abelian-groupsabstract-algebraordered-groupsordered-ringsring-theory
Let $\Lambda$ be an ordered abelian group, (there is a total order on $\Lambda$ which is compatible with addition). Is there a multiplication map on $\Lambda$ that turns it into an ordered ring?
I have tried using classification results, such as $\Lambda$ is a subgroup of $\mathbb{R}^\Omega$ for some set $\Omega$ (Hahn embedding theorem) and the classification of subgroups of $\mathbb{Q}$.
Partial results are also welcome.
EDIT: I thought it would be appropriate, given the perhaps unexpected descriptive set-theoretic nature of this answer, to give a foreword explaining the focus on the Baire property (BP for short). A subset of a topological space has the BP if it differs from an open set by a meager set (a set contained in the union of countably many nowhere dense sets). In the setting of a Polish space (such as $\mathbb{Z}_p$ or the real numbers), the BP sets contain the Borel sets (and continuous images of Borel sets) and satisfy some nice regularity properties. In this context BP sets should be considered analogous to measurable sets, with meager sets serving as analogs of the null sets. For example, the Polish space itself is not meager, and the Kuratowski-Ulam theorem asserts that a subset of the plane is meager if and only if only a meager set of vertical sections are nonmeager (this is basically a Fubini theorem).
One particularly noteworthy fact is that it's consistent with the axioms $\mathtt{ZF+DC}$ that every subset of a Polish space has the BP. Recall that $\mathtt{ZF}$ is the standard axiomatization of set theory without choice, and $\mathtt{DC}$ is the axiom of dependent choice. Intuitively, $\mathtt{DC}$ gives you the freedom to make a countable sequence of choices, where each choice is allowed to refer to properties of your previous choices (so they aren't "independent"). Dependent choice is enough to do almost all of common mathematics: you can perform most of analysis, carry out typical inductive constructions, Borel sets behave reasonably, the first uncountable cardinal is not a countable union of countable sets, etc. Some contexts in which $\mathtt{DC}$ doesn't bestow the full power of $\mathtt{AC}$ include performing wildly nonconstructive acts like building a Vitali set or choosing bases from huge vector spaces. So, I think that $\mathtt{ZF+DC}$ is a reasonable framework in which to carry out your request for an "explicit" linear order of $(\mathbb{Z}_p,+)$. Once we rule out the existence of such a linear order with the property of Baire, we're therefore forced to concede that no argument producing this order may be carried out in $\mathtt{ZF+DC}$, dashing our hopes of an explicit construction.
By the way, I focus on $(\mathbb{Z}_p, +)$ rather than $(\mathbb{Q}_p,+)$ simply for convenience. It should be clear that any order of the latter induces an order of the former, so if anything the problem is harder for $\mathbb{Z}_p$.
There is no linear order of the additive group $(\mathbb{Z}_2,+)$ of $2$-adic integers which has the property of Baire (with respect to the usual Polish topology). In particular, it is consistent with $\mathtt{ZF+DC}$ that no such order exists at all, so a large fragment of the axiom of choice is indeed required to build such an order. Analogous arguments will work for all $\mathbb{Z}_p$, with slightly more obnoxious notation.
From here on we will identify elements of $\mathbb{Z}_2$ with infinite binary strings, that is, elements of $2^\omega$ with the product topology. We define an equivalence relation $E_0$ on $2^\omega$ by setting two strings equivalent iff they differ in finitely many coordinates. This $E_0$ has a nice interpretation in $\mathbb{Z}_2$: $x$ and $y$ are $E_0$ related iff their difference is a (standard) integer. More precisely, $x E_0 y$ iff for some $n$, $x + 1 + 1 + \cdots + 1 (n \mbox{ times}) = y$ or vice-versa, where $1$ denotes the standard integer $1$ (i.e., the sequence $10000\ldots$).
(That last part isn't literally true, since the constant $1$ sequence plus $1$ equals the constant $0$ sequence. But it is true off of the eventually constant sequences, which is enough to make the below argument go through (since there are only countably many eventually constant sequences).)
We use without proof two standard facts about $E_0$:
Now, given a putative order $<$ with the BP, we can partition $2^\omega$ into three $E_0$-invariant BP pieces:
(here $0$ is the identity element of the group: the constant $0$ sequence). So $X_-$ is the union of the $E_0$-classes which are entirely negative, $X_+$ the union of those entirely positive, and $X_0$ the union of those which are sometimes positive and sometimes negative.
(Technically, these pieces might not have the BP, but by Kuratowski-Ulam there's some element in $2^\omega$ we can use in place of $0$ to make the pieces have the BP. For ease of notation, let's assume $0$ works.)
