Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 18I (Galois Theory)

If a monic polynomial $f$ has roots ${\alpha }_1,\dots ,{\alpha }_n$ in a splitting field, its polynomial discriminant isIt is a symmetric polynomial in the roots, so it belongs to the base field; it is nonzero exactly when $f$ is a separable polynomial.

Putthe Vandermonde determinant. A root permutation $\sigma$ sends $D$ to $\mathrm{sgn}(\sigma )D$. Hence the Galois group of a polynomial is contained in the alternating group $A_n$ exactly when every Galois automorphism fixes $D$, which by the fixed-field property is equivalent to $D\in K$. This implies that $\mathrm{disc}(f)=D^2$ is a square in $K$. Conversely, if $\mathrm{disc}(f)=d^2$ for $d\in K$, then $D^2=d^2$, so $D=\pm d\in K$. The Galois group therefore fixes $D$ and consists of even permutations. This proves the discriminant criterion for an alternating Galois group.

Now letThe rational root theorem shows that none of $\pm 1,\pm 2$ is a root, so this cubic is irreducible over $\mathbb{Q}$. The discriminant of a depressed cubic isThis is not a square in $\mathbb{Q}$, so the Galois group of an irreducible cubic is $S_3$ over $\mathbb{Q}$.

Over $K=\mathbb{Q}(\sqrt{-19})$, the discriminant is the squareThe cubic remains irreducible: a root in the degree-two extension $K/\mathbb{Q}$ would generate over $\mathbb{Q}$ both a degree-three field, by irreducibility, and a subfield of a degree-two field, contradicting the tower law for field extensions. Its Galois group over $K$ is consequently $A_3\cong C_3$.

In characteristic two, the formal derivative in positive characteristic ofisA polynomial is separable exactly when it and its derivative have no common root. If $r$ is a root of $f^{\prime }$, thenThus $f$ and $f^{\prime }$ have a common root exactly when $b=0$. Therefore $f$ is a separable polynomial if and only if $b\ne 0$.

Let $x_1,x_2,x_3\in L$ be the distinct roots of $f$, and define the two cyclic sumsVieta's relations in characteristic two areA direct symmetric expansion using these identities givesConsequently $r$ and $s$ are the roots of the characteristic-two cubic resolventThey are distinct because $r+s=b\ne 0$. Thus $g$ splits into distinct linear factors in $L[T]$.

A three-cycle of the roots fixes each cyclic sum, whereas any transposition interchanges $r$ and $s$. The induced action on $\{r,s\}$ is therefore exactly the sign action of the symmetric group $S_3$ on $S_3/A_3$. If the Galois group $G$ is contained in $A_3$, it fixes $r$ and $s$, so both lie in the fixed field $K$ and $g$ splits over $K$. Conversely, if $g$ splits in $K[T]$, its distinct roots $r,s$ belong to $K$ and every element of $G$ fixes them individually. No element of $G$ can then act as an odd root permutation, so $G\subseteq A_3$.