TITLE: Number fields with all degrees equal to a power of three
QUESTION [7 upvotes]: Say  that a number field $\mathbb K$ is $3$-powerful if the degree (over $\mathbb Q$) of every non-rational element of ${\mathbb K}$ is a power of $3$. By Zorn’s lemma, the field $\cal A$ of all algebraic numbers contains maximal $3$-powerful fields. We can also see that there are several different maximal $3$-powerful fields, as for example a maximal $3$-powerful field containing ${\mathbb Q}(2^{\frac{1}{3}})$ will never coincide with a maximal $3$-powerful field containing ${\mathbb Q}(e^{\frac{2\pi i}{3}}2^{\frac{1}{3}})$.
Are any two maximal $3$-powerful fields isomorphic ?
 UPDATE 26/01/2012  If $\mathbb M$ is any maximal $3$-powerful field, then any irreducible $P\in{\mathbb Q}[X]$ of degree $3$ has a root in $\mathbb M$ (otherwise we could extend $\mathbb M$ by adding a root of $P$). 
And for any $n$, we can find a sequence of irreducible polynomials $P_1,P_2, \ldots, P_n \in {\mathbb Q}[X]$, all of degree three, such that if $\alpha_k$ is a root of $P_k$ for each $k$, then $P_k$ is irreducible on ${\mathbb Q}[\alpha_1,\alpha_2, \ldots ,\alpha_{k-1}]$. Then any maximal $\mathbb M$ will contain a subfield isomorphic to ${\mathbb Q}[\alpha_1,\alpha_2, \ldots ,\alpha_{n}]$. 
This already shows that any two maximal $3$-powerful fields ${\mathbb M}_1$ and ${\mathbb M}_2$ contain arbitrarily large isomorphic subfields.
Sub-question : is an isomorphism between $3$-powerful fields  necessarily unique ? (the fields may have  finite or infinite degree over $\mathbb Q$ ; if true for the finite case, it will also hold in the infinite case)

REPLY [2 votes]: Elaborating on Hurkyl’s idea, I believe I have found a proof.
Let ${\sf P}=(P_t)_{t\lt \alpha}$ be a well-ordered family of polynomials with rational coefficients, indexed by an ordinal $\alpha$ (for the application we make here, we shall only need $\alpha$ finite or $\alpha=\omega$ ). 
If a family ${\sf L}=(\lambda_t)_{t\lt \alpha}$ is such that $\lambda_t$ is a root of $P_t$ for every $t \lt \alpha$, I say that $\sf L$ is associated with $\sf P$. I also say that the field ${\mathbb Q}((\lambda_t)_{t\lt \alpha})$ is associated with $\sf P$. Now, I say that $\sf P$ is rigid if for any $\beta \lt \alpha$, $P_{\beta}$ is irreducible over any field associated with $(P_t)_{t \lt \beta}$. Clearly, we have :
 Remark 1.  If ${\sf L}=(\lambda_t)_{t\lt \alpha}$ ${\sf M}=(\mu_t)_{t\lt \alpha}$
and are two different families associated with a rigid family $\sf P$ (with the same index set $\alpha$), then there is a field automorphism $\sigma$ of $\mathbb C$ sending $\sf L$ to $\sf M$ (i.e. $\sigma(\lambda_t)=\mu_t$ for every $t$). 
 Corollary 1.  Let $\sf P$ be a rigid family, and let $Q$ be a polynomial in ${\mathbb Q}[X]$. If $Q$ is irreducible over one field associated with $\sf P$, it is irreducible over all the fields associated with $\sf P$. 
 Corollary 2.  Let $\sf P$ be a rigid family, and let $Q\in {\mathbb Q}[X]$ be a polynomial such that $Q$ is irreducible over all  the fields associated with $\sf P$, as in corollary 1. For a field $\mathbb K$, consider the property : there is a root $\nu$ of $Q$ such that ${\mathbb K}(\nu)$ is $3$-powerful. If this property holds for  one field associated with $\sf P$, it holds for all the fields associated with $\sf P$. 
Now, let ${\cal B}=(B_n)_{n\geq 1}$ be an enumeration of all the non-constant polynomials in ${\mathbb Q}[X]$ whose degree is a power of $3$. Imagine an initially empty box and consider all the $A_k$ in increasing order, one by one. At each step, insert $A_n$ in the box iff the new well-ordered family thus obtained is rigid, and any field associated with it is $3$-powerful, as in corollary 2.
This uniquely defines an infinite subsequence ${\cal B}’$ of $\cal B$. Then, by construction, any maximal $3$-powerful field is associated with ${\cal B}’$, so any two 3-maximal fields are isomorphic by Remark 1.