We first observe that $X_0$ is meager. Note that the set $\{x : 0 \leq x < 1\}$ (which is meager by Fact 1) hits each $X_0$ class in exactly one point, so $X_0$ is the union of countably many homeomorphic translations (namely, the standard integer shifts) of a meager set, thus is meager.
Now let $f: 2^\omega \to 2^\omega$ denote the bitflipping homeomorphism, so $f(01001110\ldots) = 10110001\ldots$. We note that $x \in A_-$ iff $f(x) \in A_+$, since $x + f(x) + 1 = 0$ for all $x$. This means that $A_-$ cannot be comeager, else $A_+ = f[A_-]$ would be a disjoint comeager set. But then by Fact 2, $A_-$ is meager, thus so is $A_+ = f[A_-]$, and consequently we've written $2^\omega$ as the union of three meager sets. So we've hit a contradiction.
(*) By request, here is a reference for Fact 2: Theorem 3.2 of G. Hjorth: Classification and Orbit Equivalence Relations, Mathematical Surveys and Monographs, 75, American Mathematical Society, Providence, RI, 2000. Although actually this theorem is overkill for this special case -- here's a sketch of a more elementary argument that works here.
Suppose that $B \subseteq 2^\omega$ is nonmeager; we want to show that $[B]_{E_0} = \{x: \exists y \in B\ (x E_0 y)\}$ is comeager. By localization, there's a basic open set $U$ such that $B \cap U$ is comeager in $U$. We can find a finite binary string $s$ let's say of length $n$ such that $U$ contains all elements of $2^\omega$ beginning with $s$. Now look at the $2^n$ homeomorphisms of $2^\omega$ which flip some subset of the first $n$ bits of a string and leave the rest unchanged. These maps send each $x$ to something $E_0$-related to $x$, so it follows that $[B]_{E_0}$ is comeager in the union of $U$'s images under these maps. But the union of these images is all of $2^\omega$!
Accordingly to this page: "Order Preserving Isomorphism", it is sufficient to infer the Archimedean property (for every $a,b\in G$, if $a>0$, then there is a $n\in\mathbb{N}$ such that $na\geq b$).
Let $a>0$ be an element of $G$. Let $H=\{x\in G:\exists n\in\mathbb{N}:-na<x<na\}$. Then $-a<0<a$, so $0\in H$. For $x,y\in H$, there are $m,n\in\mathbb{N}$ such that $-ma<x<ma$ and $-na<y<na$, so $-(m+n)a<x-y<(m+n)a$, so $x-y\in H$. Also, for $x,y\in G$, if $0\leq y\leq x$ and $x\in H$ then there is $n\in\mathbb{N}$ such that $0\leq x<na$, so $0\leq y\leq x<na$, so $y\in H$. Therefore, either $H$ is an isolated subgroup (then $H=\{0\}$) or $H=G$, but $-2a<a<2a$, so $a\in H$, so $H\neq\{0\}$, so $H=G$. Therefore, $G$ is Archimedean.
Best Answer
No, and the classification of subgroups of $\mathbb{Q}$ can be used to find a counterexample. Consider the subgroup $A$ generated by $\frac{1}{p}$ as $p$ runs over all primes (or more generally infinitely many primes). I claim that $A$ cannot be the additive group of any ring (ordered or otherwise). To see this, first note that by partial fraction decomposition $A$ consists exactly of the set of fractions $\frac{a}{b}$ such that, when written in lowest terms, $b$ is squarefree.
Now suppose by contradiction that $\star : A \times A \to A$ is a multiplication for a ring structure on $A$, and let $e = \frac{a}{b}$ be its multiplicative identity. Let $p$ be a prime not dividing $b$; then $\frac{e}{p} = \frac{a}{pb} \in A$ satisfies
$$\underbrace{\frac{e}{p} + \dots + \frac{e}{p}}_{p \text{ times}} = e$$
(I am not writing this as $p \frac{e}{p}$ because we should carefully distinguish between multiplication of fractions and multiplication in $A$) and hence its square $\frac{e}{p} \star \frac{e}{p}$ satisfies
$$\underbrace{ \frac{e}{p} \star \frac{e}{p} + \dots + \frac{e}{p} \star \frac{e}{p} }_{p \text{ times}} = \frac{e}{p}.$$
It follows that $\frac{e}{p} \star \frac{e}{p} = \frac{e}{p^2} = \frac{a}{p^2 b}$, which is not in $A$; contradiction. So there is no such multiplication $\star$.