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Define $\overline{\mathbb{Q}} \subset \mathbb{C}$ to be the subset consisting of all complex numbers which are algebraic over $\mathbb{Q}$. We know that $\overline{\mathbb{Q}}$ is a countable field and that is algebraically closed.

  1. Show that there exists a sequence of finite extensions $E_{0}=Q \subset E_{1} \subset \ldots \subset E_{n} \subset \ldots \overline{\mathbb{Q}}$, i.e. each $E_{i}/E_{i-1}$ is a finite exntesion and $\overline{\mathbb{Q}} = \cup_{n} E_{n}$.
  2. (Using the above) show that for any prime $p$, the $p$-adic absolute value extends to an absolute value on $\overline{\mathbb{Q}}$.

So, I proved 1 (just define $E_{i} = \mathbb{Q} (a_{1}, a_{2},\ldots, a_{i}$, where $\overline{\mathbb{Q}} = \left\{a_{1}, a_{2},\ldots\right\}$ ), but I don't know how to formalize $2$. Of course, using the fact that every nonarchimedean absolute value on a field, extends in at least one way to every finite extension, we get an extension of the p-adic valuation on each $E_{n}$, but I don't see how to end $2$. Perhaps, a continuity argument?


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As you say, you get an extension of the $p$-adic valuation to each $E_n$, and every element of Q-bar is in $E_n$ for some $n$, so you have extended the $p$-adic valuation to Q-bar, haven't you?

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Thanks. I must have been really sleeping :-). – TJIF May 6 '11 at 7:12
The extension of the $p$-adic valuation to $E_n$ is not unique, so there is something still to be prove (i.e. that you can choose the extensions compatibly). – Matt E May 6 '11 at 12:51
@Matt E, sorry, I don't get it. Pick an extension, any extension, from $E_0$ to $E_1$. Now that you have it defined on $E_1$, pick any extension of it from $E_1$ to $E_2$. Now that gives an extension from $E_0$ to $E_2$. Now you've got it on $E_2$, extend it any way you like from $E_2$ to $E_3$. Etc. I figured compatibility would be ensured, since each step is an extension of the previous. – Gerry Myerson May 6 '11 at 13:04
Dear Gerry, You are correct, the maps are surjective! (I was ---stupidly --- thinking of extending from $\mathbb Q$ to $E_n$ each time, rather than from $E_n$ to $E_{n+1}$, hence my insistence on checking compatibility.) Sorry about that! Best wishes, – Matt E May 6 '11 at 13:25

Note that the extension of the $p$-adic valuation to $E_n$ is typically not unique; the possible extensions are in bijection with the number of primes in the ring of integers of $E_n$ lying over the prime $p$.

So for each $n$ there is a (non-empty!) finite set $S_n$ of extensions of the $p$-adic valuation to $E_n$. Restricting an extension from $E_{n+1}$ to $E_n$ gives a map $S_{n+1} \to S_n$. So you have a projective sequence of finite sets $$ \cdots \to S_{n+1} \to S_n \to \cdots \to S_1 \to S_0$$ and you are trying to choose an element from each in a compatible fashion, i.e. you are trying to choose an element in the projective limit. This will be possible if (and only if!) the projective limit is non-empty.

Added: As Gerry points out in (the comments to) his answer, these maps are surjective, and so one can just successively choose an element of each $S_n$ that maps to the previous choice in $S_{n-1}$; this gives the desired extensino.

Earlier unnecessary discussion:

General fact: the projective limit of a projective sequence of non-empty finite sets is always non-empty.

Proof: A special case of a more general fact about projective limits called the Mittag--Leffler property, which is also pretty easy to prove directly (although not completely trivial if you've never thought about this kind of things before).

Added: When the maps $S_{n+1} \to S_n$ are surjective (as they are in our context), this really is trivial; as was already noted above in this special case, just choose a point in each $S_{n+1}$ mapping to the previous choice in $S_n$!

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