# Galois theory

In mathematics, **Galois theory** is that branch of abstract algebra which studies the symmetries of the roots of polynomials. In other words, the Galois theory is the study of solutions to polynomials and how the different solutions are related to each other. Symmetries are usually expressed in terms of symmetry groups, and in fact the very notion of a group was invented by Evariste Galois to describe symmetries of roots. Galois connections describe special relations between partially ordered sets.

## Application to classical problems

*"Which regular**polygons can be constructed in the classical way with straight edge and compass?"*

*"Why can't all angles be trisected?"*

- ''"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?"

## Symmetries of roots

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.

## Example of a quadratic equation

As a very simple example, consider any quadratic polynomial

*f(x) = ax*.^{2}+bx+c

*y=f(x)*, we get a parabola with a vertical axis of symmetry across the line

*y=-b/2a*. Reflecting across that line permutes the roots of the polynomial because they are equidistant from the line. Since there are only two roots, it is easy to see that this is the only symmetry of

*f*, so the group of symmetries of

*f*, that is, the Galois group of

*f*, is

**Z**/2

**Z**. Note that the roots may be complex, but they will still be symmetric across

*y=-b/2a*.

## Another example

For a trickier example, consider the polynomial

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 permutations:

- (
*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*)

## Modern approach by field theory

In the modern approach, the setting 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**.

## Solvable groups and solution by radicals

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.

## Inverse problems

It is easy to construct field extensions with a given Galois group: Choose a field *K* and a group *G*. Cayley's theorem says that *G* is (up to isomorphism) a subgroup of the symmetric group *S* on the elements of *G*. Choose indeterminates {*x _{α}*}, one for each element

*α*of

*G*, and adjoin them to

*K*to get the field

*K({x*. Contained within

_{α}})*K({x*is the field

_{α}})*L*of symmetric rational functions in the {

*x*}. The Galois group of

_{α}*L*over

*K*is

*S*, by a basic result of Artin.

*G*acts on

*L*by mapping

*x*to

_{α}x_{β}*x*, and if the fixed field of this action is

_{αβ}*M*, then, by the Fundamental Theorem of Galois Theory, the Galois group of

*L*over

*M*is

*G*.

It is an open problem (in general) how to construct field extensions of a *fixed* ground field with a given Galois group. This is called the *inverse Galois problem*, and is usually posed for extensions of the rational number field **Q**. There is a great deal of detailed information in particular cases.