# How to Simplify $\mathbb{Q}(\zeta_4,\zeta_8 , \zeta_{12},\zeta_6 )$

If $$mdc(m,n)=1$$ then $$\mathbb{Q}(\zeta_m,\zeta_n )=\mathbb{Q}(\zeta_{mn} )$$,but what if the degree of the roots are divisors and multiples of each other?

I guess that $$\mathbb{Q}(\zeta_4,\zeta_6, \zeta_{12} )=\mathbb{Q}(\zeta_{12})$$

but what about $$\mathbb{Q}(\zeta_{12},\zeta_8 )$$?

• What exactly is your question? Are you wondering whether something in general can be said that’s analogous to or implies the $gcd(m,n) = 1$ case? – Dionel Jaime Oct 31 '18 at 2:43
• Hint: $\zeta_{12}=\zeta_{24}^2$, $\zeta_8=\zeta_{24}^3$. Because $2$ and $3$ are coprime, by Bezout's identity, there exist integers $u, v$ such that $2u+3v=1$. Consequently $$\zeta_{12}^u\cdot\zeta_8^v=\zeta_{24}^{2u}\cdot\zeta_{24}^{3v}=\zeta_{24}^{2u+3v}=???$$ And therefore ????? – Jyrki Lahtonen Oct 31 '18 at 4:52

Start with just adjoining the primitive 4th root of unity and 6th root of unity. If you multiply these two primitive roots together, what power is required to bring you back to $$1$$? What can you conclude?