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Let $\omega_n$ denote a primitive $n^{th}$ root of unity. If $m$ and $n$ are positive integers with $lcm(m,n)=k$, show that $\mathbb{Q}(\omega_n,\omega_m)=\mathbb{Q}(\omega_k)$.

To start, I am aware that $(\omega_n\omega_m)^k=1$, and so $o(\omega_n\omega_m)|k$, I am working towards showing that $o(\omega_n\omega_m)$ is in fact equal to $l$ (although I don't know this to be true at this point). This would show that $\mathbb{Q}(\omega_n,\omega_m)$ contains a primitive $k-th$ root of unity and thus all of $\mathbb{Q}(\omega_k)$. Any help would be appreciated, I think I am over thinking things (especially regarding the other direction of the inclusion).

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This may become obvious if you write everything additively instead of multiplicatively.

That is, you know the $k$-th roots of unity are a cyclic group under multiplication of order $k$, and so it is isomorphic to the integers modulo $k$ under addition. Use the isomorphism to translate the original problem into a question about integers modulo $k$.

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  • $\begingroup$ Okay I see from this how to show the inclusion that I did not mention, it becomes quite trivial once we have identified <omega_k> with Z/kZ, since it then contains a subgroup of order n and m. The opposite inclusion (the one I did some work on above) is still bugging me though. Btw thanks! $\endgroup$
    – CWsl2
    Nov 5, 2013 at 4:33
  • $\begingroup$ I guess since $<\omega_n,\omega_m>$ is a finite subgroup of the multiplicative group of a field, we know that is necessarily cyclic (and thus abelian), so since it contains an element of order n and an element of order m, it can be shown (I believe I have seen this proof before) that it necessarily contains an element of order $lcm(m,n)=k$, and thus it contains a primitive $k-th$ root of unity, and so this should give us the opposite inclusion $\endgroup$
    – CWsl2
    Nov 5, 2013 at 5:37

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