Fundamental theorem of Galois theory

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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 which is finite and Galois, there is a one-to-one correspondence between its intermediate fields (fields K satisfying F ⊆ K ⊆ E; also called subextensions of E/F) and subgroups of its Galois group.

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[edit] Proof

The proof of the fundamental theorem is not trivial. The crux in the usual treatment is a rather delicate result of Emil Artin which allows one to control the dimension of the intermediate field fixed by a given group of automorphisms. See linear independence of automorphisms of a field.

There is also a fairly simple proof using the primitive element theorem. This proof seems to be ignored by most modern treatments, possibly because it requires a separate (but easier) proof in the case of finite fields. See [1]

In terms of its abstract structure, there is a Galois connection; most of its properties are fairly formal, but the actual isomorphism of the posets requires some work.

[edit] 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 field, usually 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 just Aut(E/K), that is, the set of those automorphisms in Gal(E/F) which fix every element of K.

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).

[edit] 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 EH1 ⊇ EH2 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 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.

[edit] Example

Consider the field K = Q(√2, √3) = Q(√2)(√3). Since K is first determined by adjoining √2, then √3, a typical element of K can be written as:

(a+b\sqrt{2})+(c+d\sqrt{2})\sqrt{3},

where a, b, c, d are rational numbers. Its Galois group G = Gal (K/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. Suppose that f exchanges √2 and −√2, so

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

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:

(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}.
diagram of subgroups and subfields
diagram of subgroups and subfields

Therefore

G = \left\{e, 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 Q.
  • The two-element subgroup {1, f} corresponds to the subfield Q(√3), since f fixes √3.
  • The two-element subgroup {1, g} corresponds to the subfield Q(√2), again since g fixes √2.
  • The two-element subgroup {1, fg} corresponds to the subfield Q(√6), since fg fixes √6.

[edit] Example

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

Consider the splitting field K of the polynomial x3−2 over Q; that is, K = Q (θ, ω), 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

\omega = \frac{-1}2 + i\frac{\sqrt3}2.

It can be shown that the Galois group G = Gal (K/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,

f(\theta) = \omega \theta, \quad f(\omega) = \omega,
g(\theta) = \theta, \quad g(\omega) = \omega^2,

and then

diagram of subgroups and subfields
diagram of subgroups and subfields
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 Q, and the trivial group {1} corresponds to the whole field K.
  • There is a unique subgroup of order 3, namely {1, f, f2}. The corresponding subfield is Q(ω), which has degree two over Q (the minimal polynomial of ω is x2 + 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 Q.
  • There are three subgroups of order 2, namely {1, g}, {1, gf} and {1, gf2}, corresponding respectively to the three subfields Q(θ), Q(ωθ), Q(ω2θ). These subfields have degree three over 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 Q. For example, Q(θ) contains only a single root of the polynomial x3−2, so it cannot be normal over Q.


[edit] Applications

The theorem converts the difficult-sounding problem of classifying the intermediate fields of E/F into the more tractable problem of listing the subgroups of a certain finite group.

For example, to prove that the general quintic equation is not solvable by radicals (see Abel-Ruffini theorem), one first restates the problem in terms 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 convert this statement into a problem about groups. That can then be attacked directly.

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

[edit] Infinite case

There is also a version of the fundamental theorem that applies to infinite algebraic extensions, which are normal and separable. It involves defining a certain topological structure, the Krull topology, on the Galois group; only subgroups that are also closed sets are relevant in the correspondence.

  1. ^ Marcus, Daniel: "Number Fields", Appendix 2. Springer-Verlag, 1977.