Exercise 6.4.4

Let τ ∈ S n . Prove that f ↦ τ ⋅ f is a ring isomorphism from F [ x 1 , … , x n ] to itself.

Answers

Proof. We know (Exercise 6.4.3 (a)) that φ : f ↦ τ ⋅ f is a ring homomorphism. As τ ∈ S n , τ is bijective and so τ − 1 exists. Let ψ : f ↦ τ − 1 ⋅ f . Then for all f ∈ F [ x 1 , … , x n ] , by Exercise 6.4.3 (b)

( ψ ∘ φ ) ( f ) = τ − 1 ⋅ ( τ ⋅ f ) = ( τ − 1 τ ) ⋅ f = 1 [ 1 , n ] ⋅ f = f ( φ ∘ ψ ) ( f ) = τ ⋅ ( τ − 1 ⋅ f ) = ( τ τ − 1 ) ⋅ f = 1 [ 1 , n ] ⋅ f = f

Therefore ψ ∘ φ = φ ∘ ψ = 1 F [ x 1 , … , x n ] , so φ is a bijection.

Conclusion: φ is a ring isomorphism. □

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2022-07-19 00:00
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