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I've come to know that you can't define an order relation over the field of p-adic numbers that is compatible with the addition and multiplication according to the ordered field axioms. I was wondering if there actually was a subset which contains an isomorphic copy of $\mathbb Q$ and can be totally ordered (compatibly with addition and multiplication).

[Edit from comment below] In particular, is there a way to totally order some subring $T$ of the $p$-adics, with $\mathbb Q \subset T$, so that the order respects addition and multiplication, and there is some $\omega \in T$ greater than all the rationals in the order?

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Well, there are plenty of subrings of the $p$-adic numbers which are isomorphic to subrings of the reals which contains the rationals. For example, if $d$ is a positive integer which is a square mod $p$, then the ring $\mathbb{Q}[\sqrt{d}]$ is isomorphic to subrings of both $\mathbb{R}$ and $\mathbb{Q}_p$. –  Thomas Andrews Oct 27 '11 at 13:40
    
Note, this is essentially just trickery. In the same way, you can make the ring of polynomials $\mathbb{Q}[x]$ be totally ordered, by sending $x$ to a real transcendental number (say, $\pi$.) This ordering on $\mathbb{Q}[x]$ is arbitrary, because any different transcendental on would yield a different order. Similarly, for the case of $\sqrt{d}$, it is a choice of which square root to send to the real positive square root. –  Thomas Andrews Oct 27 '11 at 14:01
    
Thanks. That's an excellent answer, so I'm trying to be more precise. My original aim was to find an ordered subset$A$ so that $\mathbb Z \subset A \subset \mathbb Q_p $ with _"infinities"_ in it. I say $\omega \in A $ is an _infinity_ iff $\forall n \in \mathbb Z$ I get $\omega >n$. With $\mathbb Z$ I mean an isomorphic copy of the well known set of integers and $>$ is the order relation over $A$. –  Lorenzo Cavallina Oct 27 '11 at 14:07
    
@HenningMakholm you are right. I asked a question yesterday (without logging up) and today I wasn't able to add comments. So I decided to create an account. Do you think I can have access to the reputation I earned yesterday? Or I should just have the other "account" deleted? –  Lorenzo Cavallina Oct 27 '11 at 14:11
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@Lorenzo: I've merged your accounts. –  Zev Chonoles Oct 27 '11 at 18:19
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In particular, then, you want a total ordering on $\mathbb{Q}[\omega]$ for some $p$-adic $\omega$, with $\omega$ greater than the elements of $\mathbb Q$ in the order.

If $\omega$ is transcendental over $\mathbb Q$ in $\mathbb Q_p$, then $\mathbb Q[\omega] \cong \mathbb Q[x]$, where $x$ is an indeterminate, so you'd want a total order on the ring of polynomials over $\mathbb{Q}$ in which $x>r$ for all rational $r$

But if the total order is compatible with multiplication and addition, we have the following:

$$\forall n\geq 0, r\in \mathbb Q: x^{n+1}>rx^n$$

From this, we can deduce that a polynomial $p(x)$ is greater than zero if it's first coefficient is greater than zero.

That entirely defines the total order on $\mathbb Q[x]$. $p(x)>q(x)$ if the leading coefficient of $p(x)-q(x)$ is positive.

If $\omega$ is not transcendental, then you can show there can be no such ordering, basically by the same argument that $\omega^n$ must be strictly greater than any linear combination of smaller powers of $\omega.$

So what this shows is that the only numbers that you can consider "infinites" are transcendentals, but that any attempt to understand infinite values this way is more a study of $\mathbb{Q}[x]$ rather than the $p$-adics.

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I really appreciate your answer. I still don't understand why such $\omega \in \mathbb Q_p $ exists. How can I be sure there is at least one transcendental element (over $\mathbb Q$) in $\mathbb Q_p$? (In my defence I've discovered p-adic numbers yesterday evening) –  Lorenzo Cavallina Oct 27 '11 at 16:15
    
I think it is probably misguided to look for answers about "infinite" values in $p$-adics. I can understand the urge, because when written out as infinite sums of powers of $p$, they naturally look "infinite," but that appearance is deceptive. The whole point of $p$-adic values is that powers of $p$ are reinterpreted as "small" - that's essentially the reason that they can't be totally ordered in a way that's compatible with $(\mathbb Q,<)$. –  Thomas Andrews Oct 27 '11 at 16:23
    
The standard argument is to show that $\mathbb Q_p$ is uncountable, and the set of algebraic elements of $\mathbb Q_p$ is countable, so there must exist non-algebraic (aka transcendental) elements of $\mathbb Q_p$. –  Thomas Andrews Oct 27 '11 at 16:25
    
But more specifically, you are missing the point if you think such a representative $\omega$ says anything about infinity. For example, you can similarly totally order $\mathbb{Q}[\pi]\subset \mathbb R$, but we don't think of $\pi$ as an infinite value. This is still just trickery. For example, if $\omega$ is transcendental, then so is $-\omega$, but my construction allows you to choose arbitrarily which one is "positive" infinity. It's not intrinsic to $\omega$, and we can find total orders on $\mathbb{Q}[\omega]$ where $\omega$ is finite, too. –  Thomas Andrews Oct 27 '11 at 16:30
    
OK. I'm convinced about the existence of a transcendental number. Then, of course, it's all crystal-clear about how trickery it is. There is nothing like infinity in $\mathbb Q_p$, just things that behave like infinities under some circumstances (but, on the other hand, also the set of real numbers contains such elements, and as you noticed $\mathbb R [x]\cong\mathbb R[\pi]$). So I think you are right when you say that "any attempt to understand infinite values this way is more a study of $\mathbb Q[x]$ rather than the p-adics" –  Lorenzo Cavallina Oct 27 '11 at 16:52
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