# Finding a minimal polynomial of a root of unity over a field extension

I'm trying to find the minimal polynomial of the seventh root of unity over the field $$Q(i\sqrt{7})$$. I know how to do this over the rationals and have proceeded to finding that $$(x-1)(x^6 +x^5 +x^4 +x^3 +x^2 +x + 1)=0$$ has the seventh root of unity as a root. If I was just doing this over the rationals I'd be done since I can fairly easily prove that the polynomial of degree 6 is irreducible. However, I don't know if it's reducible over this field extension.

• Did you mean $x\color{red}-1$? May 21, 2020 at 4:28
• Yes I did. Thanks for noticing. I've edited the post. May 21, 2020 at 5:37

Let $$\zeta=\exp(2\pi i/7)$$ The Galois group of $$K=\Bbb Q(\zeta)$$ over $$\Bbb Q$$ is cyclic of order $$6$$. It is generated by $$\sigma_3$$ which takes $$\zeta$$ to $$\zeta^3$$. So $$K$$ has a quadratic subfield; the fixed field of $$\left<\sigma^2_3\right>$$. This includes $$\zeta+\sigma_3^2(\zeta)+\sigma_3^4(\zeta)=\zeta+\zeta^2+\zeta^4$$ which turns out to be $$\frac12(-1+i\sqrt7)$$. Therefore $$K\supseteq L=\Bbb Q(i\sqrt7)$$.
The Galois group of $$K/L$$ is of order $$3$$ and is generated by $$\sigma_3$$. The conjugates of $$\zeta$$ are $$\zeta^2$$ and $$\zeta^4$$, and so its minimal polynomial is \begin{align} (X-\zeta)(X-\zeta^2)(X-\zeta^4)&=X^3-(\zeta+\zeta^2+\zeta^4)X^2+ (\zeta^3+\zeta^5+\zeta^6)X-\zeta^7\\ &=X^3-\frac12(-1+i\sqrt7)X^2+\frac12(-1-i\sqrt7)X-1. \end{align}