# Galois theory

The Galois theory is a branch of algebra . From a classical perspective, Galois theory deals with the symmetries of the zeros of polynomials . These symmetries can in principle be described by groups of permutations , that is to say subgroups of the symmetrical group . Évariste Galois discovered that these symmetries allow statements about the solvability of the equation. In a modern perspective, body extensions are examined with the help of their Galois group .

The Galois theory has many applications in classical problems, such as “Which regular polygons can be constructed with compasses and rulers ?”, “Why can't an angle be divided into three?” (Again only with compasses and ruler), “Why can a cube be made the side of a cube with double the volume cannot be constructed? "and" Why is there no closed formula for calculating the zeros of polynomials of the fifth or higher degree that only uses the four basic arithmetic operations and the extraction of roots? "( Abel-Ruffini's theorem ) .

## Classic approach

A “symmetry of the zeros of polynomials” is a permutation of the zeros , so that every algebraic equation over these zeros is still valid even after the zeros have been exchanged using the permutation. These permutations form a group . Depending on the coefficients that are allowed in the algebraic equations, different Galois groups result.

Galois himself described a method with which a single equation satisfied by the zeros can be constructed (the so-called Galois resolvent ) so that the Galois group consists of the symmetries of this one equation.

### example

The Galois group of the polynomial should be determined over the field of the rational numbers . By pulling the roots twice, we get - together with the relationship - the zeros: ${\ displaystyle \ left (x ^ {2} -5 \ right) ^ {2} -24}$${\ displaystyle 5 \ pm 2 {\ sqrt {6}} = ({\ sqrt {2}} \ pm {\ sqrt {3}}) ^ {2}}$

${\ displaystyle x_ {1} = {\ sqrt {2}} + {\ sqrt {3}}}$,
${\ displaystyle x_ {2} = {\ sqrt {2}} - {\ sqrt {3}}}$,
${\ displaystyle x_ {3} = - {\ sqrt {2}} + {\ sqrt {3}}}$,
${\ displaystyle x_ {4} = - {\ sqrt {2}} - {\ sqrt {3}}}$.

There are ways to permute (swap) these four zeros : ${\ displaystyle 4! = 24}$

${\ displaystyle \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right) \ mapsto ...}$
No. permutation No. permutation No. permutation No. permutation
1 ${\ displaystyle \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right)}$ 7th ${\ displaystyle \ left (x_ {2}, x_ {1}, x_ {3}, x_ {4} \ right)}$ 13 ${\ displaystyle \ left (x_ {3}, x_ {1}, x_ {2}, x_ {4} \ right)}$ 19th ${\ displaystyle \ left (x_ {4}, x_ {1}, x_ {2}, x_ {3} \ right)}$
2 ${\ displaystyle \ left (x_ {1}, x_ {2}, x_ {4}, x_ {3} \ right)}$ 8th ${\ displaystyle \ left (x_ {2}, x_ {1}, x_ {4}, x_ {3} \ right)}$ 14th ${\ displaystyle \ left (x_ {3}, x_ {1}, x_ {4}, x_ {2} \ right)}$ 20th ${\ displaystyle \ left (x_ {4}, x_ {1}, x_ {3}, x_ {2} \ right)}$
3 ${\ displaystyle \ left (x_ {1}, x_ {3}, x_ {2}, x_ {4} \ right)}$ 9 ${\ displaystyle \ left (x_ {2}, x_ {3}, x_ {1}, x_ {4} \ right)}$ 15th ${\ displaystyle \ left (x_ {3}, x_ {2}, x_ {1}, x_ {4} \ right)}$ 21st ${\ displaystyle \ left (x_ {4}, x_ {2}, x_ {1}, x_ {3} \ right)}$
4th ${\ displaystyle \ left (x_ {1}, x_ {3}, x_ {4}, x_ {2} \ right)}$ 10 ${\ displaystyle \ left (x_ {2}, x_ {3}, x_ {4}, x_ {1} \ right)}$ 16 ${\ displaystyle \ left (x_ {3}, x_ {2}, x_ {4}, x_ {1} \ right)}$ 22nd ${\ displaystyle \ left (x_ {4}, x_ {2}, x_ {3}, x_ {1} \ right)}$
5 ${\ displaystyle \ left (x_ {1}, x_ {4}, x_ {2}, x_ {3} \ right)}$ 11 ${\ displaystyle \ left (x_ {2}, x_ {4}, x_ {1}, x_ {3} \ right)}$ 17th ${\ displaystyle \ left (x_ {3}, x_ {4}, x_ {1}, x_ {2} \ right)}$ 23 ${\ displaystyle \ left (x_ {4}, x_ {3}, x_ {1}, x_ {2} \ right)}$
6th ${\ displaystyle \ left (x_ {1}, x_ {4}, x_ {3}, x_ {2} \ right)}$ 12 ${\ displaystyle \ left (x_ {2}, x_ {4}, x_ {3}, x_ {1} \ right)}$ 18th ${\ displaystyle \ left (x_ {3}, x_ {4}, x_ {2}, x_ {1} \ right)}$ 24 ${\ displaystyle \ left (x_ {4}, x_ {3}, x_ {2}, x_ {1} \ right)}$

