Let a be algebraic of degree n over F. Show from Corollary 48 .5 that there are at most n different isomorphisms of F(α) onto a subfield of F and leaving F fixed.

Data from Corollary 48.5

Let α be algebraic over a field F. Every isomorphism ψ mapping F(α) onto a subfield
of F̅ such that ψ(a) = a for a ∈ F maps α onto a conjugate β of α over F. Conversely,
for each conjugate β of α over F, there exists exactly one isomorphism ψ_α,β of F(α) onto a subfield of F̅ mapping α onto β and mapping each a ∈ F onto itself.

Proof Let ψ be an isomorphism of F(α) onto a subfield of F̅ such that ψ(a) = a for a ∈ F.
Let irr(α, F) = a₀ + a₁x + + a_nxⁿ. Then a₀ + a₁α + · · ·a_nαⁿ = 0, so 0 ψ(a₀ + a₁α + · · ·a_nαⁿ )= a₀ +a₁ψ(α) +· · · + a_nψ(α)ⁿ, and β = ψ(α) is a conjugate of α. 

Conversely, for each conjugate β of α over F, the conjugation isomorphism ψ_α,β of Theorem 48.3 is an isomorphism with the desired properties. That ψ_α,β is the only such isomorphism follows from the fact that an isomorphism of F(α) is completely determined by its values on elements of F and its value on α.