I have read that the value group of the $p$-adic complex numbers ($\mathbb{C}_p$) and the algebraic closure of $\mathbb{Q}_p$ is $\mathbb{Q}$. I understand why the value group of $\mathbb{Q}_p$ is $\mathbb{Z}$, but fail to see how to relate this to the other two fields. More generally what is the relationship between the value group of a Non-Archimedean (NA) field and its algebraic closure.
Value group of the $p$-adic complex numbers
algebraic-number-theorynumber theoryp-adic-number-theoryvaluation-theory
Related Solutions
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.
The algebraic closure has countably infinite dimension over $\mathbb Q_p$, and therefore (by the Baire category theorem) is not metrically complete. (Except the case $\mathbb Q_\infty = \mathbb R$, where the algebraic closure has finite dimension, and is metrically complete.)
explanation
In $\mathbb Q_p$, let $x_n$ be a solution of $X^{n}=p$. Then $\{x_2,x_3,x_4,\dots\}$ is linearly independent over $\mathbb Q_p$. But we still need a proof that the algebraic closure does not have uncountable dimension.
Torsten Schoenberg provided the missing part:
Countable dimension follows from Krasner / Hensel and compactness of $\mathbb Z_p$ which shows for each $d \in \mathbb N$
, $\mathbb Q_p$
only has only countably many (actually finitely many) extension of degree $d$ in $\overline{\mathbb Q_p}$. I doubt there is a more elementary argument for that.
former explanation
We see in Incompleteness of algebraic closure of p-adic number field, from $9$ years ago question that it is incorrect...
How about an example? In $\mathbb Q_2$, the partial sums of the series $$ \sum_{n=1}^\infty 2^{n+1/n} $$ belong to $\overline{\mathbb Q_2}$, but the sum of the series does not. The partial sums form a Cauchy sequence with no limit in $\overline{\mathbb Q_2}$.
Why does this sum not exist in $\overline{\mathbb Q_2}$ ?
It is not trivial, but interesting: any $x$ which is algebraic of degree $n$ over $\mathbb Q_2$ has a unique series expansion $$ x = \sum 2^{u_j} $$ where $u_j \to \infty$ (unless it is a finite sum) and all $u_j$ are rationals with denominator that divides $n!$. (Maybe divides $n$ in fact?) But the series expansion in this example has arbitrarily large denominators.
Best Answer
To elaborate on stillconfused's comment, note that for any finite extension $K/\mathbb Q_p$, the value group is $\frac1{e(K)}\mathbb Z$, where $e(K)$ is the ramification degree. Thus, since the algebraic closure $\overline{\mathbb Q}_p$ is the union $\bigcup_{K/\mathbb Q_p}K$ where $K/\mathbb Q_p$ runs over all the finite extensions, the value group for $\overline{\mathbb Q}_p$ is $\bigcup_{K/\mathbb Q_p}\frac1{e(K)}\mathbb Z$. Thus, it suffices to check for each integer $n>0$ there is some field $K/\mathbb Q_p$ with $e(K)=n$. Note that $K=\mathbb Q_p(p^{1/n})$ is such an example.
Now, since $\mathbb C_p$ is the $p$-adic completion of $\overline{\mathbb{Q}}_p$, its character group is also $\mathbb Q$.