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

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?

• Consider the real subfield of the fields obtained by by adding a primitive root of unity to $Q$. What is its group? – Mariano Suárez-Álvarez Mar 26 '17 at 18:54
• @MarianoSuárez-Álvarez but primitive roots of unity are not generally contained in $\Bbb R$ – Omnomnomnom Mar 26 '17 at 19:03
• Let $\sigma \in Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ be the complex conjugate, $R_n$ is the fixed field of $\{Id, \sigma\}$ so it is cyclic ? – reuns Mar 26 '17 at 19:09
• @Omnomnomnom Hence, real subfield. – Starfall Mar 26 '17 at 19:10
• @Omnomnomnom, that is precisely why I said "consider the real subfield of the fields obtained by adding a primitive root of unity" and not "consider the fields obtained by adding a primitive root of unity", maybe ;-) – Mariano Suárez-Álvarez Mar 26 '17 at 19:21

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 $$G':=\text{Gal}(E/\mathbb{Q}) = G/H.$$ 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 $$\text{Gal}(F/\mathbb{Q}) = G'/H' \cong \mathbb{Z}/n\mathbb{Z}.$$ Thanks to Mariano for the hint, and to Watson for linking to this very helpful answer.

• Yes, $G'$ is a quotient of a cyclic group, and is thus itself cyclic (and hence abelian). – Austin Christian Mar 26 '17 at 20:45
• $\mathbb{Q}(\zeta_{n})$ is cyclic iff $\mathbb{Z}_{n}^\times$ is cyclic, that is when $n = 2p^k$ or $n = 4$ – reuns Mar 26 '17 at 21:19
• For the idea that tower of Galois extensions means the Galois group is the direct product of the Galois groups of each extension, see math.uconn.edu/~kconrad/blurbs/galoistheory/galoiscorrthms.pdf . And for any $k$ need to find a $n$ such that $\mathbb{Z}_n^\times/ \mathbb{Z}_2$ has a cyclic subgroup of order $k$ – reuns Mar 26 '17 at 21:25
• You can get by without using Dirichlet's theorem if you use the fact that the structure of the group of units is known. It is not difficult to find for each n an m such that the group of units in Z/mZ contains a cyclic group of order n. – Mariano Suárez-Álvarez Mar 26 '17 at 21:31
• @MarianoSuárez-Álvarez Right, we can use the Chinese Remainder Theorem to write the group of units as a product of groups of units. I'm willing to accept Dirichlet's big gun for the sake of brevity in this case though. – Austin Christian Mar 26 '17 at 21:54

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).

• One of the nicest answers I've seen yet - concise, simple, and easy to follow. The one thing I don't understand is why $(\mathbf Z/qp^r \mathbf Z)^{*} \cong (\mathbf Z/q \mathbf Z)^{*} \times (\mathbf Z/q \mathbf Z)$ – P-S.D Jan 6 '18 at 3:15
• For example if p=3,q=3,r=2, then the order of $(\mathbf Z/qp^r\mathbf Z)^*$ is 18, but on the right hand side we get the order is 2 x 3 = 6. Obviously, 6 does not equal 18 so where have I gone wrong? – P-S.D Jan 6 '18 at 3:51
• Misprint. Should read $Z/p^rZ$. I edit that. – nguyen quang do Jan 6 '18 at 8:45