I am studying Stillwell's Elements of Algebra. He mentions the following exercise:
If $\gcd(m, n) = 1$ and the regular $m$-gon and the regular $n$-gon are constructible, show that the regular $mn$-gon is constructilbe.
My solution:
We've constructed a regular $k$-gon if know that the real and imaginary components of any of the primitive $k$-th roots of unity are constructible. Let $\zeta_{mn}$ be an $mn$-th primitive root of unity. Then it is easily seen that $\zeta_n := \zeta_{mn}^m$ and $\zeta_m :=\zeta_{mn}^n$ are primitive $m$-th and $n$-th roots of unity.
Now, since $\gcd(m, n) = 1$, there exist integers $M$, $N$ such that $Mm + Nn = 1$. Hence $\zeta_{mn} = \zeta_n^M\;\zeta_m^N$. (Since none of $\zeta$'s are zero, we can raise them with (possibly) negative integers.)
Now observing that the product of constructible complex numbers is constructible, and inverse of a constructible nonzero complex number is also constructible, we conclude that $\zeta_n^M \; \zeta_m^N$, and hence $\zeta_{mn}$ is constructible.
Question: Is my solution correct?