Galois group of a polynomial

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This term is related to: Galois theory
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This article gives a basic definition in the following area: Galois theory
View other basic definitions in Galois theory |View 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

Particular 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.

Over finite fields

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 $\mathrm {lcm} (n_{1},\dots ,n_{r})$ .

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Results helping with the computation of a Galois group

One has that a separable polynomial of degree $n$ has its Galois group a subgroup of the symmetric group $S_{n}$ .

The Galois group of a polynomial is a transitive subgroup of $S_{n}$ if and only if the polynomial is irreducible.

It is natural to ask whether or not the Galois group is a subgroup of the alternating group $A_{n}$ , i.e. whether it contains an odd permutation or not. The following result turns out to hold: suppose $f$ is a separable polynomial of degree $n$ in a field $K$ , with the characteristic of $K$ not equal to $2$ . Then $\mathrm {Gal} (f/K)$ is a subgroup of $A_{n}$ if and only if the discriminant of $f$ is a square in $K$ .

"Reduction mod p": Suppose we want to find the Galois group of a polynomial $f\in \mathbb {Z} [X]$ over $\mathbb {Q}$ . For $p$ a prime number, we may define ${\overline {f}}\in \mathbb {F} _{p}[X]$ to be the polynomial in the finite field with $p$ elements such that $f\equiv {\overline {f}}$ modulo $p$ . Then, it turns out that $\mathrm {Gal} ({\overline {f}}/\mathbb {F} _{p})\leq \mathrm {Gal} (f/\mathbb {Q} )$ . This turns out to be quite helpful - as remarked in the examples section on this page, the Galois group of a polynomial over the field $\mathbb {F} _{p}$ has a particular structure (a cyclic group generated by a permutation with a cycle type corresponding to the irreducible factors' degrees) allowing for its computation. One may thus iterate through the primes to gain information on what cycle types are present in the Galois group over $\mathbb {Q}$ , which may determine what the group is.

Applications

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.