So I understand that there is no order on the field of p-adic numbers $\mathbb{Q_p}$ that makes it into an ordered field (i.e.) compatible with both addition and multiplication.

Now, from the responses to a couple of my previous questions,

  1. $\mathbb{Q_p}$ is a divisible abelian group under addition (being a field of characteristic $0$).

  2. $\mathbb{Q_p}$ is torsion free.

  3. It admits an order compatible with the group operation (addition), since every torsion free abelian group is orderable.

My question is, can I write an explicit ordering of $\mathbb{Q_p}$ compatible with the group operation? By "explicit", I mean an ordering in which, given two p-adic numbers, I can decide which is greater.

P.S.: I was not sure how to classify this problem, so please feel free to change the tags.

  • $\begingroup$ Brittany, just a comment. The algebra tag is used for more elementary precalculus type algebra questions. You should instead use the abstract algebra tag. You can read that in the description of the algebra tag. $\endgroup$ – Adrián Barquero Jul 7 '11 at 0:55
  • $\begingroup$ @Adrian: Thanks. I actually did mean to say abstract algebra here, but I wrote algebra and chose the wrong item from the drop down menu. $\endgroup$ – B M Jul 7 '11 at 1:06
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    $\begingroup$ For what it's worth, G Rangan, On orderability of topologicaL groups, Internat J Math Math Sci 8 (1985) 747-754, proves that ${\bf Q}_p$ has no order compatible with both its topology and addition, although it has orders compatible with each separately. $\endgroup$ – Gerry Myerson Jul 7 '11 at 3:26
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    $\begingroup$ If I understand the discussion of your earlier question, math.stackexchange.com/questions/49150, the first step in constructing the order you want is finding a basis for ${\bf Q}_p$ as a vector space over the rationals. But maybe this can't be done without invoking Choice. If so, "explicit" seems unlikely. $\endgroup$ – Gerry Myerson Jul 7 '11 at 3:48
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    $\begingroup$ @Gerry: I see. However, we don't necessarily have to use that construction. It just shows that a compatible order exists. Perhaps there is a construction some other way, but as you said, seems unlikely. $\endgroup$ – B M Jul 7 '11 at 3:53

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 $\mathtt{Z}_p$.

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 $E_0$:

  1. If $A \subseteq 2^\omega$ has the Baire property (from now on abbreviated BP) and meets each $E_0$-class in at most one point, then $A$ is meager (this is essentially the Vitali argument);
  2. If $A \subseteq 2^\omega$ has the BP and is $E_0$-invariant (i.e., $x \in A$ and $x E_0 y$ implies $y \in A$), then $A$ is either meager or comeager (this is a form of ergodicity) (*).

Now, given a putative order $<$ with the BP, we can partition $2^\omega$ into three $E_0$-invariant BP pieces:

  • $X_- = \{x \in 2^\omega : \forall y (y E_0 x \implies y < 0)\}$;
  • $X_+ = \{x \in 2^\omega : \forall y (y E_0 x \implies y > 0)\}$;
  • $X_0 = 2^\omega \backslash (X_- \cup X_+)$

(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 $2^\omega$!

  • $\begingroup$ Thanks for your elaborate answer, though I don't fully understand it yet. Although, more importantly, I don't see why the order is required to satisfy the Baire property? I just need a linear order that is compatible with addition. $\endgroup$ – B M Jul 11 '11 at 14:54
  • $\begingroup$ There are two reasons I looked at orders with the Baire property. First, you ask for an "explicit" order, which suggests some sort of definability constraint. BP sets include Borel sets, analytic sets, and so on, so this argument rules out anything "explicit" in that sense. Second, and more important, it is consistent with the axioms of ZF that every subset of $2^\omega$ has the Baire property. So this argument shows that any argument that there is an order of the sort you want necessarily uses the axiom of choice. $\endgroup$ – user83827 Jul 11 '11 at 15:34
  • $\begingroup$ Thanks for the explanation. So is it true that your proposition does not hold for the real numbers, since we can produce an order compatible with addition (and multiplication) explicitly? $\endgroup$ – B M Jul 11 '11 at 16:07
  • $\begingroup$ Yes, this very much uses the particular structure of addition in the $p$-adic integers. For instance, the subgroup generated by $1$ is dense in the $p$-adics whereas of course no cyclic subgroup is dense in the reals. $\endgroup$ – user83827 Jul 11 '11 at 16:15
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    $\begingroup$ I'm giving this +1 because it is the first answer which shows complete understanding of the question! (Whether or not it is actually correct is beyond my expertise -- I don't know any descriptive set theory, or whatever it is that is going into this answer. But at this point I consider that to be a bonus that others here will be qualified to judge.) I think it might be helpful to include a link to this "property of Baire" business. I conjecture that a lot of people interested in the question will not know what that is. The conjecture is true for me, at least. $\endgroup$ – Pete L. Clark Jul 11 '11 at 16:22

$1$ is either positive or negative. Say it is positive (otherwise, reverse the order). Then $0<1<2<3<...$. But there is a subsequence of these that converges to $0$. Impossible.

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    $\begingroup$ $@$GEdgar: I don't understand what you're trying to do here. As discussed above, since the additive group of $\mathbb{Q}_p$ is torsionfree, there does exist an ordering compatible with the group structure. The question is whether this can be made (in some precise sense) explicit. You seem to be trying to show that a putative ordering is in some sense not compatible with the topology...but that's not the question. $\endgroup$ – Pete L. Clark Jul 11 '11 at 4:46
  • $\begingroup$ @Pete: You are correct, this does not solve that problem. $\endgroup$ – GEdgar Jul 11 '11 at 13:11

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