= Square discriminants in characteristic two
{title2=$\operatorname{char}F=2\quad\Longrightarrow\quad\operatorname{disc}(f)\in F^2$}
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.
Back to article page
