Exercise 12.1.1

Let 𝜃 ( x ) be the resolvent polynomial defined in (12.3). Use the second bullet following (12.1) to show that 𝜃 ( x ) ∈ K [ x ] .

Answers

Proof. Let σ be any permutation of S n , and

𝜃 ( x ) = ∏ i = 1 r ( x − φ i ) ,

where φ 1 = φ , φ 2 , … φ r are the distinct conjugates of φ :

A : = { τ ⋅ φ | τ ∈ S n } = { φ 1 , … , φ r } .

As in the proof of Theorem 7.1.1, we first show that σ permute the φ i .

If φ i ∈ A , then φ i = τ ⋅ φ for some τ ∈ S n . Therefore

σ ⋅ φ i = σ ⋅ ( τφ ) = ( στ ) ⋅ φ .

Since σ ′ = στ ∈ S n , σ ′ ⋅ φ = ( στ ) ⋅ φ ∈ A , therefore

{ σ ⋅ φ 1 , … , σ ⋅ φ r } ⊂ { φ 1 , … , φ r } .

Since the map ψ ↦ σ ⋅ ψ is injective, the elements σ ⋅ φ i are distinct, and A is finite, thus we can conclude that

{ σ ⋅ φ 1 , … , σ ⋅ φ r } = { φ 1 , … , φ r } .

Hence the exists some τ ∈ S r such that,

σ ⋅ φ i = φ τ ( i ) , 1 ≤ i ≤ r .

σ ⋅ 𝜃 ( x ) = σ ⋅ ∏ i = 1 r ( x − φ i ) = ∏ i = 1 r σ ⋅ ( x − φ i ) = ∏ i = 1 r ( x − σ ⋅ φ i ) = ∏ i = 1 r ( x − φ τ ( i ) ) = ∏ j = 1 r ( x − φ j ) ( j = τ ( i ) ) = 𝜃 ( x ) .

Since σ ⋅ 𝜃 ( x ) = 𝜃 ( x ) for all σ ∈ S n , the coefficients of 𝜃 are in K by theorem 2.2.7, so 𝜃 ( x ) ∈ K [ x ] . □

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2022-07-19 00:00
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