Exercise 12.1.17

In Theorem 12.1.9, we used Galois correspondence to show that rational functions φ and ψ are similar if and only if K ( φ ) = K ( ψ ) . Give another proof of this result that uses only Theorem 12.1.6.

Answers

Proof. If φ , ψ ∈ F ( x 1 , … , x n ) are similar, then H ( φ ) = H ( ψ ) . So σ ⋅ ψ = ψ for every σ ∈ H ( φ ) . By Theorem 12.1.6, ψ ∈ K ( φ ) . Exchanging φ and ψ , we obtain similarly φ ∈ K ( ψ ) . Therefore

K ( φ ) = K ( φ , ψ ) = K ( ψ , φ ) = K ( ψ ) .

Conversely, if K ( φ ) = K ( ψ ) , then ψ ∈ K ( φ ) , so ψ ( x 1 , … , x n ) = f ( φ ( x 1 , … , x n ) ) , where f ∈ K ( x ) . Therefore, for all σ ∈ H ( φ ) ,

σ ⋅ ψ = f ( φ ( x σ ( 1 ) , … , x σ ( n ) ) ) = f ( φ ( x 1 , … , x n ) ) = ψ .

So H ( φ ) ⊂ H ( ψ ) , and similarly H ( ψ ) ⊂ H ( φ ) , thus H ( φ ) = H ( ψ ) . □

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2022-07-19 00:00
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