the quotient field of $A\,$ if $\,B\,$ contains an isomorphic image $\bar A$ of $A$ such that every $\,0\neq a\in A\,$ maps to a unit $\, \bar a = g(a)\,$ in $B,\,$ and every $\,q\in B\,$ is a fraction over $\bar A,\,$ i.e. $\,q = \bar a_1 \bar a_2^{-1}\,$ for some $\,a_i\in A,\, a_2\neq 0$.

Remark $ $ More generally an analogous corollary holds for the total ring of fractions of a commutative ring - which inverts every $\,s\in S = $ all regular elements (non-zero-divisors).

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