But not all of these permutations also belong to the Galois group. This is because all algebraic equations with rational coefficients only that the variables , , and included, must retain their validity under the permutations of the Galois group. If you look at, for example ${\ displaystyle x_ {1}}$${\ displaystyle x_ {2}}$${\ displaystyle x_ {3}}$${\ displaystyle x_ {4}}$

${\ displaystyle x_ {1} + x_ {4} = 0}$,

so this equation is not fulfilled for all exchanges of the zeros. Under the permutation, which leaves and the same and and interchanges, the equation produces a false statement, because is unequal . Therefore this permutation (No. 2) does not belong to the Galois group. The same applies to permutations no. 4, 5, 6, 7, 9, 10, 12, 13, 15, 16, 18, 19, 20, 21, 23 in the table, because the sum of two of the four zeros is only the equations and correct. ${\ displaystyle x_ {1}}$${\ displaystyle x_ {2}}$${\ displaystyle x_ {3}}$${\ displaystyle x_ {4}}$${\ displaystyle x_ {1} + x_ {3}}$${\ displaystyle 0}$${\ displaystyle x_ {2} + x_ {3} = x_ {3} + x_ {2} = 0}$${\ displaystyle x_ {4} + x_ {1} = 0}$

Another algebraic equation with rational coefficients satisfying the zeros is

${\ displaystyle (x_ {1} + x_ {2}) ^ {2} -8 = 0}$.

Therefore, in addition the permutations Nos. 3, 11, 14 and 22 excluded, because it is , , and . ${\ displaystyle (x_ {1} + x_ {3}) ^ {2} -8 \ neq 0}$${\ displaystyle (x_ {2} + x_ {4}) ^ {2} -8 \ neq 0}$${\ displaystyle (x_ {3} + x_ {1}) ^ {2} -8 \ neq 0}$${\ displaystyle (x_ {4} + x_ {2}) ^ {2} -8 \ neq 0}$

There remain four permutations: No. 1, 8, 17 and 24. Since the polynomial is an irreducible polynomial of the 4th degree, the Galois group consists of at least four elements. So these four permutations form the Galois group of the polynomial : ${\ displaystyle \ left (x ^ {2} -5 \ right) ^ {2} -24}$${\ displaystyle \ mathbb {Q}}$ ${\ displaystyle \ left (x ^ {2} -5 \ right) ^ {2} -24}$

${\ displaystyle \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right) \ mapsto \ left (x_ {1}, x_ {2}, x_ {3}, x_ { 4} \ right)}$
${\ displaystyle \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right) \ mapsto \ left (x_ {2}, x_ {1}, x_ {4}, x_ { 3} \ right)}$
${\ displaystyle \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right) \ mapsto \ left (x_ {3}, x_ {4}, x_ {1}, x_ { 2} \ right)}$
${\ displaystyle \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right) \ mapsto \ left (x_ {4}, x_ {3}, x_ {2}, x_ { 1} \ right)}$

or in cycle notation :

${\ displaystyle \ operatorname {id}}$( Identity ) , and .${\ displaystyle (x_ {1} x_ {2}) (x_ {3} x_ {4})}$${\ displaystyle (x_ {1} x_ {3}) (x_ {2} x_ {4})}$${\ displaystyle (x_ {1} x_ {4}) (x_ {2} x_ {3})}$

This group is isomorphic to Klein's group of four .

