# The Unapologetic Mathematician

## Archimedean Groups and the Largest Archimedean Field

Okay, I’d promised to get back to the fact that the real numbers form the “largest” Archimedean field. More precisely, any Archimedean field is order-isomorphic to a subfield of $\mathbb{R}$.

There’s an interesting side note here. I was thinking about this and couldn’t quite see my way forward. So I started asking around Tulane’s math department and seeing if anyone knew. Someone pointed me towards Mike Mislove, and when I asked him, he suggested we ask Laszlo Fuchs around the corner from him. Dr. Fuchs, it turned out, did know the answer, and it was in a book he’d written himself: Partially Ordered Algebraic Systems. It’s an interesting little volume, which I may come back and mine later for more topics.

Anyhow, we’ll do this a little more generally. First let’s talk about Archimedean ordered groups a bit. In a totally-ordered group $G$ we’ll say two elements $a$ and $b$ are “Archimedean equivalent” ($A\sim B$) if there are natural numbers $m$ and $n$ so that $|a|<|b|^m$ and $|b|<|a|^n$ (here I’m using the absolute value that comes with any totally-ordered group). That is, neither one is infinitesimal with respect to the other. This can be shown to be an equivalence relation, so it chops the elements of $G$ into equivalence classes. There are always at least two in any nontrivial group because the identity element is infinitesimal with respect to everything else. We say a group is Archimedean if there are only two Archimedean equivalence classes. That is, for any $a$ and $b$ other than the identity, there is a natural number $n$ with $|a|<|b|^n$.

Now we have a theorem of Hölder which says that any Archimedean group is order-isomorphic to a subgroup of the real numbers with addition. In particular, we will see that any Archimedean group is commutative.

Now either $G$ has a least positive element $g$ or it doesn’t. If it does, then $e\leq x implies that $x=e$ ($e$ is the identity of the group). By the Archimedean property, any element $a$ has an integer $n$ so that $g^n\leq a. Then we can multiply by $g^{-n}$ to find that $e\leq g^{-n}a, so $g^{-n}a=e$. Every element is thus some power of $g$, and the group is isomorphic to the integers $\mathbb{Z}\subseteq\mathbb{R}$.

On the other hand, what if given a positive $x$ we can always find a positive $y$ with $y? In this case, $y^2$ may be greater than $x$, but in this case we can show that $(xy^{-1})^2\leq x$, and $xy^{-1}$ itself is less than $x$, so in either case we have an element $z$ with $e and $z^2\leq x$.

Now if two positive elements $a$ and $b$ fail to commute then without loss of generality we can assume $ba. Then we pick $x=aba^{-1}b^{-1}>e$ and choose a $z$ to go with this $x$. By the Archimedean property we’ll have numbers $m$ and $n$ with $z^m\leq a and $z^n\leq b. Thus we find that $x, which contradicts how we picked $z$. And thus $G$ is commutative.

So we can pick some positive element $a\in G$ and just set $f(a)=1\in\mathbb{R}$. Now we need to find where to send every other element. To do this, note that for any $b\in G$ and any rational number $\frac{m}{n}\in\mathbb{Q}$ we’ll either have $a^m\leq b^n$ or $a^m\geq b^n$, and both of these situations must arise by the Archimedean property. This separates the rational numbers into two nonempty collections — a cut! So we define $f(b)$ to be the real number specified by this cut. It’s straightforward now to show that $f(bc)=f(b)+f(c)$, and thus establish the order isomorphism.

So all Archimedean groups are just subgroups of $\mathbb{R}$ with addition as its operation. In fact, homomorphisms of such groups are just as simple.

Say that we have a nontrivial Archimedean group $A\subseteq\mathbb{R}$, a (possibly trivial) Archimedean group $B\subseteq\mathbb{R}$, and a homomorphism $f:A\rightarrow B$. If $f(a)=0$ for some positive $a\in A$ then this is just the trivial homomorphism sending everything to zero, since for any positive $x$ there is a natural number $n$ so that $x. In this case the homomorphism is “multiply by ${0}$“.

On the other hand, take any two positive elements $a_1,a_2\in A$ and consider the quotients (in $\mathbb{R}$) $\frac{a_1}{a_2}$ and $\frac{f(a_1)}{f(a_2)}$. If they’re different (say, $\frac{f(a_1)}{f(a_2)}<\frac{a_1}{a_2}$) then we can pick a rational number $\frac{m}{n}$ between them. Then $nf(a_1), while $ma_2, which contradicts the order-preserving property of the isomorphism! Thus we find the ratio $\frac{f(a)}{a}$ must be a constant $r>0$, and the homomorphism is “multiply by $r$“.

Now let’s move up to Archimedean rings, whose definition is the same as that for Archimedean fields. In this case, either the product of any two elements is ${0}$ (we have a “zero ring”) and the additive group is order-isomorphic to a subgroup of $\mathbb{R}$, or the ring is order-isomorphic to a subring of $\mathbb{R}$. If we have a zero ring, then the only data left is an Archimedean group, which the above discussion handles, so we’ll just assume that we have some nonzero product and show that we have an order-isomorphism with a subring of $\mathbb{R}$.

So we’ve got some Archimedean ring $R$ and its additive group $R_+$. By the theorem above, $R_+$ is order-isomorphic to a subgroup of $\mathbb{R}$. We also know that for any positive $a\in R$ the operation $\lambda_a(x)=a\cdot x$ (the dot will denote the product in $R$) is an order-homomorphism from $R_+$ to itself. Thus there is some non-negative real number $r_a$ so that $\lambda_a(x)=r_ax$. If we define $r_{-a}=-r_a$ then the assignment $a\mapsto r_a$ gives us an order-homomorphism from $R_+$ to some group $S_+\subseteq\mathbb{R}$.

Again, we must have $r_a=sa$ for some non-negative real number $s$. If $s=0$ then all multiplications in $R$ would give zero, so $0, and so the assignment is invertible. Now we see that $a\cdot b=r_ab=sab$. Similarly, we have $r_{a\cdot b}=s(a\cdot b)=(sa)(sb)=r_ar_b$, and so the function $a\mapsto r_a$ is an order-isomorphism of rings.

In particular, a field $\mathbb{F}$ can’t be a zero ring, and so there must be an injective order-homomorphism $\mathbb{F}\rightarrow\mathbb{R}$. In fact, there can be only one, for if there were more than one the images would be related by multiplication by some positive $r\in\mathbb{R}$: $\phi_1(a)=r\phi_2(a)$. But then $r\phi_2(a)\phi_2(b)=r\phi_2(a\cdot b)=\phi_1(a\cdot b)=\phi_1(a)\phi_1(b)=r^2\phi_2(a)\phi_2(b)$, and so $r=1$.

We can sum this up by saying that the real numbers $\mathbb{R}$ are a terminal object in the category of Archimedean fields.

December 17, 2007 -

1. Here’s a little exercise I thought of while reading this post: prove that all totally ordered groups are abelian, or if this is false, give a counterexample. (A totally ordered group means of course a group G equipped with a total order < such that for all x, y, z in G, y &lt z; implies xy < xz and yx < zx.)

I don’t know if this will be considered hard or easy, but I thought it was sort of fun.

Comment by Todd Trimble | December 18, 2007 | Reply

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