 Solution

For the root-discriminant convention in the requested identity, defineThe Vandermonde determinant is $\det V(\alpha )={\prod }_{i<j}({\alpha }_j-{\alpha }_i)$. Squaring removes the signs, provingIt is nonzero because the roots are distinct. It lies in $F$, since its root expression is fixed by every Galois automorphism and $K^G=F$.

If $f$ is not monic and has leading coefficient $c$, the usual polynomial discriminant is $c^{2n-2}{\prod }_{i<j}({\alpha }_i-{\alpha }_j{)}^2$. Thus the printed Vandermonde equality uses the discriminant of the monic normalization $f/c$. The extra factor is itself a nonzero square in $F$, so it does not alter the square-class criterion in the next part.