Past exam of the galois theory course of the University of Oxford / 2023 / 2

The Galois group acts transitively on the $n$ roots because $f$ is irreducible and separable. The orbit-stabilizer theorem givesso $n$ divides $|G|$. Equivalently, the tower law gives $[K:F]=[K:F({\alpha }_1)]n$, while the finite Galois extension has group order $[K:F]$.

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.

First assume $\mathrm{char}F\ne 2$. Set $d=\det V(\alpha )\ne 0$. 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 $\pi$,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 $K$, givingThis is the discriminant criterion for an alternating Galois group.

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.

Denote the three stated pair sums by $r_1,r_2,r_3$. They are pairwise distinct: for example,and the other differences factor similarly. Every Galois automorphism fixes each $r_i\in F$, and hence fixes each of the three corresponding partitions of four roots into two pairs. The kernel of the permutation action $S_4\to S_3$ on these partitions isThus $G$ embeds in the Klein four-group. By part (a), $4$ divides $|G|$, and the containing group has order four. ConsequentlyThis Split resolvent criterion for a Klein four Galois group uses the cubic resolvent of a quartic. The pair-sum differences remain nonzero in characteristic two, so this proof does not need the missing characteristic assumption from part (c).