# Subject to some conditions, is it possible to conclude a subfield of an abelian extension generated by a unit is a cyclic extension

My research is mostly in the area of modular categories. In the course of my research I came across a constraining set of number theoretic conditions that I'd like to exploit. It has been pointed out that several of the conditions seem a bit odd from a number theoretic point of view (which is perhaps why my attempts to find help in the literature have been fruitless), never-the-less they are what I have to work with. Discussing the source of the conditions would, I feel, take me to far afield from the question but if you're interested the major sources are

On the Classification of Modular Tensor Categories

On Formal Codegrees of Fusion Categories

If $\mathbb{K}:=\mathbb{Q}\left(d_{1}, d_{2},\ldots, d_{n}\right)$ is an abelian extension of $\mathbb{Q}$ with Galois group $G=Gal\left(\mathbb{K}/\mathbb{Q}\right)$ and ring of integers $\mathcal{O}_{\mathbb{K}}$ such that

1. $G$ is an abelian subgroup of $\mathfrak{S}_{n}$, the symmetric group on $n$-letters.
2. $d_{i}\in\mathcal{O}_{\mathbb{K}}$
3. $\frac{d_{i}}{\sigma\left(d_{i}\right)}$ is a unit in $\mathcal{O}_{\mathbb{K}}$ $\forall \sigma\in G$
4. $d_{1}$ is a unit in $\mathcal{O}_{\mathbb{K}}$
5. There is an element $\tau\in G$ such that

a. $\tau\left(d_{1}\right)\neq d_{1}$.

b. $\displaystyle{\prod_{1\leq a\leq ord\left(\tau\right)}}\tau^{a}\left(d_{1}\right)=\pm1$

c. $\tau$ induces a permutation $\hat{\tau}\in\mathfrak{S}_{n}$ such that $d_{1}\tau\left(d_{i}\right)=\pm d_{\hat{\tau}\left(i\right)}$.

I'd really like to understand $\mathbb{Q}\left(d_{1}\right)$ in some reasonable way.

The thing that jumped out at me was that if $\mathbb{Q}\left(d_{1}\right)$ was a cyclic extension of $\mathbb{Q}$ with Galois group $\langle\tau\rangle$ then $$\displaystyle{\prod_{1\leq a\leq ord\left(\tau\right)}}\tau^{a}\left(d_{1}\right)=\pm1$$ would be exactly the condition that $d_{1}$ is a unit in $\mathcal{O}_{\mathbb{Q}\left(d_{1}\right)}^{\times}$.

In light of this, I would really like to conclude that $\mathbb{Q}\left(d_{1}\right)$ is a cyclic extension of $\mathbb{Q}$ with Galois group $\langle \tau\rangle$. I haven't been able to find a counter example in the context of modular categories but perhaps from a number theoretic standpoint this is asking to much. If one cannot conclude that $Gal\left(\mathbb{Q}\left(d_{1}\right)/\mathbb{Q}\right)=\langle\tau\rangle$, what can one say?

As I mentioned above, the number-theory/field theory literature hasn't been very helpful. This could simply be a symptom of not having the correct vocabulary to search it efficiently. For instance $\displaystyle{\prod_{1\leq a\leq ord\left(\tau\right)}}\tau^{a}\left(d_{1}\right)$ looks an awful lot like a norm, but that doesn't seem to be quite what it is, and I'm not really sure what to call it.

