# Every ordered field has a subfield isomorphic to $\mathbb Q$?

I'm going through the first chapter in a text on real analysis, which contains preliminaries on ordered fields, the real numbers, etc. Supposedly I had learned about such things already, in calculus, but I thought it wouldn't hurt to go over it again.

Up until the paragraph which is the subject of my question, everything is thoroughly proven and examples are provided. However the following excerpt just goes through facts, whose proof I cannot conceive:

Although $\mathbb Q$ is an archimedean field, these properties cannot be used to define $\mathbb Q$ since $\\$ there are many Archimedean ordered fields. What distinguishes $\mathbb Q$ from the other $\\$Archimedean ordered fields is that $(\mathbb Q,<)$ is the $smallest$ ordered field in the following sense: $\\$ if $(X,\prec)$ is an ordered field, then $X$ contains a sub-field which is (field) isomorphic to $\mathbb Q$.$\\$ Furthermore such an isomorphism preserves the order relation.

I've concatenated the paragraph a bit, so the text isn't idem from the book (if anyone wishes to know it's Phillips, An Introduction to Analysis and Integration Theory, Dover Publications). Also $\mathbb Q$ contains no proper subfields, and this can be verified by the apparently easy-to-prove fact that, if a field has characteristic zero, then it contains a subfield isomorphic to $\mathbb Q$. Then it remains to prove the preservation of order.

I'm guessing that some concepts from abstract algebra or field theory would easily suffice, but at the moment such topics are a bit over my head, so all I can think of doing is actually constructing the isomorphism $\phi:\mathbb Q\to\hat{\mathbb Q}$, where $\hat{\mathbb Q}$ is a certain subfield of $X$ in such a way that field operations are preserved, but I really can't come up with anything. So the question is: how do I construct this isomorphism, or, if there's a better way of proving this, how is it done? Thanks for any help.

Edit: Assume the existence of $(\mathbb Q,<)$ as an archimedean ordered field.

• That "apparently easy-to-prove fact" is quite easy to prove. Given a characteristic $0$ field, there is a $1$, and it's closed under addidion and additive inverse, so it has to contain $\Bbb Z$. It's also closed under multiplicative inverses, so it contains the field of fractions of $\Bbb Z$, which is $\Bbb Q$. There are some details that needs to be ironed out, but that's the gist of it. – Arthur Jan 25 '14 at 13:25
• @Arthur Just to be sure, when you say "contain" this is a sort of abuse of language is it not? Surely the term would be "is isomorphic to"? Also thank you I will work upon that – GPerez Jan 25 '14 at 14:43
• When I say a certain field "contains" $\Bbb Q$, I mean there's an injective homomorphism from $\Bbb Q$ into the field. It is standard terminology, although somewhat abusive, yes. – Arthur Jan 25 '14 at 15:24
• If you’re asking how to get the unique map from $\mathbb Q$ into a given field $K$ of characteristic zero, then it’s standard, but tedious to go through the details. First you use the existence of $1\in K$ and induction to get a map of $\mathbb N$ into $K$, then extend it by the existence of additive inverse, to get your unique map of $\mathbb Z$ into $K$. Finally, extend this to a map of $\mathbb Q$ into $K$ by using the existence of reciprocals in $K$. Tedious, as I say, and you have to check at each step that you’re forced to do what you’ve done. – Lubin Jan 27 '14 at 23:38

Any ordered field $F$ has characteristic $0$, so it contains a copy of $\mathbb{Z}$; by the universal property of the quotient field, the ring monomorphism $\mathbb{Z}\to F$ lifts to a monomorphism $\mathbb{Q}\to F$. We can identify $\mathbb{Q}$ with its image, so it's not restrictive to assume that $\mathbb{Q}\subseteq F$.
It's not really difficult: if $m/n\in\mathbb{Q}$, then we send it to $$\frac{f(m)}{f(n)}\in F$$ where $f\colon \mathbb{Z}\to F$ is the (unique) monomorphism. Is this a field homomorphism? Just a check.
Now we come to the order. First of all, positive integers are positive in $(F,\prec)$: if $n>0$, then $$n=\underbrace{1+1+\dots+1}_{\text{n times}}$$ and therefore $0\prec n$. Conversely, if $n<0$, then $$n=-(\,\underbrace{\,1+1\dots+1}_{\text{-n times}}\,)$$ and so $n\prec0$.
Any element of $\mathbb{Q}$ can be represented as $m/n$ with $n>0$, because $a/b=(-a)/(-b)$, where $a,b\in F$, $b\ne0$. So, let $0\prec m/n$ in the ordering of $F$, with $n>0$. Then, by the properties of ordered fields, $$0\prec n\cdot\frac{m}{n}=m$$ and therefore $m>0$. So a rational which is positive in $(F,\prec)$ is also positive in the usual order. A rational which is negative in $(F,\prec)$ is the opposite of a positive rational (in both orders).