Suppose E is an extension of the field F, and consider the set of all field automorphisms of E which fix F pointwise. This set of automorphisms forms a group G. If there are no elements of E \ F which are fixed by all members of G, then the extension E/F is called a Galois extension, and G is the Galois group of the extension and is usually denoted Gal(E/F).
It can be shown that E is algebraic over F if and only if the Galois group is profinite.
Fundamental theorem of Galois theory. Let E be a finite Galois extension of the field F with Galois group G. For every subgroup H of G, let E^{H} denote the subfield of E consisting of all elements which are fixed by all elements of H. Then the function
Proof: coming up soon
Search Encyclopedia

Featured Article
