We have seen that for $F=\Q$ there is a specific fifth degree polynomial that is not solvable by radicals. This implies that there can be no formula for the roots of polynomials of degree greater than or equal to five, where by "formula'' we mean an expression using the four simple arithmetic operations and arbitrary $k$th roots.

We can show directly that no such formula is possible, for any $F$ of characteristic 0. Consider the "general'' polynomial $f(t)=t^n-a_1t^{n-1}+ \cdots+(-1)^n a_n$, $n\ge 5$. A formula for a root of $f(t)$ would be an expression containing the symbols $a_i$ as placeholders for elements of $F$. We may instead interpret $f(t)$ as a specific polynomial in $F(a_1,\ldots,a_n)$, the field of rational functions in the symbols $a_i$. Now the question is whether $f(t)$ is solvable by radicals over $F(a_1,\ldots,a_n)$.

Let $E$ be a splitting field for $f(t)$, so in $E$ $$\prod_{i=1}^n (t-x_i)=f(t).$$ Then $E=F(x_1,\ldots,x_n)$ and the coefficients $a_i$ are the elementary symmetric functions in the $x_i$. As we saw in theorem 1.2.3, $G(E/F(\overline a))\cong S_n$. Since $S_n$ is not solvable, $f(t)$ is not solvable by radicals over $F(\overline a)$, so there can be no formula for any root of $f(t)$.