3.7. Solvability of equations.

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Let f be a polynomial. We say the equation f(X) = 0 is solvable (by extracting radicals)

if there is a tower

F = F0 ⊂ F1 ⊂ F2 ⊂ ・ ・ ・ ⊂ Fm

such that

(a) Fi = Fi−1[αi], αmi

i

∈ Fi−1;

(b) Fm contains a splitting field for f.

Theorem 3.25. (Galois, 1832) Let F be a field of characteristic zero. The equation f = 0

is solvable if and only if the Galois group of f is solvable.

We shall prove this later. Also we shall exhibit polynomials f(X) ∈ Q[X] with Galois

group Sn, which therefore are not solvable when n ≥ 5.

Remark 3.26. If F has characteristic p, then the theorem fails for two reasons:

(i) f may not be separable, and so not have a Galois group;

(ii) Xp − X − a is not solvable by radicals.

If the definition of solvable is changed to allow extensions of the type in (ii) in the chain,

and f is required to be separable then the theorem becomes true in characteristic p.

28 J.S. MILNE

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