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In mathematics, more specifically in the area of modern algebra known as Galois theory, the Galois group of a certain type of field extension is a specific group associated with the field extension. The study of field extensions (and polynomials which give rise to them) via Galois groups is called Galois theory, so named in honor of Évariste Galois who first discovered them.

For a more elementary discussion of Galois groups in terms of permutation groups, see the article on Galois theory.

Definition
Suppose that E is an extension of the field F (written as E/F and read E over F). An automorphism of E/F is defined to be an automorphism of E that fixes F pointwise. In other words, an automorphism of E/F is an isomorphism α from E to E such that α(x) = x for each x in F. The set of all automorphisms of E/F forms a group with the operation of function composition. This group is sometimes denoted by Aut(E/F).

If E/F is a Galois extension, then Aut(E/F) is called the Galois group of (the extension) E over F, and is usually denoted by Gal(E/F).

Examples
In the following examples F is a field, and C, R, Q are the fields of complex, real, and rational numbers, respectively. The notation F(a) indicates the field extension obtained by adjoining an element a to the field F.


 * Gal(F/F) is the trivial group that has a single element, namely the identity automorphism.
 * Gal(C/R) has two elements, the identity automorphism and the complex conjugation automorphism.
 * Aut(R/Q) is trivial. Indeed it can be shown that any automorphism of R must preserve the ordering of the real numbers and hence must be the identity.
 * Aut(C/Q) is an infinite group.
 * Gal(Q(√2)/Q) has two elements, the identity automorphism and the automorphism which exchanges √2 and &minus;√2.
 * Consider the field K = Q(³√2). The group Aut(K/Q) contains only the identity automorphism. This is because K is not a normal extension, since the other two cube roots of 2 (both complex) are missing from the extension &mdash; in other words K is not a splitting field.
 * Consider now L = Q(³√2, ω), where ω is a primitive third root of unity. The group Gal(L/Q) is isomorphic to S3, the dihedral group of order 6, and L is in fact the splitting field of x3 &minus; 2 over Q.
 * If q is a prime power, and if F = GF(q) and E = GF(qn) denote the Galois fields of order q and qn respectively, then Gal(E/F) is cyclic of order n.
 * If f is an irreducible polynomial of prime degree p with rational coefficients and exactly two non-real roots, then the Galois group of f is the full symmetric group Sp.

Computing Galois Groups
Below is a list of useful basic properties of Galois groups.


 * If σ is an element of Aut(E/F), then σ(α) is a root of the minimal polynomial of α over F.
 * If E is a finite separable extension of F, then there exists a primitive element α such that E = F(α). Furthermore, any σ in Aut(E/F) is uniquely determined by the value of σ(α).
 * If E = F(α) is finite field extension.
 * If E is a finite normal extension of F, then |Aut(E/F)| = [E:F], where the latter denotes the dimension of E as an F-vector space.
 * In particular, if E = F(α) is finite field extension, and β is a root of the minimal polynomial of α. Then the function σ that maps α to β while fixing everything else is a field isomorphism. If β is in E, then σ is an automorphism.

Example 1
Below is an example to illustrate how to use the properties of Galois group to compute the Galois group.

let E = Q(√2) and F = Q. Since the minimal polynomal of √2 over Q is x2 &minus; 2, we know that E is a two dimensional vector space over Q with basis 1, √2.

If σ is an element of Aut(E/F), and α + β√2 is an element of E, then σ(α + β√2) = ασ(1) + βσ(√2) = α + βσ(√2) because σ(1) is fixed and σ preserves field operations. So σ is uniquely determined by the value of σ(√2).

We know that √22 &minus; 2 = 0, so σ2(√2) &minus; σ(2) = σ(0), hence σ(√2) is a root of x2 &minus; 2.

Lastly, by the fundamental homomorphism of rings, we can confirm that the map which exchange √2 for &minus;√2 is an automorphism.

Example 2
Let E = Q(√2, √3) and F = Q. We can use the fact that Q(√2, √3) is the same as Q(√2 + √3), then we can appeal to the fact that the minimal polynomial of √2 + √3 must be of degree 4, since [Q(√2):Q] is a degree 2 extension, and [Q(√3):Q(√2)] is a degree 2 extension. So the minimal polynomial must be of the form
 * $$(x - \sqrt{2} - \sqrt{3})(x - \sqrt{2} + \sqrt{3})(x + \sqrt{2} - \sqrt{3})(x + \sqrt{2} + \sqrt{3})

$$.

Therefore, the 4 automorphisms are:
 * the idenity map,
 * the map that exchanges √2 for -√2,
 * the map that exchanges √3 for -√3,
 * the map that exchanges √2 for -√2 and √3 for √3.

Aut[E/F] is isomorphic to the [Kline four-group].

Fundamental Theorem of Galois Theory
The significance of an extension being Galois is that it obeys the fundamental theorem of Galois theory: the closed (with respect to the Krull topology) subgroups of the Galois group correspond to the intermediate fields of the field extension.

If E/F is a Galois extension, then Gal(E/F) can be given a topology, called the Krull topology, that makes it into a profinite group.

In the finite case, this means there is a 1-1 correspondence between the subgroup of the Galois group, and the intermediate fields of the field extensions.

For example, in the case of $$ E = \mathbb{Q}(\sqrt{3}, \sqrt{2}) $$ and $$ F = \mathbb{Q} $$. The subgroups of Gal(E/F) viewed as the Kline four group are:
 * {(0,0)},
 * {(0,0), (1,0)},
 * {(0,0), (0,1)},
 * ((0,0), (1,1)} and
 * {(0,0), (1,0), (0,1), (1,1)}.

The corresponding intermediate fields are: Q, Q(√2), Q(√3), Q(√6), Q(√2, √3).

Solvability by radicals
An important result in Galois theory is that for a field $$F $$ of characteristics 0, if a polynomial $$f(x) \in F[x]$$ is solvable by radicals, then the Galois group of $$ f(x) $$ is solvable.

As a consequence, polynomials of degree 5 or higher are generally not solvable by radicals. In other words, there are no formulas like the quadratic formula for solving roots.

For the complete proof, see Abel–Ruffini theorem.

Inverse Galois problem
A famous open problem in mathematics is whether any finite group can be realized as a Galois group of a Galois extension of $Q$. More precisely, given a finite group G, does there exists a Galois extension K of $Q$ such that Gal(K/$Q$) is isomorphic to G?

For example, for groups of order 4, Kline four is isomorphic to Gal(Q(√2, √3)/Q), and the cyclic group of order four can be obtained by adjoining Q}] with all the roots of $$x^4 -2 $$.