Abstract Algebra and Algebraic Number Theory

3.5 Galois Extension
Let E be a field and F ⊂ E. Then
Aut(E) = {σ : E ↦→ E : σ is an automorphism }
forms a group with respect to composition of maps. and
Aut(E/F ) = {σ : E ↦→ E : σ is F -automorphism i.e. σ(a) = a∀a ∈ F }
will be a subgroup of the Aut(E).
Note. Prime subfield P of E is generated by {0, 1}. Since any automorphism
σ takes 1 to 1 and 0 to 0, Aut(E) = Aut(E/P ).
Proposition 3.5.1. Let E/F be a field extension. Aut(K) permutes the
roots of irreducible polynomials in F (x) i.e., if α ∈ E is a root of an irreducible
polynomial f(x) in F (x), then σ(α) is also a root of f(x) for all
σ ∈ Aut(E).
Example 3.5.1. Let Q( √ 2)/Q, if τ ∈ Aut(Q( √ 2)) so τ( √ 2) = ± √ 2, as
there are two roots ± √ 2 of the min √ 2,Q (x) = x2 −2. Since Q( √ 2) is a vector
space over Q with basis {1, √ 2}, Aut(Q( √ 2)) = {I, τ}, where τ( √ 2) = − √ 2
and I is identity automorphism. Since Q is a prime subfield of Q( √ 2).
Aut(Q( √ 2)) = Aut(Q( √ 2)/Q) = {I, τ}.
Size of automorphism group in splitting filed Let f(x) ∈ F [x]
and E be splitting field of F . Theorem 3.2.3 shows that any isomorphism
ϕ : F ↦→ ¯F extends to an isomorphism σ : E ↦→ Ē, where Ē is splitting field
ϕ(f(x)).
σ : E −→ Ē
↿
τ : F (α) −→ ¯F (β)
↿
ϕ : F −→ ¯F
Using induction on [E : F ], it can be shown that number of such extensions
is at most [E : F ], with equality if f(x) is separable over F .
In particular case when F = ¯F , ϕ is an identity map and isomorphism
σ : E ↦→ Ē, becomes F -automorphism and we have a theorem:
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