Exercise 13.3.3

In the proof of proposition 13.3.2, we asserted that

φ ( α 1 , . . . , α n ) = φ ( α τ ( 1 ) , . . . , α τ ( n ) )

follows from β = φ ( α 1 , . . . , α n ) F and τ G f . Prove this.

Proof. Let σ Gal ( L F ) corresponding to τ G f , so that σ ( α i ) = α τ ( i ) , i = 1 , , n .

Since F is fixed for σ , and β = φ ( α 1 , . . . , α n ) F , β = σ ( β ) , so that

φ ( α 1 , . . . , α n ) = σ ( φ ( α 1 , . . . , α n ) ) .

Moreover

σ ( φ ( α 1 , . . . , α n ) ) = φ ( σ ( α 1 ) , , σ ( α n ) = φ ( α τ ( 1 ) , . . . , α τ ( n ) ) .

Therefore, φ ( α 1 , . . . , α n ) = φ ( α τ ( 1 ) , . . . , α τ ( n ) ) for φ ( α 1 , . . . , α n ) F and τ G f . □

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2022-07-19 00:00
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