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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!!

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    $\begingroup$ $x^3-2x+1$ is irreducible, it has a real root $\alpha_1$, its discriminant $D=-4(-2)^3-27 \ 1^2= 5 = (\prod_{i < j} (\alpha_i-\alpha_j))^2$, its splitting field is $\Bbb{Q}(\alpha_1,\sqrt{D}) \subset \Bbb{R}$ thus the cubic field $\Bbb{Q}(\alpha_1)$ has 3 real embeddings. For $x^3-3x+1$ it is $D = 81$ thus $\Bbb{Q}(\alpha_1)$ is Galois and totally real. $\endgroup$
    – reuns
    Aug 14, 2019 at 21:47
  • $\begingroup$ See this thread. Close to being a duplicate of that actually. $\endgroup$ Aug 15, 2019 at 4:42

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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 $n$:

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Given a number field of degree $n$ then the number of real embeddings and pairs of complex embeddings $r_1,r_2$ can be anything prior that $r_1+2r_2 = n$.

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.

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