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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?

Bill Dubuque
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BigD4J
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    $\mathbb R$ is its own field of fractions, since it is already a field. – Matt Samuel Apr 05 '15 at 21:13
  • Not sure how that answers my question. Also, aren't $\mathbb{Q}$ and $\mathbb{C}$ also fields? – BigD4J Apr 05 '15 at 21:29
  • Whats your definition of $\mathbb{Q}$? Usually $\mathbb{Q}$ is definded to be the fraction field of $\mathbb{Z}$. – abc Apr 05 '15 at 21:29
  • $\mathbb{Q}$ is the set of rational numbers here. – BigD4J Apr 05 '15 at 21:29
  • Sure, but whats your definition of "rational number", the usual one is "an element in the fraction field of $\mathbb{Z}$". So there is nothing to prove. – abc Apr 05 '15 at 21:31
  • Ah I'm running in circles. So the fraction field of $\mathbb{R}$ is not $\mathbb{C}$ but instead $\mathbb{R}$ itself? Why? – BigD4J Apr 05 '15 at 21:33
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    @BigD4J: A fraction field for an integral domain $R$ is the smallest field containing $R$ as a subring, or put differently, it is the smallest extension of the ring $R$ to a ring $F(R)$ that has all the multiplicative inverses $x^{-1}$ for every $x\in R\setminus{0}$. – String Apr 05 '15 at 21:37
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    And in $\mathbb R$, every element $x\in\mathbb R\setminus{0}$ already have their inverses in $\mathbb R$. Thus $\mathbb R$ is already its own fraction field. The simple way to say this is to say, that $\mathbb R$ is already a field (meaning that all non-zero elements have inverses). – String Apr 05 '15 at 21:40
  • Regarding $\mathbb C$, it is the algebraic closure of $\mathbb R$, which is something quite different. Both $\mathbb C$ and $\mathbb R$ are fields, the former being an extension field of the latter. – String Apr 05 '15 at 21:41
  • Maybe I'm confusing field extensions with fraction fields. I know $\mathbb{C}$ is a field extension of $\mathbb{R}$. I thought this translated to $\mathbb{C}$ being the field of fractions of $\mathbb{R}$, but I must have that mixed up. – BigD4J Apr 05 '15 at 21:45
  • OK, that makes sense then :) – String Apr 05 '15 at 21:46
  • Isn't it trivial that $\mathbb{C}$ is its own field of fractions? Is the case for $\mathbb{R}$ non-trivial? – BigD4J Apr 05 '15 at 21:47
  • @BigD4J by the definition above it is trivial to show that every field is its own field of fractions. – Matt Samuel Apr 05 '15 at 22:18
  • @Matt Samuel that's what I'm confused about as well. If $\mathbb{Z}$ is its own field of fractions, how can $\mathbb{Q}$ also be the field of fractions of $\mathbb{Z}$? – BigD4J Apr 05 '15 at 23:22
  • @BigD4J $\mathbb Z$ is not a field. – Matt Samuel Apr 05 '15 at 23:33

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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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Bill Dubuque
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  • The universal property of the fraction field $Q$ of $A$ is: every embedding of $A$ into a field $K$ uniquely extends to an embedding of $Q$ into $K$ (i.e. $Q$ is the most general (universal) field-image of $A).$ As above, can deduce this from the more general universal property of localizatations ("partial" rings of fractions) – Bill Dubuque Apr 13 '23 at 18:55