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In characteristic $0$ , a polynomial solvable by radicals has a solvable Galois group

Statement

Let $F$ be a field of characteristic $0$ , and let $f \in F [x]$ be nonzero. If $f$ is solvable by radicals, then the Galois group of its splitting field over $F$ is a solvable group.

Facts & Assumptions

Given: A polynomial $f \in F [x]$ of characteristic $0$ that is solvable by radicals, with splitting field $E / F$ .

[F1]

Solvable by radicals means that $E$ lies inside some radical extension of $F$ (A polynomial is solvable by radicals when its splitting field lies in a radical extension).

[L1]

The normal closure of a radical extension is radical (The normal closure of a radical extension is again radical).

[L2]

Adjoining roots of unity to a finite Galois extension adds an abelian kernel and preserves solvability of the Galois group (Adjoining roots of unity to a finite Galois extension adds an abelian kernel and preserves solvability).

[L3]

Subgroups, quotients, and extensions of solvable groups are solvable (Subgroups and quotients of solvable groups are solvable, Extensions and finite direct products of solvable groups are solvable).

Proof

technique · direct

1.1

By [F1], choose a finite radical extension $L / F$ with $E \subseteq L$ . Replacing $L$ by its normal closure over $F$ , [L1] lets us assume from the start that $L / F$ is finite Galois and radical.

F1L1choose

2.1

Let the radical tower for $L / F$ use exponents $m_{1}, \dots, m_{r}$ , and let $N = m_{1} \dots m_{r}$ . Adjoin $μ_{N}$ to $L$ . By [L2], solvability of $Gal (L / F)$ is equivalent to solvability of $Gal (L (μ_{N}) / F)$ . So it is enough to prove the latter solvable.

step 1.1L2

3.1

After adjoining $μ_{N}$ , every step of the radical tower becomes a finite Galois extension with cyclic Galois group: if $F_{i} = F_{i - 1} (α_{i})$ with $α_{i}^{m_{i}} \in F_{i - 1}$ , then the enlarged lower field already contains $μ_{m_{i}}$ , so every root of $x^{m_{i}} - α_{i}^{m_{i}}$ is $ζ α_{i}$ with $ζ \in μ_{m_{i}}$ . Thus $F_{i} (μ_{N}) / F_{i - 1} (μ_{N})$ is the splitting field of a separable polynomial, and every automorphism is determined by $α_{i} \mapsto ζ α_{i}$ , so its Galois group embeds in the cyclic group $μ_{m_{i}}$ . In particular each step has solvable Galois group, and repeated use of [L3] up the tower makes $Gal (L (μ_{N}) / F (μ_{N}))$ solvable.

step 2.1L3algebra

4.1

The cyclotomic extension $F (μ_{N}) / F$ has abelian Galois group, hence solvable. Applying [L3] once more to the tower $F \subseteq F (μ_{N}) \subseteq L (μ_{N})$ shows that $Gal (L (μ_{N}) / F)$ is solvable. By step 2.1 the same is true of $Gal (L / F)$ .

step 3.1L2L3

5.1

The splitting field $E$ is an intermediate field of the finite Galois extension $L / F$ , so $Gal (E / F)$ is a quotient of a subgroup of $Gal (L / F)$ . Therefore [L3] makes $Gal (E / F)$ solvable.

step 4.1L3∎

Depends on

Used by

For n≥5 over a characteristic-zero base, the general polynomial of degree n is not solvable by radicals Corollary

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Fields and Galois Theory, v5.10, Theorem 5.34 (standard reference, not scraped)
J. Ash, Basic Abstract Algebra, Sections 6.8.2-6.8.4 (standard reference, not scraped)

In characteristic 00, a polynomial solvable by radicals has a solvable Galois group