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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?

Pearl
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  • An elementary proof can also be found at this question, which is originally for $d=1$, but also discusses the general case. – Dietrich Burde Oct 04 '18 at 14:50
  • @Dietrich Burde I think that is special case for $d=1$. Because totient function is not multiplicative when $(m,n)\neq 1$. Could you give me generalization of the link? – Pearl Oct 05 '18 at 00:03

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I am first guessing here that $\zeta_m$ 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 $\zeta_d \in \Bbb Q(\zeta_m)$ and $\zeta_d \in \Bbb Q(\zeta_n)$ to do $\supseteq$ in the claim. But, those two are analogous to each other, hence I do $\zeta_d \in\Bbb Q(\zeta_m)$, 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, $\Bbb Q(\zeta_m) \cap \Bbb Q(\zeta_n)$ is extension of $\Bbb Q$ by a single element. Let $L$ be that intersection. Then obviously $\Bbb Q \leq L \leq\Bbb Q(\zeta_m)$ and $\Bbb Q \leq L \leq \Bbb Q(\zeta_n)$. Applying chain rules to those two we see that $|L:\Bbb Q|\mid\gcd(m,n)$. Now, if we take $\Bbb Q(\zeta_\ell)$ where $\ell < \gcd(m,n)$ then the intersection is larger then $L$. Hence $L$ has to be exactly $\Bbb Q(\zeta_d)$.

ultralegend5385
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nikola
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  • Thanks for help. But I can't understand that because of $d=gcd(m,n)$, L is extension of $\mathbb Q$ by a single element. – Pearl Oct 04 '18 at 14:34
  • No, $L$ is extension by one element because it is intersection of two extensions of that type – nikola Oct 04 '18 at 14:58
  • I think it looks obvious. But could you give me a concrete proof about L is extension of $\mathbb Q$ by a single element? I can't do that. – Pearl Oct 04 '18 at 23:18
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    @nicola Furthermore I think $[L:\mathbb Q] | gcd(\phi(m),\phi(n))$, not $gcd(m,n)$ – Pearl Oct 04 '18 at 23:32