Worked solution: Abel–Ruffini theorem and the Galois solvability criterion (1824)

Step 2 of 8: Sharpening a radical tower into a chain of cyclic Galois extensions

In plain words

A staircase built one radical at a time is uneven and hard to reason about symmetrically, unless you first bolt a handrail onto every step: adjoining enough roots of unity (numbers like $\zeta_n$ with $\zeta_n^n=1$ ) makes each step of the staircase a normal, symmetric extension whose automorphism group is as simple as a clock face — cyclic.

F = K_0 \subseteq K_0(\zeta_{n_1}) \subseteq K_1' \subseteq \cdots \subseteq K_m', \qquad \mathrm{Gal}(K_i'/K_{i-1}') \cong \mathbb{Z}/n_i\mathbb{Z}

Detailed analysis

Following the strategy Galois and later expositors use (see e.g. the field-theoretic reformulation summarised on Wikipedia's Abel–Ruffini theorem article), a radical tower is not yet normal, so before applying Galois theory one inserts, at each stage, a primitive $n_i$ -th root of unity $\zeta_{n_i}$ (if it is missing) and takes the normal closure. The resulting refined tower still ends at a field containing all the roots of $f$ , but now each single step $K_i'/K_{i-1}'$ is a normal extension whose Galois group is cyclic of order dividing $n_i$ — this is the classical fact that adjoining an $n$ -th root of an element, once $n$ -th roots of unity are already present, produces a cyclic automorphism group (the automorphisms just multiply the new root by a root of unity).

Terms in this step

Root of unity: A number $\zeta$ with $\zeta^n=1$ for some positive integer $n$ ; a primitive $n$ -th root of unity has $n$ as its smallest such exponent.
Normal extension: A field extension that contains every root of every irreducible polynomial it contains at least one root of — no root is left behind.
Cyclic group: A group generated by a single element, so every element is that generator raised to some power — as simple in structure as the hours on a clock.

Knowledge used in this step

Galois theory