3

Let $p$ be an odd prime and let $p^*=(-1)^{\frac{p-1}{2}}p$. Let $n\geqslant 1$. Show that $\mathbb{Q}(\sqrt{p^*})$ is the only quadratic extension of $\mathbb{Q}$ contained in $\mathbb{Q}(\zeta_{p^n})$. (Hint. Consider the Gauss sum $g=\sum_{a=1}^{p-1}\left(\frac{a}{p}\right)\zeta_p^a$, where $\left(\frac{a}{p}\right)$ is the Legendre symbol.

My try: Since $g^2=\left(\frac{-1}{p}\right)p=p^*$, we have $\sqrt{p^*}=g\in\mathbb{Q}(\zeta_p, \zeta_p^2, \dotsc, \zeta_p^{p-1})=\mathbb{Q}(\zeta_p)$. I want to know what does $\mathbb{Q}(\zeta_{p^n})$ do and how to prove the uniqueness?

Any hints are appreciated, thanks in advance.

Ryze
  • 1,136
  • For $n=1$ see here. – Dietrich Burde Oct 14 '19 at 08:13
  • Is there an essential difference between "case $\mathbb{Q}(\zeta_{p})$" and "case $\mathbb{Q}(\zeta_{p^n})$"? – Ryze Oct 14 '19 at 08:40
  • Maybe it is a printing mistake. Could you please give a more detailed explanation? – Ryze Oct 14 '19 at 09:20
  • Thanks for your help! I still need some time... – Ryze Oct 14 '19 at 09:52
  • 1
    @DietrichBurde I think we only get other 2-torsion in the Galois group of $\Bbb{Q}(\zeta_{m})$ iff $m$ has more odd prime factors or is a power of two $>4$. When $p$ is an odd prime $Gal(\Bbb{Q}(\zeta_{p^n})/\Bbb{Q})$ is cyclic of order $(p-1)p^n$, so has a unique element of order two. And, more relevantly, a unique subgroup of index two. – Jyrki Lahtonen Oct 14 '19 at 12:34

2 Answers2

4

One could note that there is a counterexample for $p=2$ and, for example, $n=4$. Then $\Bbb Q(\zeta_{p^n})$ contains three different quadratic subfields:

Determining subfields of cyclotomic field extension

Otherwise the above answer and the comment by Jyrki show that the claim is true for $p>2$ and $n\ge 1$.

Dietrich Burde
  • 130,978
4

The Galois group of $\mathbb{Q}(\zeta_{p^n})/ \mathbb{Q}$ is isomorphic to $(\mathbb{Z}/p^n\mathbb{Z})^\times$, which is cyclic for all $n\geq 1$ since $p$ is odd.

In particular, it has a unique subgroup of order $\varphi(p^n)/2$, hence the Galois extension $\mathbb{Q}(\zeta_{p^n})/\mathbb{Q}$ has a unique quadratic subextension, by Galois correspondence.

GreginGre
  • 15,028