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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.

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  • $\begingroup$ Did you mean $x\color{red}-1$? $\endgroup$ – J. W. Tanner May 21 '20 at 4:28
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    $\begingroup$ Yes I did. Thanks for noticing. I've edited the post. $\endgroup$ – HicorySauce May 21 '20 at 5:37
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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}

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