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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

So without further ado...

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
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2 Answers

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
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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
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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
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