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.

  • 1
    $\begingroup$ Just to warn you that you refer to GAP 3.4.4 manual (1997). Unless you use GAP 3.4.4, the most recent version is at gap-system.org/Doc/manuals.html and is also supplied with GAP (and searchable from GAP command line). $\endgroup$ Oct 19, 2017 at 14:52
  • $\begingroup$ @AlexanderKonovalov Thanks. $\endgroup$
    – usermath
    Oct 20, 2017 at 4:01

3 Answers 3


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.

  • $\begingroup$ Thanks a lot for the example and link to the theorem. $\endgroup$
    – usermath
    Oct 19, 2017 at 2:42
  • 1
    $\begingroup$ (+1) To add a specific instance of your two examples (for @usermath's amusement): $\sqrt{2} = 2 \cos \pi/4 = \zeta + \zeta^{-1}$, where $\zeta$ is the $8$-th primitive root of unity in the first quadrant. $\endgroup$
    – peter a g
    Oct 19, 2017 at 2:48
  • $\begingroup$ @peterag Great. Thanks. I was reading the Kronecker-Weber theorem and it seems to me that what you said holds in general. $\endgroup$
    – usermath
    Oct 19, 2017 at 2:51

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 $2n-2 > n$.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.