Alternatively, the Galois group can also be determined with the help of a primitive element . This is a special case, because the zero point x 1 is - like the zero point x 2 , x 3 or x 4 - already such a primitive element. With

${\ displaystyle x_ {1} ^ {2} = 5 + 2 {\ sqrt {6}}}$, and${\ displaystyle \ quad x_ {1} ^ {3} = 11 {\ sqrt {2}} + 9 {\ sqrt {3}} \ quad}$${\ displaystyle \ quad x_ {1} ^ {4} = 49 + 20 {\ sqrt {6}}}$

one obtains the equations:

${\ displaystyle x_ {1} ^ {3} -9x_ {1} = 2 {\ sqrt {2}} \ quad}$ and .${\ displaystyle \ quad x_ {1} ^ {3} -11x_ {1} = - 2 {\ sqrt {3}}}$

This allows us to replace and as a polynomial with the variable x 1 : ${\ displaystyle \ textstyle {\ sqrt {2}}}$${\ displaystyle \ textstyle {\ sqrt {3}}}$

${\ displaystyle {\ sqrt {2}} = {\ tfrac {1} {2}} (x_ {1} ^ {3} -9x_ {1})}$and .${\ displaystyle {\ sqrt {3}} = - {\ tfrac {1} {2}} (x_ {1} ^ {3} -11x_ {1})}$

Thus the four zeros result as polynomials p 1 , p 2 , p 3 , p 4 with the variable x 1 :

${\ displaystyle x_ {1} = p_ {1} (x_ {1}) = x_ {1}}$,
${\ displaystyle x_ {2} = p_ {2} (x_ {1}) = x_ {1} ^ {3} -10x_ {1}}$,
${\ displaystyle x_ {3} = p_ {3} (x_ {1}) = - x_ {1} ^ {3} + 10x_ {1}}$,
${\ displaystyle x_ {4} = p_ {4} (x_ {1}) = - x_ {1}}$.

In the general case, the associated minimal polynomial and its further zeros must be determined for the primitive element . In this example, however, the minimal polynomial of x 1 is the initial polynomial with the other zeros x 2 , x 3 and x 4 that are already known . (For the general procedure: see example for the theorem of the primitive element .) If one now replaces the variable x 1 in the polynomials p 1 , ... p 4 with x 2 , x 3 or x 4 , the zeros x 1 result again , x 2 , x 3 , x 4 of the initial polynomial, but in a different order. These permutations of the zeros form the Galois group. Inserting x 1 provides the identity, the other relationships result from recalculation:

${\ displaystyle p_ {1} (x_ {1}) = x_ {1}, \ quad p_ {2} (x_ {1}) = x_ {2}, \ quad p_ {3} (x_ {1}) = x_ {3}, \ quad p_ {4} (x_ {1}) = x_ {4}, \ quad \ Longrightarrow \ quad \ sigma _ {1}: \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right) \ mapsto \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right)}$,
${\ displaystyle p_ {1} (x_ {2}) = x_ {2}, \ quad p_ {2} (x_ {2}) = x_ {1}, \ quad p_ {3} (x_ {2}) = x_ {4}, \ quad p_ {4} (x_ {2}) = x_ {3}, \ quad \ Longrightarrow \ quad \ sigma _ {2}: \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right) \ mapsto \ left (x_ {2}, x_ {1}, x_ {4}, x_ {3} \ right)}$,
${\ displaystyle p_ {1} (x_ {3}) = x_ {3}, \ quad p_ {2} (x_ {3}) = x_ {4}, \ quad p_ {3} (x_ {3}) = x_ {1}, \ quad p_ {4} (x_ {3}) = x_ {2}, \ quad \ Longrightarrow \ quad \ sigma _ {3}: \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right) \ mapsto \ left (x_ {3}, x_ {4}, x_ {1}, x_ {2} \ right)}$,
${\ displaystyle p_ {1} (x_ {4}) = x_ {4}, \ quad p_ {2} (x_ {4}) = x_ {3}, \ quad p_ {3} (x_ {4}) = x_ {2}, \ quad p_ {4} (x_ {4}) = x_ {1}, \ quad \ Longrightarrow \ quad \ sigma _ {4}: \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right) \ mapsto \ left (x_ {4}, x_ {3}, x_ {2}, x_ {1} \ right)}$.

