Constructing Quartic number fields

Question

Let $Qrt(n)$ be the $n$th quartic field. For $n = 5$, the polynomial $x^4+x^3+x^2+x+1$ defines the quartic field $Qrt(5)$. What is the polynmial $P(x)$ of degree $4$ that defines the $13$th quartic field? In other words, for $Qrt(13)$, what is the smallest degree $4$ polynomial which defines that field? The discriminant of $P(x)$ should be $13^3$. Thanks for help.

Answer 1

I am following LStU's interpretation, about the polynomial generating the quartic subfield of $\mathbb{Q}(\zeta_{13})$. Let $\omega=\exp\left(\frac{2\pi i}{13}\right)$. Since $2$ is a generator for $\mathbb{Z}/(13\mathbb{Z})^*$ and the quartic residues in $\mathbb{Z}/(13\mathbb{Z})^*$ are $\{1,3,9\}$, $$ \alpha = \omega+\omega^3+\omega^9 $$ $$ \beta = \omega^2+\omega^6+\omega^5 $$ $$ \gamma = \omega^4+\omega^{12}+\omega^{10} $$ $$ \delta = \omega^8+\omega^{11}+\omega^{7} $$ are algebraic conjugates and the quartic subfield of $\mathbb{Q}(\omega)$ is generated by the minimal polynomial of $\alpha$, namely $\color{blue}{x^4+x^3+2x^2-4x+3}$. The coefficients of this polynomial are simply found by evaluating the elementary symmetric polynomials of $\alpha,\beta,\gamma,\delta$.