 Square discriminants in characteristic two

For a separable polynomial $f\in F[T]$ in characteristic two, the Vandermonde determinant of its roots is fixed by every Galois automorphism: root permutations change it only by a sign, and $-1=1$ in $F$. It therefore lies in $F$. Its square, multiplied by the square of the appropriate leading-coefficient power for a nonmonic polynomial, is the polynomial discriminant. Thus being a square does not detect containment of the Galois group in the alternating group. For example $T^2+T+1$ over ${\mathbb{F}}_2$ has discriminant one and Galois group $C_2$, while $A_2$ is trivial. The discriminant criterion for an alternating Galois group requires characteristic different from two.