Galois group

The Galois group (after Évariste Galois ) is a group with the help of which field extensions can be examined in algebra .

The intermediate bodies of a body extension can be assigned to certain subgroups of the Galois group. This means that structural studies of body extensions can be linked with group theoretical studies. Since finite Galois groups belong to finite-dimensional field extensions, such structural investigations can often be greatly simplified.

It was historically significant that the classic questions of constructibility - with compasses and ruler - of certain algebraic numbers could thus be translated into a group-theoretical formulation. For details on the classic question of constructibility with compass and ruler , examples and their modern solution, see → Constructible polygon .

definition

Be (read: " over ") an extension of the body. That means: and are bodies and the body is contained in as a sub-ring . Thus, at the same time (not necessarily finite) - vector space . ${\ displaystyle F / K}$${\ displaystyle F}$${\ displaystyle K}$${\ displaystyle K}$${\ displaystyle F}$${\ displaystyle K}$${\ displaystyle F}$${\ displaystyle F}$${\ displaystyle K}$

In this situation the group of all body automorphisms of the extension body , which fix the base body element by element, is called the Galois group from over and is referred to as, formal ${\ displaystyle F}$${\ displaystyle K}$${\ displaystyle F}$${\ displaystyle K}$${\ displaystyle \ mathrm {Gal} (F / K)}$

${\ displaystyle \ mathrm {Gal} (F / K) = \ lbrace \ varphi \ in \ mathrm {Aut} (F) \ mid \ forall k \ in K: \ varphi (k) = k \ rbrace}$.

This can also be formulated like this: The Galois group of over consists exactly of the body automorphisms of , which are at the same time vector space endomorphisms of as -vector space. ${\ displaystyle F}$${\ displaystyle K}$${\ displaystyle F}$${\ displaystyle F}$${\ displaystyle K}$

Galois group of a polynomial

Be a body. The Galois group of the polynomial in the polynomial ring is the group where is a decay field of the polynomial . One speaks in this case of the disintegration body, since disintegration bodies - and thus the Galois group of a polynomial - are uniquely determined except for isomorphism. ${\ displaystyle K}$ ${\ displaystyle f}$ ${\ displaystyle K [x]}$${\ displaystyle \ mathrm {Gal} (F / K)}$${\ displaystyle F}$${\ displaystyle f}$

The decay field of a polynomial is normal above the base field . In this case the - here finite-dimensional - body expansion is Galois if the irreducible factors are separable from . The article Galois theory deals with the concept of the Galois group of a polynomial, in this case the first version of the main clause mentioned below is sufficient - the main clause for finite Galois extensions. ${\ displaystyle F}$${\ displaystyle K}$${\ displaystyle F / K}$${\ displaystyle K}$${\ displaystyle f}$

Different meanings of the term

The Galois group is particularly useful when the body extension is a Galois extension (see below). In the literature, the term “Galois group” is often only used in this case. The group of automorphisms of used in this article is then denoted by. ${\ displaystyle F / K}$${\ displaystyle K}$${\ displaystyle F}$${\ displaystyle \ mathrm {Aut} _ {K} (F)}$

