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In an answer to another question I used the fact that $\mathbb{Q}(\zeta_m)\subseteq \mathbb{Q}(\zeta_n)$ if and only if $m$ divides $n$ (here $\zeta_n$ stands for a primitive $n$th root of unity, Edit: and neither $m$ nor $n$ is twice an odd number; see KCd comments below).

More generally, one can show that $\mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\zeta_m)=\mathbb{Q}$ when $\gcd(n,m)=1$. The only proof of this fact that comes to mind uses facts about discriminants of cyclotomic extensions, and the fact that every non-trivial number field extension over $\mathbb{Q}$ ramifies at least at one prime (see, for instance, Washington, "Introduction to Cyclotomic Fields", Chapter 2, Proposition 2.4).

Since the original question that I was trying to answer was somewhat elementary, I was left wondering if there are more elementary proofs of the fact $$\mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\zeta_m)=\mathbb{Q}, \text{ when } \gcd(n,m)=1. $$ By "elementary proof" here I mean some proof that does not involve algebraic number theory results about discriminants, or ramification of primes in rings of integers of number fields.

Can anyone think of an elementary proof?

Thanks!

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  • $\begingroup$ The fact stated in the first sentence is incorrect due to problems at 2: try $m=2$ and $n=1$. You need to use the standard indexing convention that $m$ and $n$ are not two times an odd number. $\endgroup$
    – KCd
    Mar 16, 2012 at 1:27
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    $\begingroup$ A more general formula is ${\mathbf Q}(\zeta_m) \cap {\mathbf Q}(\zeta_n) = {\mathbf Q}(\zeta_{(m,n)})$. This is a lot more subtle than the corresponding composite field formula ${\mathbf Q}(\zeta_{m}){\mathbf Q}(\zeta_n) = {\mathbf Q}(\zeta_{mn})$ because the second formula is actually true over any base field in place of ${\mathbf Q}$ (well, one over which those primitive roots of unity exist) but the first formula is false: $F(\zeta_m) \cap F(\zeta_n)$ need not be $F(\zeta_{(m,n)})$. For example, ${\mathbf F}_3(\zeta_5) \cap {\mathbf F}_3(\zeta_7)$ is ${\mathbf F}_9$, not ${\mathbf F}_3$. $\endgroup$
    – KCd
    Mar 16, 2012 at 1:31
  • $\begingroup$ That composite field formula in my previous comment should have been ${\mathbf Q}(\zeta_m){\mathbf Q}(\zeta_n) = {\mathbf Q}(\zeta_{[m,n]})$. $\endgroup$
    – KCd
    Mar 16, 2012 at 1:39
  • $\begingroup$ @KCd, thanks! I've changed the first paragraph accordingly. $\endgroup$
    – Álvaro Lozano-Robledo
    Mar 16, 2012 at 2:05
  • $\begingroup$ The examples I gave using finite fields, to show $F(\zeta_5) \cap F(\zeta_7)$ need not equal $F$, can be bootstrapped to characteristic 0 using $p$-adics: let $F$ be ${\mathbf Q}_3$ instead of ${\mathbf F}_3$. $\endgroup$
    – KCd
    Mar 16, 2012 at 3:16

2 Answers 2

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$\newcommand{\Q}{\mathbf Q}$This answer assumes that we're willing to use $[\Q(\zeta_n) : \Q] = \varphi(n)$, which is not obvious. A freely available reference is Milne's notes, Lemma 5.9 and Theorem 5.10.

Since $n$ and $m$ are coprime, the proof you gave shows that $\mathbf Q(\zeta_{nm}) = \mathbf Q(\zeta_n, \zeta_m)$, and the totient function satisfies $\varphi(nm) = \varphi(n)\varphi(m)$. Now \[ [\mathbf Q(\zeta_{nm}) : \mathbf Q] = [\mathbf Q(\zeta_n, \zeta_m) : \mathbf Q(\zeta_n)][\mathbf Q(\zeta_n) : \mathbf Q]. \] If we had $\mathbf Q(\zeta_n) \cap \mathbf Q(\zeta_m) \neq \mathbf Q$ then the degree of $\zeta_m$ over $\mathbf Q(\zeta_n)$ would be less than $\varphi(m)$.

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  • $\begingroup$ This is very nice, but it is worth pointing out that here you are using the fact that the degree of $\mathbb{Q}(\zeta_n)$ is $\varphi(n)$, which is not trivial by any means. $\endgroup$
    – Álvaro Lozano-Robledo
    Dec 23, 2011 at 16:02
  • $\begingroup$ @ÁlvaroLozano-Robledo True! But that fact seems necessary to talk about these fields at all. I will add a reference. $\endgroup$
    – Dylan Moreland
    Dec 23, 2011 at 16:13
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    $\begingroup$ @ÁlvaroLozano-Robledo: Dear Alvaro, I don't think it is unreasonable to use the fact about degrees to prove this. As KCd points out in his comment above, the fact you are asking about is not true if you replace $\mathbb Q$ by other fields. He gives counterexamples involving finite fields and $p$-adic fields, but here is another counterexample involving a number field: If $F = \mathbb Q(\sqrt{3})$, then $F(\zeta_4) = F(\zeta_3) \neq F$. If we tried to apply Dylan's argument, it would break down at the point where we had to compute $[F(\zeta_{12}):F]$; we have that ... $\endgroup$
    – Matt E
    Mar 16, 2012 at 3:55
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    $\begingroup$ ... $F(\zeta_{12}) = F(\zeta_4) = F(\zeta_3)$ is of degree $2$ over $F$ rather than of degree $4 = \varphi(12)$. This suggests that Dylan's proof is capturing the key point: when $F = \mathbb Q$, the different cyclotomic extension are "as independent as possible", and we can see this because their degree over $\mathbb Q$ is as large as possible (i.e. the cyclotomic polynomials are irreducible over $\mathbb Q$, whereas e.g. the $12$ cyclotomic polynomial is not irreducible over $\mathbb Q(\sqrt{3})$). Regards, $\endgroup$
    – Matt E
    Mar 16, 2012 at 3:58
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    $\begingroup$ @DylanMoreland Im sorry but I really don't see how you've concluded that the degree is less than $φ(m)$ and I don't see the relevance of the bit of the tower law that you've mentioned $\endgroup$
    – user366818
    May 7, 2018 at 22:04
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This is a complement to Dylan's answer: we prove that $\mathbb{Q}(\zeta_{mn})=\mathbb{Q}(\zeta_m,\zeta_n)$.

We have $(\zeta_m)^{mn}=1^n=1$ and $(\zeta_n)^{nm}=1^m=1$, and so the inclusion $\mathbb{Q}(\zeta_{mn})\supseteq\mathbb{Q}(\zeta_m,\zeta_n)$ is clear. Conversely, $\zeta_{mn}^m$ is a primitive $n$-th root of unity, and $\zeta_{mn}^n$ is a primitive $m$-th root of unity, which gives the reverse inclusion.

I believe this is as elementary as it gets.

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  • $\begingroup$ Yes, I agree that his is elementary. The simplicity of this step is what got me wondering whether there is also such an elementary proof of $\mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\zeta_m)=\mathbb{Q}$, when $\gcd(n,m)=1$. $\endgroup$
    – Álvaro Lozano-Robledo
    Dec 23, 2011 at 19:23

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