Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 4 / 18F (Galois Theory)

Artin fixed-field theorem says that for a finite group $G\le \mathrm{Aut}L$, $L/L^G$ is Galois with group $G$ and degree $|G|$. Let any finite $G$ act faithfully by left translation on variables $X_g$ in $L=\mathbb{C}(X_g:g\in G)$. Artin's theorem then realizes $G$ as $\mathrm{Gal}(L/L^G)$.

$s_i$ is the sum of all square-free degree-$i$ monomials in the $X_j$. The symmetric group $S_n$ permutes variables and fixes exactly $K=\mathbb{C}(s_1,\dots ,s_n)$, so Artin gives $[L:K]=n!$. The Vandermonde $f={\prod }_{i<j}(X_i-X_j)$ is not symmetric, but $f^2$ is symmetric and hence belongs to $K$.

The relations are ${\sigma }^n={\tau }^2=1$ and $\tau \sigma \tau ={\sigma }^{-1}$, so the generated group is $D_{2n}$. The functions $XY$ and $X^n+Y^n$ are fixed. Conversely $X^n,Y^n$ are roots of $T^2-(X^n+Y^n)T+(XY{)}^n$, and adjoining $X$ then $Y$ gives degree at most $2n$. Artin gives degree exactly $|G|=2n$, proving $M^G=\mathbb{C}(X^n+Y^n,XY)$.