Subfields of a cyclotomic field

Question

I was doing some study on the programming language GAP and I came to know from here (in the very fist line) that " $\mathbb Q(\sqrt{5})$ is a number field that is not cyclotomic but contained in the cyclotomic field $\mathbb {Q}_5 = \mathbb Q(e^{\frac{2\pi i}{5}})$".

So I think it is an example that says that in general not all subfields of a cyclotomic field are cyclotomic. But a question came across in my mind from here, that I want to ask.

The above example deals with the $5$-th root of unity and $5$ is an odd prime. But I was thinking if we take $\mathbb Q(\theta)$ where $\theta$ is a primitive $2^n$-th root of unity for some $n>1$, will then same statement holds? In other words, I want to know

Are all subfields of $\mathbb Q(\theta)$ cyclotomic where $\theta$ is a primitive $2^n$-th root of unity for some $n>1$ ?

I am not at all good in algebraic number theory, so I am really sorry that I can not show much work from my side. I might be completely wrong or missing something trivial. Sorry again.

I will be really grateful if someone helps me to find an answer to this.

Thanks in advance.

Answer 1

It's certainly not the case that a subfield of $\Bbb Q(\theta)$ where $\theta=\exp(2\pi i/2^n)$ is a primitive $2^n$-th root of unity must be a cyclotomic field. For instance it contains the subfield $\Bbb Q(\cos(2\pi i/2^n))$ which is contained in $\Bbb R$.

By the Kronecker-Weber theorem, the subfields of cyclotomic fields are precisely the finite extensions of $\Bbb Q$ whose Galois group is Abelian. In particular, all quadratic fields $\Bbb Q(\sqrt m)$ for $m\in\Bbb Z$ are contained in cyclotomic fields.

Answer 2

Let $f(x)$ be a polynomial with integer coefficients. For a root of unity $\zeta$, the subfields $\mathbf{Q}[f(\zeta +\bar\zeta )]\subset \mathbf{Q}[\zeta]$ give for most polynomials $f$ give examples non-cyclotomic fields.(of course two different choices of $f$ may lead to the same subfield). These are all reals.

When $\zeta$ is a $p$-th root of unity, for a prime number $p$, using Quadratic Gauss sums Gauss showed (generalizing your example $p=5$) that when $p-1$ is a multiple of $4$, the real quadratic field $\mathbf{Q}[\sqrt p]$ is a subfield of the $p$th cyclotomic field.

Answer 3

Let $\zeta_m$ denote a primitive $m$th root of unity. When $n \geq 3$, you can conclude that there exist noncyclotomic subfields of $\mathbb{Q}(\zeta_{2^n})$ without computing any examples.

The Galois group of $\mathbb{Q}(\zeta_{2^n})$ over $\mathbb{Q}$ is isomorphic to $(\mathbb{Z}/2^n\mathbb{Z})^{\ast} \cong \mathbb{Z}/2^{n-2}\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. The number of subgroups of this last group is the same as the number of subfields of $\mathbb{Q}(\zeta_{2^n})$.

How many subfields of $\mathbb{Q}(\zeta_{2^n})$ are cyclotomic? The only roots of unity in $\mathbb{Q}(\zeta_{2^n})$ are $\pm \zeta_{2^n}^j, j = 0, ... , n-1$. (see A Classical Introduction to Modern Number Theory, Chapter 14, Section 5, Lemma 1). So the only cyclotomic subfields are $$\mathbb{Q} = \mathbb{Q}(\zeta_2), \mathbb{Q}(\zeta_4) = \mathbb{Q}(i), ... , \mathbb{Q}(\zeta_{2^n})$$

$n$ in all. But there are more than $n$ subgroups of $\mathbb{Z}/2^{n-2}\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. There are $n-1$ subgroups of $\mathbb{Z}/2^{n-2}\mathbb{Z}$, and for each such subgroup $H$, you have two subgroups $H \times \{0\}$ and $H \times \mathbb{Z}/2\mathbb{Z}$ of $\mathbb{Z}/2^{n-2}\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. So this gives you at least

$$2(n-1) = 2n -2$$

subfields of $\mathbb{Q}(\zeta_{2^n})$. Since $n > 2 $, we have $2n-2 > n$.