Is the algebraic closure of $\mathbb Q$ in the field $\mathbb Q_5$ a number field?

Question

Is the algebraic closure of $\mathbb Q$ in the field $\mathbb Q_5$ of $5$-adic numbers a number field, if yes what is the degree ?

To be honest I don't understand the question, what does it mean to be the algebraic closure of $\mathbb Q$ in the field $\mathbb Q_5$? Maybe $\bar{\mathbb Q}\cap\mathbb Q_5$ ?

on Wikipedia I saw also;

''The algebraic closure of a field $K$ has the same cardinality as $K$ if $K$ is infinite, and is countably infinite if $K$ is finite.''

Therefore If $\mathbb Q$ is not entirely contained in $\mathbb Q_5$, $\bar{\mathbb Q}\cap\mathbb Q_5$ cannot be a finite extension of $\mathbb Q$, is that correct ?

Answer 1

Firstly, to address your concern about being an algebraic closure in an extension, you are correct, $\overline{\Bbb Q}\cap\Bbb Q_5\subseteq\overline{\Bbb Q}_5$ is exactly the set you are looking for.

To see that we are not dealing with a number field, we appeal to the theory of quadratic extensions of $\Bbb Q$. Since $\Bbb Q_5$ is a completion at an odd prime, we know by Hensel's lemma and quadratic reciprocity that for any prime $q\equiv 1\mod 5$ we have that $\sqrt{q}\in\Bbb Q_5$ since

$$\left({q\over 5}\right)=\left({1\over 5}\right)=1.$$

Then Dirichlet's theorem on infinitely many primes in arithmetic progressions implies there are infinitely many distinct primes $q_1,q_2,\ldots $ so that $q_i=5k+1$. Hence $\Bbb Q(\sqrt{q_i})\subseteq \Bbb Q_5$. But then the algebraic closure of $\Bbb Q$ in $\Bbb Q_5$ contains the infinite dimensional extension $\Bbb Q(q_1,q_2,\ldots)$, hence cannot be a number field, since--by definition--number fields are finite, algebraic extensions of $\Bbb Q$.

Answer 2

I think Adam Hughes gave a great answer to your initial question. Let me try to say a bit more about what the algebraic closure of $\mathbb{Q}$ in $\mathbb{Q}_5$ is using algebraic number theory.

Let $p$ be any prime, then we can embed $\mathbb{Q} \subseteq \mathbb{Q}_p$. The algebraic closure $N$ of $\mathbb{Q}$ in $\mathbb{Q}_p$ is then the largest subextension $\mathbb{Q}_p|N|\mathbb{Q}$ that is algebraic over $\mathbb{Q}$. If we fix an algebraic closure $\overline{\mathbb{Q}}_p$ of $\mathbb{Q}_p$, this must contain an algebraic closure of $\mathbb{Q}$ and we can say $\overline{\mathbb{Q}} \subseteq \overline{\mathbb{Q}}_p$. So algebraic extensions of $\mathbb{Q}_p$ give rise to algebraic extensions of $\mathbb{Q}$ (or rather of $N$). But we can go the other way round:

Let $K|\mathbb{Q}$ be any number field and choose a prime $\mathfrak{p}$ of the ring of integers of $K$ over $p$. Then the completion of $K$ at $\mathfrak{p}$ will be a finite extension of $\mathbb{Q}_p$. Different primes may produce different extensions, but if $K$ is a Galois number field, those will be isomorphic.

More precisely, if $K|\mathbb{Q}$ has Galois group $G$, then the corresponding extension $K_\mathfrak{p}|\mathbb{Q}_p$ will be Galois and its Galois group can be identified with the decomposition group of $\mathfrak{p}$, that is $D_\mathfrak{p} = \{\sigma \in G| \sigma \mathfrak{p} = \mathfrak{p}\}$. In the limit, we can therefore identify $Gal(\overline{\mathbb{Q}}_p|\mathbb{Q}_p)$ with the decomposition group of a prime of $\overline{\mathbb{Z}}$ (the algebraic closure of $\mathbb{Z}$ in $\overline{\mathbb{Q}}$). So the choice of an algebraic closure in the beginning was actually a choice of a prime $\mathfrak{P}\subseteq \overline{\mathbb{Z}}$.

A bit of work shows that if $e(\mathfrak{p}|p)$ is the ramification index and $f(\mathfrak{p}|p)$ is the residue degree, then: $$[K_\mathfrak{p}:\mathbb{Q}_p] = e(\mathfrak{p}|p)f(\mathfrak{p}|p).$$

So we can give an explicit description of $N$, the algebraic closure of $\mathbb{Q}$ in $\mathbb{Q}_p$: It is the compositum of all number fields $K$ in which the prime lying under $\mathfrak{P}$ has trivial ramification index and residue degree in $K|\mathbb{Q}$

In particular, $N$ contains all number fields in which $p$ is completely split.

As Adam Hughes pointed out much more straightforward then I did, this includes all quadratic extensions $\mathbb{Q}(\sqrt{d})|\mathbb{Q}$ for which $d\equiv \pm 1 \bmod 5$, and $N$ cannot be a number field.

Answer 3

As if the excellent answers of Adam Hughes and @benh weren’t enough, I’ll give another argument something like that of Adam, but not using Hensel.

For simplicity, let’s take $p\ne2$. Look at the Binomial expansion of $(1+x)^{1/2}=1+\frac12x- \frac18x^2 + +\frac1{16}x^3 - \frac5{128}x^4 +\cdots\>$. You’ll notice that the only denominators are powers of $2$, which are “units” in $\Bbb Q_p$, that is, $|2|_p=1$, so the denominators have nothing to do with the size of the coefficient. The numerators, whatever they are, are ordinary integers, so their $p$-adic absolute values are all $\le1$. Now, if for $x$ you substitute something that’s small, like $pm$ for any integer $m$, you get a Cauchy series out of $(1+x)^{1/2}$, so that the square root of $1+pm$ will always be in $\Bbb Q_p$.

Two remarks and I’ll quit. First, if $q$ is a prime, the series $(1+x)^{1/q}$ has only powers of $q$ in its denominators, so for any $p\ne q$, you’ll have $q$-th roots of $1+pm$. So in $\Bbb Q_2$, you have lots of cube roots of integers. Second, you can also show that $(1+4x)^{1/2}$ has ordinary everyday integers for coefficients, so that can give you what you want also, for $\Bbb Q_2$.