Solvability by Radicals

A polynomial is solvable by radicals if and only if its Galois group is solvable.

The idea

Theorem (Solvability by radicals).

A polynomial over a field of characteristic $0$ is solvable by radicals if and only if its Galois group is a solvable group.

The theorem is a translation, and the translation is the whole content. On one side is a question about formulas: starting from the coefficients, can the roots be reached in finitely many additions, subtractions, multiplications, divisions and extractions of $n$-th roots? A formula of that shape is nothing but a tower of fields, each got from the previous one by adjoining a single radical $\sqrt[n]{a}$.

On the other side is a question about groups, and the bridge is that one radical step is a very tame extension. Once the $n$-th roots of unity are available, an automorphism fixing the ground field can only send $\sqrt[n]{a}$ to $\zeta^{j} \sqrt[n]{a}$, and the exponents $j$ add modulo $n$ — so the step has cyclic Galois group. Under the Galois correspondence the radical tower $F = K_0 \subset K_1 \subset \cdots \subset K_m$ becomes a chain of subgroups $G = G_0 \trianglerighteq G_1 \trianglerighteq \cdots \trianglerighteq 1$, and every step cyclic becomes every successive quotient abelian. That is precisely the definition of a solvable group.

So a radical formula for $f$ manufactures a solvability chain for $\operatorname{Gal}(f)$, and the converse holds too. The payoff is that ruling out a formula — which as a statement about formulas looks impossible to prove, since there are infinitely many to rule out — becomes the finite job of examining one group.

Ways to work on it

• Walkthrough. Radical towers, solvable groups, and why the quintic breaks.
• Practice. Decide whether a given Galois group is solvable.
• Hardest. Show an explicit quintic has Galois group S_5, hence no radical formula.

Not sure where to start? Take the ten-question placement test.