**Galois theory**is that branch of abstract algebra which studies the symmetries of the roots of polynomials. Symmetries are usually expressed in terms of symmetry groups, and in fact the abstract notion of group was invented by Evariste Galois for the very purpose of describing symmetries of roots.

Galois theory has many applications to such classic mathematical problems as * "Which regular polygons can be constructed in the classical way with straight edge and compass?" * and * "Why can't an angle be trisected?" * (again, in the classical way with straight edge and compass) and "Why isn't there a formula for the roots of a 5th or higher degree polynomial in terms of the usual algebraic operations (+,-,*,/) and the taking of n-th roots?" (The Abel-Ruffini theorem.)

What exactly do we mean when we say "symmetries of roots of polynomials"? Such a symmetry is a permutation of the roots such that any algebraic equation which is satisfied by the roots is still satisfied after the roots have been permuted. These permutations form a group. Depending on the coefficients we allow in the algebraic equations, one gets thus different Galois groups.

For example, consider the polynomial (*x*^{2}-5)^{2}-24; we want to describe the Galois group of this polynomial "over the field of rational numbers" (i.e. allowing only rational numbers as coefficients in the invariant algebraic equations). The roots of the polynomial are

*a*= √2 + √3,*b*= √2 - √3,*c*= -√2 + √3,*d*= -√2 - √3.

*a*,

*b*,

*c*and

*d*and rational numbers. One such identity is

*a*+

*d*= 0. Therefore the permutation

*a*→

*a*,

*b*→

*b*,

*c*→

*d*and

*d*→

*c*is not permitted, as

*a*maps to

*a*and

*d*maps to

*c*, but

*a*+

*c*is not zero. A less obvious fact is that (

*a*+

*b*)

^{2}= 8. Therefore, we could send (

*a*,

*b*) to (

*c*,

*d*), as we also have (

*c*+

*d*)

^{2}= 8, but we could not send (

*a*,

*b*) to (

*a*,

*c*) as (

*a*+

*c*)

^{2}= 12. On the other hand, we can send (

*a*,

*b*) to (

*c*,

*d*), despite the fact that

*a*+

*b*= 2√2 and

*c*+

*d*= -2√2. This is because the identity

*a*+

*b*= 2√2 contains an irrational number, and so we don't require the Galois group to preserve it. Putting all this together, we see that the Galois group contains only the following four permuations:

- (
*a*,*b*,*c*,*d*) → (*a*,*b*,*c*,*d*) - (
*a*,*b*,*c*,*d*) → (*c*,*d*,*a*,*b*) - (
*a*,*b*,*c*,*d*) → (*b*,*a*,*d*,*c*) - (
*a*,*b*,*c*,*d*) → (*d*,*c*,*b*,*a*)

In the modern approach, the formalism is changed somewhat, in order to achieve a precise and more general definition: one starts with a field extension *L*/*K* and defines its Galois group as the group of all field automorphisms of *L* which keep all elements of *K* fixed. In the example above, we computed the Galois group of the field extension **Q**(*a*,*b*,*c*,*d*)/**Q**.

The notion of a solvable group in group theory allows us to determine whether or not a polynomial is solvable in the radicals, depending on whether or not its Galois group has the property of solvability. In essence, each field extension *L*/*K* corresponds to a factor group in a composition series of the Galois group. If a factor group in the composition series is cyclic of order *n*, then the corresponding field extension is a radical extension, and the elements of *L* can then be expressed using the *n*th root of some element of *K*.

If all the factor groups in its composition series are cyclic, the Galois group is called *solvable*, and all of the elements of the corresponding field can be found by repeatedly taking roots, products, and sums of elements from the base field (usually **Q**).

One of the great triumphs of Galois Theory was the proof that for every *n* > 4, there exist polynomials of degree *n* which are not solvable by radicals. This is due to the fact that for *n* > 4 the symmetric group *S*_{n} contains a simple, non-cyclic, normal subgroup.