Create an order relation over the field of p-adic numbers 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?
 A: 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.
