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The proof that every field has an algebraic closure is known to require at least a weak form of AC, the boolean prime ideal theorem. But I recall reading somewhere that for concrete, sufficiently small (say, countable) fields, it is not necessary to invoke choice. Now $\Bbb Q_p$ is not countable, but $2^{\aleph_0}$ is at least manageably small and I'm told that $\overline{\Bbb Q_p}$ should be no larger than this, so perhaps it is possible to show the existence of an algebraically closed field extension to $\Bbb Q_p$.

On a related note, is it possible to define the algebraically closed field extension to $\Bbb Q_p$? That is, is there some definable class and operations that form an appropriate extension of $\Bbb Q_p$? The usual (ZFC) proof shows that there exists a field extension, but this extension is not defined uniquely in the set-theoretic sense but is only unique up to some isomorphism fixed on $\Bbb Q_p$. But perhaps it is possible to collect all these extensions up into a set and work with the equivalence classes instead to get a single object that acts like an extension.

Note that you don't need proper classes or unbounded cardinalities in the above because when extending a field $F$ you can work within the "universe" $\Bbb N\times F^{<\omega}$ since every algebraic extension should have all its elements algebraic over $F$ and hence be of the form "root $k$ of the polynomial $\sum_n^da_nx^n$" where $k\in\Bbb N$ and $(a_1,\dots,a_d)\in F^{<\omega}$.

