---
tag: Z7R10WW158
kind: definition
name: Solvable by radicals
msc: [12F10, "12E12"]
defines: J7YS0DA3HQ
uses: [V74WMGQQ9D, 02NZZ1J0FH, 0QA8CHH8MG, 0P3KZK0FTV]
sources:
  - url: https://en.wikipedia.org/wiki/Solution_in_radicals
---

## Definition

Let $K$ be a [[field|0QA8CHH8MG]] and let $f \in K[x]$ be a nonconstant polynomial.
The polynomial $f$ is *solvable by radicals* over $K$ if there is a [[radical extension|V74WMGQQ9D]] $R/K$ that contains a [[splitting field|02NZZ1J0FH]] of $f$ over $K$.
The equation $f(x) = 0$ is then also called solvable by radicals.

## Remarks

Equivalently, every root of $f$ lies in some radical extension of $K$, since the compositum of finitely many radical extensions inside a common field is radical.
Example: $f = x^3 - 2$ over $\Q$ splits in $R = \Q(\omega, \sqrt[3]{2})$, with $\omega = e^{2\pi i/3}$.
This $R$ is radical, since $\omega^3 = 1 \in \Q$ and $(\sqrt[3]{2})^3 = 2 \in \Q(\omega)$, so $f$ is solvable by radicals.
Non-example: $f = x^5 - 4x + 2$ over $\Q$ is not solvable by radicals, because its Galois group is $S_5$, which is not [[solvable|0P3KZK0FTV]]; see the [[quintic equation|S1WS18QEGJ]].
In characteristic $0$, [[Galois' theorem on solvability|VHCWCFA3MX]] says that $f$ is solvable by radicals if and only if its Galois group is solvable.