Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


Let $\zeta$ and $\eta$ be $m$th and $n$th primitive roots of unity, respectively, with $(n,m)=1$. I want to prove that $\mathbb{Q}(\eta)\cap{\mathbb{Q}(\zeta)}=\mathbb{Q}$.

I give an answer below

share|cite|improve this question
up vote 4 down vote accepted

Let $\zeta$ and $\eta$ be $m$th and $n$th primitive roots of unity, respectively, with $(n,m)=1$. We may assume $m<n.$

Since $(m,n)=1$, $\zeta\eta$ has order $mn$ in $\langle\mathbb{C},\cdot\rangle$, i.e., $\zeta\eta$ is an $nm$th root of unity, in consequence $[\mathbb{Q}(\zeta\eta):\mathbb{Q}]=\varphi(mn)$.

We have that $(\zeta\eta)^{m}=\eta^m$, but $(n,m)=1$, then $\eta^m$ is a $n$th primitive root of unity and $\eta^m\in{\mathbb{Q}(\zeta\eta)}$, but $\mathbb{Q}(\zeta\eta)$ is an splitting field over $\mathbb{Q}$,then since $\mathbb{Q}(\zeta\eta)$ contains a root of $\Phi_n(x)$; namely $\eta^m$, it follows that $\mathbb{Q}(\zeta\eta)$ contains all roots of $\Phi_n(x)$ in $\bar{\mathbb{Q}}$, in particular $\eta\in{\mathbb{Q}(\zeta\eta)}$.

This proves that $\mathbb{Q}(\eta)\subseteq{\mathbb{Q}(\zeta\eta)}$ and $\mathbb{Q}(\zeta\eta)=\mathbb{Q}(\zeta,\eta)$

Then $\varphi(nm)=[\mathbb{Q}(\zeta,\eta):\mathbb{Q}]=[\mathbb{Q}(\zeta,\eta):\mathbb{Q}(\eta)][\mathbb{Q}(\eta):\mathbb{Q}]=[\mathbb{Q}(\zeta,\eta):\mathbb{Q}(\eta)]\varphi(n),$ but since $\varphi(mn)=\varphi(m)\varphi(n)$, it follows that $$[\mathbb{Q}(\zeta,\eta):\mathbb{Q}(\eta)]=\varphi(m),$$

but $\zeta$ is a root of $\Phi_m(x)$ and $deg{\Phi_m(x)}=\varphi(m)$, thus $\Phi_m(x)$ is irreducible over $\mathbb{Q}(\eta).$

Let $K=\mathbb{Q}(\zeta)\cap{\mathbb{Q}(\eta)}$, since $\Phi_m(x)$ is irreducible over $\mathbb{Q}(\eta)$, $\Phi_m(x)$ is irreducible over $K$, thus $[K(\zeta):K]=\varphi(m)$, but since $\mathbb{Q}\subseteq{K}\subseteq{\mathbb{Q}(\eta)}$, it follows that $K(\eta)=\mathbb{Q}(\eta)$, then

$$\varphi(n)=[\mathbb{Q}(\eta):\mathbb{Q}]=[\mathbb{Q}(\eta):K][K:\mathbb{Q}]=[K(\eta):K][K:\mathbb{Q}]=\varphi(n)[K:\mathbb{Q}],$$ then we must have that $[K:\mathbb{Q}]=1$, i.e, $K=\mathbb{Q}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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