Let $L \mid \mathbb{Q}$ be Galois with Galois group $S_3$. I want to show that if there are no real embeddings $L \to \mathbb{C}$, then there is some unit in the ring of integers whose non-zero powers are not real.

Should I try to look at the unit group of the ring of integers of the three degree 3 subextensions ? The unit ring for the extension of degree 2 has order 6, and consists of the 6th roots of unity. I assume the unit group for the degree 3 cases are going to be free of rank 1.

For the integer units of L, by Dirichlet's unit theorem the unit group has torsion-free rank 2. Can I use this to generate such and element?

I would prefer a simple elementary approach over one that uses Dirichlet's unit theorem.

  1. Let $K$ be a sub-extension of degree $3$. There must be a real place of $K$, so we assume that $K=\Bbb{Q}(\alpha)$ is a subfield of $\Bbb{R}$;
  2. Let $f$ be the minimal polynomial of $\alpha$. If $K$ is totally real, then all the roots of $f$ should be real, and hence the discriminant $D(f)=\prod (x_i-x_j)^2>0$. On the other hand, we have $L=K(\sqrt{D})$, so it indicates that $L$ is also totally real, a contradiction.
  3. Therefore $K$ can only possess one real place. Now by Dirichlet's theorem and some knowledge of the free abelian group, there is a non-torsion unit $u$ inside $L\backslash K$ that is linearly independent of any non-torsion unit $v$ in $K$(means $u^mv^n=1$ if and only if $m=n=0$).
  • $\begingroup$ Why does this particular choice of $u$ make the last statement absurd? We cannot conclude from this that $u$ lies in $K$ can we? $\endgroup$ – Improve Dec 11 '17 at 11:52
  • $\begingroup$ @Improve The non-torsion units in $\mathcal{O}_L$ constitute a free group of rank two, so the powers of $u$ will not intersect with $K$. $\endgroup$ – Phil. Z Dec 11 '17 at 12:33
  • $\begingroup$ How do you make sure there is no $n$ such that $u^n$ generates a free group below? $\endgroup$ – Improve Dec 12 '17 at 14:12
  • $\begingroup$ @Improve ok, I get your point. It is not clear that $u^m\notin K$ for all units $u\in L\backslash K$. I've edited the third part of the proof. Thank you for catching that. $\endgroup$ – Phil. Z Dec 13 '17 at 1:42

A same idea of Phil.Z.
By tracing the proof of Dirichlet's unit Theorem , we have for a Galois extension $L$, there is a unit $u$ such that $\{u^\sigma|\sigma \in \mathrm{Gal}(L/\mathbb{Q})/<\tau>\}$ generate a finite index subgroup in the unit group. Here $\tau$ is the complex conjugation and $\sigma$ runs over the representative elements of $\mathrm{Gal}(L/\mathbb{Q})/<\tau>$.

In your case, choose a such $u$, then $\mathbb{Q}(u,u^\sigma)=L$, since any nontrivial subfield does not admit a unit group with rank $2$. As $L$ is complex, $u,u^\sigma$ can't be both real.


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