# On the existence of an algebraically closed field containing other fields

This question arose while I was reading a paper I found in the web. It might be very simple, but I don't know the answer. Let $$\mathbb{R}$$ be the set of real numbers and $$\mathbb{Q}_p$$ the set of all $$p$$-adic numbers.

My question is: how can I construct (or at least guarantee the existence of) an algebraically closed field $$\Omega$$ of characteristic $$0$$ containing both $$\mathbb{R}$$ and $$\mathbb{Q}_p$$ for all primes $$p$$?

More generally, given a finite or infinite family of fields with the same characteristic (and possibly a common subfield), can I prove the existence of such a field? If not, under which conditions does it hold?

Edit: about my background, my level is basic; that is, I know what an algebraically closed field is and basic facts about Field Theory from a basic Galois theory course

It is possible to embed the algebraic closure of $$\Bbb Q_p$$ into $$\Bbb C$$, if you want. We also can consider the completion $$\Bbb C_p$$ of $$\overline{\Bbb Q_p}$$. This field is called the field of $$p$$-adic complex numbers.

The details have been already discussed at this site, e.g., here:

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

The embedding is guaranteed by the axiom of choice.

• Can you explain in more details this construction? (A reference would be appreciated) I mean, once you have embedded each $\mathbb{Q}_p$ in $\mathbb{C}_p$, how can I "glue" them to get an algebrically closed field containig them all? (I think in the same way I can proceed for $\mathbb{C}$. – LBJFS Feb 22 at 13:05
• Just embed each $\Bbb Q_p$ into $\Bbb C$ and you are done. – Dietrich Burde Feb 22 at 13:08
• Sorry, but how is the embedding defined? Once it is defined, does it mean that $\Omega =\mathbb{C}$? – LBJFS Feb 22 at 13:11
• The embedding is given by the axiom of choice. – Dietrich Burde Feb 22 at 13:18
• Yes, there are several standard references. On MO for example, the proof is given in the first six lines of this post. But of course, the problem for many people here is, that it relies on the axiom of choice (and not on much more). – Dietrich Burde Feb 22 at 13:29