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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?

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2 Answers2

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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.

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  • Yes, $G'$ is a quotient of a cyclic group, and is thus itself cyclic (and hence abelian). – 211792 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 http://www.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
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    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
  • @user1952009 The group of units $(\mathbb{Z}/n\mathbb{Z})^\times$ is also cyclic when $n$ is prime. – 211792 Mar 26 '17 at 21:50
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    @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. – 211792 Mar 26 '17 at 21:54
  • Yes sorry, I was trying to put together each argument. Now I agree with your answer – reuns Mar 26 '17 at 22:24
  • @AustinChristian, your question is really an elementary one, so using a big theorem to solve it is somewhat weird. – Mariano Suárez-Álvarez Mar 27 '17 at 07:32
  • For what it's worth, while Dirichlet's theorem is something of a nuclear weapon here, I don't think it's weird to use it, since it is only being used in a significantly weakened capacity. Indeed, the statement that there are infinitely many primes of the form $km+1$ for any integer $m$ is elementary. An outline of the proof of this weakened form which only uses basic theory of cyclotomic extensions is provided in exercises 14-17 of Section 13.6 of Dummit and Foote. – Alex Wertheim Jun 06 '17 at 04:20
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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).

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