properties

• The Galois group is a subgroup of the automorphism group of .${\ displaystyle F}$
• Is the body enlargement finite, i.e. H. is finite dimensional over , the group order is less than or equal to the degree of expansion . In this case the minimal polynomial of over exists for every element of the body . If is a finite Galois expansion, then we have .${\ displaystyle F / K}$${\ displaystyle F}$${\ displaystyle K}$${\ displaystyle \ mathrm {Gal} (F / K)}$ ${\ displaystyle [F: K]}$${\ displaystyle \ alpha \ in F}$ ${\ displaystyle f = m_ {K, \ alpha}}$${\ displaystyle \ alpha}$${\ displaystyle K}$${\ displaystyle F / K}$${\ displaystyle | \ mathrm {Gal} (F / K) | = [F: K]}$
• Let be a decay field of the polynomial over . Every automorphism from the Galois group of the polynomial maps a zero from back to a zero. The Galois group operates on the set of zeros in the body as a permutation group and is therefore isomorphic to a subgroup of the symmetrical group . For a separables about irreducible polynomial this operation is even transitive, that is, to two distinct zeros there is an element of the Galois group, which on maps: .${\ displaystyle F}$${\ displaystyle f}$${\ displaystyle K}$${\ displaystyle \ mathrm {Gal} (F / K)}$${\ displaystyle f}$${\ displaystyle f}$${\ displaystyle N = \ {u_ {1}, u_ {2}, \ ldots u_ {n} \}}$${\ displaystyle f}$${\ displaystyle F}$ ${\ displaystyle S_ {n}}$${\ displaystyle K}$${\ displaystyle f}$${\ displaystyle u_ {j} \ neq u_ {k}}$${\ displaystyle \ varphi}$${\ displaystyle u_ {j}}$${\ displaystyle u_ {k}}$${\ displaystyle \ varphi (u_ {j}) = u_ {k}}$

Galois Correspondence, Completed Subgroups and Intermediate Bodies

One can assign the subgroup of the Galois group to each intermediate body of the extension , the elements of which remain fixed element by element, and vice versa to each subgroup of the intermediate body it fixes. According to Hungerford (1981) , the "priming notation" is used here for both assignments, both of which are also referred to as Galois correspondence : ${\ displaystyle L}$${\ displaystyle F / K}$${\ displaystyle G = \ mathrm {Gal} (F / K)}$${\ displaystyle L}$${\ displaystyle H}$${\ displaystyle \ mathrm {Gal} (F / K)}$

${\ displaystyle L ^ {\ prime}: = \ mathrm {Gal} (F / L): = \ {\ varphi \ in \ mathrm {Gal} (F / K) \, | \, \ forall l \ in L : \ varphi (l) = l \}}$
${\ displaystyle H ^ {\ prime}: = \ {f \ in F \, | \, \ forall \ eta \ in H: \ eta (f) = f \}}$

The following relationships apply to intermediate bodies and the extension, subgroups and from : ${\ displaystyle L}$${\ displaystyle M}$${\ displaystyle H}$${\ displaystyle J}$${\ displaystyle G}$

1. ${\ displaystyle F ^ {\ prime} = 1}$and ,${\ displaystyle K ^ {\ prime} = G}$
2. ${\ displaystyle 1 ^ {\ prime} = F}$,
3. ${\ displaystyle L \ subseteq M \ Rightarrow M ^ {\ prime} \ leq L ^ {\ prime}}$,
4. ${\ displaystyle H \ leq J \ Rightarrow J ^ {\ prime} \ subseteq H ^ {\ prime}}$,
5. ${\ displaystyle L \ subseteq L ^ {\ prime \ prime}}$and ,${\ displaystyle H \ leq H ^ {\ prime \ prime}}$
6. ${\ displaystyle L ^ {\ prime} = L ^ {\ prime \ prime \ prime}}$and .${\ displaystyle H ^ {\ prime} = H ^ {\ prime \ prime \ prime}}$

The expansion of the body is called Galois expansion if it is normal and separable. This is exactly the case if applies, i.e. if the Galois group does not fix any other elements of apart from the basic body . Since holds in all cases , the expansion is Galois if and only if is. The same condition applies to intermediate fields : The extension is a Galois extension if and only if applies. The terms normal and separable are defined in the article body enlargement regardless of the classifications used here. There, in the Galois extension section, the same is defined for the case that the extension is algebraic. According to Emil Artin and Hungerford (1981) , the definition used here does not allow algebraic extensions. ${\ displaystyle F / K}$${\ displaystyle G ^ {\ prime} = K}$${\ displaystyle F}$${\ displaystyle K ^ {\ prime} = G}$${\ displaystyle K = K ^ {\ prime \ prime}}$${\ displaystyle L}$${\ displaystyle F / L}$${\ displaystyle L = L ^ {\ prime \ prime}}$

