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

Statement

Let $F$ be a field of characteristic $0$ . For every $n \geq 5$ , the general polynomial of degree $n$ over the rational function field $F (e_{1}, \dots, e_{n})$ is not solvable by radicals.

Facts & Assumptions

Given: A field $F$ of characteristic $0$ , an integer $n \geq 5$ , and the general polynomial of degree $n$ of the previous theorem.

[L1]

The general polynomial of degree $n$ has Galois group $S_{n}$ (The general polynomial of degree $n$ has Galois group $S_{n}$ ).

[L2]

The group $S_{n}$ is not solvable for $n \geq 5$ ( $A_{5}$ and $S_{n}$ for $n \geq 5$ are not solvable).

[L3]

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

Proof

technique · direct

1.1

By [L1], the Galois group of the general polynomial is $S_{n}$ , and [L2] says that group is not solvable for $n \geq 5$ .

L1L2

2.1

If the polynomial were solvable by radicals, [L3] would force its Galois group to be solvable, contradicting step 1.1. Therefore it is not solvable by radicals.

step 1.1L3∎

Depends on

Used by

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Dependency tree · two levels

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Sources

J. S. Milne, Fields and Galois Theory, v5.10, Theorem 5.40 (standard reference, not scraped)
J. Ash, Basic Abstract Algebra, Section 6.8 (standard reference, not scraped)

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