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Let $l:=\text{lcd}(n,m)$. Show $\mathbb Q(\zeta_n,\zeta_m)=\mathbb Q(\zeta_l)$.

What I got:

$\subseteq$: Let $\text{lcm}(n,m)=r\cdot n=s\cdot m$. Then $$\zeta_l^r=\exp\left(\frac{2\pi i r}{l}\right)=\exp\left(\frac{2\pi i}{n}\right)=\zeta_n$$ and $$\zeta_l^s=\exp\left(\frac{2\pi i s}{l}\right)=\exp\left(\frac{2\pi i}{m}\right)=\zeta_m.$$

$\supseteq$: Bezout's theorem guarantees the existence of $a,b\in\mathbb Z$ with $n\cdot a+m\cdot b=\gcd(n,m)$ so we have $\text{lcm}(n,m)=\frac{n\cdot m}{n\cdot a+m\cdot b}$, and thus $$\displaystyle{\zeta_n^b\cdot\zeta_m^a}=\exp\left(\frac{2\pi i}{n}b\right)\exp\left(\frac{2\pi i}{m}a\right)=\exp\left(2\pi i\frac{bm+an}{nm}\right)=\exp\left(\frac{2\pi i}{l}\right)=\zeta_l.$$

Is my proof correct? And if so, is there anything else do add?

user26857
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Buh
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1 Answers1

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Your proof is fine. Here is a slightly bigger picture:

Let $\Omega_k = \{ z \in \mathbb C : z^k = 1 \} = \langle \zeta_k \rangle$, a subgroup of $\mathbb C^\times$. The inclusion lattice of these subgroups is isomorphic to $\mathbb Z$ under divisibility. In particular, $\Omega_n \vee \Omega_m= \Omega_{\text{lcm}(n,m)}$.

lhf
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