Abstract Algebra Theory and Applications - Computer Science ...

21.1 FIELD AUTOMORPHISMS 365Proof. If σ and τ are automorphisms of E, then so are στ and σ −1 . Theidentity is certainly an automorphism; hence, the set of all automorphismsof a field F is indeed a group.□Proposition 21.2 Let E be a field extension of F . Then the set of allautomorphisms of E that fix F elementwise is a group; that is, the set of allautomorphisms σ : E → E such that σ(α) = α for all α ∈ F is a group.Proof. We need only show that the set of automorphisms of E that fix Felementwise is a subgroup of the group of all automorphisms of E. Let σand τ be two automorphisms of E such that σ(α) = α and τ(α) = α for allα ∈ F . Then στ(α) = σ(α) = α and σ −1 (α) = α. Since the identity fixesevery element of E, the set of automorphisms of E that leave elements of Ffixed is a subgroup of the entire group of automorphisms of E. □Let E be a field extension of F . We will denote the full group of automorphismsof E by Aut(E). We define the Galois group of E over F tobe the group of automorphisms of E that fix F elementwise; that is,G(E/F ) = {σ ∈ Aut(E) : σ(α) = α for all α ∈ F }.If f(x) is a polynomial in F [x] and E is the splitting field of f(x) over F ,then we define the Galois group of f(x) to be G(E/F ).Example 1. Complex conjugation, defined by σ : a + bi ↦→ a − bi, is anautomorphism of the complex numbers. Sinceσ(a) = σ(a + 0i) = a − 0i = a,the automorphism defined by complex conjugation must be in G(C/R).Example 2. Consider the fields Q ⊂ Q( √ 5 ) ⊂ Q( √ 3, √ 5 ). Then fora, b ∈ Q( √ 5 ),σ(a + b √ 3 ) = a − b √ 3is an automorphism of Q( √ 3, √ 5 ) leaving Q( √ 5 ) fixed. Similarly,τ(a + b √ 5 ) = a − b √ 5is an automorphism of Q( √ 3, √ 5 ) leaving Q( √ 3 ) fixed. The automorphismµ = στ moves both √ 3 and √ 5. It will soon be clear that {id, σ, τ, µ} is