= Solution
First assume $\operatorname{char}F\ne2$. Set $d=\det V(\alpha)\ne0$. If $\Delta_f=b^2$ with $b\in F$, then $(d-b)(d+b)=0$ in the <field> $K$, so $d=\pm b\in F$. For an automorphism inducing the root permutation $\pi$,
$$
\sigma(d)=\operatorname{sgn}(\pi)d.
$$
Since $\sigma$ fixes $d$ and $1\ne-1$ in $F$, every induced permutation has positive sign. The root action is faithful because the roots generate $K$, giving
$$
\boxed{G\hookrightarrow A_n\qquad(\operatorname{char}F\ne2)}.
$$
This is the <discriminant criterion for an alternating Galois group>.
\b[As printed for arbitrary fields, the assertion is false in characteristic two.] Over $\mathbb F_2$, the polynomial $T^2+T+1$ is irreducible and separable, with roots $\alpha,\alpha+1$ in $\mathbb F_4$. Their difference is one, so the discriminant is the square $1$. Nevertheless its <Galois group> is $C_2$, whereas $A_2$ is trivial. In characteristic two the permutation signs become indistinguishable as field elements; <square discriminants in characteristic two> therefore cannot detect the alternating subgroup in this way.
Back to article page