Seclusion

According to Hungerford (1981) a subgroup of the Galois group or an intermediate body of the extension is called closed , if this applies. ${\ displaystyle X}$${\ displaystyle X}$${\ displaystyle X = X ^ {\ prime \ prime}}$

• All objects that appear as images of the correspondence described above are completed (after 6.).${\ displaystyle X = Y ^ {\ prime}}$
• The trivial subgroup 1, and are closed.${\ displaystyle G}$${\ displaystyle F}$
• The expansion is a Galois expansion if and only if is complete.${\ displaystyle F / K}$${\ displaystyle K}$

With the terms agreed at the beginning of the section, the following applies:

• If is complete and is finite, then is complete and it applies .${\ displaystyle L}$${\ displaystyle [M: L]}$${\ displaystyle M}$${\ displaystyle [L ^ {\ prime}: M ^ {\ prime}] = [M: L]}$
• If is complete and is finite, then is complete and .${\ displaystyle H}$${\ displaystyle [J: H]}$${\ displaystyle J}$${\ displaystyle [H ^ {\ prime}: J ^ {\ prime}] = [J: H]}$
• Especially applies (for ): Every finite subgroup of the Galois group is closed.${\ displaystyle H = 1}$
• If is a finite-dimensional Galois extension of , then all intermediate fields and all subgroups of the Galois group are closed and the Galois group has the order .${\ displaystyle F}$${\ displaystyle K}$${\ displaystyle [F: K]}$

Laws of Galois Theory

Finite-dimensional body expansion

If there is a finite-dimensional Galois extension of , then the Galois correspondence mediates a bijection between the set of intermediate fields and the set of subgroups of the Galois group. This correspondence maps the subset lattice of the intermediate bodies (with the order ) to the lattice of the subgroups (with the order ) in an orderly manner, whereby the subset relationship is reversed. The following applies: ${\ displaystyle F}$${\ displaystyle K}$${\ displaystyle \ subset}$${\ displaystyle>}$

1. The relative dimension of two intermediate bodies is equal to the relative index of the corresponding subgroups.
2. ${\ displaystyle F}$is Galois over every intermediate body . The Galois group coincides with the subgroup .${\ displaystyle L}$${\ displaystyle \ mathrm {Gal} (F / L)}$${\ displaystyle L ^ {\ prime}}$
3. An intermediate field is Galois if and only if the corresponding subgroup is a normal subgroup of the Galois group . In this case, which is the factor group isomorphic to the Galois group of the body over .${\ displaystyle L}$${\ displaystyle K}$${\ displaystyle L ^ {\ prime}}$${\ displaystyle G = \ mathrm {Gal} (F / K)}$ ${\ displaystyle G / L ^ {\ prime}}$${\ displaystyle \ mathrm {Gal} (L / K)}$${\ displaystyle L}$${\ displaystyle K}$

Infinite dimensional algebraic expansion

If an algebraic, not necessarily finite-dimensional Galois extension of , then the Galois correspondence mediates a bijection between the set of all intermediate fields and the set of closed subgroups of the Galois group. This correspondence maps the subset lattice of the intermediate bodies (with the order ) to the lattice of the closed subgroups (with the order ) in an orderly manner, whereby the subset relationship is reversed. The following applies: ${\ displaystyle F}$${\ displaystyle K}$${\ displaystyle \ subset}$${\ displaystyle>}$

1. ${\ displaystyle F}$is Galois over every intermediate body . The Galois group coincides with the subgroup .${\ displaystyle L}$${\ displaystyle \ mathrm {Gal} (F / L)}$${\ displaystyle L ^ {\ prime}}$
2. An intermediate field is Galois if and only if the corresponding subgroup is a normal subgroup of the Galois group . In this case, the factor group is isomorphic to the Galois group of the body over .${\ displaystyle L}$${\ displaystyle K}$${\ displaystyle L ^ {\ prime}}$${\ displaystyle G = \ mathrm {Gal} (F / K)}$${\ displaystyle G / L ^ {\ prime}}$${\ displaystyle \ mathrm {Gal} (L / K)}$${\ displaystyle L}$${\ displaystyle K}$

