Exercise 8.17

(continuation) By Exercise 8.7, J ( χ , χ ) = χ ( 1 ) J ( χ , ρ ) . Let π = J ( χ , ρ ) . Show that

(a)
N = p 3 6 Re π if p 1 ( mod 8 ) .
(b)
N = p + 1 2 Re π if p 5 ( mod 8 ) .

Answers

Proof. Let g a generator in 𝔽 p . As ( g ( p 1 ) 2 ) 2 = 1 and g ( p 1 ) 2 1 , then g ( p 1 ) 2 = 1 . As in Exercise 8.16, write χ ( g ) = e qiπ 2 , with q odd.

Then 1 is a fourth power in 𝔽 p iff (see Exercice 8.16)

δ 4 ( 1 ) = 1 χ ( 1 ) = 1 χ ( g ( p 1 ) 2 ) = 1 e q ( ( p 1 ) 2 ) 2 = 1 4 q ( p 1 ) 2 4 ( p 1 ) 2 p 1 ( mod 8 ) .

By Exercise 8.7, as χ is a character of order 4,

J ( χ , χ ) = χ ( 1 ) J ( χ , ρ ) .

If p 1 [ 8 ] ,

χ ( 1 ) = 1 , so J ( χ , χ ) = J ( χ , ρ ) , and δ 4 ( 1 ) = 1 .

N = p + 1 δ 4 ( 1 ) 4 + 2 Re J ( χ , χ ) + 4 Re J ( χ , ρ ) = p 3 + 6 Re J ( χ , ρ ) = p 3 6 Re π , where π = J ( χ , ρ ) .

If p 5 [ 8 ] ,

χ ( 1 ) = 1 , so J ( χ , χ ) = J ( χ , ρ ) , and δ 4 ( 1 ) = 0 .

N = p + 1 δ 4 ( 1 ) 4 + 2 Re J ( χ , χ ) + 4 Re J ( χ , ρ ) = p + 1 + 2 Re J ( χ , ρ ) = p + 1 2 Re π .
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2022-07-19 00:00
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