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What is the context where this problem is actually arising? What you list seems like a peculiar list of conditions to string together. Do you intend that the $d_i$'s are a full set of ${\mathbf Q}$-conjugates? (If so, then $G$ being abelian implies $K = {\mathbf Q}(d_1)$ since ${\mathbf Q}(d_1)$ is Galois over ${\mathbf Q}$ and thus contains the other $d_i$'s.) There is no content in listing the elements of $G$ as $\sigma_1,\dots,\sigma_m$ since that is never used later. Is $\sigma_1$ just supposed to be some nonidentity element of $G$? –  KConrad Apr 9 '12 at 2:44
@KConrad I realize that the set of of conditions is rather peculiar. The setting I'm working in is that of modular categories. The problem seemed to boil down to a field theory question, a field I'm by no means an expert in, so after some fruitless research I thought I'd check in with some experts. To address your questions: 1. The $d_{i}$ need not be a full set of $\mathbb{Q}$-conjugates. In some situations they are, though in those cases $\mathbb{Q}\left(d_{1}\right)$ is a cyclic extension. 2. $\sigma_{1}$ should be a non-identity element. I'll edit the post to address your concerns. –  Paul Apr 9 '12 at 3:36
I do not quite understand you question: do you ask if a number field satisfying 1.-5. exists or you want to assume it exists and try to understand $\mathbb{Q}(d_1)$? Also, is n a fixed parameter or you can play with it? –  Filippo Alberto Edoardo Apr 9 '12 at 4:26
@Filippo Alberto Edoardo: Sorry for the confusion, this is my first time posting here. The field $\mathbb{K}$ should be taken to exist, I can provide examples if you'd like, though the ones that come to mind are secretly $\mathbb{Q}(d_{1})$. Additionally, $n$ is a fixed parameter. However, if you have thoughts that only work for certain types of $n$, e.g. prime, I'd like to hear them. –  Paul Apr 9 '12 at 5:33
I do not have thoughts, yet – but what puzzles me is that $n=n$: when you say that $G$ is a subgroup of $\mathfrak{S}_n$, I guess you mean that the map sending $\sigma\in G$ to the corresponding permutation of $n$ chosen generators, identifies $G$ with a subgroup. Then $[\mathbb{K}:\mathbb{Q}]$ is a divisor of $n$ (can we say it is $n$, i.e. assume $n$ is minimal?). But then either all the $d_i$'s are conjugated and $\mathbb{K}=\mathbb{Q}(d_1)$ or I do not understand what is their role. They seem to be just random integers, they do not intervene in 1.-5... –  Filippo Alberto Edoardo Apr 9 '12 at 5:48

I'm having a lot of trouble following all of the details, but the following obeys all conditions except 5c, and as I commented above something is wrong with 5c.

Let $K$ be a totally real field with Galois group $\mathbb{Z}/4$. To be concrete, let $\zeta$ be a $17$th root of unity and take the subfield of $\mathbb{Q}(\zeta)$ generated by $\alpha:=\zeta^{1} + \zeta^{4} + \zeta^{-1} + \zeta^{-4}$. Write $\sigma$ for the generator of $\mathbb{Z}/4$: say $\sigma: \zeta \mapsto \zeta^3$. Our $\tau$ will be $\sigma^2$.

Let $L$ be the quadratic subfield of $K$. In our concrete example, $L = \mathbb{Q}(\sqrt{17})$. The unit groups of $K$ and $L$ are $\{ \pm 1 \} \times \mathbb{Z}^3$ and $\{ \pm 1 \} \times \mathbb{Z}$.

Take $u$ a unit of $K$ such that neither $u$ nor $u^2$ is in $L$. Set $d_1 = u/\tau(u)$. By construction, $d_1 \tau(d_1) =1$.

However, I claim that $\mathbb{Q}(d_1) = K$, which is cyclic of order $4$, not of order $2$. The only intermediate subfield is $L$. Suppose for the sake of contradiction that $d_1 \in L$. Then $\tau(d_1) = d_1$ so $d_1^2 =1$ and $d_1 = \pm 1$. But then $u = \pm \tau(u)$ and $u^2 = \tau(u^2)$, contradicting that $u^2 \not \in K$.

So we have now achieved that $d_1 \tau(d_1) = 1$ and that $Gal(\mathbb{Q}(d_1),\mathbb{Q})$ is not $\langle \tau \rangle$. I now just have to add additional $d$'s to make the rest of the conditions hold. Taking $d_1$, $d_2$, $d_3$ and $d_4$ to be the $\sigma$ orbit of $d_1$ works.