Examples

• The complex numbers are a body and contain the body of the real numbers . So is a body enlargement. Since a vector space of dimension 2 is over ( is a basis ), we have . The Galois group contains the identity and the complex conjugation . The set of roots of the minimal polynomial is . The identity maps these two elements back onto itself as they are permuted by the complex conjugation . So the Galois group is restricted to the set of roots isomorphic to the symmetric group .${\ displaystyle \ mathbb {C} / \ mathbb {R}}$${\ displaystyle \ mathbb {C}}$${\ displaystyle \ mathbb {R}}$${\ displaystyle (1, i)}$${\ displaystyle [\ mathbb {C}: \ mathbb {R}] = 2}$${\ displaystyle f = X ^ {2} +1}$${\ displaystyle \ lbrace i, -i \ rbrace}$${\ displaystyle S_ {2}}$
• More detailed examples for the calculation of a Galois group:
• Be the body of the rational functions above . Then for every number the mapping defined by is an -automorphism. If the body is infinite, there are infinitely many of these automorphisms and the Galois group is an infinite group. If the element itself is not a root of unity, then the subgroup of generated by the automorphism is not closed.${\ displaystyle F = K (x)}$${\ displaystyle \ rho}$${\ displaystyle K}$${\ displaystyle a \ in K \ setminus \ {0 \}}$${\ displaystyle \ varphi _ {a} \ colon \ rho (x) \ mapsto \ rho (ax)}$${\ displaystyle K}$${\ displaystyle K}$${\ displaystyle G = \ mathrm {Gal} (F / K)}$${\ displaystyle a \ neq 0}$${\ displaystyle \ varphi _ {a}}$${\ displaystyle G}$
• The field of real numbers does not allow any nontrivial automorphisms, because its arrangement is an algebraic invariant: It is for two real numbers if and only if there is a square. Therefore the field of real numbers is not Galois over any of its proper subfields; the same is true for the field of real algebraic numbers.${\ displaystyle r \ leq s}$${\ displaystyle sr}$
• More generally this applies to all Euclidean fields : the Galois group of a Euclidean field over one of its subfields is always the trivial group .

Galois group of a cubic polynomial

The following detailed example shows , using the polynomial , how intermediate fields can be determined with the aid of the Galois group. ${\ displaystyle f (x) = x ^ {3} -2}$

The real number of the above generated number field has the Galois group 1 because no other zeros of the minimal polynomial of (real!) In number fields are. So this extension is not Galois. Its degree is 3 because it is isomorphic to the factor ring (see factor ring ). The same applies to the two number fields and , which are generated by the two non-real roots and by over . All three fields are isomorphic intermediate fields of the decay field of the polynomial . ${\ displaystyle \ xi _ {1} = {\ sqrt [{3}] {2}}}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle L_ {1} = \ mathbb {Q} \ left ({\ sqrt [{3}] {2}} \ right)}$${\ displaystyle f (x) = x ^ {3} -2}$${\ displaystyle \ xi _ {1}}$${\ displaystyle L_ {1}}$${\ displaystyle L_ {1}}$${\ displaystyle \ mathbb {Q} (x) / (f)}$${\ displaystyle L_ {2} = \ mathbb {Q} (\ xi _ {2})}$${\ displaystyle L_ {3} = \ mathbb {Q} (\ xi _ {3})}$${\ displaystyle \ xi _ {2} = {\ sqrt [{3}] {2}} \ cdot \ exp \ left ({\ frac {2 \ pi i} {3}} \ right)}$${\ displaystyle \ xi _ {3} = {\ sqrt [{3}] {2}} \ cdot \ exp \ left ({\ frac {4 \ pi i} {3}} \ right)}$${\ displaystyle f}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle F}$${\ displaystyle f}$

