I recently posted a question on MathOverflow (if you're interested it can be found here). While some answers were quickly produced there were a few points that I found confusing. I requested some clarification, but sadly the MathOverflow community has been a bit reticent, so I thought I would try a different forum that seems more amenable to requests for explanations regarding solutions.
I should probably preface this post with the statement that while I'm quite interested in number theory I'm still learning, so I apologize in advance if this is simply some trivial matter that I'm being dense about.
I'd be satisfied either with clarification of the supplied explanation or an alternative way to think about the problem.
The question I asked essentially boiled down to the following situation:
Consider a totally real abelian extension $\mathbb{K}$ of $\mathbb{Q}$ with Galois group $G$, ring of integers denoted by $\mathcal{O}_{\mathbb{K}}$, and unit group denoted by $\mathcal{O}_{\mathbb{K}}^{\times}$. Suppose that $d_{1}\in\mathcal{O}_{\mathbb{K}}^{\times}$ and $\exists \tau\in G$ such that $\displaystyle{\prod_{a=1}^{ord(\tau)}}\tau^{a}(d_{1})=\pm1$. How can this occur if $\mathbb{Q}\left(d_{1}\right)\neq\langle\tau\rangle$? (As an aside, I'm particularly interested in characterizing when, if at all, $\mathbb{Q}\left(d_{1}\right)$ can fail to be a cyclic extension).
David Speyer was kind enough to produce the following novel explanation:
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) $|G|-|G|/|H|$.
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.
I believe that $H=\langle\tau\rangle$. Furthermore, I think that in the last paragraph the subspace being referred to is the kernel of the projection onto the $H$-invariant subspace not the image. The reason being that $d_{1}$ has trivial projection onto the $H$-invariant subspace and so must have stabilizer $G$ on this space.
There are a few things that I don't understand here.
Why is $U$ the regular representation of $G$ modulo the trivial? If I'm understanding things correctly, then I can see how $G$ acts by permutation of the logarithmic embedding of $\mathcal{O}_{\mathbb{K}}^{\times}$. It certainly bears a striking resembelence to the regular representation but I don't see why this is what it must be. The claim is that this fact follows from the proof of Dirichlet's unit theorem, but when I go through that proof I fail to make this particular connection.(sorted out)Why is the dimension of $U$ given by $|G|-|H|$. I feel like this may be tied to a class equation or the orbit-stabilizer theorem but I haven't made much headway there.It should be $|G|-|G|/|H|$, thanks for the clarification David.- I get truly lost when I come to the last two sentences. If there is a subspace on which $d_{1}$ has trivial stabilizer and the action of $G$ commutes with the projection onto this space then I believe I can produce a contradiction. However, I don't have any idea why either of these things should happen.
Assuming these points can be clarified, this seems to imply that if $\mathbb{Q}(d_{1})$ is a cyclic extension with Galois group $\langle\tau\rangle$ then $G=\langle\tau\rangle$. This seems incredibly unlikely to me, though perhaps I just don't have the proper number theoretic intuition. Alternatively, it could be that I've completely missed the point of the last paragraph in the above explanation.
Thanks in advance.