---
tag: VHCWCFA3MX
kind: statement
role: theorem
name: Galois' theorem on solvability
msc: [12F10, 20D10]
uses: [J7YS0DA3HQ, 0JZY3R908H, 0P3KZK0FTV, V74WMGQQ9D, 1ZB13N78XW]
about: [J7YS0DA3HQ, 0P3KZK0FTV]
sources:
  - url: https://en.wikipedia.org/wiki/Solvable_group
---

## Statement

Let $K$ be a field of [[characteristic|1ZB13N78XW]] $0$, and let $f \in K[x]$ be a nonconstant polynomial.
Then $f$ is [[solvable by radicals|J7YS0DA3HQ]] over $K$ if and only if the [[Galois group of $f$|0JZY3R908H]] over $K$ is a [[solvable group|0P3KZK0FTV]].

## Remarks

For one direction, put a radical extension containing the roots of $f$ inside a Galois radical extension that first adjoins enough roots of unity.
Each later step then adjoins an $n$-th root in the presence of the $n$-th roots of unity, which gives an abelian extension.
So the big Galois group is solvable, and so is its quotient, the Galois group of $f$.
For the other direction, a solvable group has a chain of subgroups with cyclic quotients of prime order.
After adjoining roots of unity, each cyclic step of prime degree $q$ is generated by a $q$-th root, found with a [[Lagrange resolvent|2GDFFH7Y7T]].
Example: $x^5 - 4x + 2$ over $\Q$ has Galois group $S_5$, which is not solvable, so its roots cannot be written with radicals.
The characteristic hypothesis matters: in characteristic $p$, the [[Artin–Schreier|7YNMNRGH3T]] extensions are cyclic of degree $p$ but in general not radical.

## Explanations

### L1 human

The theorem answers an old question: when can the roots of a polynomial be written with $+$, $-$, $\times$, $\div$ and $n$-th roots?
Galois' answer is to look at the group of permutations of the roots that respect all algebraic relations.
The roots have such a formula exactly when this group can be broken down into commutative pieces.

### L2 human

The theorem reduces a question about formulas to a question about finite groups.
Since $S_n$ is solvable for $n \le 4$, every polynomial of degree at most $4$ is solvable by radicals, which matches the [[cubic formula|FB6J5WAT5V]] and the [[quartic formula|4CDFMKKZ01]].
Since [[$S_n$ is not solvable for $n \ge 5$|0J53NE674D]], the [[Abel–Ruffini theorem|163A2WQQJX]] follows once a polynomial with Galois group $S_n$ is found.
The proof uses the [[fundamental theorem of Galois theory|0T1Q8D0FW8]] together with [[Kummer theory|XZQENYV15Y]] for the cyclic steps.