# Existence of totally real number fields of any degree [duplicate]

Are there simple/canonical examples for totally real number fields of degree $n$ for any natural number $n \in \mathbb{N}$? Of course, the question can be reformulated by asking if there exists an irreducible polynomial of degree $n$ with real roots only, i.e. all roots in the algebraic closure $\mathbb{C}$ are real.

Currently, I just started to read about field theory, but in the standard books (and soucres) I could not find an answer to my question.

My question arose from Geometry of Numbers/Analytic Number Theory: Any totally real number field gives rise for an 'admissible lattice' $\Lambda$, that means $\inf_{(m_1,\ldots,m_n) \in \Lambda \setminus \{0\}} |m_1 \cdots m_d| >0$ in standard coordinates, as follows. Under the map $\mathbb{K} \rightarrow \mathbb{R}^n$, $x \mapsto (\sigma_1(x),\ldots,\sigma_n(n))$, where $\sigma_1,\ldots,\sigma_n$ denote all embeddings (which are real since $\mathbb{K}$ is totally real) of the number field $\mathbb{K}$, the image of the ring of integers $\mathcal{O}_{\mathbb{K}}$ is an admissible lattice.

Skriganov has shown in Constructions of Uniform Distributions in Terms of Geometry of Numbers, 1994, that 'admissible lattices' are very uniformly distributed in parallelepipeds. (The remainder term is only of type $\log^{d-1}{r}$.)

Let $p\equiv1\pmod{2n}$ be prime (existence guaranteed by a weak form of Dirichlet's theorem). The field $\Bbb Q(\cos(2\pi/p))$ has degree $\frac12(p-1)$ and is cyclic over $\Bbb Q$. So it has a degree $n$ subextension. This is totally real (and also cyclic over $\Bbb Q$).
• I was looking for an example like that - thanks for your answer! My missing point was that every subfield of a totally real subfield is also totally real. (Let $\mathbb{F} \subset \mathbb{K}$ be a subfield. Any embedding $\sigma \colon \mathbb{F} \rightarrow \mathbb{C}$ can be extended (by the Isomorphism extension theorem) to an embedding $\tilde{\sigma} \colon \mathbb{K} \rightarrow \mathbb{C}$. Thus, if $\mathbb{K}$ is totally real, then also $\mathbb{F}$.) – p4sch Feb 11 '18 at 22:47