# Fundamental theorem of Galois theory

In mathematics, the fundamental theorem of Galois theory is a result that describes the structure of certain types of field extensions.

In its most basic form, the theorem asserts that given a field extension E/F that is finite and Galois, there is a one-to-one correspondence between its intermediate fields and subgroups of its Galois group. (Intermediate fields are fields K satisfying FKE; they are also called subextensions of E/F.)

## Explicit description of the correspondence

For finite extensions, the correspondence can be described explicitly as follows.

• For any subgroup H of Gal(E/F), the corresponding fixed field, denoted EH, is the set of those elements of E which are fixed by every automorphism in H.
• For any intermediate field K of E/F, the corresponding subgroup is Aut(E/K), that is, the set of those automorphisms in Gal(E/F) which fix every element of K.

The fundamental theorem says that this correspondence is a one-to-one correspondence if (and only if) E/F is a Galois extension. For example, the topmost field E corresponds to the trivial subgroup of Gal(E/F), and the base field F corresponds to the whole group Gal(E/F).

The notation Gal(E/F) is only used for Galois extensions. If E/F is Galois, then Gal(E/F) = Aut(E/F). If E/F is not Galois, then the "correspondence" gives only an injective (but not surjective) map from ${\displaystyle \{}$ subgroups of Aut(E/F)${\displaystyle \}}$  to ${\displaystyle \{}$ subfields of E/F${\displaystyle \}}$ , and a surjective (but not injective) map in the reverse direction. In particular, if E/F is not Galois, then F is not the fixed field of any subgroup of Aut(E/F).

## Properties of the correspondence

The correspondence has the following useful properties.

• It is inclusion-reversing. The inclusion of subgroups H1 ⊆ H2 holds if and only if the inclusion of fields EH1EH2 holds.
• Degrees of extensions are related to orders of groups, in a manner consistent with the inclusion-reversing property. Specifically, if H is a subgroup of Gal(E/F), then |H| = [E:EH] and |Gal(E/F)|/|H| = [EH:F].
• The field EH is a normal extension of F (or, equivalently, Galois extension, since any subextension of a separable extension is separable) if and only if H is a normal subgroup of Gal(E/F). In this case, the restriction of the elements of Gal(E/F) to EH induces an isomorphism between Gal(EH/F) and the quotient group Gal(E/F)/H.

## Example 1

Lattice of subgroups and subfields

Consider the field

${\displaystyle K=\mathbb {Q} \left({\sqrt {2}},{\sqrt {3}}\right)=\left[\mathbb {Q} ({\sqrt {2}})\right]({\sqrt {3}}).}$

Since K is first determined by adjoining 2, then 3, each element of K can be written as:

${\displaystyle \left(a+b{\sqrt {2}}\right)+\left(c+d{\sqrt {2}}\right){\sqrt {3}},\qquad a,b,c,d\in \mathbb {Q} .}$

Its Galois group ${\displaystyle G={\text{Gal}}(K/\mathbb {Q} )}$  can be determined by examining the automorphisms of K which fix a. Each such automorphism must send 2 to either 2 or 2, and must send 3 to either 3 or 3 since the permutations in a Galois group can only permute the roots of an irreducible polynomial. Suppose that f exchanges 2 and 2, so

${\displaystyle f\left((a+b{\sqrt {2}})+(c+d{\sqrt {2}}){\sqrt {3}}\right)=(a-b{\sqrt {2}})+(c-d{\sqrt {2}}){\sqrt {3}}=a-b{\sqrt {2}}+c{\sqrt {3}}-d{\sqrt {6}},}$

and g exchanges 3 and 3, so

${\displaystyle g\left((a+b{\sqrt {2}})+(c+d{\sqrt {2}}){\sqrt {3}}\right)=(a+b{\sqrt {2}})-(c+d{\sqrt {2}}){\sqrt {3}}=a+b{\sqrt {2}}-c{\sqrt {3}}-d{\sqrt {6}}.}$

These are clearly automorphisms of K. There is also the identity automorphism e which does not change anything, and the composition of f and g which changes the signs on both radicals:

${\displaystyle (fg)\left((a+b{\sqrt {2}})+(c+d{\sqrt {2}}){\sqrt {3}}\right)=(a-b{\sqrt {2}})-(c-d{\sqrt {2}}){\sqrt {3}}=a-b{\sqrt {2}}-c{\sqrt {3}}+d{\sqrt {6}}.}$

Therefore,

${\displaystyle G=\left\{1,f,g,fg\right\},}$

and G is isomorphic to the Klein four-group. It has five subgroups, each of which correspond via the theorem to a subfield of K.

