 Solution

The conjugates of $\omega$ are $\omega ,ldots,{\omega }^{p-1}$. Fromwe obtain, at every nontrivial $p$th root $\alpha$,The product of the roots is one and ${\prod }_{a=1}^{p-1}(1-{\omega }^a)={\Phi }_p(1)=p$. The derivative formula for a polynomial discriminant therefore yields the discriminant of a prime cyclotomic power basis:

Since $p\equiv 1(\mathrm{mod}4)$, the sign is positive. The Vandermonde determinantbelongs to $L$ and satisfies ${\Delta }^2=p^{p-2}$. Hence