Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 4 / 18F / b / Solution

$s_i$ 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 $K$.