Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

  1. 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)
  2. 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.
  3. 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.

share|cite|improve this question
There seems to a typo on line 14: you write "$\mathbb Q(d_1) \neq \langle \tau \rangle$". One side of this purported inequality is a field, the other is a Galois group, so they are certainly not equal, but this is probably not what you meant. Did you mean to write $\mathbb K^{\langle \tau \rangle}$ rather than just $\langle \tau \rangle$? Regards, – Matt E Apr 16 '12 at 4:54

First of all, a correction -- the dimension of the kernel of $\sum \tau^k$ is $|G|-|G/H|$. Sorry.

Might an example help? I'll work with $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$. The Galois group $G$ is $(\mathbb{Z}/2)^2$. Let $\sigma$ be the generator that negates $\sqrt{2}$ and fixes $\sqrt{3}$; let $\tau$ be the generator that fixes $\sqrt{2}$ and negates $\sqrt{3}$.

I think that the unit group is generated by $\pm 1$, $u_1=1+\sqrt{2}$, $u_2=2 + \sqrt{3}$ and $u_3=\sqrt{2} + \sqrt{3}$. It's possible that this is a finite index subgroup of the unit group, but that won't cause trouble even if its true.

Let $\ell_i$ be the image of $u_i$ in $U$. We write $U$ additively.

I claim that $\sigma$ and $\tau$ act on $\ell_1$, $\ell_2$ and $\ell_3$ by the characters $(-1,1)$, $(1,-1)$ and $(-1,-1)$ of $G$ respectively. For example, $\tau(u_2) = u_2^{-1}$ so $\tau(\ell_2) = - \ell_2$. Slightly more subtly, $(\sigma \tau)(u_3) = -u_3$ and the map to $U$ kills the multiplicative factor of $-1$, so $(\sigma \tau) \ell_3 = \ell_3$. You can go through and check all the other cases.

So, as promised, $U$ is a three dimensional representation of $G$ whose character is the regular rep modulo the trivial rep. It is isomorphic to the group of functions $f:G \to \mathbb{Q}$ with $\sum_{g \in G} f(g)=0$. The elements $\ell_i$ correspond to the nontrivial characters of $G$.

Let's look for units with $v \tau(v)=1$. So we want $\ell \in U$ with $(1+\tau) \ell=0$. Thinking in terms of functions $G \to \mathbb{Q}$, we want $f(e) + f(\tau)=f(\sigma) + f(\sigma \tau) = 0$. So this is a two dimensional vector space. A basis for this vector space is $\ell_2$ and $\ell_3$.

If $a_2$ and $a_3$ are nonzero integers, then $a_2 \ell_2 + a_3 \ell_3$ has trivial stabilizer. This is what happens in general. We have a $|G|-|G/H|$ dimensional subspace of $U$ and, except on lower dimensional subspaces of it, the stabilizer is trivial.

Now, if life were simple, this would mean that $v:=u_2^{a_2} u_3^{a_3}$ would have $v \tau(v)=1$ and would have trivial stabilizer, when $a_2$ and $a_3$ are both nonzero. It is true that $v$ has trivial stabilizer, since its image in $U$ does. But we might have $v \tau(v) = -1$ instead, since working down in $U$ we can't see torsion.

However, we can always fix this by passing to a finite index subgroup of the unit group. In this case, we have $v \tau(v)=1$ as long as $a_3$ is even.

To sum up: For any $(a_2, a_3)$ with $a_3$ even, the element $v=(2+\sqrt{3})^{a_2} (\sqrt{2}+\sqrt{3})^{a_3}$ has $v \tau(v)=-1$. But, as long of $a_2$ and $a_3$ are both nonzero, $v$ has trivial stabilizer.

share|cite|improve this answer
Thanks, this does clarify some things. There is still one sticking point for me, though perhaps it's simply the way the argument is worded. The last paragraph of your proof in my post seems to indicate that if $Gal(\mathbb{Q}(d_{1})/\mathbb{Q})=H$ then $G=H$ which is not true e.g., in your example here take $d_{1}=(2+\sqrt{3})$. It seems that there is some assumption about $d_{1}$ being generic ($a_{2},a_{3}\neq0$ in your example). This all seems to indicate that essentially nothing can be said about $Gal(\mathbb{Q}(d_{1})/\mathbb{Q})$ just from the fact that $\prod_{h\in H}h(d_{1})=\pm1$. – Paul Apr 16 '12 at 1:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.