Real Cyclic Extensions of $\mathbb Q$ of given degree

Question

An old qual problem asks us to

Show that for every positive integer $n$, there exists a cyclic extension of $\mathbb{Q}$ of degree $n$ which is contained in $\mathbb{R}$.

A first thought might be towards Kummer theory: we could adjoin an $n^\text{th}$ root of, say, a prime number. But when $n>2$, $\mathbb{Q}$ lacks the full cohort of roots of unity that would make this work. If $n$ is a power of $2$ we can get what we want by adjoining ($\mathbb{Q}$-linearly independent) square roots to $\mathbb{Q}$, and I think some casus irreducibilis things can be done in other degrees ( at least $n=3$ and $n=5$) but a more general $n$ has me stumped.

Could I get a nudge in the right direction on this problem?

Answer 1

This solution uses the hint provided by Mariano Suarez-Alvarez.

Fix a positive integer $n$, and use Dirichlet's theorem on arithmetic progressions to select a prime $p$ with the property that $p\equiv 1(\text{mod } 2n)$. Let $\zeta_p$ be a primitive $p^\text{th}$ root of unity, so that $K=\mathbb{Q}(\zeta_p)$ is a Galois extension with Galois group $G=(\mathbb{Z}/p\mathbb{Z})^\times\cong \mathbb{Z}/(p-1)\mathbb{Z}$. Notice that complex conjugation $\sigma\colon K\to K$ is a $\mathbb{Q}$-automorphism of $K$, so $G$ has an order two subgroup $H\leq G$ generated by $\sigma$. We let $E\subset K$ be the fixed subfield of $H$. Notice that $E\subset \mathbb{R}$, since no non-real element of $K$ is fixed by conjugation. Also, since $G$ is abelian, $H$ is a normal subgroup, so $E/\mathbb{Q}$ is Galois and \begin{equation} G':=\text{Gal}(E/\mathbb{Q}) = G/H. \end{equation} Because $G$ is cyclic, so is this quotient. Now $|G'|=(p-1)/2$ is divisible by $n$, and thus contains a subgroup $H'\leq G'$ of index $n$ (necessarily normal, since $G'$ is abelian). We finally let $F\subset E$ be the fixed field of $H'$ and have \begin{equation} \text{Gal}(F/\mathbb{Q}) = G'/H' \cong \mathbb{Z}/n\mathbb{Z}. \end{equation} Thanks to Mariano for the hint, and to Watson for linking to this very helpful answer.

Answer 2

First decompose $n$ into a product of prime powers, say $n= p_1 ^{r_1}...p_m ^{r_m}$. Then the (additive) cyclic group $\mathbf Z/n\mathbf Z$ is isomorphic to the direct product of the cyclic groups $\mathbf Z/p_i ^{r_i}\mathbf Z$, so it suffices to deal with the particular case $n=p ^{r}$, where $p$ is a prime. For convenience, put $q=p$ if $p$ is odd, $4$ if $p=2$, and introduce the cyclotomic field $F_r=\mathbf Q(\zeta_{qp^r})$ for $r\ge 0$. Since $Gal(F_r/\mathbf Q) \cong (\mathbf Z/qp^r \mathbf Z)^{*} \cong (\mathbf Z/q \mathbf Z)^{*} \times (\mathbf Z/p^r \mathbf Z)$, the subfield $\mathbf B_r$ fixed by $(\mathbf Z/q \mathbf Z)^{*}$ is cyclic of degree $p^r$ over $\mathbf Q = \mathbf B_0$.

Note that the field $\mathbf B_{cyc} :=\cup \mathbf B_r$ is infinite Galois above $\mathbf Q$, with Galois group isomorphic to $\mathbf Z_p $, the additive group of the ring of $p$-adic integers. An extension with such a Galois group is called a $\mathbf Z_p $-extension in (the algebraic part of) Iwasawa theory, see e.g. Washington's "Introduction cyclotomic fields", chapter 13. Any number field $K$ admits a $\mathbf Z_p $-extension, which is $K.\mathbf B_{cyc}$. The celebrated Leopoldt conjecture states that $K$ admits exactly $(1+r_2)$ "independent" (in an obvious sense) $\mathbf Z_p $-extensions. Up to now, it has been proved only for abelian number fields (Brumer's theorem).