Galois group of a polynomial

This article defines a natural context where a group occurs, or is associated, with another algebraic, topological, analytic or discrete structure
View other occurrences of groups

This term is related to: Galois theory
View other terms related to Galois theory | View facts related to Galois theory

Definition

For $K$ a field, let $f(X)\in K[X]$ be a separable polynomial with coefficients in $K$, and $L$ a splitting field for $f$ over $K$. Then the Galois group (in the sense of field extensions) of $L$ over $K$ acting on the roots $\alpha _{1},\dots ,\alpha _{n}$ of $f$ determines an injective group homomorphism $\iota :G\to S_{n}$. (Here, $S_{n}$ is the symmetric group on $n$ letters).

Then the image of $\iota$ is a subgroup of $S_{n}$, and is termed the Galois group of the polynomial $f(X)$.

Examples

• The polynomial $f(X)=(X^{2}-2)(X^{2}-3)\in \mathbb {Q} [X]$ has splitting field $\mathbb {Q} ({\sqrt {2}},{\sqrt {3}})$ over $\mathbb {Q}$. It can be shown that this field extension has Galois group isomorphic to the Klein four-group.

• The polynomial $X^{n}-X-1\in \mathbb {Q} [X]$ has Galois group the symmetric group on $n$ letters.

• Let $p$ be a prime, and $\mathbb {F} _{p}$ denote the finite field with $p$ elements. Let $f\in \mathbb {F} _{p}[X]$ be a separable polynomial that factors into irreducible factors of degrees $n_{1},\dots n_{r}$, and denote the degree of $f$ as $n:=\sum _{i}n_{i}$. Then the Galois group of $f$ over $\mathbb {F} _{p}$ viewed as a subgroup of the symmetric group $S_{n}$ is generated by a single permutation with cycle type $(n_{1},\dots ,n_{r})$. Thus, it is cyclic of order $\mathrm {lcm} (n_{1},\dots ,n_{r})$.

PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.

Applications and results

Insolvability of the quintic

It can be shown that an irreducible polynomial $f$ is solvable in terms of radicals if and only if its Galois group is a solvable group. But there exist degree five polynomials with Galois group alternating group:A5 and symmetric group:S5, which are not solvable groups. Thus in general, the quintic is not solvable in radicals.

A particular polynomial over $\mathbb {Q}$ that is not solvable in terms of radicals is $X^{5}-X+1$, which has Galois group symmetric group:S5.

Every subgroup of symmetric group:S4 is solvable, which tells us that the quadratic formula, the cubic formula and the quartic formula do indeed exist, i.e. such polynomials are always solvable in radicals.