Question: We consider the field E = Q(2, 3, 5). It can be shown that [E : Q] = 8. In the notation of Theorem 48.3,

We consider the field E = Q(√2, √3, √5). It can be shown that [E : Q] = 8. In the notation of Theorem 48.3, we have the following conjugation isomorphisms (which are here automorphisms of E):

2.-2 : (Q(3, 5))(2) (Q(3, 5))(-2), V3.-3: (Q(2, 5))(3) (Q(2, 5))(-3), 5.-5


For shorter notation, let τ2 = ψ2.-√2, τ3 = ψ√3 -√3, and , τ5 = ψ√5.-√5· Compute the indicated element of E. 

τ2(√2 +√5)

Data from Theorem 48.3:

Let F be a field, and let α and β be algebraic over F with deg(α, F) = n. The map ψα :F(α) → F(β) defined by

: (Q(2, 3))(5) (Q(2, 3))(-5).

for c∈ F is an isomorphism of F(α) onto F(β) if and only if a and are conjugate over F.

2.-2 : (Q(3, 5))(2) (Q(3, 5))(-2), V3.-3: (Q(2, 5))(3) (Q(2, 5))(-3), 5.-5 : (Q(2, 3))(5) (Q(2, 3))(-5).

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