Fundamental Theorem of Galois Theory - Example

Example

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

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 since the permutations in a Galois group can only permute the roots of an irreducible polynomial. Suppose that f exchanges √2 and −√2, so

and g exchanges √3 and −√3, so

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:

Therefore

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.