Since the basic field is perfect as a field with the characteristic 0 , the desired decay field is a Galois extension of and the Galois group must operate transitively on the zeros of . The only real subgroup of the symmetric group that operates transitively on is the normal subgroup of , generated by the 3-cycle , the alternating group . Since we have already identified three real intermediate bodies and which have no real subgroups, it cannot yet be the full Galois group. So this can only be the full symmetrical group, so it applies ${\ displaystyle \ mathbb {Q}}$${\ displaystyle F = \ mathbb {Q} (\ xi _ {1}, \ xi _ {2}, \ xi _ {3})}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle G}$${\ displaystyle f}$${\ displaystyle S_ {3}}$${\ displaystyle \ {1,2,3 \}}$${\ displaystyle (1,2,3)}$${\ displaystyle S_ {3}}$ ${\ displaystyle A_ {3}}$${\ displaystyle A_ {3}}$

${\ displaystyle \ operatorname {Gal} (F / \ mathbb {Q}) = S_ {3}}$.

In addition to the intermediate bodies that we have already identified, there must also be a normal intermediate body that is two-dimensional over (index of ). This remains fixed with cyclical interchanges of the zeros, this only applies to the circle division field of the third unit root, which is generated by the unit root . All results are shown in the graph below. ${\ displaystyle E}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle A_ {3}}$${\ displaystyle \ omega = \ exp \ left ({\ frac {2 \ pi i} {3}} \ right) = {\ frac {\ xi _ {2}} {\ xi _ {1}}} = { \ frac {\ xi _ {3}} {\ xi _ {2}}} = {\ frac {\ xi _ {1}} {\ xi _ {3}}}}$

Subgroup association of the Galois group and interbody association of the body extension F in the example. The arrows in the left diagram can be read as "is a subgroup of" (thin) or "is normal divisor of" (thick), in the right diagram as "is an extension of" (thin) or "is a Galois extension of" (thick) . The numbers on the arrows indicate relative indices in the left diagram and the relative dimension of the expansion in the right diagram. If the two graphs are pushed over one another, the objects that correspond to one another in the Galois correspondence come to rest on one another. So z. B. the real field L1 is fixed by the group <(2,3)>, the generating automorphism, which swaps the two non-real roots of f , is the restriction of the complex conjugation on F.

The intermediate bodies can now be used, among other things, to obtain various representations of the disintegration body:

• ${\ displaystyle F = \ mathbb {Q} (\ xi _ {1}, \ xi _ {2}, \ xi _ {3})}$, this follows - entirely without Galois theory - from its definition as a disintegrating body.
• ${\ displaystyle F = \ mathbb {Q} (\ xi _ {1}, \ xi _ {2})}$: That two zeros are sufficient for generation, follows from the fact that there are no other bodies between the bodies that are generated by a zero .${\ displaystyle F}$
• ${\ displaystyle F = E (\ xi _ {1}) = \ mathbb {Q} (\ omega, \ xi _ {1})}$: Here the (in this case the only maximum) subnormal series of the Galois group is reproduced (in the graphic the path on the far right). The relative extensions in are all Galois and their Galois groups are simple Abelian groups.${\ displaystyle \ mathbb {Q} \ subset E \ subset E (\ xi _ {1})}$
• ${\ displaystyle F}$can also be represented as a simple body extension: is certainly an element of and is not fixed by any nontrivial element of the Galois group. Hence is .${\ displaystyle \ omega + \ xi _ {1}}$${\ displaystyle F}$${\ displaystyle F = \ mathbb {Q} (\ omega + \ xi _ {1})}$

Of course, the zeros can be exchanged in any of the above representations . ${\ displaystyle \ xi _ {k}}$