Prove that $\mathbb Q(\zeta_m)\cap\mathbb Q(\zeta_n)=\mathbb Q(\zeta_d)$ where $d=\gcd(m,n)$.
I want to solve this problem without Galois theory.
I know only about field extension. For example, algebraic extension, cyclotomic extension, splitting field and algebraic closure.
Can I solve it without Galois theory?
I am first guessing here thet $\zeta_m$ means prime root of $1$ of order $m$, i.e. a generator of set of solutions of $x^m-1=0$. That being said, let's see what we have here. Left side is intersection of fields. Intersection of fields is a field again. Right side is a field. So what we need to prove is equality of fields
Now, field on the right is completely determined by $\zeta_d$ hence we need to show $\zeta_d \in Q(\zeta_m)$ and $\zeta_d \in Q(\zeta_n)$ to do $\supseteq$ in the claim. But, those two are analogous to each other, hence I do $\zeta_d \in Q(\zeta_m)$, where $d = gcd(m,n)$ or $d|m$, which is enough, because there is some $k$ such that $dk = m$ hence $\zeta_m^k = \zeta_d$ hence that's it.
About the converse. Here, it is important that $d = gcd(m,n)$. So, $Q(\zeta_m) \bigcap Q(\zeta_n)$ is extension of $Q$ by a single element. Let $L$ be that intersection. Then obviously $Q \leq L \leq Q(\zeta_m)$ and $Q \leq L \leq Q(\zeta_n)$. Applying chain rules to those two we see that $|L:Q| | gcd(m,n)$. Now, if we take $Q(\zeta_l)$ where $l < gcd(m,n)$ then the intersection is larger then $L$> Hence $L$ has to be exactly $Q(\zeta_d)$