In our advanced algebra course, we just learned about the notion of the conductor of certain abelian extensions of $\mathbb{Q}$. In the specific example I had in mind, the conductor is simply the smallest value of $m$ such that $\mathbb{Q}(\sqrt{3}) \subset \mathbb{Q}(\zeta_m)$. It turns out that this is equal to the discrimant of $K = \mathbb{Q}(\sqrt{3})$, which, by https://en.wikipedia.org/wiki/Discriminant_of_an_algebraic_number_field, is equal to $12$.
But, I'm wondering how one can rigorously show that this really is the smallest such desired value. I understand that, using one of the elementary cases that motivated Kronecker-Weber Theorem, $\sqrt{3}$ can be written as a sum of two powers of primitive $12$th roots of unity, since $\sqrt{3} = e^{\frac{2 \pi i}{12}} - e^{\frac{10 \pi i}{12}}$. But I don't understand why $12$ is the smallest value $m$ such that we can write $\sqrt{3}$ as a sum of powers of $m$th roots of unity. Is there a robust way to show this ?
Thanks!