 Solution

Let $x_1,\dots ,x_5$ be algebraically independent indeterminates and set $a_i=e_i(x_1,\dots ,x_5)$, the elementary symmetric polynomials. ThenThe Fundamental theorem of symmetric polynomials identifies the fixed field of the natural $S_5$ action on $\mathbb{Q}(x_1,\dots ,x_5)$ with $\mathbb{Q}(a_1,\dots ,a_5)=K$. Every permutation of the roots therefore gives a $K$-automorphism, and the Galois group of a polynomial isA characteristic-zero polynomial is solvable by radicals exactly when its Galois group is a solvable group. Part (c) shows that $S_5$ is not solvable, so