---
tag: SDV692Z915
kind: statement
role: theorem
name: Shafarevich's theorem on solvable Galois groups
msc: [12F12, 11R32]
uses: [0P3KZK0FTV, 0DS2F7GK05, 0XVR72M51M]
about: [YJS6X8JN7D, 0P3KZK0FTV]
sources:
  - url: https://en.wikipedia.org/wiki/Inverse_Galois_problem
---

## Statement

Let $G$ be a finite [[solvable group|0P3KZK0FTV]].
Then there is a [[Galois extension|0DS2F7GK05]] $L/\Q$ whose [[Galois group|0XVR72M51M]] $\Gal(L/\Q)$ is isomorphic to $G$.

## Remarks

The proof is long and uses class field theory and embedding problems over number fields.
It builds the extension in steps, and each step solves an embedding problem.
Given a normal subgroup $N$ of $G$ and an extension with group $G/N$, one must find a larger extension with group $G$ that contains it.
Example: the symmetric group $S_4$ is solvable, so it is a Galois group over $\Q$; the splitting field of $x^4 - x - 1$ is one such extension.
The theorem answers the [[inverse Galois problem|YJS6X8JN7D]] for solvable groups only; nonsolvable groups need other methods.

## Explanations

### L1 human

The inverse Galois problem asks which finite groups arise as symmetry groups of rational polynomials.
Shafarevich's theorem says that all solvable groups do.
These are the groups that can be built from commutative pieces, the same groups that correspond to equations solvable by radicals.

### L2 human

Abelian groups are easy to realise inside cyclotomic fields, but solvable groups need many successive extensions.
The difficulty lies in the embedding problem: an extension with a given quotient group need not extend to one with the larger group.
The theorem holds over every number field, not only over $\Q$.