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 relation to groups. It was proved by Évariste Galois in his development of Galois theory.

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 F ⊆ K ⊆ E; they are also called subextensions of E/F.)

For any subgroup H of Gal(E/F), the corresponding , denoted E^H, is the set of those elements of E which are fixed by every automorphism in H.

fixed field

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.

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

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 $\{{\text{subgroups of Aut}}(E/F)\}$ to $\{{\text{subfields of }}E/F\}$ , 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).

It is inclusion-reversing. The inclusion of subgroups H₁ ⊆ H₂ holds if and only if the inclusion of fields E^H₁ ⊇ E^H₂ 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:E^H] and |Gal(E/F)|/|H| = [E^H:F].

The field E^H is a 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 E^H induces an isomorphism between Gal(E^H/F) and the quotient group Gal(E/F)/H.

normal extension

The correspondence has the following useful properties.

The trivial subgroup ${1}$ corresponds to the entire extension field $K$ .

The entire group $G$ corresponds to the base field $\mathbb {Q} .$

The subgroup ${1, f}$ corresponds to the subfield $\mathbb {Q} ({\sqrt {3}}),$ since $f$ fixes $\sqrt 3$ .

The subgroup ${1, g}$ corresponds to the subfield $\mathbb {Q} ({\sqrt {2}}),$ since $g$ fixes $\sqrt 2$ .

The subgroup ${1, fg}$ corresponds to the subfield $\mathbb {Q} ({\sqrt {6}}),$ since $fg$ fixes $\sqrt 6$ .

Consider the field

Since $K$ is constructed from the base field $\mathbb {Q}$ by adjoining $\sqrt 2$ , then $\sqrt 3$ , each element of $K$ can be written as:

Its Galois group $G={\text{Gal}}(K/\mathbb {Q} )$ comprises the automorphisms of $K$ which fix $a$ . Such automorphisms must send $\sqrt 2$ to $\sqrt 2$ or $- \sqrt 2$ , and send $\sqrt 3$ to $\sqrt 3$ or $- \sqrt 3$ , since they permute the roots of any irreducible polynomial. Suppose that $f$ exchanges $\sqrt 2$ and $- \sqrt 2$ , so

and $g$ exchanges $\sqrt 3$ and $- \sqrt 3$ , so

These are clearly automorphisms of $K$ , respecting its addition and multiplication. There is also the identity automorphism $e$ which fixes each element, and the composition of $f$ and $g$ which changes the signs on both radicals:

Since the order of the Galois group is equal to the degree of the field extension, $|G|=[K:\mathbb {Q} ]=4$ , there can be no further automorphisms:

which is isomorphic to the Klein four-group. Its five subgroups correspond to the fields intermediate between the base $\mathbb {Q}$ and the extension $K$ .

As always, the trivial group {1} corresponds to the whole field K, while the entire group G to the base field $\mathbb {Q}$ .

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

Consider the splitting field K of the irreducible polynomial $x^{3}-2$ over $\mathbb {Q}$ ; that is, $K=\mathbb {Q} (\theta ,\omega )$ where θ is a cube root of 2, and ω is a cube root of 1 (but not 1 itself). If we consider K inside the complex numbers, we may take $\theta ={\sqrt[{3}]{2}}$ , the real cube root of 2, and $\omega =-{\tfrac {1}{2}}+i{\tfrac {\sqrt {3}}{2}}.$ Since ω has minimal polynomial $x^{2}+x+1$ , the extension $\mathbb {Q} \subset K$ has degree:

Since there are only 3! = 6 such permutations, G must be isomorphic to the symmetric group of all permutations of three objects. The group can be generated by two automorphisms f and g defined by:

and $G=\left\{1,f,f^{2},g,gf,gf^{2}\right\}$ , obeying the relations $f^{3}=g^{2}=(gf)^{2}=1$ . Their effect as permutations of $\alpha _{1},\alpha _{2},\alpha _{3}$ is (in cycle notation): $f=(123),g=(23)$ . Also, g can be considered as the complex conjugation mapping.

The subgroups of G and corresponding subfields are as follows:

Applications[edit]

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.

Galois connection

Milne, J. S. (2022). . Kea Books, Ann Arbor, MI. ISBN 979-8-218-07399-2.

Fields and Galois Theory

Media related to Fundamental theorem of Galois theory at Wikimedia Commons

at PlanetMath.

proof of fundamental theorem of Galois theory

The Stacks Project authors. .

"Theorem 9.21.7 (Fundamental theorem of Galois theory)"