One can show that the statement that every field has an algebraic closure requires the axiom of choice.

However, for almost all "everyday" fields, it seems that one can actually produce an algebraic closure quite explicitly without using AC (by which I mean: one can produce an algebraic extension of the base field, and prove that it is algebraically closed, without using AC). One cannot necessarily prove the uniqueness of the algebraic closure, but one can nevertheless construct one. For instance:

  1. Finite fields;
  2. Fields of formal Laurent series over an algebraically closed field (the algebraic closure being given by the field of Puiseux series);
  3. The field of rational functions over an algebraically closed field: it embeds in the field of formal Laurent series, so one can take its algebraic closure in the field of Puiseux series. More generally, any subfield of a field whose algebraic closure exists in ZF also has an algebraic closure in ZF.
  4. The real numbers. This one I am only 95% sure about. One can obviously construct $\mathbf C$ from $\mathbf R$ without choice, but all of the proofs that $\mathbf C$ is algebraically closed use analysis in some way, and perhaps things might go wrong here. (The analytic input in Legendre's proof that $\mathbf C$ is algebraically closed seems to be only the intermediate value theorem, which doesn't depend on AC. Nevertheless, I'd be happy if an expert could chime in!)
  5. $\mathbf Q$. If (5) is indeed true, then one can construct the algebraic closure of $\mathbf Q$ without AC by taking the algebraic closure of $\mathbf Q$ in $\mathbf C$. Alternatively, and more simply, one can also do this directly by writing down an enumeration of $\mathbf Q[x]$.
  6. $\mathbf Q_p$. This one I am not completely sure about either. One can use Krasner's lemma to prove that $\mathbf Q_p$ has finitely many extensions of a given degree up to isomorphism, and one should be able to use this to write down a countable subset of $\mathbf Q_p[x]$ whose splitting field is an algebraic closure of $\mathbf Q_p$. But perhaps I am missing a subtle point.

Question: Does there exist an explicit example of a field the existence of whose algebraic closure can be proven to depend on AC? A natural candidate might be a field of rational functions in a very large number of variables. However, the less "degenerate" the example is, the better!

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    $\begingroup$ Your first sentence is plain wrong. One cannot show that, because the existence (and uniqueness) of algebraic closures follows from the Boolean Prime Ideal theorem which is weaker than the axiom of choice. $\endgroup$
    – Asaf Karagila
    May 20, 2014 at 23:31
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    $\begingroup$ This is just a thought, and might be totally off base as I know very little about when the axiom of choice is really needed for something, but in the proof that $\mathbf{Q}_p$ has only finitely many totally ramified extensions of a given degree in a fixed algebraic closure, I think one makes use of the compactness of $\mathbf{Z}_p$, which maybe needs the axiom of choice (in the form "a product of compact spaces is compact"). $\endgroup$ May 20, 2014 at 23:37
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    $\begingroup$ @Keenan: The spaces we take product for defining $\Bbb Z_p$ are uniformly well-orderable (these are $\Bbb{Z/p^n Z}$ if I understand the construction), so their product is compact. The proof is the same proof that $2^\Bbb N$ is compact in $\sf ZF$. $\endgroup$
    – Asaf Karagila
    May 20, 2014 at 23:40
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    $\begingroup$ Another subtle point to the 6th item is that the countable union of finite sets need not be countable without the axiom of choice. (Note that for $\Bbb Q[x]$ we have an effective enumeration, so that's fine, but in general it need not be the case.) Whether or not this is true in this case, I don't know, since I don't know enough about $p$-adic number theory. $\endgroup$
    – Asaf Karagila
    May 21, 2014 at 0:48
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    $\begingroup$ In any case, I have to wake up in roughly 3.5 hours, and I have a long day tomorrow. So I'll leave this one for now. But for what it's worth you got me searching thoroughly, and it doesn't seem to have an answer in the literature (at least the $p$-adic case, 1,3,4,5 are indeed true; I don't know enough to say something intelligible about 2). $\endgroup$
    – Asaf Karagila
    May 21, 2014 at 0:50

1 Answer 1


Re: (4). As you are only "95% sure".....In ZF, without analysis, we have

(i). If $X$ is a compact space and $f:X\to [0,\infty)$ is continuous and $y=\inf_{x\in X}f(x)$ then $\exists x\in X\;(f(x)=y).$ Otherwise $\{f^{-1}(y+r,\infty):r>0\}$ is an open cover of $X$ with no finite sub-cover.

