Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 4 / 17F / ii / 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 $\boxed{\alpha =\Delta \notin K,{\alpha }^2\in K}$.