Totally real Galois extension of given degree

Solution 1:

Fix $n$, and choose a prime $p$ such that $p\equiv 1$ (mod $2n$). Dirichlet's theorem on primes in arithmetic progressions guarantees the existence of infinitely many such primes.

If $K=\mathbb{Q}(\zeta_p)$, then $K$ is a Galois extension of $\mathbb{Q}$ with $\mathrm{Gal}(K/\mathbb{Q})$ cyclic of order $p-1$. If $F=\mathbb{Q}(\zeta_p+\zeta_p^{-1})$ then $F$ is totally real and $[K:F]=2$, hence $F$ is Galois over $\mathbb{Q}$ with $\mathrm{Gal}(F/\mathbb{Q})$ cyclic of order $\frac{p-1}{2}$.

Now $n$ divides $\frac{p-1}{2}$ by our choice of $p$, hence $G=\mathrm{Gal}(F/\mathbb{Q})$ has a subgroup $H$ of index $n$. Let $E=F^H$, then $E$ is Galois over $\mathbb{Q}$, is totally real, and $[E:\mathbb{Q}]=[G:H]=n$.