If L is the splitting field of $t^{13}-1$, then obviously $L=\mathbb Q(\zeta_{13})$. Now an intermediate field $M=\mathbb Q(\zeta_{13}^{12}+\zeta_{13})$. How to express L in the form $M(\sqrt{d})$, where d is in M?
============ I found $[\mathbb Q(\zeta_{13}):\mathbb Q(\zeta_{13}^{12}+\zeta_{13})]=2$. So $\zeta_3$ must satisfy a quadratic equation over $\mathbb Q(\zeta_3^{12}+\zeta_{13})$.But I got stuck here. How can I find such quadratic equation and thus express L in that form?
Let $\beta = \zeta_{13} + \zeta_{13}^{-1}$. Then $\zeta_{13}$ is a root of the equation
$x^2-\beta x + 1 = 0$
Notice then that the discriminant $\beta^2-4$ is an element of the real subfield whose square root generates the cyclotomic field.