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
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


*

*$G$ is an abelian subgroup of $\mathfrak{S}_{n}$, the symmetric group on $n$-letters. 

*$d_{i}\in\mathcal{O}_{\mathbb{K}}$

*$\frac{d_{i}}{\sigma\left(d_{i}\right)}$ is a unit in $\mathcal{O}_{\mathbb{K}}$ $\forall \sigma\in G$

*$d_{1}$ is a unit in $\mathcal{O}_{\mathbb{K}}$

*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.
 A: 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.
A: 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.
