A few days ago I was recalling some facts about the p-adic numbers, for example the fact that the p-adic metric is an ultrametric implies very strongly that there is no order on $\mathbb{Q}_p$, as any number in the interior of an open ball is in fact its center.
This argument is not correct. For instance why does it not apply to $\mathbb{Q}$ with the $p$-adic metric? In fact any field which admits an ordering also admits a nontrivial non-Archimedean metric.
It is true though that $\mathbb{Q}_p$ cannot be ordered. By the Artin-Schreier theorem, this is equivalent to the fact that $-1$ is a sum of squares. Using Hensel's Lemma and a little quadratic form theory it is not hard to show that $-1$ is a sum of four squares in $\mathbb{Q}_p$.
I know that if you take the completion of the algebraic close of the p-adic completion you get something which is isomorphic to $\mathbb{C}$ (this result was very surprising until I studied model theory, then it became obvious).
I don't mean to pick, but I am familiar with basic model theory and I don't see how it helps to establish this result. Rather it is basic field theory: any two algebraically closed fields of equal characteristic and absolute transcendence degree are isomorphic. (From this the completeness of the theory of algebraically closed fields of any given characteristic follows easily, by Vaught's test.)
So I was thinking, is there a $p$-adic number whose square equals 2? 3? 2011? For which prime numbers $p$?
All of these answers depend on $p$. The general situation is as follows: for any odd $p$, the group of square classes $\mathbb{Q}_p^{\times}/\mathbb{Q}_p^{\times 2}$ -- which parameterizes quadratic extensions -- has order $4$, meaning there are exactly three quadratic extensions of $\mathbb{Q}_p$ inside any algebraic closure. If $u$ is any integer which is not a square modulo $p$, then these three extensions are given by adjoinging $\sqrt{p}$, $\sqrt{u}$ and $\sqrt{up}$. When $p = 2$ the group of square classes has cardinality $8$, meaning there are $7$ quadratic extensions.
How far down the rabbit hole of algebraic numbers can you go inside the p-adic numbers? Are there general results connecting the choice (or rather properties) of $p$ to the "amount" of algebraic closure it gives?
I don't know exactly what you are looking for as an answer here. The absolute Galois group of $\mathbb{Q}_p$ is in some sense rather well understood: it is an infinite profinite group but it is "small" in the technical sense that there are only finitely many open subgroups of any given index. Also every finite extension of $\mathbb{Q}_p$ is solvable. All in all it is vague -- but fair -- to say that the fields $\mathbb{Q}_p$ are "much closer to being algebraically closed" than the field $\mathbb{Q}$ but "not as close to being algebraically closed" as the finite field $\mathbb{F}_p$. This can be made precise in various ways.
If you are interested in the $p$-adic numbers you should read intermediate level number theory texts on local fields. For instance this page collects notes from a course on (in part) local fields that I taught last spring. I also highly recommend books called Local Fields: one by Cassels and one by Serre.
Added: see in particular Sections 5.4 and 5.5 of this set of notes for information about the number of $n$th power classes and the number of field extensions of a given degree.
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 $E_0$:
- 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);
- 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 $2^\omega$!
Best Answer
Note that every such isomorphism of $\Bbb C_p\to\Bbb C$ is actually a $\Bbb Q$-automorphism of $\Bbb C$.
It is consistent that without the axiom of choice there are only two automorphisms of $\Bbb C$, the identity and conjugation. Obviously if $\Bbb C_p$ is a $p$-adic field, such automorphism is neither of the two. Therefore its existence relies on the axiom of choice, and cannot be written explicitly.
Of course if such an embedding of $\Bbb C_p$ does not exist without using the axiom of choice to begin with, then we cannot embed $\Bbb Q_p$ into $\Bbb C$. Otherwise we could have taken the intersection of all algebraically closed subfields of $\Bbb C$ which contain the embedded $\Bbb Q_p$.