# 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)

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I think you mean a cube root of unity rather than $j$? $j \sqrt[3]{2}$ is degree 6. –  Hurkyl Jan 26 '13 at 18:51
@Hurkyl : Corrected, thanks. –  Ewan Delanoy Jan 26 '13 at 21:07
I think this boils down to some group theory : if $G$ is a finite group, if $H_1$ and $H_2$ are subgroups of $G$ whose indices are powers of $3$, and if they are not conjugated, can we deduce that the index of $H_1 \cap H_2$ in $G$ is again a power of $3$ ? I feel this shouldn't be true. –  mercio Jan 26 '13 at 22:23
@mercio : but all Galois groups appearing here will be $3$-groups (or profinite pro-$3$-groups in the infinite degree case), so all indices will be powers of $3$ ? –  Ewan Delanoy Jan 27 '13 at 8:55
I don't get how you show that if $M$ is $3$-powerful and doesn't contain $\alpha_1$, $M(\alpha_1)$ is still $3$-powerful. What prevents $P_1$ from getting factored over $M$ into smaller factors whose degrees aren't powers of $3$ ? –  mercio Jan 27 '13 at 11:39

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.

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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.

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It seems to me you’re on the right track. One should try to reformulate the lemma in “finite-degree” terms. –  Ewan Delanoy Jan 27 '13 at 18:10
My plan was to choose a sequence of polynomials that, by adjoining their roots, yields a sequence of fields $\mathbb{M}_{1,n}$ whose union is $\mathbb{M}_1$. Then construct a tower of fields $\mathbb{M}_{2,n} \subseteq \mathbb{M}_2$ by adjoining roots of those same polynomials. Then we can use an inductive argument to show $\mathbb{M}_{1,n} \cong \mathbb{M}_{2,n}$ for all $n$. This requires that the chosen $f_{n+1}$ to be irreducible over $\mathbb{M}_{1,n}$ and to have a root in $\mathbb{M}_2$. I don't think multiple roots would cause a problem, but I'm not sure. –  Hurkyl Jan 27 '13 at 18:16
@EwanDelanoy We can of course simply adjoin the coefficients of $p(t)$ to $\mathbb Q$ and get a finite degree extension over which $p(t)$ must be irreducible. –  JSchlather Jan 27 '13 at 18:26
@Hurkyl Yes, so picking your $f_{n+1}$ is in some sense equivalent to the step I'm stuck at. Perhaps in your construction though it's simpler since you have more control over your tower than I do my maximal element. –  JSchlather Jan 27 '13 at 18:29

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.

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A rigid family of rational polynomials is an interesting thought: I had always considered each polynomial to have coefficients in the previous field. Anyways, I think this argument still suffers from two things related to what's been giving me problems: –  Hurkyl Jan 27 '13 at 20:41
The first is that it's not clear that an arbitrary maximal 3-powerful field is associated to $\mathcal{B}'$: a particular polynomial could potentially wind up being reducible, but not have any roots. My idea to proceed past this is to argue in such a case, it would have to have an irreducible factor of degree $3^j$. Then reorganize the argument to do all degree-3 extensions first, then all degree-9 extensions, then all degree-27 extensions, and so forth. –  Hurkyl Jan 27 '13 at 20:43
The second is that it's not clear that a field associated with $\mathcal{B}$' is actually maximal 3-powerful: it may be the case that there exist irreducible polynomials of degree $3^k$ over the field, but there don't exist any that are rational. –  Hurkyl Jan 27 '13 at 20:44
@Hurkyl : thanks for those two very informative comments. –  Ewan Delanoy Jan 28 '13 at 5:37
@Hurkyl, Ewan I posted a version of my other question on mathoverflow here where a counterexample was found, essentially for $7$-maximal fields. –  JSchlather Feb 5 '13 at 21:50

As you already noted, there exists one such field in which $X^2-2$ has a root and one in which $X^3-2$ has no root ...

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I don’t quite follow you. No $3$-powerful field can contain $\sqrt{2}$, because the degree of $\sqrt{2}$ is $2$ which is not a power of $3$. –  Ewan Delanoy Jan 25 '13 at 15:41