I am trying to find the possible subgroups of the unit group in the ring of integers, $\mathcal{O}_L$. While using Dirichlet’a unit theorem, I started to doubt myself whether such an extension is possible. It is clear to me that complex embeddings come in pair, but I am not sure if I can have a number field, say of degree 3, where all the extensions are real.
I would appreciate any hint or examples.
Thanks!!
On
Argument borrowed from a related question.
Let $n$ be greater than $1$. Then by Dirichlet's theorem on arithmetic progressions, there exists a smallest prime $p(n)$ such that $n$ divides $p(n)-1$.
Now consider the cyclotomic extension $\mathbb{Q}(\zeta_{p(n)})$. We know that the Galois group $G$ of $\mathbb{Q}(\zeta_{p(n)})$ over $\mathbb{Q}$ is cyclic of order $p(n)-1$. Let $H$ be the subgroup of order $\tfrac{p(n)-1}{n}$. Then the fixed field $K_n\stackrel{\text{def}}{=}\mathbb{Q}(\zeta_{p(n)})^H$ has degree $n$ and Galois group $G/H$ cyclic of order $n$.
But Galois extensions of odd degree are totally real. Therefore if $n$ is odd, then $K_n$ has exactly $n$ real embeddings.
Here are minimal defining polynomials for the first few odd $n$:
Let $f\in \Bbb{Z}[x]_{monic}$ be irreducible in $\Bbb{F}_p[x]$, let $g \in \Bbb{Z}[x]$ of same degree with $r_2$ pairs of complex roots and with $r_1$ distinct real roots, then $f+mp g$ is irreducible and since $g+\frac{f}{mp} \to g$ locally uniformly, for $m$ large enough it has $r_1$ real roots and $r_2$ pairs of complex roots. With $\alpha$ one of its roots then $\Bbb{Q}(\alpha)$ has $r_1$ real embeddings and $r_2$ pairs of complex embeddings.