• The trivial subgroup (containing only the identity element) corresponds to all of K.
• The entire group G corresponds to the base field ${\displaystyle \mathbb {Q} .}$
• The two-element subgroup {1, f} corresponds to the subfield ${\displaystyle \mathbb {Q} ({\sqrt {3}}),}$  since f fixes 3.
• The two-element subgroup {1, g} corresponds to the subfield ${\displaystyle \mathbb {Q} ({\sqrt {2}}),}$  again since g fixes 2.
• The two-element subgroup {1, fg} corresponds to the subfield ${\displaystyle \mathbb {Q} ({\sqrt {6}}),}$  since fg fixes 6.

## Example 2

Lattice of subgroups and subfields

The following is the simplest case where the Galois group is not abelian.

Consider the splitting field K of the polynomial ${\displaystyle x^{3}-2}$  over ${\displaystyle \mathbb {Q} ;}$  that is, ${\displaystyle K=\mathbb {Q} (\omega ,\theta )}$  where θ is a cube root of 2, and ω is a cube root of 1 (but not 1 itself). For example, if we imagine K to be inside the field of complex numbers, we may take θ to be the real cube root of 2, and ω to be

${\displaystyle \omega ={\frac {-1}{2}}+i{\frac {\sqrt {3}}{2}}.}$

It can be shown that the Galois group ${\displaystyle G={\text{Gal}}(K/\mathbb {Q} )}$  has six elements, and is isomorphic to the group of permutations of three objects. It is generated by (for example) two automorphisms, say f and g, which are determined by their effect on θ and ω,

${\displaystyle f(\theta )=\omega \theta ,\quad f(\omega )=\omega ,}$
${\displaystyle g(\theta )=\theta ,\quad g(\omega )=\omega ^{2},}$

and then

${\displaystyle G=\left\{1,f,f^{2},g,gf,gf^{2}\right\}.}$

The subgroups of G and corresponding subfields are as follows:

• As usual, the entire group G corresponds to the base field ${\displaystyle \mathbb {Q} }$  and the trivial group {1} corresponds to the whole field K.
• There is a unique subgroup of order 3, namely ${\displaystyle \{1,f,f^{2}\}.}$  The corresponding subfield is ${\displaystyle \mathbb {Q} (\omega ),}$  which has degree 2 over ${\displaystyle \mathbb {Q} }$  (the minimal polynomial of ω is ${\displaystyle x^{2}+x+1}$ ), corresponding to the fact that the subgroup has index two in G. Also, this subgroup is normal, corresponding to the fact that the subfield is normal over ${\displaystyle \mathbb {Q} .}$
• There are three subgroups of order 2, namely ${\displaystyle \{1,g\},\{1,gf\}}$  and ${\displaystyle \{1,gf^{2}\},}$  corresponding respectively to the three subfields ${\displaystyle \mathbb {Q} (\theta ),\mathbb {Q} (\omega \theta ),\mathbb {Q} (\omega ^{2}\theta ).}$  These subfields have degree 3 over ${\displaystyle \mathbb {Q} ,}$  again corresponding to the subgroups having index 3 in G. Note that the subgroups are not normal in G, and this corresponds to the fact that the subfields are not Galois over ${\displaystyle \mathbb {Q} .}$  For example, ${\displaystyle \mathbb {Q} (\theta )}$  contains only a single root of the polynomial ${\displaystyle x^{3}-2,}$  so it cannot be normal over ${\displaystyle \mathbb {Q} .}$

## Example 3

Let ${\displaystyle E=\mathbb {Q} (\lambda )}$  be the field of rational functions in ${\displaystyle \lambda }$  and let

${\displaystyle G=\left\lbrace {\lambda ,{\frac {1}{1-\lambda }},{\frac {\lambda -1}{\lambda }},{\frac {1}{\lambda }},{\frac {\lambda }{\lambda -1}},1-\lambda }\right\rbrace \subset {\rm {Aut}}(E)}$

which is a group under composition, isomorphic to ${\displaystyle S_{3}}$  (see: six cross-ratios). Let ${\displaystyle F}$  be the fixed field of ${\displaystyle G}$ , then ${\displaystyle {\rm {Gal}}(E/F)=G}$ .

