Number fields with all degrees equal to a power of three

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)

For the subquestion, isomorphisms between 3-powerful fields need not be unique. For example, any abelian extension of $\mathbb{Q}$ of degree $3^k$ for $k > 0$ would give a counter-example, as its automorphism group provides more than one isomorphism from it to itself.

A concrete example is $\mathbb{Q}(\zeta_7 + \overline{\zeta_7})$ where $\zeta_7$ is a primitive seventh root of unity. This field is the real subfield of the cyclotomic field $\mathbb{Q}(\zeta_7)$, and is thus an abelian extension of $\mathbb{Q}$ of degree 3.

This is more of a comment than an answer, but it's a little bit too long for the comment box.

Let $\Omega$ be the set of automorphisms of $\mathcal A$ and $\mathbb M_1, \mathbb M_2$ two $3$-maximal fields embedded in $\mathcal A$. We may extend any isomorphism of subfields of $\mathcal A$ to an automorphism of $\mathcal A$. In particular the following set is non-empty

$$\mathcal I=\{(\sigma,E,F) \mid E \subset \mathbb M_1\;,F \subset\mathbb M_2,\sigma \in \Omega, \sigma(E)=F\},$$

that is the tuples of automorphisms of $\mathcal A$ that restrict to isomorphims between subfields of $\mathbb M_1$ and $\mathbb M_2$. Order $\mathcal I$ by $(\sigma,E,F) \leq (\sigma^\prime,E^\prime,F^\prime)$ if $\sigma^\prime|_E=\sigma$, $E \subset E^\prime$ and $F \subset F^\prime$. By Zorn's lemma we can find a maximal element $(\varphi,K,L)$. Suppose that $K \neq \mathbb M_1$ then there exists some element $\alpha \in \mathbb M_1 \setminus K$ so $p(t)=\min_{ K}(\alpha,t)$ is irreducible over $K$ and its image under $\varphi$, $q(t)=\varphi(p(t))$ is also irreducible over $L$. Then we may extend $\varphi$ to an automorphism of $K(\alpha)$ to $L(\alpha)$. Now we just need to fit $L(\alpha)$ into $\mathbb M_2$. Which is where I'm stuck, notice that if we can do this it will force $L=\mathbb M_2$ otherwise $\mathbb M_1$ would not be maximal. Inevitably $\mathbb M_2$ contains a $\mathbb Z$-conjugate of $\alpha$, but it's not clear that this $\mathbb Z$-conjugate is also a $K$-conjugate. The question is boils down to is whether or not $q(t)$ is irreducible over $\mathbb M_2$. It seems to me that the following lemma would provide an affirmative or negative answer to this question:

Let $K$ be a subfield of a $3$-maximal field $\mathbb M$. Can a $p(t) \in K[t]$ exist such that $p(t)$ is irreducible and every root of $p(t)$ has a power of $3$ over $\mathbb Q$. Or equivalently can we show that if $p(t) \in K[t]$ and every root of $p(t)$ has order a power of $3$ over $\mathbb Q$ then $p(t)$ has a root in $\mathbb M$.

Edit: There's also the other question. Let $p(t) \in \mathbb Q[t]$ be irreducible if $p(t)$ takes a root in a $3$-maximal field does it take a root in every $3$-maximal field. While this is true for degree $3$ polynomials we know by this question of Ewan's that we can't necessarily construct a $3$-maximal field by a tower of degree $3$ extensions.

Some thoughts on this. Let $p(t) \in \mathbb Q[t]$ be a irreducible polynomial of degree $3^t$ where $t>1$ and $\mathbb M$ a $3$-maximal field containing a root of $p(t)$. Suppose that $\mathbb M_2$ is a $3$-maximal field not containing a root of $p(t)$, then $\mathbb M_2(\alpha)$ where $\alpha$ is any root of $p(t)$ is not $3$-maximal. In particular we can find $\beta \in \mathbb M_2(\alpha)$ whose minimum polynomial is not a power of $3$. Let $L=\mathbb Q(\alpha_1,\dots,\alpha_n)$ where $\alpha_n=\alpha$ and for $i<n$ $\alpha_i \in \mathbb M_2$ be a subfield of $\mathbb M_2(\alpha)$ containing $\beta$. Then the minimum polynomial of $\alpha_n=\alpha$ over $\mathbb Q(\alpha_1,\dots,\alpha_{n-1})$ has degree $>1$, is not a power of $3$ and $p(t)$ takes no roots in $\mathbb Q(\alpha_1,\dots,\alpha_{n-1})$. I feel like this shouldn't be possible, but I'm not quite sure. I've asked another question concerning this last bit since I think it might get more attention on its own.

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.