{ } is thus the Galois group of the polynomial . ${\ displaystyle \ textstyle \ sigma _ {1}, \ sigma _ {2}, \ sigma _ {3}, \ sigma _ {4}}$${\ displaystyle \ left (x ^ {2} -5 \ right) ^ {2} -24}$

## Modern approach

The modern approach, which goes back to Richard Dedekind , formulates the Galois theory in the language of algebraic structures : Based on a body extension , the Galois group is defined as the group of all body automorphisms of , which hold the elements of individually. ${\ displaystyle L / K}$${\ displaystyle L}$${\ displaystyle K}$

Here is a decay field of the given polynomial, i.e. a smallest expansion field of , in which the polynomial decays into linear factors. It is called the normal or Galois extension body of . The Galois group , consisting of those automorphisms of , which leave the sub-body fixed element-wise, necessarily also leaves every term fixed whose value is an element from . ${\ displaystyle L}$${\ displaystyle K}$${\ displaystyle K}$${\ displaystyle L}$${\ displaystyle K}$${\ displaystyle K}$

The reference to the classical procedure of Galois arises when one applies an automorphism of the Galois group to a zero of the corresponding polynomial : ${\ displaystyle \ sigma}$${\ displaystyle \ alpha}$${\ displaystyle p (x) = x ^ {n} + a_ {n-1} x ^ {n-1} + \ dots + a_ {0}}$

${\ displaystyle p (\ alpha) = \ alpha ^ {n} + a_ {n-1} \ alpha ^ {n-1} + \ dots + a_ {0} = 0}$.
${\ displaystyle p (\ sigma (\ alpha)) = (\ sigma (\ alpha)) ^ {n} + a_ {n-1} (\ sigma (\ alpha)) ^ {n-1} + \ dots + a_ {0}}$.

Because it is a body homomorphism and also leaves the coefficients of the polynomial fixed as elements of the body , we get: ${\ displaystyle \ sigma}$${\ displaystyle K}$

${\ displaystyle p (\ sigma (\ alpha)) = \ sigma (\ alpha ^ {n} + a_ {n-1} \ alpha ^ {n-1} + \ dots + a_ {0}) = \ sigma ( p (\ alpha)) = \ sigma (0) = 0}$.

So there is also a zero of the polynomial . This means that the automorphism swaps the zeros. The Galois group thus operates on the set of zeros of the polynomial and acts there as a permutation group. ${\ displaystyle \ sigma (\ alpha)}$${\ displaystyle p}$${\ displaystyle \ sigma}$

Knowledge of solvable groups in group theory allows us to find out whether a polynomial is solvable by radicals , depending on whether its Galois group is solvable or not. Every enlargement of the body belongs to a factor group of the main sequence of the Galois group. If a factor group of the main series is cyclic of the order , the associated body extension is a radical extension, and the elements from can be understood as the -th roots of an element . ${\ displaystyle L / K}$${\ displaystyle n}$${\ displaystyle L}$${\ displaystyle n}$${\ displaystyle K}$

If all the factor groups of the main series are cyclic, the Galois group is said to be solvable, and all elements of the associated field can be obtained from the elements of the basic field (usually ) by successive root extraction, product formation and summation . ${\ displaystyle \ mathbb {Q}}$

One of the greatest triumphs of Galois theory was the proof that for each there is a polynomial with degree that cannot be resolved by radicals. This is due to the fact that the symmetric group contains a simple non-cyclic normal divisor . ${\ displaystyle n> 4}$${\ displaystyle n}$${\ displaystyle n> 4}$ ${\ displaystyle S_ {n}}$

### Law of Galois Theory

If is a finite Galois extension of the body , and the corresponding Galois group , then is Galois over every intermediate body , and there is an inclusion-reversing bijection ${\ displaystyle L}$${\ displaystyle K}$${\ displaystyle \ operatorname {Gal} (L / K)}$${\ displaystyle L}$ ${\ displaystyle Z}$

