$s_i$ is the sum of all square-free degree-$i$ monomials in the $X_j$. The symmetric <group> $S_n$ permutes variables and fixes exactly $K=\mathbb C(s_1,\ldots,s_n)$, 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 $K$.

Solution

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

 Solution (source code)