Simplified Galois proof?

Question

I have learned about Galois epoch-making proof that any polynomial of the fifth degree has no solution representable in terms of its coefficients.

Can his proof be simplified and clarified in modern language? If yes, please show that.

Answer 1

More accurately, the result is that the general fifth degree equation \begin{equation} f(x) = x^5 + a_4 x^4 + \cdots + a_1x + a_0 = 0 \end{equation}with coefficients $a_i\in \mathbb{C}$ isn't solvable by radicals. (Some equations, like $x^5 - 1 = 0$ and the less obvious $x^5 + 20x^3 + 20x^2 + 30x + 10$, are easy to solve, and certain special functions can parametrize the solutions.) That is, there isn't a formula combining via coefficients $a_0, \dots, a_4$ by addition, multiplication, and rational powers that gives the zeros of the quintic $f$. Contrast this result to the quadratic formula or the more complicated results in degree $3$ and $4$.

More formally, the idea is as follows. For simplicity, I'm going to take $a_i\in \mathbb{Q}$. The quintic above is said to be solvable by radicals if there exists a tower of fields $K_0 = \mathbb{Q}\subset K_1 \subset \cdots \subset K_n$ such that $K_{i+1} = K_i(\theta_i^{1/{n_i}})$ for some $\theta_i\in K_i$ and integer $n_i > 0$. By standard Galois theory machinery, this condition is equivalent to $G = \text{Gal}(K_n/K_0) = \text{Gal}(L/\mathbb{Q})$ being a solvable group, where $L$ is the splitting field of $f$. The group $G$ acts faithfully by permuting the roots of $f$ and thus embeds in $S_d$ for $d = \deg f$. The groups $S_1, \dots, S_4$ are solvable, but $S_5$ is not. We thus only need to write down some $f$ with $G = S_5$, which can be done with a bit of computation; $f(x) = x^5 - x + 1$ works, I think.

The argument above is the basic proof you'd see in any first Galois theory class, although the original proof preceded Galois by a decade or so. Here's what looks like a description of that proof, though I've only skimmed it myself. There are more modern proofs; Arnol'd, for example, has one with a topological bent, but it boils down (in more concrete terms) to the fact that $S_5$ is not a solvable group.