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.

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

1 Answer 1


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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.