If ${\displaystyle H}$  is a subgroup of ${\displaystyle G}$  then the coefficients of the following polynomial

${\displaystyle P(T):=\prod _{h\in H}(T-h)\in E[T]}$

generate the fixed field of ${\displaystyle H}$ . Galois correspondence means that every subfield of ${\displaystyle E/F}$  can be constructed this way. For example, if ${\displaystyle H=\{\lambda ,1-\lambda \}}$  then the fixed field is ${\displaystyle \mathbb {Q} (\lambda (1-\lambda ))}$  and if ${\displaystyle H=\{\lambda ,1/\lambda \}}$  then the fixed field is ${\displaystyle \mathbb {Q} (\lambda +1/\lambda )}$ . Likewise, one can write ${\displaystyle F}$ , the fixed field of ${\displaystyle G}$ , as ${\displaystyle \mathbb {Q} (j),}$  where j is the j-invariant.

Similar examples can be constructed for each of the symmetry groups of the platonic solids as these also have faithful actions on the projective line ${\displaystyle \mathbb {P} ^{1}(\mathbb {C} )}$  and hence on ${\displaystyle \mathbb {C} (x)}$ .

## Applications

The theorem classifies the intermediate fields of E/F in terms of group theory. This translation between intermediate fields and subgroups is key to showing that the general quintic equation is not solvable by radicals (see Abel–Ruffini theorem). One first determines the Galois groups of radical extensions (extensions of the form F(α) where α is an n-th root of some element of F), and then uses the fundamental theorem to show that solvable extensions correspond to solvable groups.

Theories such as Kummer theory and class field theory are predicated on the fundamental theorem.

## Infinite case

Given an infinite algebraic extension we can still defined it to be Galois if it is normal and seperable. The problem that one encounters in the infinite case is that the bijection in the fundamental theorem does not hold as we get too many subgroups generally. More precisely if we just take every subgroup we can in general find two different subgroups that fix the same interemidiate field. Therefore we amend this by introducing a topology on the Galois group.

Let ${\displaystyle E/F}$  be a Galois extension (possible infinite) and let ${\displaystyle G={\text{Gal}}(E/F)}$  be the Galois group of the extension. Let

${\displaystyle {\text{Int}}_{\text{F}}(E/F)=\{G_{i}={\text{Gal}}(L_{i}/F)~|~L_{i}/F{\text{ is a finite Galois extension and }}L_{i}\subseteq E\}}$

be the set of the Galois groups of all finite intermidiate Galois extension. Note that for all ${\displaystyle i\in I}$  we can define the maps ${\displaystyle \varphi _{i}:G\rightarrow G_{i}}$  by ${\displaystyle \sigma \mapsto \sigma _{|L_{i}}}$ . We then define the Krull Topology on ${\displaystyle G}$  to be weakest topology such that for all ${\displaystyle i\in I}$  the maps ${\displaystyle \varphi _{i}:G\rightarrow G_{i}}$  are continuous, where we endow each ${\displaystyle G_{i}}$  with the discrete topology. Stated differently ${\displaystyle G\cong \varprojlim G_{i}}$  as an inverse limit of topological groups (where again each ${\displaystyle G_{i}}$  is endowed with the discrete topology). This makes ${\displaystyle G}$  a profinite group (in fact every profinite group can be realised as the Galois group of an Galois extension, see for example [1]). Note that when ${\displaystyle E/F}$  is finite, the Krull topology is the discrete topology.

Now that we have defined a topology on the Galois group we can restate the fundamental theorem for infinite Galois extension.

Let ${\displaystyle {\mathcal {F}}}$  denote the set of all finite intermediate field extension of ${\displaystyle E/F}$  and let ${\displaystyle {\mathcal {C}}}$  denote the set of all closed subgroups of ${\displaystyle G={\text{Gal}}(E/F)}$  endowed with the Krull topology. Then there exists a bijection between ${\displaystyle {\mathcal {F}}}$  and ${\displaystyle {\mathcal {C}}}$  given by the map

${\displaystyle \Phi :{\mathcal {F}}(E/F)\rightarrow {\mathcal {C}}(G)}$

defined by ${\displaystyle L\mapsto {\text{Gal}}(E/L)}$  and the map
${\displaystyle \Gamma :{\mathcal {C}}(G)\rightarrow {\mathcal {F}}(E/F)}$

defined by ${\displaystyle N\mapsto {\text{Fix}}_{E}(N):=\{a\in E~|~\sigma (a)=a{\text{ for all }}\sigma \in N\}}$ . One important thing one need to check is that ${\displaystyle \Phi }$  is a well-defined map, that is that ${\displaystyle \Phi (L)}$  is a closed subgroup of ${\displaystyle G}$  for all intermediate. For a proof see for example [1]

## References

1. ^ a b Ribes, Zalesskii (2010). Profinite groups. Springer. ISBN 978-3-642-01641-7.