Exercise 7.21

Let F be a field with q = p n elements. For α ∈ F set f ( x ) = ( x − α ) ( x − α p ) ( x − α p 2 ) ⋯ ( x − α p n − 1 ) . Show that f ( x ) ∈ ℤ ∕ pℤ [ x ] . In particular, α + α p + ⋯ + α p n − 1 and α α p α p 2 ⋯ α p n − 1 are in Z ∕ pℤ .

Answers

Proof. Let F : { 𝔽 q → 𝔽 q x ↦ x p .

As the characteristic of 𝔽 q is p , ( x + y ) p = x p + y p et ( xy ) p = x p y p , and each homomorphism of field is injective, so F is a field automorphism (Frobenius automorphism).

For every automorphism H in 𝔽 q , and every polynomial p ( x ) = ∑ ⁡ a i x i ∈ 𝔽 q [ x ] , write ( H ⋅ p ) ( x ) = ∑ ⁡ i H ( a i ) x i . Then for all ( p , q ) ∈ 𝔽 q [ x ] 2 , H ⋅ ( pq ) = ( H ⋅ p ) ( H ⋅ q ) .

With this notation,

f ( x ) = ( x − α ) ( x − Fα ) ( x − F 2 α ) ⋯ ( x − F n − 1 α ) ,

( F ⋅ f ) ( x ) = ( x − Fα ) ( x − F 2 α ) ( x − F 3 α ) ⋯ ( x − F n α ) .

Since α ∈ 𝔽 p n , F n α = α p n = α , thus

F ⋅ f = f .

In other words, if f ( x ) = ∑ ⁡ i a i x i , then for all i , F ( a i ) = a i , so a i p = a i , thus a i ∈ 𝔽 p , and f ∈ 𝔽 p [ x ] . In particular, the coefficients a n − 1 = α + α p + ⋯ + α p n − 1 , a 0 = α α p α p 2 ⋯ α p n − 1 are in 𝔽 p . □

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2022-07-19 00:00
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