(ii).(Edited). $[-r,r]^2$ is compact for positive real $r$. We derive this from (iii) below. [I don't want to re-number all the references in the second half of this A.]

(iii). A closed bounded real interval is compact.

(iv). (Elementary). For $r>0$ the set $R(r)=\{z\in \mathbb C: \max (|Re (z)|, |Im(z)|)\leq r\}$ is homeomorphic to $[-r,r]^2.$ By (ii) and (iii), $R(r)$ is compact.

(v). (Elementary). If $p:\mathbb C\to \mathbb C$ is a non-constant polynomial then $|p(z)|\to \infty$ uniformly as $|z|\to \infty$. And by elementary algebraic means, $|p|$ is a continuous function.

Theorem. A non-constant polynomial $p:\mathbb C\to \mathbb C$ has a zero. PROOF: The case deg $(p)=1$ is trivial, so assume deg$(p)>1.$

By (v) there exists $r>0$ such that $\forall z \in \mathbb C$ \ $R(r)\;( |p(z)|>|p(0)|).$

Let $R(r)$ be as in (iv). Now $R(r)$ is compact so there exists $z_0\in R(r)$ with $|p(z_0)|=\min \{|p(z)|:z\in R(r)\}.$

We have $|p(z_0)|\leq |p(0)|.$ Also $z\not \in R(r)\implies |p(z)|>p(0)|\geq |p(z_0)|.$ Therefore $$|p(z_0)|=\min \{|p(z)|:z\in \mathbb C\}.$$

We now assume $p(z_0)\ne 0$ and obtain a contradiction.

Since deg$(p)>1$ there exists (unique) $k\in \mathbb N$ and $A\ne 0$ and polynomial $q$ such that $$\forall z\;(p(z)=p(z_0)+A(z-z_0)^k(1+(z-z_0)q(z-z_0)).$$ There exists $y$ such that $Ay^k/p(z_0)$ is a negative real number. For such $y,$ we have $A(ys)^k/p(z_0)<0$ for all $s>0.$ Now there exists (small ) $s>0$ such that $$A(ys)^k=-tP(z_0)\text { for some } t\in(0,1)$$ $$\text { and }\quad |ys\cdot q(ys)|<1/2$$ . $$\text {Then }\quad |p(z_0+ys)|=|(1-t)p(z_0)-tp(z_0)(ys\cdot q(ys)|\leq$$ $$ \leq (1-t)|p(z_0)|+t|p(z_0)|/2=$$ $$=(1-t/2)|p(z_0)|<|p(z_0)|$$ contrary to the minimality of $|p(z_0)|.$

The long list of preliminaries was to ensure that AC hadn't been assumed somewhere.

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    $\begingroup$ Heine–Borel gives you sequential compactness. This is not necessarily the same as compactness without countable choice. However, there are direct proofs of compactness of closed intervals on the line. $\endgroup$
    – Asaf Karagila
    Jan 7, 2017 at 9:13
  • $\begingroup$ @AsafKaragila It just occurred to me that $A\times B$ is compact when A,B are compact, without AC , might be a mistake on my part. However in this Q we only need compactness of $[-r,r]$ and $[-r,r]^2$ which we can do without AC. ... For the method for $[-r,r]$, I just mis-named what I was thinking of. I'm about to edit. $\endgroup$ Jan 7, 2017 at 9:35
  • $\begingroup$ Ugh, fine. I went the extra mile and found you those links. math.stackexchange.com/a/750499/622 for the compactness of $[0,1]$ and math.stackexchange.com/a/765018/622 for the product of two compact sets being compact. $\endgroup$
    – Asaf Karagila
    Jan 7, 2017 at 10:30
  • $\begingroup$ @AsafKaragila.Thank you $\endgroup$ Jan 7, 2017 at 10:34
  • $\begingroup$ @AsafKaragila Thank you .I have no problem showing that a closed square region in $\mathbb C$ is closed without AC. The general case ($A\times B$) is interesting and without AC, harder. $\endgroup$ Jan 7, 2017 at 10:38

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