It is well known that there is only one quadratic subfield of $\mathbb{Q}(\zeta)$, where $\zeta$ is a primitive $p$th root of unity for some prime $p$. I wonder if the same is true for $\mathbb{Q}(\alpha)$ with $\alpha$ a $n$th primitive root of unity (not necessarily prime).
Asked
Active
Viewed 1,196 times
3 Answers
10
It's not true. Consider $\Bbb Q[\zeta_{12}]$. It contains both $\Bbb Q[i]$ and $\Bbb Q[\zeta_3]$, each of which have degree $2$ over $\Bbb Q$.
Robert Shore
- 26,056
- 3
- 21
- 49
2
When $\zeta_p$ is a primitive $p$-th root of unity for an odd prime $p$, the cyclotomic field $K_p=\mathbf Q(\zeta_p)$ admits only one quadratic subfield because $Gal(K_p/\mathbf Q)$ is cyclic (this remains true for $K_4$). But for $K_n$ with composite $n$, you know that $Gal(K_n/\mathbf Q)\cong (\mathbf Z/n)^*$, which is not cyclic in general. To catch the quadratic subfields of $K_n$, you have just to decompose $(\mathbf Z/n)^*$ into a direct product of cyclic components.
nguyen quang do
- 15,189
2
Not true. As I explained here, $\Bbb{Q}(\zeta_n)$ contains $\Bbb{Q}(\zeta_{p^\ell})$ as a subfield for any odd prime power divisor $p^\ell$ of $n$. And that field contains the quadratic subfield $\Bbb{Q}(\sqrt{\pm p})$ with the sign determined by the residue class of $p$ modulo $4$.
And those quadratic subfields are distinct for different primes $p$.
The even prime is, again, odd. If $n$ is divisible by a power of two, we get quadratic subfields $\Bbb{Q}(i)$ as well as both $\Bbb{Q}(\sqrt{\pm2})$ according to whether the power is four or at least $8$ respectively.
The quadratic subfield is unique if and only if $n=4$, a power of an odd prime, or twice a power of an odd prime.
Jyrki Lahtonen
- 140,891
-
2so, as another answer indicated, it's unique when $(\mathbb Z/n)^\times$ is cyclic – J. W. Tanner Nov 23 '19 at 23:59