# If $K_1$ and $K_2$ are the splitting fields of $x^n-1$ and $x^m-1$ over $\mathbb{Q}$;(m,n)=1 then $K_1\cap{K_2}=\mathbb{Q}$

Greets

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

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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, $K=\mathbb{Q}$.

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