Mario Carneiro
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    I laughed when I read "but $2^{\aleph_0}$ is at least manageably small". – Asaf Karagila Nov 21 '14 at 20:06
  • @Asaf It's all a matter of perspective. Sure it can't be proven to be less than any (specific) $\aleph_\alpha$, but at least it's not of that mysterious "arbitrarily large" cardinality that always seems to cause problems. Something like "$g_{64}$ is closer to zero than infinity". – Mario Carneiro Nov 21 '14 at 20:09
  • "Much to learn you still have, my old padawan." – Asaf Karagila Nov 21 '14 at 20:12
  • In other news, this is a good question, and I've been wondering about it myself for some time now. One thing I'm almost certain of is that "the algebraic closure" is probably going to be impossible to define. Maybe an algebraic closure is possible. But if $\Bbb Q$ cannot have a unique algebraic closure in $\sf ZF$, I wouldn't expect any field whose closure has infinite degree to have a unique algebraic closure in $\sf ZF$. And that includes $\Bbb Q_p$. – Asaf Karagila Nov 21 '14 at 20:18
  • @Asaf Well, I'm not looking for it to have a unique closure, just a definable one. $\Bbb Q$ has a definable algebraic closure, namely the set of algebraic numbers in $\Bbb C$, but I'm sure that's just cause we got lucky with a simple field. Since we seem to be forced into this abstract construction with $\Bbb Q_p$, you may be right that since we are forced to rely on some kind of uniqueness if we want to pick one definably, it is probably not possible. – Mario Carneiro Nov 21 '14 at 20:24
  • I don't know about you, but when you write "the" in front of an object, it suggests uniqueness. – Asaf Karagila Nov 21 '14 at 20:25
  • @Asaf Fair enough. But consider this interpretation: I want to make a definition $\overline{\Bbb Q_p}=\dots$ and then just use it freely the same way algebraic number theorists do. It is quite the same situation as with $\Bbb R$ - it is not unique in the set-theoretic sense, yet we still call it the real numbers since it is unique up to isomorphism, and we have a definable representative so that we can work with it as if it were a single specific object. – Mario Carneiro Nov 21 '14 at 20:29
  • Also, I'm willing to trade that Yoda quote with "You know nothing, Jon Snow." if you're more comfortable with being a bastard (supposedly, if you reject R+L=J) than you are being a Sith Lord. – Asaf Karagila Nov 21 '14 at 20:29
  • But $\Bbb R$ is in fact unique. The real numbers is not some arbitrary model of $\sf RCF$. It is the unique Dedekind-complete ordered field. You can't say "I want to have a definable algebraic closure, but I'll call it "the algebraic closure" even if there is a completely different one out there". It does not compute. – Asaf Karagila Nov 21 '14 at 20:31
  • @Asaf But that still means unique up to isomorphism, which is just as true for $\overline{\Bbb Q_p}$ - any two representatives of $\overline{\Bbb Q_p}$ are isomorphic. $$$$ What is the issue with just building $F_{n+1}=$ the set of roots of polynomials from $F_n[x]$ and repeat this $\omega$ times starting from $\Bbb Q_p$? – Mario Carneiro Nov 21 '14 at 20:37
  • Ugh. I can't edit my comment. But wasn't it obvious that I meant "up to isomorphism"? Please don't read what I say, read what I meant to say. And clearly I had meant to say "unique up to isomorphism". – Asaf Karagila Nov 21 '14 at 20:44
  • $\mathbb Q_p$ has only finitely many extensions of a given finite degree, and the proof of this fact seems reasonably constructive. My guess is thus that a $\overline{\mathbb Q_p}$ can be constructed without invoking choice. – user8268 Nov 21 '14 at 21:07
  • @user8268: That's new to me. Thanks! (But let me just add that a Russell set, which can be split into countably many pairs but has no countably infinite subset, can be written as the increasing union of finite sets. "Constructive enough" is not necessarily "constructive". Doesn't mean none in this case, since I'm vague on the details, and can't give a proper solution. But just to make my point clear. I hope.) – Asaf Karagila Nov 21 '14 at 21:14
  • @user8268 Do you have a reference to a proof of this fact? Perhaps it will give some clues. – Mario Carneiro Nov 24 '14 at 00:22
  • @Asaf Doesn't the fact that $\Bbb Q_p$ has finitely many extensions of any given finite degree imply that you can construct $\overline{\Bbb Q_p}$ using only dependent choice over finite sets? You just let your relation be proper algebraic field extension, and at each stage you just have to choose one of the finitely many field extensions of index $1$, and then take the union of an infinite path in this relation to be your $\overline{\Bbb Q_p}$. – Mario Carneiro Nov 24 '14 at 07:20
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    @MarioCarneiro It follows from Krasner's lemma, see e.g. http://www.jmilne.org/math/CourseNotes/ANT.pdf the "Krasner lemma and its applications" section – user8268 Nov 24 '14 at 07:37
  • @Mario: If you take into consideration that all sets involved in the process have size continuum, it might be helpful for monkeying around with finite sets. Of course you might still have to do some choosing here and there, though. But this fact can be helpful. – Asaf Karagila Nov 24 '14 at 09:16
  • @Asaf I am short on time ATM, but I think I can see a proof that I've attempted to outline in my answer, by choosing the $\epsilon$-net described in proposition 7.46 carefully and directly plugging it into Krasner's lemma. As long as you don't use the "compact" predicate to forget about your choices, I think it can be made completely constructive. – Mario Carneiro Nov 24 '14 at 09:40
  • @Asaf The original proof is constructive, and so it is possible to construct $\overline{\Bbb Q_p}$. See my proof. – Mario Carneiro Nov 24 '14 at 14:44

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user8268 has supplied the proof (in Algebraic Number Theory, Remark 7.65) that there are finitely many extensions $(K_{n,i}/\Bbb Q_p)_{i=1,\dots,d_n}$ with degree $n$. From this, you can prove the existence of $\overline{\Bbb Q_p}$: Consider the subset relation on finite extensions of $\Bbb Q_p$. This is a directed set, because we can take the union of two finite extensions to get another finite extension. Then the union of all extensions of a given degree $K_n=\bigcup_iK_{n,i}$ is also a finite extension, which contains all roots of $n$-degree polynomials over $\Bbb Q_p$, and $\bigcup_nK_n:=\overline{\Bbb Q_p}$ then contains all roots of polynomials over $\Bbb Q_p$ (and I believe that a theorem then gives that it must also contain roots of polynomials over $\overline{\Bbb Q_p}$.)

This relies on $K_{n,i}$ being a definable set, or at least $\{K_{n,i}:i=1,\dots,d_n\}$ being a definable finite set. Otherwise you need countable choice to finish this one.