${\ displaystyle \ {\ mathrm {Zwischenk {\ ddot {o}} rper} \} \ to \ {{\ text {Subgroups of}} \ operatorname {Gal} (L / K) \}}$
${\ displaystyle Z \ mapsto \ operatorname {Gal} (L / Z)}$

Under this bijection, normal body extensions correspond to normal divisions of . ${\ displaystyle M / K}$${\ displaystyle \ operatorname {Gal} (L / K)}$

• ${\ displaystyle [Z \ colon K] = {\ frac {| \ operatorname {Gal} (L / K) |} {| \ operatorname {Gal} (L / Z) |}}}$
• ${\ displaystyle Z \ subset Z '\ \ Rightarrow \ \ operatorname {Gal} (L / Z') \ subset \ operatorname {Gal} (L / Z)}$

A somewhat more general formulation is explained in the article Galois group .

### example

For the example given above, the elements of the Galois group are now to be determined as body automorphisms. The roots of the polynomial are ${\ displaystyle \ textstyle \ left (x ^ {2} -5 \ right) ^ {2} -24}$

${\ displaystyle x_ {1} = {\ sqrt {2}} + {\ sqrt {3}}}$,
${\ displaystyle x_ {2} = {\ sqrt {2}} - {\ sqrt {3}}}$,
${\ displaystyle x_ {3} = - {\ sqrt {2}} + {\ sqrt {3}}}$,
${\ displaystyle x_ {4} = - {\ sqrt {2}} - {\ sqrt {3}}}$.

The disintegration body is thus . A basis for as vector space over is , i.e. H. each element of is of the form with from . It is therefore an algebraic field expansion of degree 4 above . The order of the Galois group corresponds to the degree of the body expansion, its elements permute - as shown above - the zeros of the polynomial as follows: ${\ displaystyle \ mathbb {Q} (x_ {1}, x_ {2}, x_ {3}, x_ {4}) = \ mathbb {Q} ({\ sqrt {2}}, {\ sqrt {3} })}$${\ displaystyle \ mathbb {Q} ({\ sqrt {2}}, {\ sqrt {3}})}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle \ {1, {\ sqrt {2}}, {\ sqrt {3}}, {\ sqrt {6}} \}}$${\ displaystyle \ mathbb {Q} ({\ sqrt {2}}, {\ sqrt {3}})}$${\ displaystyle a + b {\ sqrt {2}} + c {\ sqrt {3}} + d {\ sqrt {6}}}$${\ displaystyle a, b, c, d}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle \ mathbb {Q} ({\ sqrt {2}}, {\ sqrt {3}})}$${\ displaystyle \ mathbb {Q}}$

${\ displaystyle \ sigma _ {1}: \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right) \ mapsto \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right)}$
${\ displaystyle \ sigma _ {2}: \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right) \ mapsto \ left (x_ {2}, x_ {1}, x_ {4}, x_ {3} \ right)}$
${\ displaystyle \ sigma _ {3}: \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right) \ mapsto \ left (x_ {3}, x_ {4}, x_ {1}, x_ {2} \ right)}$
${\ displaystyle \ sigma _ {4}: \ left (x_ {1}, x_ {2}, x_ {3}, x_ {4} \ right) \ mapsto \ left (x_ {4}, x_ {3}, x_ {2}, x_ {1} \ right)}$

${\ displaystyle \ sigma _ {1}}$(as a permutation) the identity remains, but now becomes a body automorphism of : ${\ displaystyle \ sigma _ {1} '}$${\ displaystyle \ mathbb {Q} ({\ sqrt {2}}, {\ sqrt {3}})}$

${\ displaystyle \ sigma _ {1} ': a + b {\ sqrt {2}} + c {\ sqrt {3}} + d {\ sqrt {6}} \ mapsto a + b {\ sqrt {2} } + c {\ sqrt {3}} + d {\ sqrt {6}}}$.

