Is there an explicit embedding from the various fields of p-adic numbers $\mathbb{Q}_p$ into $\mathbb{C}$?

For any field of p-adic numbers $\mathbb{Q}_p$, one can construct the field $\mathbb{C}_p$, the metric completion of one of its algebraic completions. By the axiom of choice, we can prove this to be isomorphic to the usual field of complex numbers $\mathbb{C}$. Therefore, since $\mathbb{Q}_p$ embeds into $\mathbb{C}_p$, there must be an embedding of $\mathbb{Q}_p$ into $\mathbb{C}$.

Is there any way to explicitly construct such an embedding, so that given an arbitrary p-adic number, we can rewrite it as a complex number to arbitrary precision?

I'm hoping that this sheds some light on what the (algebraic, non-topological) tensor products of things like $\mathbb{Q}_p \otimes_\mathbb{Q} \mathbb{Q}_q$ and $\mathbb{R} \otimes_\mathbb{Q} \mathbb{Q}_p$ and so on might look like.

(The above post was a lot longer, but it was confusing everyone, so I ditched it and wrote my question much more simply.)

Solution 1:

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 $\Bbb Q_p$.

Solution 2:

If you don't care about topology, then any field of characteristic $0$ and cardinality not greater than that of $\Bbb C$ is a subfield of $\Bbb C$.

In fact, using the Axiom of Choice, for each cardinal $\kappa > \aleph_0$, there is "only one" algebraically closed field of cardinal $\kappa$. (while there are many countable algebrically closed fields of characteristic $0$ : $\overline{\Bbb Q},\overline{\Bbb Q(X)},\overline{\Bbb Q(X,Y)}$ and so on, which can also be realised as subfields of $\Bbb C$).

So if $|K| \le |\Bbb C|$ then $K$ is a subfield of $\overline{K}$, which is either isomorphic to $\Bbb C$ or to one of the countable algebraically closed fields.