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For a dominant integer tuple $\lambda$ and $\delta =(m-1,\dots ,0)$, the rational Schur character is $\det (x_i^{{\lambda }_j+m-j})/\det (x_i^{m-j})$. For partitions it is a symmetric polynomial; for general tuples it is Laurent on the diagonal torus. The formula follows by comparing the trace of a permuted tensor power with the Frobenius alternant character formula and using independence of symmetric-group characters. Apparent poles at coincident nonzero eigenvalues are removable.

For the permutation action on $n$ coordinates with $n\ge 2$, every alternating-group invariant is a symmetric polynomial plus the Vandermonde product $\Delta$ times a symmetric polynomial. Thus the invariant ring is $\mathbb{C}[e_1,\dots ,e_n,Z]/(Z^2-D(e_1,\dots ,e_n))$, where $D$ is the discriminant polynomial. The two summands are independent over the symmetric polynomial ring.

Over a field of characteristic zero, an alternating polynomial changes sign when two variables are interchanged. It vanishes when two coordinates agree, so each factor $x_i-x_j$ divides it. These distinct linear factors are relatively prime in the polynomial ring, hence their product, the Vandermonde determinant, divides it. The quotient is a symmetric polynomial.

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 $\boxed{\alpha =\Delta \notin K,{\alpha }^2\in K}$.