It can be seen that under with the permutation of the four zeros always and are swapped. The corresponding body automorphism is thus: ${\ displaystyle \ sigma _ {2}}$${\ displaystyle {\ sqrt {3}}}$${\ displaystyle - {\ sqrt {3}}}$${\ displaystyle \ sigma _ {2} '}$

${\ displaystyle \ sigma _ {2} ': a + b {\ sqrt {2}} + c {\ sqrt {3}} + d {\ sqrt {6}} \ mapsto a + b {\ sqrt {2} } -c {\ sqrt {3}} - d {\ sqrt {6}}}$.

The body remains fixed element by element. The same applies to for and . The sign changes for both roots below . The corresponding body automorphisms are: ${\ displaystyle \ mathbb {Q} ({\ sqrt {2}})}$${\ displaystyle \ sigma _ {3}}$${\ displaystyle {\ sqrt {2}}}$${\ displaystyle - {\ sqrt {2}}}$${\ displaystyle \ sigma _ {4}}$

${\ displaystyle \ sigma _ {3} ': a + b {\ sqrt {2}} + c {\ sqrt {3}} + d {\ sqrt {6}} \ mapsto from {\ sqrt {2}} + c {\ sqrt {3}} - d {\ sqrt {6}}}$with the fixed body and${\ displaystyle \ mathbb {Q} ({\ sqrt {3}})}$
${\ displaystyle \ sigma _ {4} ': a + b {\ sqrt {2}} + c {\ sqrt {3}} + d {\ sqrt {6}} \ mapsto from {\ sqrt {2}} - c {\ sqrt {3}} + d {\ sqrt {6}}}$with the fixed body .${\ displaystyle \ mathbb {Q} ({\ sqrt {6}})}$

${\ displaystyle \ sigma _ {2} '}$, and are inverse to themselves, so together with identity they each form a subgroup of the Galois group. There are no more real subgroups, because the addition of a further element would already create the entire Galois group. Executing and in succession results in that the Galois group is isomorphic to Klein's group of four and, in particular, is commutative. Therefore all subgroups of the Galois group are also normal divisors. So, according to the main theorem of Galois theory , and are the only intermediate bodies of the body extension . The intermediate bodies themselves are body extensions from degree 2 above . ${\ displaystyle \ sigma _ {3} '}$${\ displaystyle \ sigma _ {4} '}$${\ displaystyle \ sigma _ {2} '}$${\ displaystyle \ sigma _ {3} '}$${\ displaystyle \ sigma _ {4} '}$${\ displaystyle \ mathbb {Q} ({\ sqrt {2}})}$${\ displaystyle \ mathbb {Q} ({\ sqrt {3}})}$${\ displaystyle \ mathbb {Q} ({\ sqrt {6}})}$${\ displaystyle \ mathbb {Q} ({\ sqrt {2}}, {\ sqrt {3}})}$${\ displaystyle \ mathbb {Q}}$

### Kronecker's theorem

The Kronecker set to Galois extensions of the field of rational numbers is one of the classic sets the mathematician Leopold Kronecker and is considered one of the best sets of algebraic number theory . The sentence says:

Every Galois extension with an Abelian Galois group is contained in one of the fields of the circle .${\ displaystyle L / {\ mathbb {Q}}}$ ${\ displaystyle \ mathrm {Gal} (L / {\ mathbb {Q}})}$ ${\ displaystyle \ mathbb {Q} (\ zeta _ {n}) \; (n \ in \ mathbb {N})}$

## Generalizations

In the case of an infinite expansion , the automorphism group can be provided with the so-called Krull topology (according to W. Krull ). If separable and normal (i.e. a Galois extension), there is then a natural bijection between partial extensions and closed subgroups of . ${\ displaystyle L / K}$${\ displaystyle \ mathrm {Aut} (L / K)}$${\ displaystyle L / K}$ ${\ displaystyle K \ subseteq Z \ subseteq L}$${\ displaystyle \ operatorname {Gal} (L / K)}$

If an infinite expansion is not necessarily algebraic, then there is no longer such a general theory: if, for example, a perfect field of the characteristic is through ${\ displaystyle L / K}$${\ displaystyle L}$${\ displaystyle p> 0}$

