Exercise 4.2.3

Use part (a) of Exercise 2 with a = 1 to give another proof of Proposition 4.2.5.

Answers

Proof. Lemma: If p is prime, then for all k , 0 ≤ k ≤ p − 1 ,

( p − 1 k ) ≡ ( − 1 ) k ( mod p ) .

Proof by induction on k .

∙ If k = 0 , ( p − 1 0 ) = 1 = ( − 1 ) 0 .

∙ Suppose that this property is true for k − 1 ( 1 ≤ k ≤ p − 1 ):

( p − 1 k − 1 ) ≡ ( − 1 ) k − 1 ( mod p )

Then, as 1 ≤ k ≤ p − 1 , we know that ( p k ) ≡ 0 ( mod p ) , thus from Pascal’s formula,

( p − 1 k ) = ( p k ) − ( p − 1 k − 1 ) ≡ 0 − ( − 1 ) k − 1 ≡ ( − 1 ) k ( mod p ) ,

which concludes the induction. □

If p = 2 , Φ 2 = 1 + x is irreducible. Suppose now that p is an odd prime.

Applying the lemma, we obtain

Φ p ( x ) − ( x − 1 ) p − 1 = ∑ k = 0 p − 1 x k − ∑ k = 0 p − 1 ( − 1 ) p − 1 − k ( p − 1 k ) x k = ∑ k = 0 p − 1 [ 1 − ( − 1 ) p − 1 − k ( p − 1 k ) ] x k = ∑ k = 0 p − 1 [ 1 − ( − 1 ) k ( p − 1 k ) ] x k = p ∑ k = 0 p − 1 a k x k ( a k ∈ ℤ )

since every coefficient [ 1 − ( − 1 ) k ( p − 1 k ) ] is divisible by p , of the form p a k , a k ∈ ℤ .

Consequently

Φ p ( x ) = ( x − 1 ) p − 1 + pF ( x ) , F ( x ) = ∑ k = 0 p − 1 a k x k ∈ ℤ [ x ] , deg ⁡ ( F ) ≤ p − 1 .

Moreover

F ( 1 ) = ∑ k = 0 p − 1 a k = ∑ k = 0 p − 1 1 − ( − 1 ) k ( p − 1 k ) p = 1 − 1 p ∑ k = 0 p − 1 ( − 1 ) k ( p − 1 k ) = 1 − 1 p ( 1 − 1 ) p − 1 = 1 .

F ( 1 ) ≢ 0 ( mod p ) . By Exercise 2, Φ p is irreducible. □

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2022-07-19 00:00
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