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Algebraic elements in $\mathbb{Q}_p$ are well studied and there was a question on mathoverflow about information regarding this.

I want to know following information:

Let $K_p=\overline{\mathbb{Q}}\cap \mathbb{Q}_p$. So we have extensions

$$\mathbb{Q} \subset K_p \subset \mathbb{Q}_p.$$ The first one is algebraic extension (of $\mathbb{Q}$) and next one is transcendental extension (of $K_p$).

Q. What is known about degrees of these extensions: $[K_p:\mathbb{Q}]$ and $[\mathbb{Q}_p:K_p]$?

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Beginner
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    Since $K_p$ is countable, the latter degree is uncountably infinite. I believe the former is countably infinite. – Wojowu Apr 03 '19 at 09:16
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    Some of the answers to my (very) old question here might be also relevant. – Asaf Karagila Apr 03 '19 at 09:19
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    Also, just by the fact that $K_p$ has no algebraic extensions in $\Bbb Q_p$, and by cardinality arguments ($K_p$ is countable and $\Bbb Q_p$ has the cardinality of $\Bbb R$), the second extension has infinite degree. – Asaf Karagila Apr 03 '19 at 09:21
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    @Wojowu: Since $\overline{\Bbb Q}$ is countable, $[K_p:\Bbb Q]$ is at most countably infinite. On the other hand, e.g. for $n$ odd and not divisible by $p$, $\Bbb Q_p$ and hence $K_p$ contains an $n$-th root of $1+p^2$, and for infinitely many such $n$, this root has degree $n$ over $\Bbb Q$, which implies that $[K_p:\Bbb Q]$ cannot be finite. So your belief is correct. – Torsten Schoeneberg Apr 04 '19 at 17:28
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    @Torsten: This fact (On the other hand ...) is interesting; roots of unity in $Q_p$ are $p-1$-th roots only, but roots of other elements can be of arbitrary height (in some sense). May I get to know some reference for (proof of) this fact? – Beginner Apr 05 '19 at 15:18
  • See my answer. I originally wrote $1+p^2$ because I wanted to argue with exp and log and include the case $p=2$, but I've noticed now that's all not needed and we even have $n$-th roots of $1+p$ (and of course many other integers) for an unbounded set of $n$'s. – Torsten Schoeneberg Apr 08 '19 at 00:06

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To summarise some comments into an answer:

A. $[K_p:\Bbb Q]$ is countably infinite.

Namely, since the polynomial ring $\Bbb Q[x]$ is countable, so is $\overline{\Bbb Q}$ and so is $K_p$, so $[K_p:\Bbb Q]$ is at most countably infinite.

To see that it's not finite, notice the following: For $x \in \Bbb Z_p^\times$, $\Bbb Q_p$ contains an $n$-th root of $x$ for all $n$ which are coprime to $p(p-1)$. (E.g. by Hensel's lemma; it is maybe noteworthy that this characterises the set $\Bbb Z_p^\times$ within $\Bbb Q_p$, which can be used to prove e.g. that all automorphisms of $\Bbb Q_p$ are continuous hence trivial.)

Now let's for example look at the element $\alpha:=1+p \in \Bbb Z_p^\times \cap \Bbb Z$. Notice that $\alpha$ can be a perfect $\ell$-th power $k^\ell$ with $\ell$ prime and $k \in \Bbb Q$ (necessarily $k \in \Bbb N$), for at most finitely many primes $\ell_1, ..., \ell_r$, and hence for all $n$ with $gcd(n, 2\ell_1 ... \ell_r) =1$ the polynomial $x^n - \alpha$ is irreducible over $\Bbb Q$. So, for all $n$ with $gcd(n, 2\ell_1 ... \ell_r p(p-1)) =1$, the subextension $\Bbb Q(\sqrt[n]{\alpha}) \vert \Bbb Q$ of $K_p\vert \Bbb Q$ has degree $n$, where $\sqrt[n]{\alpha}$ denotes the $n$-th root of $\alpha$ existing in $\Bbb Q_p$ according to the previous paragraph. Since such $n$ can be arbitrarily high, $[K_p:\Bbb Q]$ must indeed be (countable) infinite.

B. $[\Bbb Q_p: K_p]$ is uncountable, more precisely, has the cardinality of $\Bbb Q_p$ which is $2^{\aleph_0}$.

This is just a cardinality argument using again that $K_p$ is countable, and so would be any countable-dimensional extension of it.

Torsten Schoeneberg
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  • What does mean by "algebraic closure of $\mathbb Q$ in $\mathbb Q_p$" ? Isn't just the set $\bar{\mathbb Q} \cap \mathbb Q_p$ ? – MAS Apr 12 '23 at 23:05
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    @ANG: That's how it is written in the question, although I do not like that (there is no natural superset in which one can take that intersection). I would just define it as ${x \in \mathbb Q_p: x \text{ is algebraic over } \mathbb Q}$. – Torsten Schoeneberg Apr 13 '23 at 02:45
  • Isn't $\bar{\mathbb Q} \subset \mathbb C_p \supset \mathbb Q_p$ ? – MAS Apr 13 '23 at 04:28
  • Depends. How is $\overline{\mathbb Q} \subset \mathbb C_p$? Unless one defines $\overline{\mathbb Q} :={x \in \mathbb C_p: x \text{ is algebraic over } \mathbb Q}$, in which case the definition in my previous comment has less redundancy. Some people define $\overline{\mathbb Q}$ as the algebraic closure of $\mathbb Q$ in $\mathbb C$, which is not naturally identified with $\mathbb C_p$. Or abstractly via polynomials and some kind of axiom of choice. – Torsten Schoeneberg Apr 13 '23 at 04:32
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    Yes, I understand it depends on situation. I found something here – MAS Apr 13 '23 at 04:34