Lectures on the Algebraic Theory of Fields - Tata Institute of ...

7. Galois extensions 39Proof. Let k be the fixed field of the group G(K/k) of k-automorphismsof K. Letω∈K. Letω 1 (=ω),...ω n be all the distinct conjugates ofωthat lie in K. Consider the polynomialf (x)=(x−ω 1 )···(x−ω n )Ifσis an element of G(K/k),σ permutesω 1 ,...,ω n so thatσleavesthe polynomial f (x) unaltered. The coefficients of this polynomial arefixed under all elements of G and hence since k is the fixed field G, f (x)∈k[x]. Hence the minimum polynomial roots of the minimum polynomialofω, sinceω 1 ,...ω n are conjugates. Thus f (x)/φ(x). ThereforeK splitting field ofφ(x),φ(x) has all roots distinct. Thus K/k is normaland separable.□Suppose now K/k is normal and separable. Consider the group 46G(K/k) of k-automorphisms of K. Letα∈K. Since K/k is separable,all conjugates ofαare distinct. Also since K/k is normal K contains allthe conjugates. Ifαis fixed under allσ∈G(K/k), thenαis a purelyinseparable element of K and hence is in k.Our theorem is thus proved.We thus see that galois extensions are identical with extension fieldswhich are both normal and separable.Examples of Galois extensions are the splitting fields of polynomialsover perfect fields.Let k be a field of characteristic 2 and let K= k( √ α) forα∈kand √ αk.( √ α) 2 =α∈k. Every element of K is uniquely of the forma+ √ α·b, a, b∈k. Ifσis an automorphism of K which is trivial on k,then its effect on K is determined by its effect on √ α. Nowα=σ { ( √ α) 2} =σ( √ α).σ( √ α)or thatσ( √ α)/ √ α=λ is such thatλ 2 = 1. Sinceλ∈K,λ=±1. Thusσ is either th identity or the automorphismσ( √ α)=− √ αThus G(K/k) is a group of order 2. K/k is normal and separable.We shall obtain some important properties of galois extensions.