# Is there a particular field that cannot be proven to have an algebraic closure in ZF?

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.

• 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 supports Monica Mar 17 '16 at 16:46
• 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
• 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
• Seems like mathoverflow.net/questions/46566/… is closelY related this question. – Benjamin Steinberg Mar 17 '16 at 17:15
• @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 0:26