${\ displaystyle F_ {p} \ colon L \ to L, \ quad x \ mapsto x ^ {p}}$

defines a body automorphism, the so-called Frobenius homomorphism . The subgroup of produced by is generally "much" smaller than the group of automorphisms produced by , but it holds . If is an algebraic closure of , then the subgroup generated by Frobenius automorphism lies close in , that is, its closure is equal to the Galois group. ${\ displaystyle F_ {p}}$${\ displaystyle H}$${\ displaystyle \ mathrm {Aut} \ left (L / \ mathbb {F} _ {p} \ right)}$${\ displaystyle L}$${\ displaystyle L ^ {H} = \ mathbb {F} _ {p}}$${\ displaystyle L}$${\ displaystyle \ mathbb {F} _ {p}}$${\ displaystyle \ mathrm {Gal} \ left ({\ bar {\ mathbb {F} _ {p}}} / \ mathbb {F} _ {p} \ right)}$

If, however, a field extension is with (this does not imply that L / K is algebraic and therefore in particular not Galois), the following still applies: and are mutually inverse, inclusion-reversing bjections between the set of compact subgroups of and the set of intermediate fields where Galois is over . ${\ displaystyle L / K}$${\ displaystyle L ^ {\ mathrm {Gal} (L / K)} = K}$${\ displaystyle H \ mapsto L ^ {H}}$${\ displaystyle M \ mapsto \ mathrm {Gal} (L / M)}$${\ displaystyle \ mathrm {Gal} (L / K)}$${\ displaystyle K \ subseteq M \ subseteq L}$${\ displaystyle L}$${\ displaystyle M}$

There is also a generalization of Galois theory for ring extensions instead of body extensions.

## The inverse problem of Galois theory

It is easy to construct field extensions with any given finite group as a Galois group if one does not specify the basic field. All finite groups therefore appear as Galois groups.

To do this, you choose a body and a finite group . According to Cayley's theorem, is isomorphic to a subgroup of the symmetric group on the elements of . If one chooses variables for each element of and adjoins them , one obtains . In containing the body of symmetric rational functions in the . Then is , and the fixed field from below has Galois group according to the main theorem of Galois theory. ${\ displaystyle K}$${\ displaystyle G}$${\ displaystyle G}$${\ displaystyle G}$${\ displaystyle \ {X_ {a} \} _ {a \ in G}}$${\ displaystyle a}$${\ displaystyle G}$${\ displaystyle K}$${\ displaystyle F = K \ left (\ left \ {X_ {a} \ right \} \ right)}$${\ displaystyle F}$${\ displaystyle L}$${\ displaystyle \ {X_ {a} \}}$${\ displaystyle \ operatorname {Gal} (F / L) = S_ {| G |}}$${\ displaystyle M = F ^ {G}}$${\ displaystyle F}$${\ displaystyle G}$${\ displaystyle G = \ operatorname {Gal} (F / M)}$

The outlined procedure represents the strategy of Emmy Noether (1918) for the solution of the inverse Galois problem, whereby she considered the rational numbers as the basic field . If the fixed field M is a rational function field over the rational numbers, one can, according to Noether, construct a Galois field extension of with Galois group using Hilbert's irreducibility theorem . A counterexample for their strategy was found in 1969 by Richard Swan . It is a generally unsolved problem how and whether such a construction can be carried out for a solid base body, for example . ${\ displaystyle K = \ mathbb {Q}}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle G}$${\ displaystyle \ mathbb {Q}}$

The inverse problem of Galois theory is generally unsolved and asks for a given field K and especially (the rational numbers) whether every finite group can be realized as a Galois group of a field extension of K. If K is a finite field, this is not the case, since in this case the Galois group is cyclic. But the inverse problem can be solved for every finite group for the case of the function field in a variable over the complex numbers or more generally over algebraically closed fields with characteristic 0. Even in the case of the rational numbers there are only partial results. For finite Abelian groups over there has already been solved in the 19th century ( Leopold Kronecker , Heinrich Weber ) and it is solved for finite resolvable groups ( Igor Shafarevich ). The problem is for the sporadic groups via solved except Mathieugruppe M23 (for Mathieugruppen Heinrich Matzat , for the monster group John G. Thompson , which at the same time, most cases of sporadic groups were done). ${\ displaystyle K = \ mathbb {Q}}$${\ displaystyle G}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle \ mathbb {Q}}$