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

Statement

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

Facts & Assumptions

Given: A polynomial $f \in F [x]$ with splitting field $E / F$ and solvable Galois group $G = Gal (E / F)$ .

[F1]

A polynomial is solvable by radicals when its splitting field is contained in a radical extension of the base field (A polynomial is solvable by radicals when its splitting field lies in a radical extension).

[L1]

A finite solvable group has a composition series whose factors are cyclic of prime order (A finite group is solvable if and only if all its composition factors are cyclic of prime order).

[L2]

Normal subgroups in a finite Galois group correspond to Galois intermediate fields, and quotient groups give the corresponding Galois groups (Normal subgroups, conjugate fields, and quotient groups in the Galois correspondence).

[L3]

Over a base containing the needed roots of unity, a cyclic degree- $d$ extension is generated by adjoining one $d$ -th root (If $μ_{n} \subseteq F$ and $char F ∤ n$ , then a degree- $n$ extension is cyclic exactly when it is $F (α)$ with $α^{n} \in F$ and $x^{n} - α^{n}$ irreducible).

Proof

technique · direct

1.1L1L2

By [L1], choose a composition series $G = G_{0} ▹ G_{1} ▹ \dots ▹ G_{r} = 1$ whose factors $G_{i - 1} / G_{i}$ have prime order $p_{i}$ . Put $E_{i} : = E^{G_{i}}$ . Then $F = E_{0} \subseteq E_{1} \subseteq \dots \subseteq E_{r} = E,$ and [L2] makes each step $E_{i} / E_{i - 1}$ a cyclic Galois extension of degree $p_{i}$ .

2.1step 1.1F1choose

We enlarge the lower fields one step at a time. Starting with $R_{0} = F$ , suppose $R_{i - 1}$ is already a radical extension of $F$ containing $E_{i - 1}$ . Adjoin a primitive $p_{i}$ -th root of unity to $R_{i - 1}$ . This is one radical step because it adjoins a root of $x^{p_{i}} - 1$ . Let the enlarged field be $R_{i - 1}^{'}$ .

3.1step 2.1L3

The compositum $E_{i} R_{i - 1}^{'} / R_{i - 1}^{'}$ is still cyclic of degree $1$ or $p_{i}$ , because it is a quotient of the cyclic group $Gal (E_{i} / E_{i - 1})$ after base change. Since $R_{i - 1}^{'}$ contains the primitive $p_{i}$ -th roots of unity, [L3] shows that $E_{i} R_{i - 1}^{'}$ is obtained from $R_{i - 1}^{'}$ by adjoining one $p_{i}$ -th root. Therefore $R_{i} : = E_{i} R_{i - 1}^{'}$ is radical over $R_{i - 1}$ , hence radical over $F$ .

4.1step 3.1F1∎

Iterating step 3.1 produces a radical extension $R_{r} / F$ with $E = E_{r} \subseteq R_{r} .$ By [F1], the polynomial $f$ is solvable by radicals.

Depends on

Used by

Every polynomial of degree at most four is solvable by radicals Corollary

Dependency tree · two levels

27 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 6.8.3-6.8.4 (standard reference, not scraped)

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