Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 18I / b

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 $G\subseteq A_3$.