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You'll need to excuse my number theoretic naivety, but I'm a bit confused by a few points. I'll have to post them in two comments to make them fit and be (I hope) cogent. First, $L$ is a subfield of $K$ and unit groups are multiplicative subgroups of the ring of integers. So if $u\in\mathcal{O}_{L}^{\times}$ then $u,u^{2}\in\mathcal{O}_{L}^{\times}$. However, $\mathcal{O}_{L}^{\times}\subset L\subset K$ and so $u,u^{2}\in K$. That is, how can one choose a unit $u\in L$ such that neither $u$ nor $u^{2}\in K$? –  Paul Apr 9 '12 at 21:22
If I take $K=\mathbb{Q}(\alpha)$ and $L=\mathbb{Q}(\sqrt{17})$ then $\alpha^3+\alpha^2-5\alpha-1=\frac{1}{2}(-1+\sqrt{17})$ is a generator for $L$. Plugging in $\alpha$ in terms of $\zeta$ and then sending $\zeta\mapsto\zeta^{9}$ (applying $\tau=\sigma^{2}$), I find that $\tau$ fixes $\frac{1}{2}(-1+\sqrt{17})$. Thus $\tau$ is trivial when restricted to $L$. Since $u\in L$, I have $\tau(u)=u$ and $d_{1}=u/\tau(u)=1$. I do find that $\sigma(\sqrt{17})=-\sqrt{17}$ but that is less helpful since $\mathbb{Q}(\frac{u}{\sigma(u)})$ will be a subfield of $L$. –  Paul Apr 9 '12 at 21:30
I think David meant to choose $u \in K$ with neither $u,u^2$ not in $L$. Then everything else seems to hold. –  Felipe Voloch Apr 10 '12 at 0:17
Thanks Felipe! Fixed now. –  David Speyer Apr 10 '12 at 12:11
Excellent, thanks for the clarification. It seems that if we take $d_{2}=\frac{\sigma(u)\sigma^{2}(u)}{u\sigma^{3}(u)}$, $d_{3}=\frac{\sigma(u)}{\sigma^{3}(u)}$, $d_{4}=1$, and $\hat{\sigma}=(1234)$ then we will also have condition 5c. –  Paul Apr 10 '12 at 17:58

It seems like your more general question is "why should $\prod_{a=1}^{\mathrm{ord} \tau} \tau^a(d_1)$ be $\pm 1$ if $\tau$ doesn't generate $\mathrm{Gal}(\mathbb{Q}(d_1))$?" Here is a way to think about that. For simplicity, let $K/\mathbb{Q}$ be totally real, I leave it to you to work out the complex case. Let $U$ be $\mathbb{R} \otimes \mathcal{O}_K^{\times}$. The proof of the Dirichlet unit theorem shows that, as a representation of $G$, $U$ is the regular representation modulo the trivial representation.

The image of $\mathcal{O}_K^{\times}$ in $U$ is a discrete lattice of full rank and the kernel of $\mathcal{O}_K^{\times}\to U$ is the torsion. Since $K$ is totally real, the torsion is just $\pm 1$. Thus, an equality between units which holds in $U$ will also hold up to sign in $\mathcal{O}_K^{\times}$. Let $u$ be the image of $d_1$ in $U$.

The condition that $\prod \tau^a(d_1) = \pm 1$ is then that the element $\sum \tau^a$ in $\mathbb{Z}[G]$ annihilates $u$. In other words, that $U$ has $0$ projection onto the $H$-trivial part of $U$. This is a subspace of $U$ of dimension $|G|-|H|$. CORRECTION This is a subspace of $U$ of dimension $|G| - |G/H|$.

The condition that $\mathrm{Gal}(\mathbb{Q}(d_1), \mathbb{Q})$ be generated by $\tau$ says that the stabilizer of $d_1$, together with $\tau$, generates $G$. Except on some lower dimensional subspaces of the subspace of $U$ above, $d_1$ has trivial stabilizer. So, unless $G = \langle \tau \rangle$, this is not going to happen.

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Thanks. This certainly is a salient question. You're answer is along very different lines than I am used to thinking and is quite novel to me. I think I follow your argument all the way until the last sentence. Two things struck me. First, assuming you mean $H=\langle\tau\rangle$, then $u$ has trivial projection onto the $\tau$-invariant subspace and hence its stabilizer there will be all of $G$, so I guess the subspace you're referring to is the kernel of this projection? Secondly, I'm likely just being dense, but why should there be a subspace of $U$ on which $d_{1}$ has trivial stabilizer? –  Paul Apr 10 '12 at 19:10