What is the smallest subfield of the complex numbers which has the property that every polynomial of odd degree has a root

Question

It can be shown using the intermediate value theorem that every polynomial of odd degree with real coefficients must have at least one real root. I was just curious, are there any other smaller fields with this property?

Since the set of algebraic numbers is algebraically closed and the real numbers form a field, it is easy to see that $\mathbb{R} \cap \mathbb{A}$ has this property, however I am not sure if this set forms a field, and I am also not sure if it is the smallest.

Does anyone have any insight into this problem?

EDIT: it has been pointed out in the comments that the intersection of two fields is a field so clearly $\mathbb{R} \cap \mathbb{A}$ is a field and thus satisfies the criteria and is in fact smaller than $\mathbb{R}$ (due to the existence of transcendental numbers). Two question still remains though, is this the smallest field with this property?

It has been pointed out that there may not exist such a "smallest" field with this property, however I am not sure how to prove nor disprove the existence of one

Answer 1

Let $\overline{\mathbb{Q}}$ be the algebraic closure of $\mathbb{Q}$, and let $\mathfrak{g} = \operatorname{Aut}(\overline{\mathbb{Q}}/\mathbb{Q})$ be its automorphism group, the absolute Galois group of $\mathbb{Q}$. Then $\mathfrak{g}$ is a profinite group, so Sylow theory applies: there is a maximal pro-$2$-subgroup $\mathfrak{s}$ of $\mathfrak{g}$ and any two such are conjugate in $\mathfrak{g}$. The fixed field $F(\mathfrak{s}) = \overline{\mathbb{Q}}^{\mathfrak{s}}$ has Galois group $\mathfrak{s}$ and thus every finite algebraic extension of $F(\mathfrak{s})$ has degree a power of $2$. This means that $F$ has the property that every odd degree polynomial has a root.

By Sylow theory and then Galois theory these fields are all Galois conjugates over $\mathbb{Q}$ and in particular are abstractly isomorphic. However I believe there will certainly be more than one of them, and I suspect uncountably many.

I need to run before thinking through the last part thoroughly, but if I am not mistaken then any field of characteristic zero in which every odd degree polynomial has a root contains $F(\mathfrak{s})$ for some $\mathfrak{s}$, so as a family these are the minimal fields.