Let $n≥1$ be an integer. I would like to prove (or disprove) the existence of a subfield $K \subset \Bbb R$ such that $K/\Bbb Q$ is Galois and has degree $n$.

It is easy to construct such a subfield for $n=2^k$: one can take $K=\Bbb Q(\sqrt 2, \sqrt 3, \sqrt 5, ..., \sqrt{p_k})\;$ where $p_k$ is the $k$-th prime number.

I found that the irreducible polynomial $x^3-6x+2 \in \Bbb Q[X]$ (Eisenstein's criterion with $p=2$) has three real roots, so its splitting field $K$ satisfies my conditions for $n=3$. Then I'm done with $n=3\cdot 2^k$ thanks to compositum of fields.

Apparently, a Galois extension of odd degree must be totally real. In all cases, I don't know how to construct, in general or in particular cases ($n$ prime for instance), such a totally real Galois extension $K/\Bbb Q$ of degree $n$.

Thank you for your help!

  • $\begingroup$ Possible duplicate of this: math.stackexchange.com/questions/36200/… $\endgroup$ Jul 22, 2016 at 19:08
  • $\begingroup$ @GregoryGrant: cyclotomic extensions are not totally real in general, or am I missing something trivial? $\endgroup$
    – Watson
    Jul 22, 2016 at 19:10
  • $\begingroup$ I thought we agreed that a Galois extension of $\Bbb Q$ of odd degree must be totally real. $\endgroup$ Jul 22, 2016 at 19:14
  • $\begingroup$ @GregoryGrant: I just linked to this page because I was wondering about a converse statement, in some sense, "can you find a totally real Galois extension of odd order $n$?" (even order is not difficult from there). $\endgroup$
    – Watson
    Jul 22, 2016 at 19:18

1 Answer 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 $[E:\mathbb{Q}]=[G:H]=n$.

  • $\begingroup$ Thank you! What argument are you using to conclude "hence $F$ is Galois over $\mathbb{Q}$"? I have an argument using the fact that $K$ is an abelian extension of $\Bbb Q$, so that every subextension (as $F$) is Galois over $\Bbb Q$. $\endgroup$
    – Watson
    Jul 22, 2016 at 19:41
  • 1
    $\begingroup$ Yes, that's it. All subgroups of an abelian group are normal, so all subfields of an abelian extension are Galois over the base field. $\endgroup$ Jul 22, 2016 at 19:42
  • $\begingroup$ Ok! I thought that it followd from "$F$ is totally real and $[K:F]=2$", this is why I asked. $\endgroup$
    – Watson
    Jul 22, 2016 at 19:46

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