Minimal polynomials and degree of field extension I have a cyclotomic field $\mathbb{Q}(\zeta_3)$, and want to know how I can find a minimal polynomial of  $\zeta_{10}$, and $\zeta_{12}$. I have determined that both the polynomials should be of degree 2. I am trying to use a basis argument for both of them. That is: 
$(x+m\zeta_3)(x+n\zeta_3)=0$; here $x=\zeta_{10}$
$(y+p\zeta_3)(y+q\zeta_3)=0$; here $y=\zeta_{12}$
and trying to find some explicit restrictions for m,n and p,q. But I am having trouble doing that.
 A: To expand from my comment, let me say:
It’s easy to see what $\Phi_{12}$ is, since $\Phi_3=X^2+X+1$ is the minimal polynomial for $\zeta_3$, the primitive sixth root $\zeta_6$ is just $-\zeta_3$, with minimal polynomial $X^2-X+1$. Either square root of that is a primitive twelfth root of unity, so $\zeta_{12}^2=\zeta_6=-\zeta_3$, your minimal polynomial for $\zeta_{12}$ over $\mathbb Q(\zeta_3)$ is thus $X^2+\zeta_3$.
For $\zeta_{10}$, the story is more complicated. Just as the minimal $\mathbb Q$-polynomial for $\zeta_5$ is $\Phi_5=X^4+X^3+X^2+X+1$, the minimal $\mathbb Q$-polynomial for $\zeta_{10}$ is $X^4-X^3+X^2-X+1$. Same argument as between the primitive cube and primitive sixth roots of unity. Now comes the advanced stuff. The field of fifth roots of unity (which is also the field of tenth roots) is ramified over $\mathbb Q$ only at $5$, and totally so. You can see this by writing down $\Phi_5(X+1)$, and noticing that it’s Eisenstein for the prime $5$.
On the other hand, the field of cube roots of unity is ramified only at $3$. Too incompatible. Another way of seeing that the tenth roots of unity are still of degree four over the field of cube roots of unity is no notice that any field containing both kinds of roots must contain the thirtieth roots of unity, but the degree of $\zeta_{30}$ over $\mathbb Q$ is $\phi(30)=\phi(2)\phi(3)\phi(5)=1\cdot2\cdot4=8$, not $4$. In short, $\Phi_{10}$ is still the minimal polynomial for $\zeta_{10}$ over the field of cube roots of unity.