But now let's analyze the proof of Remark 7.65 (A local field $K$ of characteristic 0, such as $\Bbb Q_p$ or its finite extensions, has finitely many extensions of degree $n$.) The statement follows from

  1. (Example 7.54) There is a unique unramified extension $K_m$ of $K$ for any $m$, and this extension is definable as $K(\sigma)$ where $\sigma$ is the Frobenius element defined by $\sigma\beta\equiv\beta^q\pmod{\frak p}$ for all $\beta\in B$, the discrete valuation ring in $K_n$, where $\frak p$ is the nonzero prime ideal in $B$. (Not sure I understand all that, but I'll take it as given for now.)
  2. (Proposition 7.61, Krasner's lemma) If $f\in K[X]$ is a monic irreducible polynomial, then for any monic $g$ sufficiently close to $f$ any root $\beta$ of $g$ is close to some root $\alpha$ of $f$, and for that root $K(\alpha)=K(\beta)$.
  3. (Proposition 7.64) The space $(a_1,\dots,a_n)\in{\frak p\times p}\times\dots\times A^\times\pi$ (where $A$ is the ring of integers in $K$) generates the coefficients of the Eisenstein polynomials $X^n+a_1X^{n-1}+\dots+a_n$, which together generate all the totally ramified extensions of $K$.
  4. (Proposition 7.46) The space $A$ ($\supseteq A^\times,\frak p$) is compact because it is complete and totally bounded.
  5. Every extension factors into an unramified extension and a totally ramified extension.

We want to make these statements more constructive in order to extract a definable set of candidate extensions. Now (1) is already definable, and (2) does not involve choice but invokes the existence of a neighborhood of $f$. (4) is a straight application of generalized Heine-Borel and thus involves countable choice, but if we carefully trace the linkage from (4) to (3) to (2) we can eliminate the compactness argument.

Define $T_m=\{\sum_{i=0}^ms_i\pi^i:s_0,\dots,s_m\in S\}^n\cap V$, where $V={\frak p\times p}\times\dots\times A^\times\pi$. This is the $\epsilon$-net (for $\epsilon_m=|\pi^{m+1}|$) used in the proof of Proposition 7.46, except here it is in an $n$-fold cartesian product and restricted to the subset $V\subseteq A^n$ of interest. For the same reasons as detailed there, this is a $\epsilon_m$-net of $V$ in the supremum norm. Furthermore, $T_m\subseteq T_{m+1}$ for each $m$. Now each $x\in V$ is associated with a neighborhood $U_x$ of all functions that are close enough to $X^n+a_0X^{n-1}+\dots+a_n$ to satisfy the conditions of (2).

Now we can mirror the proof of compactness of a closed interval in $\Bbb R$. Suppose $V$ is not compact (w.r.t the $U_x$ collection). Then one of the $T_0$ balls is not compact, and one of the $T_1$ balls inside that ball is not compact, and so on. There is a simple definable order on $T_m$ given an order on $S$ (which we can fix once and for all), so we can pick the first one in each case to get a sequence of points that converges to a constant sequence at some $y$ and thus $y$ is a point in one of the $T_m$ sets, which is a contradiction since it is covered by $U_y$. Therefore there is a finite subcovering of $\{U_x\}_x$ that covers $V$, and we can reduce this to say that for some $m$, $B(\epsilon_m,x)\subseteq U_x$ for each $x\in T_m$.

Thus there is a smallest such $m$, and we can define the set $\{K_{n,i}\}_i$ to be the set of all extensions $K(\sigma_d)(e_{x,k})$ where $d\mid n$, $\sigma_d$ is the Frobenius element for $K_d$ as defined in (1), and $e_{x,k}$ is the $k$-th root of the Eisenstein polynomial defined by $x\in T_m$ for that smallest $m$ (using $n/d$ instead of $n$ in the definition of $T_m$). (Do all of the roots need to be added? I feel like they would all be isomorphic assuming the root is not already in the field.)

This yields a definable $K_n$ field (from the first paragraph), and thus a definable field that is isomorphic to $\overline{\Bbb Q_p}$.

Mario Carneiro
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