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Proofs that every field has a unique (up to isomorphism) algebraic closure use some form of the axiom of choice. For uniqueness this is provably necessary: there are models of ZF in which $\mathbb{Q}$ has two non-isomorphic algebraic closures. Existence is trickier, because algebraic closures of many familiar fields can be constructed by hand.

For example, we can construct an algebraic closure of $\mathbb{Q}$ (or any number field) as the algebraic numbers inside $\mathbb{C}$. More generally, algebraic closures for any countable field can be constructed explicitly (enumerate the irreducible polynomials, adjoin a root to the first and enumerate the new field, factor the second over the new field and adjoin a root of the lexicographically first factor,...). As a further example, the function field of an algebraic curve over $\mathbb{C}$ embeds in the field of Laurent series (choose a local parameter at a point), so this field has an algebraic closure sitting inside the field of Puiseaux series.

Are there examples of specific, explicit fields for which no algebraic closure can be constructed in ZF?

There is a similar question on math.SE with no answers. I tried offering a bounty, to no avail.

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Julian Rosen
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    The main problem with "explicit examples" is that most of the examples of non-AC tend to be non-explicit. – Asaf Karagila Mar 17 '16 at 16:16
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    @AsafKaragila I suspect the OP expects an answer analoguous to $\Bbb R$ over $\Bbb Q$ or $\Bbb F_2^\infty$ over $\Bbb F_2$ as examples of explicit vector spaces without bases constructible (in the sense of proving existence) in ZF. – Wojowu Mar 17 '16 at 16:45
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    A rational function field over an unsightly set of variables, such as $\mathbb Q({x_a:a\in\mathcal P(\mathbb R)})$, might do the trick. This is pretty explicit, IMHO. – Emil Jeřábek Mar 17 '16 at 16:46
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    A side comment -- it's fairly well known, but still -- is that the ultrafilter principle or BPIT, which is strictly weaker than AC, suffices to prove existence (and uniqueness up to isomorphism) of algebraic closures. – Todd Trimble Mar 17 '16 at 16:58
  • @ToddTrimble, I am not sure why this is a "side comment". It seems like an answer to me :-) – Mikhail Katz Mar 17 '16 at 16:59
  • @katz: Because it's a side comment. It doesn't answer the question if there is an explicit field whose algebraic closure requires the axiom of choice to exist. – Asaf Karagila Mar 17 '16 at 17:00
  • @Emil: That's a good idea, actually, since it is consistent that there is an amorphous set which is a subset of $\mathcal P(\Bbb R)$. It might actually work. – Asaf Karagila Mar 17 '16 at 17:03
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    Well, to be clear, the word "require" needs some disambiguation: my point is that AC is not a necessary consequence of existence of an algebraic closure. (I would guess that BPIT is necessary as well as sufficient, but our resident expert Asaf could probably say for sure.) However, it is a side comment in the sense that the OP was very clear that his question is about working in ZF, and the comment doesn't answer his question. – Todd Trimble Mar 17 '16 at 17:04
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    Seems like http://mathoverflow.net/questions/46566/is-the-statement-that-every-field-has-an-algebraic-closure-known-to-be-equivalent is closelY related this question. – Benjamin Steinberg Mar 17 '16 at 17:15
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    Ah, so it's open whether existence of an algebraic closure implies BPIT, according to Caicedo's answer. Thanks, @BenjaminSteinberg. – Todd Trimble Mar 17 '16 at 18:00
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    I edited the title to reflect the question in the body. – Julian Rosen Mar 17 '16 at 18:59
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    @EmilJeřábek For any finite ordered set of variables, we can build an algebraically closed field of iterated Puiseax series in those variables. Enlarging the set of variables gives an extension of Puiseax series fields, so if $S$ is a totally ordered set, we can embed $\mathbb{Q}({x_s:s\in S})$ into the directed union of finite iterated Puiseax series fields, which is algebraically closed. Maybe we could choose a set on which a total order cannot be constructed in ZF – Julian Rosen Mar 18 '16 at 00:26
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    @JulianRosen It's consistent with ZF that $\mathcal{P}(\mathbb{R})$ can't be totally ordered. I suspect that's why Emil chose that example. – Jeremy Rickard Mar 18 '16 at 10:11
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    Yes, indeed, that was the reason. $\mathcal P(\mathbb R)$ looks like one of the simplest sets that can’t be proved totally ordered in ZF. – Emil Jeřábek Mar 18 '16 at 12:33
  • *slaps forehead* For some reason I was thinking of $\mathcal{P}(\mathbb{R})$ as finite subsets of $\mathbb{R}$, which can be totally ordered. – Julian Rosen Mar 18 '16 at 14:11
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    (A generalisation of @JulianRosen's comment:) For any ordered (i.e., linearly ordered) field $K$, the ordered finite extensions form a directed system. (The embeddings between them are canonical, because of the order.) Their naturally defined limit (without AC, I think) is a real closed field. Adjoin an imaginary unit $i$ to get an algebraically closed field. – Goldstern Mar 21 '16 at 00:05

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