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

Saying that a root of $f$ is 'expressible by radicals' means there is a finite tower of fields $\mathbb{Q} = K_0 \subset K_1 \subset \cdots \subset K_r$ with $K_i = K_{i-1}(\sqrt[n_i]{a_i})$ for some $a_i \in K_{i-1}$ and exponent $n_i \ge 2$, such that the root lies in $K_r$. Inside the tower the four field operations are applied freely — they never enlarge a field — so the only genuinely new elements ever created are the $n_i$-th roots. The classical formulas are exactly such towers read off in reverse: the quadratic formula uses a single layer $\mathbb{Q}(\sqrt{b^2-4ac})$, while Cardano's cubic formula needs one layer adjoining a cube root of unity and another adjoining $\sqrt[3]{-q/2+\sqrt{D}}$.

The containment formulation matters: one does not ask that $K_r$ equal the splitting field $E$ of $f$, only that $E \subseteq K_r$. That makes 'solvable by radicals' a property of field embeddings, and it converts the permutation-theoretic question Ruffini and Abel originally asked into pure field theory: does any radical tower at all have a top containing $E$?

Two subtleties already visible in the definition will matter later. First, the radicand $a_i$ ranges over all of $K_{i-1}$, not merely over $\mathbb{Q}$, so nesting is essential — that is what 'a formula with radicals inside radicals' means formally. Second, in general $K_i/K_{i-1}$ is neither normal nor Galois, so the raw tower is not yet ready for group theory; Step 1 refines it.