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Wikipedia defines the field of fractions of a domain as

The field of fractions or field of quotients of an integral domain is the "smallest" field in which it can be embedded.

What does "smallest" mean mathematically in this context? Is it possible to embed an integral domain in two different fields which have no elements in common?

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You can do that embedding, but each field will then contain a subfield isomorphic to the fraction field. –  Jason Polak Mar 3 '13 at 3:43

4 Answers 4

up vote 8 down vote accepted

"Smallest" means essentially that it satisfies the following universal property: if $R$ is a domain, and $K$ is any field for which there exists an injective ring homomorphism $f:R\to K$, then there is a unique ring homomorphism $g:\text{Frac}(R)\to K$ such that $f=g\circ i$, where $i$ is the natural inclusion of $R$ into its fraction field $\mathrm{Frac}(R)$.

As usual for a universal property, the object which satisfies it is unique up to unique isomorphism. Thus, if $L$ is a field in which $R$ embeds, and $L$ has the property that $L$ embeds in any other field in which $R$ embeds (this is the sense of "smallest"), then $L$ will necessarily be isomorphic to $\mathrm{Frac}(R)$.

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Much as I like categories, functors, adjoints, and universal properties, I feel obliged to point out that fields of fractions and the description you quoted from Wikipedia were around before categories, etc., were invented. An old-fashioned (i.e., pre-categorical) explanation of "smallest" might say that whenever a field $K$ contains (as a subring) a copy of $R$, then it must also contain a copy of the fraction field $F$ of $R$. (It might also say that this embedding of $F$ in $K$ can be chosen to extend the given embedding of $R$ in $K$, and that then the embedding is unique, but I'm not sure whether pre-category people would have included this last part.)

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This is already included in my answer. –  Martin Brandenburg Mar 3 '13 at 11:41
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@MartinBrandenburg Indeed, your answer is quite complete and thus includes mine. It is, however, at least in part, formulated in a way that is second nature nowadays, in light of category theory, but would, as far as I know, not be understood in pre-categorical times. I mean terminology like "the obvious diagram commutes" and "this embedding factors through the canonical embedding". –  Andreas Blass Mar 4 '13 at 2:01

It is meant in the sense of category theory. The field of fractions is a functor from integral domains (with injective homomorphisms) to fields (with field homomorphisms), which is left adjoint to the forgetful functor. This means: If $R$ is an integral domain, then there is a canonical injective homomorphism $R \hookrightarrow Q(R)$, and if $F$ is a field and $R \to F$ is an injective homomorphism, then there is a unique homomorphism of fields $Q(R) \to F$ such that the obvious diagram commutes. Or in other words: When $R$ is embedded into a field, then actually this embedding factors through the canonical embedding from $R$ into $Q(R)$, so that $Q(R)$ is the smallest such embedding.

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For completeness why don't we go ahead and include a proof of the universal property.

Let $R$ be an integral domain, $K$ its field of fractions and $\varphi: R \rightarrow F$ in an injective homomorphism into some field $F$. Consider the map $\tilde{\varphi}: K \rightarrow F$ given by $\tilde{\varphi}(a/b)=\varphi(a)/\varphi(b)$. We claim this map is in fact also an injective homomorphism. First note that the map is well defined because $F$ is a field so each element in $F$ has an inverse. Since $\varphi$ is a homomorphism $\tilde{\varphi}$ takes $0_K$ to $0_F$ and $1_K$ to $1_F$. To see additivity we have $\newcommand{\tv}{\tilde{\varphi}}$

$$\tv\left(\frac{a}{b}+\frac{c}{d}\right)=\tv\left(\frac{ad+bc}{bd}\right) =\frac{\varphi(ad)+\varphi(bc)}{\varphi(bd)}=\frac{\varphi(a)}{\varphi(b)}+\frac{\varphi(c)}{\varphi(d)}=\tv\left(\frac{a}{b}\right)+\tv\left(\frac{c}{d}\right).$$ Multiplication is even more straightforward $$\tv\left(\frac{ac}{bd}\right)=\frac{\varphi(a)\varphi(c)}{\varphi(b)\varphi(d)}=\tv\left(\frac{a}{b}\right)\tv\left(\frac{c}{d}\right).$$ Injectivity follows immediately because $K$ is a field. Finally $\tv$ is the unique extension of $\varphi$. Let $\psi: K \rightarrow F$ be another extension of $\varphi$, that is $\psi|_R=\varphi$. Then for $b \in R \setminus \{0\}$

$$1=\psi(b/b)=\psi(b)\psi(1/b)=\varphi(b)\psi(1/b)$$

so $\psi(1/b)=1/\varphi(b)$ and we see that $\psi(a/b)=\tv(a/b)$.

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"For completeness why don't we go ahead and include a proof of the universal property." because it is included in every textbook and otherwise it is a good exercise one should do for oneself, instead of copying it from an answer on math.SE. –  Martin Brandenburg Mar 3 '13 at 11:40

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