Galois' theorem on solvability

Remarks

For one direction, put a radical extension containing the roots of $f$ inside a Galois radical extension that first adjoins enough roots of unity. Each later step then adjoins an $n$-th root in the presence of the $n$-th roots of unity, which gives an abelian extension. So the big Galois group is solvable, and so is its quotient, the Galois group of $f$. For the other direction, a solvable group has a chain of subgroups with cyclic quotients of prime order. After adjoining roots of unity, each cyclic step of prime degree $q$ is generated by a $q$-th root, found with a Lagrange resolvent. Example: $x^5 - 4x + 2$ over $\Q$ has Galois group $S_5$, which is not solvable, so its roots cannot be written with radicals. The characteristic hypothesis matters: in characteristic $p$, the Artin–Schreier extensions are cyclic of degree $p$ but in general not radical.

Going deeper

For a student

The theorem reduces a question about formulas to a question about finite groups. Since $S_n$ is solvable for $n \le 4$, every polynomial of degree at most $4$ is solvable by radicals, which matches the cubic formula and the quartic formula. Since $S_n$ is not solvable for $n \ge 5$, the Abel–Ruffini theorem follows once a polynomial with Galois group $S_n$ is found. The proof uses the fundamental theorem of Galois theory together with Kummer theory for the cyclic steps.