Is the reals the smallest connected ordered topological ring?

The real numbers is a locally compact Tychonoff connected complete ordered topological field. I am looking at minimal collections of adjectives that can characterize the reals. The one often used to define the reals is that it is (Dedekind- or Cauchy-) complete ordered field.

Consider the real numbers among other ordered topological rings. I am wondering if the real numbers is the smallest connected ordered topological ring? Here "smallest" means that it embeds into every other connected ordered topological ring. If this is not true, do you have some minimal collection of additional adjectives (advoiding the word "complete") that can characterize the real numbers among ordered topological rings? "Minimal" means that if you drop an adjective then you there are other ordered topological rings that also fulfil your definition.

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Since the reals are characterized by being the only complete totally ordered field, any other criterion you give will have to have something with the feeling of completeness to it- I don't think you can get something from nothing, especially if the "something" is "real numbers". – Dylan Wilson Feb 9 '12 at 6:56
In lights of the comments above, I would suggest editing the title. Perhaps to something like "characterizing $\mathbb R$ as a topological ring"... – Gjergji Zaimi Feb 9 '12 at 8:47
argh, i forgot to add "connected". I meant to ask if the reals was the smallest connected ordered topological ring. – user2529 Feb 9 '12 at 13:26
this is discussed in the comments to Gjerigji's answer. – user2529 Feb 9 '12 at 13:28
Must an ordered ring be an integral domain? – Gerald Edgar Feb 9 '12 at 14:59

3 Answers

The following characterization of $\mathbb R$ and $\mathbb C$ among topological rings is due to Pontryagin and seems to be in the spirit of your question:

Theorem: If $F$ is a field with a Hausdorff ring topology which is locally compact and connected then $F$ is isomorphic as a topological field to either $\mathbb R$ or $\mathbb C$ with their usual topologies. (If you substitute "field" with "division ring" then you must add the quaternions $\mathbb H$ to the list.)

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But, as I said, you have to have some complete-ish feeling to your conditions... Locally compact Hausdorff is a nice one. (Also, en.wikipedia.org/wiki/…) – Dylan Wilson Feb 9 '12 at 7:21
I agree. The surprise here, for me, is the lack of other properties like "ordered" etc. from the statement. Locally compact Hausdorff connected seemslike too little at first sight :) – Gjergji Zaimi Feb 9 '12 at 7:35
Are there easy counterexamples if the "locally compact" condition is dropped? Now that I think about it, I really don't know a lot of connected fields... – Qiaochu Yuan Feb 9 '12 at 7:50
I can't think of any simple examples, though I read that any field with the discrete topology embeds in a connected field. It seems that it is not known whether any field embeds in a connected field... – Gjergji Zaimi Feb 9 '12 at 8:20
Waterman, Alan G., Bergman, George M., "Connected fields of arbitrary characteristic." J. Math. Kyoto Univ. 5 (1966) 177–184. Here we have the proof that an arbitrary discrete field may be imbedded in a connected field. – Gerald Edgar Feb 9 '12 at 14:57

The answer to the original question is positive, and even more is true.

Proposition: $(\mathbb R,+,\le)$ is up to isomorphism the unique nontrivial connected ordered group.

Proof:

Let $(G,+,\le)$ be such a group (not necessarily abelian despite the notation). First, observe that $G$ is archimedean, i.e., for every $a,b>0$, there is a natural number $n$ such that $b\le na$. This follows from the fact that $$\{b:\exists n\in\mathbb N\,(-na\le b\le na)\}$$ is a nontrivial convex subgroup, hence clopen.

Second, by a well-known theorem of Hölder, every archimedean group is isomorphic to a subgroup of $\mathbb R$. The argument is as follows. We fix a positive element $a$. For any $b\in G$ and $n\in\mathbb N$, there is a unique $b_n\in\mathbb Z$ such that $b_na\le nb<(b_n+1)a$. Then it is easy to show that $$f(b):=\lim_{n\to\infty}\frac{b_n}n$$ exists, and that $f$ is an ordered group embedding of $G$ into $\mathbb R$.

Now, $\mathbb R$ has no nontrivial proper connected subgroup: if $a\in\mathbb R\smallsetminus G$, then $(-\infty,a)\cap G$ and $(a,+\infty)\cap G$ form an open partition of $G$. (By the way, in general an ordered space is connected iff it is densely ordered and Dedekind complete.)

Since there is only one way how to expand $(\mathbb R,+,\le,1)$ to an ordered ring, we obtain

Corollary: $(\mathbb R,+,\cdot,\le)$ is up to isomorphism the unique nontrivial connected ordered ring.

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The real numbers is the smallest connected ordered field.

Let $F$ be an ordered field, endowed with the order topology, that is connected with this topology. (Note that the order topology makes any ordered field into a topological field).

$F$ characteristic zero and so contains a subfield isomorphic to the rational numbers.

Consider the Dedekind cuts of the field; non-empty partitions (A, B) of the field such that every element of A is less than every element of B and A has no greatest element. A is open, being a union of all negative rays from each of its elements (i.e. $A = \bigcup_{a \in A} \{x | x < a \}$) . If B has no least element then similarly it is open, and so the cut is a disconnection of $F$.

Hence by assumption for every Dedekind cut (A,B) of $F$, B has a least element.

Any Dedekind cut on the rational numbers can be extended to $F$ via the mapping (A, B) to (A', B') = ($\{x \in F | x < a, a \in A\}$, $F \setminus A'$). Thus by mapping each Dedekind cut on the rational numbers (A, B) to the least element of B' we get an embedding of the real numbers into $F$.

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Note that the last step is nonobvious. For example, let $F$ be an ultraproduct of $\mathbb Q$ over a nonprincipal ultrafilter on $\omega$. Then $F$ is an ordered field, and every Dedekind cut on $\mathbb Q$ can be separated by an element of $F$ (due to its $\aleph_1$-saturation), but there is no embedding of $\mathbb R$ into $F$. In fact, $\mathbb Q$ is relatively algebraically closed in $F$, so $F$ doesn’t even contain a square root of $2$. – Emil Jeřábek Aug 27 '14 at 15:24