Solution
= Solution
Take the <Vandermonde determinant> $\Delta=\prod_{i<j}(x_i-x_j)$. Its square is a <symmetric polynomial>, hence a polynomial in the <elementary symmetric polynomials> and belongs to $K$. It is nonzero because the variables are algebraically independent. A transposition of two variables fixes every $e_i$ and therefore fixes $K$, but sends $\Delta$ to $-\Delta\ne\Delta$ in characteristic zero. Thus $\boxed{\alpha=\Delta\notin K,\quad\alpha^2\in K}$.